CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(c)

Solve the following systems of linear inequalities graphically.
Question 1.
2x – y ≥ 0, x – 2y ≤ 0, x ≤ 2, y ≤ 2 [Hint: You may consider the point (2, 2) to determine the SR of the first two inequalities.]
Solution:
2x – y ≥ 0
x – 2y ≤ 0
x ≤ 2
y ≤ 2
Step – 1: Let us draw the lines.
2x – y = 0, x – 2y = 0, x = 2, y = 2
2x – y = 0

X 0 1
y 0 2

x – 2y = 0

X 0 2
y 0 1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c)
Step – 2: Let us consider point (1, 0) which does not line on any of these lines.
Putting x = 1, y = 0 in the inequations we get
2 ≥ 0 (True)
1 ≤ 0 (False)
1 ≤ 2 (True)
0 ≤ 2 (True)
Point (1, 0) satisfies all inequality except x – 2y < 0.
∴ Thus the shaded region is the solution region.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c)

Question 2.
x – y < 1, y – x < 1
Solution:
x – y < 1
y – x < 1
Step – 1: Let us draw the dotted lines.
x – y = 1 and y – x = 1
x – y = 1 ⇒ y = x – 1

X 1 0
y 0 -1

y – x = 1 ⇒ y = x + 1

X 0 -1
y 1 0

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 1
Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
0 < 1 (True)
0 < 1 (True)
∴ (0, 0) satisfies both the inequations.
∴ Thus the shaded region is the feasible region.

Question 3.
x – 2y + 2 < 0, x > 0
Solution:
x – 2y + 2 < 0, x > 0
Step – 1: Let us draw the dotted line x – 2y + 2 = 0
⇒ y = \(\frac{x+2}{2}\)

X -2 0
y 0 1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 2
Step – 2: Let us consider the point (1, 0) that does not lie on the lines  putting x = 0, y = 0 in the inequation, we get
2 < 0 (false)
1 > 0 (True)
⇒ (1, 0) satisfies x > 0 and does not satisfy x – 2y + 2 < 0.
∴ Thus the shaded region is the solution region.

Question 4.
x – y + 1 ≥ 0, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Solution:
x – y + 1 ≥ 0
3x + 4y ≤ 12
x ≥ 0, y ≥ 0
Step – 1: Let us draw the lines.
x – y + 1 = 0
3x + 4y = 12
Now, x – y + 1 = 0 ⇒ y = x + 1

X 0 -1
y 1 0

3x + 4y = 12

X 4 0
y 0 3

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 3
Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 ≥ 0 (True)
0 ≤ 12 (False)
∴ (0, 0) satisfies both the inequations and x > 0, y > 0 is the first quadrant.
∴ Thus the shaded region is the solution region.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c)

Question 5.
x + y > 1, 3x – y < 3, x – 3y + 3 > 0
Solution:
x + y > 1
3x – y < 3
x – 3y + 3 > 0
Step – 1: Let us draw the lines.
x + y = 1
3x – y = 3
x – 3y + 3 = 0
Now x + y = 1
⇒ y = 1 –  x

X 1 0
y 0 1

3x – y = 3
⇒ y = 3x – 3

X 1 0
y 0 -3

x – 3y + 3 = 0
⇒ y = \(\frac{x+3}{3}\)

X -3 0
y 0 1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 4
Step – 2: Let us consider the point (0, 0) that does not lie on these lines. Putting x = 0, y = 0 in the inequations we get,
0 > 1 (False)
0 < 3 (True)
3 > 0 (True)
Thus (0, 0) satisfies 3x – y < 3 and x – 3y + 3 > 0 but does not satisfy x + y > 1
∴ The shaded region is the solution region.

Question 6.
x > y, x < 1, y > 0
Solution:
x > y, x < 1, y > 0
Step – 1: Let us draw the dotted lines.
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 5
Step – 2: Let us consider a point (2, 1) that does not lie on any of the lines.
Putting x = 2, y = 2 in the inequations
we get,
2 > 1 (True)
2 < 1 (False)
1 > 0 (True)
⇒ (2, 1) satisfies x > y and y > 0 but does not satisfy x < 1.
∴ Thus the shaded region is the solution region.

Question 7.
x < y, x > 0, y < 1
Solution:
x < y
x > 0
y < 1
Step – 1: Let us draw the dotted lines.
x = y
x = 0
and y = 1
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(c) 6
Step – 2: Let us consider point (1, 0) that does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 < 0 (False)
1 > 0 (True)
0 < 1 (True)
Clearly (1, 0) satisfies x > 0, y < 1 but does not satisfy x < y.
∴ The shaded region is the solution region.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(b)

Question 1.
If Z1 and Z2 are two complex numbers then show that
\(\begin{aligned}
& \left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2 \\
& =\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)
\end{aligned}\)
Solution:
Let z1 and z2 be two complex numbers.
Let z1 = a + ib, z2 = c + id
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Question 2.
If a, b, c are complex numbers satisfying a + b + c = 0 and a2 + b2 + c2 = 0 then show that |a| = |b| = |c|
Solution:
Let a + b + c = 0 and a2 + b2 + c2 = 0
Then (a + b + c)2 = 0
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
⇒ 2(ab + bc + ca) = 0
⇒ ab + bc = – ca
⇒ b(a + c) = – ca
⇒ b(- b) = – ca [a + b + c = 0]
⇒ b2 = ca
⇒ b3 = abc
Similarly it can be shown that a3 = abc and c3 = abc
Thus a3 = b3 = c3
⇒ |a3|= |b3| = |c3|
⇒ |a|3 = |b|3 = |c|3
⇒ |a| = |b| = |c|

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Question 3.
What do the following represent?
(i) { z : |z – a| + |z + a| = 2c } where |a| < c
Solution:
{ z : |z – a| + |z + a| = 2c } where |a| < c    …….(1)
Here z is a complex number.
Let z = x + iy.
∴ (x, y) is the point corresponding to the complex number z?
Let ‘a’ and ‘ – a’ be two fixed points.
∴ Eqn. (1) implies that the sum of the distances of the point (x, y) from two points la and a’ is constant i.e. 2c
∴ The locus is an ellipse.

(ii) {z : |z – a| – |z + a| = c }
Solution:
Here {z : |z – a| – |z + a| = c } implies that, the difference of the distances of the point (x, y) from two fixed points ‘ – a’ and ‘a’ is a constant i.e. c.
So the locus is a hyperbola.

(iii) What happens in (i) |a| > c?
Solution:
In(i), if |a| > c. then there is no locus. But if |a| = c, then the locus reduces to a straight line.

Question 4.
Given cos α + cos β + cos γ = sin α + sin β + sin γ = 0 Show that cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
Solution:
Let a = cos α + i sin α,
b = cos β + i sin β
c = cos γ + i sin γ
∴ a + b + c = ( cos α + cos β + cos γ) + i ( sin α + sin β + sin γ)
= 0 + i0 = 0
∴ a3 + b3 + c3 – 3 abc
= (a + b + c )( a2 + b2 + c2 – ab – bc – ca) = 0
or, a3 + b3 + c3 = 3 abc
or, ( cos α + i sin α)3 + (cos β + i sin β)3 + ( cos γ + i sin γ)3
= 3( cos α + i sin α) (cos β + i sin β) ( cos γ + i sin γ)
or, cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ
= 3[cos (α + β + γ) + i sin (α + β + γ)]
or, (cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ)
= 3 cos (α + β + γ) + i 3 sin (a + β + γ)
∴ cos 3α + cos 3β + cos 3γ
= 3 cos (α + β + γ) and sin 3α + sin 3β + sin 3γ
= 3 sin (α + β + γ)

Question 5.
Binomial theorem for complex numbers. Show that (a+b)n = an nC1an-1b +  …..+ ncran-rbr + …..+ bn where a,b ∈ C and n, rule of multiplication of complex numbers and the relation nCr + nCr-1 = n+1Cr)
Solution:
Let a and b be two complex numbers
Let a = α1 + iβ1, b = α2 + iβ2
(a + b)1 = (α1 + iβ1 + α2 + iβ2)1
= α1 + iβ1 + α2 + iβ2
= (α1 + iβ1)1 + 1C11 + iβ1)1-11 + iβ1)1
= a1 + 1C1 a1-1 b1
∴ P1 is true
Let Pk be true
i.e., (a + b)k = ak + kC1 ak-1 b1 + … + bk
where a.b ∈ C
Now ( a + b)k+1 = (a + b)k (a + b)1
= (ak + kC1 ak-1b1 +…+ bk) (a + b)
= ak-1 + bak+ kC1 akb + kC1ak-1b2 + … + bk+1
= ak+1 + akb(kC1+ 1)+ …+ bk+1
= ak+1 + k+1C1akb1 + k+1 C2ak-1b2 +… + bk+1
∴ Pk+1 is true
∴ Pn is true for all values of n ∈ N

Question 6.
Use the Binomial theorem and De Moiver’s theorem to show
cos 3θ = 4 cos 3 θ – 3 cos θ,
sin 3θ = 3 sin θ – 4 sin 3 θ
Express cos nθ as a sum of the product of powers of sin θ and cos θ. Do the same thing for sin nθ.
Solution:
We have (cos θ + i sin θ)3
= cos 3θ + i sin 3θ       …..(1)
But by applying the Binomial theorem, we have
(cos θ + i sin θ)3
= cos 3 θ + 3C1 (cos θ)3-1 (i sin θ)1 + 3C2 (cos θ)3-2  (i sin θ)2 + (i sin θ)3
= cos3θ + 3i cos2θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ
= (cos3 θ – 3 cos θ sin2 θ) + i(3 cos2 θ sin θ – sin3 θ)
∴ cos 3 θ =cos3 θ – 3 cos θ(1 – cos2 θ)
= cos3 θ – 3 cos θ + 3 cos 3 θ
= 4 cos3 θ – 3 cos θ and
sin 3θ = 3 cos2 θ sin3 θ – sin3 θ
= 3 (1 – sin2 θ) sin θ – sin3 θ
= 3 sin θ – 3 sin3 θ – sin3 θ
= 3 sin θ – 4 sin3 θ (Proved)
Again, (cos θ + i sin θ)n
= cos nθ + i sin nθ         ….(3)
Also, (cos θ + i sin θ)n
= cosn θ + nC1 cosn-1 θ ( i sin θ) + nc2 cos n-2 θ (i sin θ)2 + …+ (i sin θ)
= cosn θ – nC2 cosn-2 θ sin2 θ + nC4 cosn-4 θ sin4 θ – …) + i (nC1 cosn-1 θ sin θ – nC3 cosn-3 θ sin 3 θ) + nC5 cosn-5 θ sin5 θ – …)    …..(4)
Equating real part and imaginary parts in (1) and (3), we have
cos nθ = cosn θ – nC2 cosn-2 θ × sin2 θ + nC4 cosn-4 θ sin4 θ …
and sin nθ = nC1 cosn-1 θ sin θ – nC3 cosn-3 θ sin3 θ + nC5 cosn-5 θ sin5 θ…

Question 7.
Find the square root of
(i) – 5 + 12 √-1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 1

(ii) – 11 – 60 √-1
Solution:
Let \(\sqrt{-11-60 \sqrt{-1}}\) = x + iy
Squaring both sides we get
– 11 – 60i = (x + iy)2
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 2

(iii) – 47 + 8 √-1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 3
As 2ab = 8 > 0, a and b have the same sign.
∴ \(\sqrt{-47+8 i}\)
\(=\pm\left(\sqrt{\frac{\sqrt{2273}-47}{2}}+i \frac{\sqrt{2273}+47}{2}\right)\)

(iv) – 8 + √-1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 4

(v) a2 – 1 +2a √-1
Solution:
a2 – 1 + 2a √-1
= a2 + i2 + 2ai = (a + i)2
∴ \(\sqrt{a^2-1+2 a \sqrt{-1}}\) = ±(a + i)

(vi) 4ab – 2 (a2 – b2) √-1
Solution:
4ab – 2 (a2 – b2) √-1
= (a + b)2 – (a – b)2 – 2(a2 – b2)i
= (a + b)2 + (a – b)2i2 – 2(a + b)(a – b)i
= (a + b) – i(a – b)2
∴ \(\sqrt{4 a b-2\left(a^2-b^2\right) \sqrt{-1}}\)
= ±[(a + b) – i (a – b)]

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Question 8.
Find the values of cos72° ….
Solution:
Let 18° = θ then 5θ = 90°
⇒ 3θ = 9θ – 2θ
⇒ cos 3θ = cos (9θ – 2θ) = sin 2θ
⇒ 4 cos3 θ – 3 cos θ = 2 sin θ cos θ
⇒ 4 cos2 0 – 3 = 2 sin 0 (∴ cos θ = cos 18° ≠ 0)
⇒ 4(1 – sin2 θ) – 3 = 2 sin θ
⇒ 4 sin2 θ + 2 sin θ – 1 = 0
⇒ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{2 \times 4}=\frac{-1 \pm \sqrt{5}}{4}\)
⇒ sin 18° = \(\frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4}\)
(∴ 18° is a cut)
Now cos 72° = cos (90° – 18°)
= sin 18° = \(\frac{\sqrt{5}-1}{4}\)
For other methods refer 148 pages of the text book.

Question 9.
Find the value of cos 36°.
Solution:
We have 36° = \(\frac{\pi^0}{5}\)
∴ cos 36° = cos \(\frac{\pi}{5}\)
Let α = cos \(\frac{\pi}{5}\) + i sin \(\frac{\pi}{5}\)
be the root of the equation x5 + 1 = 0
Again, if x5 + 1 = 0
or, x5 = – 1 = cosπ + i sinπ
or, x = (cos π + i sin π )1/5
= [cos (π + 2kπ)+ i sin (π + 2kπ)]1/5
or, x = cos \(\frac{(2 k+1) \pi}{5}+i \sin \frac{(2 k+1) \pi}{5}\)
where 2kπ is the period of sine and cosine and k = 0, 1, 2, 3, 4
∴ The eqn x5 + 1 = 0 has 5 roots out of which -1 is one root which corresponds to k = 2
Again,
x5 + 1 = (x + 1)(x4 – x3 + x2 – x + 1)
So their 4 roots will be obtained on solving the eqn.
x4 – x3 + x2 – x + 1 = 0
we have, x4 – x3 + x2 – x + 1 = 0
or, x2 – x + 1 – \(\frac{1}{x}+\frac{1}{x^2}\) = 0
(Dividing both sides by x2)
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 5
Re α i.e. cos \(\frac{\pi}{5}=\frac{1+\sqrt{5}}{5}=\frac{\sqrt{5}+1}{4}\)
and cos.108° = \(\frac{1-\sqrt{5}}{4}\)

Question 10.
Evaluate cos \(\frac{2 \pi}{17}\) using the equation x17 – 1 = 0
Solution:
x17 – 1 = 0
or, x17 = 1 = cos 0° + i sin 0°
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
or x = (cos 2kπ + i sin 2kπ)1/17
= cos \(\frac{2k \pi}{17}\) + i sin \(\frac{2k \pi}{17}\)
If k = 1, x = cos \(\frac{2 \pi}{17}\) + i sin \(\frac{2 \pi}{17}\)
If k = 0 , x = 1
As x17 – 1 = (x – 1) (x16 + x15 + …. +1)
So one root of the eqn. x17 – 1 = 0 is 1 and all other roots are the roots of the eqn.
x16 + x15 + ….+ 1 = 0
∴ The value of cos \(\frac{2 \pi}{17}\) can be found from the roots of the eqn.  (1)

Question 11.
Solve the equations.
(i) z7 = 1
Solution:
z7 = 1 = cos 0 + i sin 0
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
∴ z = (cos 2kπ + i sin 2kπ)1/7
= cos \(\frac{2k \pi}{7}\) + i sin \(\frac{2k \pi}{7}\)
where k = 0, 1, 2, 3, 4, 5, 6

(ii) z3 = i
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 6
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 7

(iii) z6 = – i
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 8

(iv) z3 = 1 + i
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 9

Question 12.
If sin α + sin β + cos γ = 0
= cos α + cos β + cos γ = 0
Show that
(i) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Solution:
Refer to Q. No. 4

(ii) sin2 α + sin2 β + sin2 γ = cos2 α + cos2 β + cos2 γ =3/2
Solution:
Let x = cos α + i sin α,
y = cos β + i sin β
z = cos β + sin β
∴ x + y + z = (cos α + cos β + cos γ) + i ( sin α + sin β + sin γ) = 0 +i0= 0
∴ xy + yz + zx = xyz (\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)) = 0
Since \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ
= (cos α + cos β + cos γ) – i (sin α + sin β + sin γ) = 0 – i0 =0
∴ (x + y + z)2 = x 2 + y2 +z2 + 2(xy + yz + zx)
= x2 + y2 + z2 + 0 = x2 + x2 + z2
or 0 = x2 + y2 + z2
∴ x2 +y2 + z2 =0
or, (cos α + i sin α)2 + (cos β + i sin β)2 + ( cos γ + i sin γ)2 = 0
or, cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0
or, (cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ) = 0
∴ cos 2α + cos 2β + cos 2γ = 0
or, cos2 α – sin2 α + cos2 β – sin2 β + cos2 γ – sin2 γ =0
or, (cos2 α + cos2 β + cos2 γ) = (sin2 α + sin2 β + sin2 γ)
But cos2 α + sin2 α + cos2 β + sin2 β + cos2 γ + sin2 γ = 1 + 1 + 1 = 3
∴ cos2 α + cos2 β + cos2 γ = sin2 α + sin2 β + sin2 γ = 3/2   (Proved)

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Question 13.
If x + \(\frac{1}{x}\) = 2 cos θ
Show that \(x^n+\frac{1}{x^n}\) = 2 cos nθ
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 10

Question 14.
xr = cos ar + i sin ar
r =1, 2, 3 and x1 + x2 + x3 = 0 Show that \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\) = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 11
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 12

Question 15.
Show that \(\left(\frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta}\right)^n\) = \(\cos \left(\frac{n \pi}{2}-n \theta\right)+i \sin \left(\frac{n \pi}{2}-n \theta\right)\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 13
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 14

Question 16.
If α and β are roots x2 – 2x + 4 = 0 then show that \(\alpha^n+\beta^n=2^{n+1} \cos \frac{n \pi}{3}\)
Solution:
we have x2 – 2x + 4 = 0
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 15
= \(2^n \times 2 \cos \frac{n \pi}{3}=2^{n+1} \cos \frac{n \pi}{3}\)

Question 17.
For a positive integer n show that
(i) (1 + i)n + (1 – i)n = \(2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 16

(ii) (1 + i√3)n + (1 – i√3)n = \(2^{n+1} \cos \frac{n \pi}{3}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 17

Question 18.
Let x + \(\frac{1}{x}\) = 2 cos α, y + \(\frac{1}{y}\) = 2 cos β, z + \(\frac{1}{z}\) = 2 cos γ. Show that
(i) 2 cos (α + β + γ) = xyz + \(\frac{1}{xyz}\)
Solution:
We can take x = cos α + i sin α
y = cos β + i sin β, z = cos γ + i sin γ
∴ xyz = (cos α + i sin α ) (cos β + i sin β ) (cos γ + i sin γ)
= cos (α + β + γ) – i sin (α + β + γ)
∴ \(\frac{1}{xyz}\) = cos (α + β + γ) – i sin(α + β + γ)
∴ xyz + \(\frac{1}{xyz}\) = 2 cos(α + β + γ)

(ii) 2 cos (pα + qβ + rγ) = \(x^p y^q z^r+\frac{1}{x^p y^q z^r}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 18

Question 19.
Solve x9 + x5 – x4 = 1
Solution:
x9 + x5 – x4 = 1
or, x5(x4 + 1) – (x4 + 1) = 0
or, (x5 – 1) (x4 + 1) = 0
x4 + 1 = 0 and x5 – 1 = 0
x4 = – 1 = cos π + i sin π
= cos (π + 2nπ) + i sin (π + 2nπ)
∴ x = [cos (2n + 1) π + 1 sin (2n + 1) π]1/4
= \(\cos \frac{2 n+1 \pi}{4}+i \sin \frac{2 n+1 \pi}{4}\)
for n = 0, 1, 2, 3
Again, x5 – 1 = 0 or, x5 = 1
or, x5 = cos 0 + i sin 0
= cos 2nπ + i sin 2nπ
or, x = (cos 2nπ + i sin 2nπ)1/5
= \(\cos \frac{2 n \pi}{5}+i \sin \frac{2 n \pi}{5}\)
Where n = 0, 1, 2, 3, 4.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Question 20.
Find the general value of θ if (cos θ + i sin θ) (cos 2θ + i sin 2θ),…..(cos nθ + i sin nθ) =1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 19

Question 21.
If z = x + iy show that |x| + |y| ≤ √2 |z|
Solution:
z = x + iy
∴ |z| = \(\sqrt{x^2+y^2}\)
∴ |z| = x2 + y2
We have (|x| – |y|)2 ≥ 0
⇒ |x|2 + |y|2 -2|x||y|> 0
⇒ 2(|x|2 + |y|2) – 2|x||y| ≥ |x|2 + |y|2
⇒ 2(|x|2 + |y|2) ≥ |x|2 + |y|2 + 2|x||y|
⇒ 2|z|2 ≥ (|x| + |y|)2
⇒ √2|z| ≥ |x| + |y|
⇒ |x| + |y| ≤ √2|z|

Question 22.
Show that
Re (Z1Z2) = Re z1, Re z2 – Im z1, Im z2
Im (Z1Z2) = Re z1, Im z2 + Re z2 Im z1
Solution:
Let z1 = a + ib, z2 = c + id
∴ z1, z2 = (a + ib) (c + id)
= ac + iad + ibc + i2bc
= (ac – bd) + i (ad + be)
∴ Re (z1, z2) = ac – bd – Re z1. Re z2
– Im z1,. Im z2
Again, Im z1, z2 = ad + be
= Re z1. Im z2 + Im z1,. Re z2

Question 23.
What is the value of arg ω + arg ω2?
Solution:
arg ω = arg ω2 = arg (ω • ω2)
= arg (ω3) = arg (1) = 2nπ
∴ The principal agrument = 0.

Question 24.
If |z1| ≤ 1, |z2| ≤ 1 show that \(\left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2=\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)\) Hence and otherwise show that. \(\left|\frac{z_1-z_2}{1-z_1 z_2}\right|<1 \text { if }\left|z_1\right|<1,\left|z_2\right|<1\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 20
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 21

Question 25.
If z12 + z22 + z32 – z1z2 – z2z3 – z3z4 = 0 Show that |z1 – z2| = |z2 – z3| = |z3 – z1|
Solution:
Let z12 + z22 + z32 – z1z2 – z2z3 – z3z4 = 0
⇒ 2z12 + 2z22 + 2z32 – 2z1z2 – 2z2z3 – 2z3z4 = 0
⇒ (z1 – z2)2 + (z2 – z3)2 + (z3 – z1)2 = 0
Put a = z1 – z2, b = z2 – z3, c = z3 – z1
Then a + b + c = 0 and a2 + b2 + c2 = 0 As in Q2 we can show that |a| = |b| = |c|
⇒ |z1 – z2| = |z2 – z3| = |z3 – z1|

Question 26.
If |a| < |c| show that there are complex numbers z satisfying |z – a| = |z + a| = 2|c|
Solution:
Let z = x + iy
∴ |z – a| + |z + a| = 2c
or, |x + iy – a| + |x + iy + a| = 2c
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 22
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 23

Question 27.
Solve \(\frac{(1-i) x+3 i}{2+i}+\frac{(3+2 i) y+i}{2-i}=-i\) where x, y, ∈ R.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 24

Question 28.
If (1 + x + x2)n = p0 + p1x + p2x2 + …..+ p2nx2n, then prove that p0 + p3 + p6 + …..+ 3n-1
Solution:
Given, (1 + x + x2)n = p0 + p1x + p2x2 + …..+ p2nx2n
putting x = ω we get
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 25

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b)

Question 29.
Find the region on the Argand plane on which z satisfying
[Hint Arg (x + iy) =\(\frac{\pi}{2}\) = 0, y>0]
(i) 1 < |z – 2i| < 3
Solution:
Let z = x + iy
The given inequality is
1 < |x + i(y – 2)| < 3
⇒ \(1<\sqrt{x^2+(y-2)^2}<3\)

(ii) arg \(\left(\frac{z}{z+i}\right)=\frac{\pi}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 26
As all are +ve we have
1 < x2 + (y – 2)2 < 9
x2 + (y – 2)2 < 9 is the region inside the circle with center (0, 2) and radius 1.
x2 + (y – 2)2 > 1 is the region outside the circle with center (0, 2) and radius.
∴ 1 < |z – 2i| < 3 is the region between two concentric circles with center (0, 2) and radius 1 and 3 which is shows below.
CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(b) 27

CHSE Odisha Class 11 Political Science Book Solutions (+2 1st Year)

CHSE Odisha 11th Class Political Science Book Solutions (+ 2 1st Year)

CHSE Odisha Class 11 Political Science Book Solutions in English Medium

Unit 1 Understanding Political Theory

Unit 2 Basic Concept

Unit 3 Indian Constitution

Unit 4 Constitution at Work-I

Unit 5 Constitution at Work-II

CHSE Odisha Class 11 Political Science Book Solutions in Odia Medium

CHSE Odisha Class 11 Political Science Syllabus (+2 1st Year)

First Year CHSE (2022-2023)
Political Science Paper-I
(Foundation of Politics and Government)

Part A: Political Theory

UNIT I Understanding Political Theory

  1. Political Theory: An Introduction-What is Politics? Nature and scope of Politics; Usages of Political Theory. (4 Periods)
  2. State: Definition; Elements of State. (4 Periods)
  3. Nature of State Activity: Individualism; Welfare State; Globalisation. (6 Periods)

UNIT II Basic Concepts

  1. Liberty: Positives and Negative Liberty; Types of Liberty (2 Periods)
  2. Equality: Meaning; Dimensions; Significance of Equality. (2 Periods)
  3. Justice: Meaning; Dimensions; Significance of Social Justice. (2 Periods)
  4. Rights: Meaning; Types; Human Rights and its significance. (4 Periods)
  5. Secularism: Meaning; Western and Indian approaches to Secularism. (2 Periods)
  6. Development: Meaning; Models of Development; Capitalistic model, Socialist model; Sustainable Development. (4 Periods)

Part B: Indian Constitution At Work

UNIT-III (Indian Constitution)

  1. Philosophy of the Constitution; Constitution- the Making; Constituent Assembly; Preamble; Basic Features; Amendment Procedure. (8 Periods)
  2. Rights in the Indian Constitution; Fundamental Rights; Directive Principles of State Policy; Relationship between Fundamental Rights & Directive Principles of State Policy; Fundamental Duties. (8 Periods)

UNIT IV Constitution at Work-I

  1. Election and Representation: Elections and Democracy; Election Commission- Composition and Functions; Challenges to Free and Fair Elections; Electoral Reforms. (8 Periods)
  2. Legislature: Parliament- Composition and Functions; State Legislatures (Odisha Vidhan Sabha) Composition and Functions. (8 Periods)

UNIT IV Constitution at Work-II

  1. Executive: President- Powers & Position; Prime Minister Functions & Role; Governor- Powers and Position; Chief MinsterFunctions & Role. (8 Periods)
  2. Judiciary: Structure of Judiciary; Supreme Court; High Court; Judicial Review; Judicial Activism. (10 Periods)

BOOK PRESCRIBED:
Bureau’s Higher Secondary (+2) Political Science, Paper-I (English & Odia) Published by Odisha State Bureau of Textbook Preparation & Production, Bhubaneswar.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(b)

Solve graphically
Question 1.
x < y
Solution:
x < y
Step – 1: Let us draw the dotted line x = y

X 0 1
y 0 1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b)
Step – 2: Let us take a point say (1, 0) which is not on the line. Putting x = 1, y = 0 in the equation we get 1 < 0 (false).
⇒ (1, 0) does not satisfy the inequality.
⇒ The solution is the half-plane that does not contain (1, 0)
Step – 3: The shaded region is the solution region.

<img src="https://bseodisha.guru/wp-content/uploads/2022/10/BSE-Odisha.png" alt="BSE Odisha" width="130" height="16" />

Question 2.
3x + 4y ≥ 12
Solution:
3x + 4y ≥ 12
Step – 1: Let us draw the line 3x + 4y = 12

X 4 0
y 0 3

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 1
Step – 2: Let us consider the point (0, 0) which is not on the line. Putting x = 0, y = 0 in the inequality we have 0 ≥ 12 (false).
∴ (0, 0) does not satisfy the inequality.
⇒ The half-plane that does not contain (0, 0) is the solution region.
Step – 3: The shaded region is the solution region.

Question 3.
x – y > 0
Solution:
x – y > 0
Step – 1: Let us draw the dotted line x = y.

X 0 1
y 0 1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 2
Step – 2: Let us consider (1, 0) which is not on the line.
Putting x = 1, y = 0 in the inequation we get 1 > 0 (True)
⇒ (1, 0) satisfies the inequation.
⇒ The half-plane containing (1, 0) is the solution region.

Question 4.
x + 2y – 5 ≤ 0
Solution:
x + 2y – 5 ≤ 0
Step – 1: Let us draw the line x + 2y – 5 = 0
⇒ y = \(\frac{5-x}{2}\)

X 5 1
y 0 2

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 3
Step – 2: Let us consider the point (0, 0) which does not lie on the line putting x = 0, y = 0 in the inequation we get – 5 < 0 (True).
⇒ The point satisfies the inequation.
⇒ The half-plane containing (0, 0) is the solution region.
Step – 3: The shaded region is the solution region.

<img src="https://bseodisha.guru/wp-content/uploads/2022/10/BSE-Odisha.png" alt="BSE Odisha" width="130" height="16" />

Question 5.
7x – 4y < 14
Solution:
7x – 4y < 14
Step – 1: Let us draw the dotted line 7x – 4y = 14

X 2 6
y 0 7

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 4
Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequality we get 0 < 14 (True).
⇒ (0, 0) satisfies the inequation.
⇒ The half-plane including (0, 0) is the solution region.
Step – 3: The solution region is the shaded region.

Question 6.
x + 8y + 10 > 0
Solution:
x + 8y + 10 > 0
Step – 1: Let us draw the dotted line x + 8y + 10 = 0

X -10 -2
y 0 -1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 5
Step – 2: Let us consider the point (0, 0) which does not lie on the line. Putting x = 0, y = 0 in the inequation we get 10 > 0 (True)
⇒ (0, 0) satisfies the inequality.
⇒ The half-plane containing origin is the solution region.
Step – 3: The solution region’ is the shaded region.

Question 7.
5x + 6y < 12
Solution:
5x + 6y < 12
Step – 1: Let us draw the dotted line 5x + 6y = 12

x -6 0 6
y 7 2 -3

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 6
Step – 2: Putting x = 0, y = 0 in the equation we get, 0 < 12 (True)
⇒ The (0, 0) satisfies the equation.
Step – 3: The shaded region is the required solution.

<img src="https://bseodisha.guru/wp-content/uploads/2022/10/BSE-Odisha.png" alt="BSE Odisha" width="130" height="16" />

Question 8.
– 3x + y > 0
Solution:
Step – 1: Let us draw the dotted line – 3x + y = 0

x 0 1 -1
y 0 3 -3

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 7
Step – 2: Putting x – 1, y = 0 in the equation we get, – 3 > 0 (false)
∴ Point (1, 0) does not satisfy the in the equation.
Step – 3: The shaded half-plane is the solution.

Question 9.
3x + 8y > 24
Solution:
Step- 1: Let us draw the dotted graph of 3x + 8y = 24

x 8 0
y 0 3

Step- 2: Putting x = 0, y- 0 we get, 0 > 24 (false)
∴ 0, (0, 0) does not satisfy the in equality.
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 8

Question 10.
x + y > 1
Solution:
Step – 1: Let us draw the graph x + y = 1

x 1 0
y 0 1

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 9
Step – 2: Putting x = 0, y = 0 in the equation we get, 0 > 1 (false)
∴ 0 (0, 0) does not satisfy the in equation
Step – 3: The shaded region is the solution.

Question 11.
x ≤ 0
Solution:
Step – 1: Let us draw the graph x = 0
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 10
Step – 2: Putting x = – 1 we get, – 1 ≤ 0 (True)
Thus, the shaded region is the solution.

<img src="https://bseodisha.guru/wp-content/uploads/2022/10/BSE-Odisha.png" alt="BSE Odisha" width="130" height="16" />

Question 12.
y > 5
Solution:
Step – 1: Let us draw the dotted graph of y = 5

x 0 1 -1
y 5 5 5

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(b) 11
Step- 2: Putting x = 0, y = 0 we have 0 > 5 (false)
we 0(0, 0) does not satisfy. the inequality.
Step – 3: The shaded region is the solution.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(a)

Question 1.
Determine whether the solution set is finite or infinite or empty:
(i) x < 1000, x ∈ N
Solution:
Finite

(ii) x < 1, x ∈ Z (set of integers)
Solution:
Infinite

(iii) x < 2, x is a positive integer.
Solution:
Finite

(iv) x < 1, x is a positive integer.
Solution:
Empty

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a)

Question 2.
Solve as directed:
(i) 5x ≤ 20 in positive integers, in integers.
Solution:
5x ≤ 20
⇒ \(\frac{5 x}{5} \leq \frac{20}{5}\)
⇒ x ≤ 4
If x is a positive integer, then the solution set is {1, 2, 3, 4}
If x is an integer, then the solution set is:
S = {x : x ∈ Z and x ≤ 4}
= { ….. -3, -2, -1, 0, 1, 2, 3, 4}

(ii) 2x + 3 > 15 in integers, in natural numbers.
Do you mark any difference in the solution sets?
Solution:
2x + 3 > 15
⇒ 2x + 3 – 3 > 15 – 3
⇒ 2x > 12
⇒ \(\frac{2 x}{2}>\frac{12}{2}\)
⇒ x > 6
If x ∈ Z, then the solution set is S = (x : x ∈ Z and x > 6}
= {7, 8, 9…… }
If x ∈ N. then the solution set is S = {x : x ∈ N and x > 6}
= {7, 8, 9…… }
Two solution sets are the same.

(iii) 5x + 7 < 32 in integers, in non-negative integers.
Solution:
5x + 7 < 32
⇒ 5x + 7 – 7 < 32 – 7
⇒ 5x < 25
⇒ \(\frac{5 x}{5}<\frac{25}{5}\)
⇒ x < 5
If x ∈ Z, then the solution set is S = { x : x ∈ Z and x < 5 }
= {…..-3, -2, -1, 0, 1, 2, 3, 4}
If x is a non-negative solution then the solution set is S = {x : x is a non-negative integer < 5}
= (0, 1,2, 3,4}

(iv) -3x – 8 > 19, in integers, in real numbers.
Solution:
– 3x – 8 > 19
⇒ – 3 x – 8 + 8 > 19 + 8
⇒ – 3x > 27
⇒ \(\frac{-3 x}{-3}<\frac{27}{-3}\)
⇒ x < – 9
If x ∈ Z, then the solution set is S = (x : x ∈ Z and x < – 9}
= { ……..- 11, – 10}
If x ∈ R then the solution set is S = {x : x ∈ R and x < – 9}
= (∞, – 9)

(v) |x – 3| < 11, in N and in R.
Solution:
|x – 3| < 11
⇒ – 1 < x – 3 < 11
⇒ – 11 + 3 < x – 3 + 3 < 11+3
⇒ – 8 < x < 14
If x ∈ N the solution set is S = {1, 2, 3, 4, 5……..12, 13}
If x ∈ R then the solution set is: S = {x : x ∈ R and – 8 < x < 14}
= (- 8, 14)

Question 3.
Solve as directed:
(i) 2x + 3 > x – 7 in R
Solution:
2x + 3 > x – 7
⇒ 2x – x > – 7 – 3
⇒  x > – 10
x ∈ R, the solution set is S = (x : x ∈ R and x > – 10} = (-10, ∞)

(ii) \(\frac{x}{2}+\frac{7}{3}\) <  3x – 1 in R
Solution:
\(\frac{x}{2}+\frac{7}{3}\) <  3x – 1
\(\frac{3 x+14}{6}\) <  3x – 1
⇒ 3x + 14 < 18x – 6
⇒ 3x – 18x < – 6 – 14
⇒ – 15x < – 20
⇒ \(\frac{-15 x}{-15}>\frac{-20}{-15}\)
⇒ x > \(\frac{4}{3}\)
If x ∈ R, the solution set is S = \(\left(\frac{4}{3}, \infty\right)=\left\{x: x \in R \text { and } x>\frac{4}{3}\right\}\)

(iii) \(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\) for non-negative real numbers.
Solution:
\(\frac{x}{2}-\frac{x}{3}+\frac{x}{5} \leq \frac{11}{3}\)
⇒ \(\frac{15 x-10 x+6 x}{30}\) ≤ \(\frac{11}{3}\)
⇒ 11x ≤ \(\frac{11}{3}\) × 30
⇒ 11x ≤ 110
⇒ x ≤ 10
If x is a non-negative real number then the solution set is S = {x : x ∈ R and 0 ≤ x ≤ 10}
= {0, 10}

(iv) 2(3x – 1) < 7x + 1 < 3 (2x + 1) for real values.
Solution:
2(3x – 1) < 7x + 1 < 3(2x + 1)
⇒ 6x – 2 < 7x + 1< 6x + 3
⇒ – 2 < x + 1 < 3
⇒ – 3 < x < 2
If x ∈ R, the solution set is S = (x : x ∈ R and -3 < x < 2}
= {-3, 2}

(v) 7(x – 3) ≤ 4 (x + 6), for non-negative integral values.
Solution:
7(x – 3) ≤ 4(x + 6)
⇒ 7x – 21 ≤ 4x + 24
⇒ 7x – 4x ≤ 24 + 21
⇒ 3x ≤ 45
⇒ x ≤ 9
If x is a non-negative integer the solution set is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(vi) Convert to linear inequality and solve for natural numbers: (x – 2) (x – 3) < (x + 3) (x – 1)
Solution:
(x – 2) (x – 3) < (x + 3) (x – 1)
⇒ x2 – 5x + 6  <  x2 + 2x – 3
⇒ – 5x + 6 < 2x – 3
⇒ – 5x – 2x < – 3 – 6
⇒ – 7x < – 9
⇒ x > \(\frac{9}{7}\)
If x ∈ N, the solution set is S = {2, 3, 4 }

(vii) Solve in R, \(\frac{x}{2}\) + 1 ≤ 2x – 5 < x. Also, find its solution in N.
Solution:
\(\frac{x}{2}\) + 1 ≤ 2x – 5 < x
⇒ \(\frac{x}{2}\) +1 ≤ 2x – 5 and 2x – 5 < x
⇒ \(\frac{x}{2}\) – 2x ≤ – 5 – 1 and x < 5
⇒ \(\frac{-3x}{2}\) ≤ – 6 and x < 5
⇒ – 3x ≤ – 12 and x < 5
⇒ x ≥ 4 and x < 5
⇒ 4 ≤ x < 5
If x ∈ R, the solution set is S = {x : x ∈ R and 4 < x < 5}
= {4, 5}
If x ∈ N, the solution set is S = { 4 }

(viii) Solve in R and also in Z: \(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)
Solution:
\(\frac{3 x+1}{5} \geq \frac{x+2}{3}-\frac{5-3 x}{5}\)
⇒ \(\frac{3 x+1}{5} \geq \frac{5 x+10-15+9 x}{15}\)
⇒ 3x + 1 ≥ \(\frac{14 x-5}{3}\)
⇒ 9x + 3 ≥ 14x – 5
⇒ 9x – 14x ≥ – 5 – 3
⇒ – 5x ≥ – 8
⇒ x ≤ \(\frac{8}{5}\)
If x ∈ R, then the solution set is S = (x : x ∈ R and x ≤ \(\frac{8}{5}\)}
= (- ∞, \(\frac{8}{5}\))
If x ∈ Z, then the solution set is S = { x : x ∈ Z and x ≤ \(\frac{8}{5}\)}
= {……. -3, -2, -1, 0, 1}

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a)

Question 4.
Solve |x – 1| >1 and represent the solution on the number line.
[Exhaustive hints: By definition of modulus function
For x – 1 ≥ 0 or x ≥ 1, |x – 1| > 1
⇔ x – 1 > 1 ⇔ x > 2 ⇔ x ∈ (2, ∞)
For x- 1 < 0 or x < 1, |x – 1| > 1
⇔ – (x – 1) > 1
⇔ x – 1 < -1 (multiplication by -1 reverses the inequality)
⇔ x < 0 ⇔ x ∈ ( -∞, 0)
∴ The solution set is the Union,
(-∞, 0) ∪ (2, ∞) Show this as two disjoint open intervals on the number line, i.e., real line.]
Solution:
|x – 1| > 1
⇒ – 1 > x – 1 > 1
⇒ 0 > x > 2
⇒ x < 0 and x > 2
∴ The solution set is S = {x : x ∈ R, x < 0 and x > 2}
= (-∞, 0) ∪ (2, ∞)
We can show this solution in the number line as
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a)

Question 5.
Solve in R and represent the solution on the number line.
(i) |x – 5| < 1
Solution:
|x – 5| < 1
⇒ – 1< x – 5 < 1
⇒ 4 < x < 6
If x ∈ R, then the solution set is S = (4, 6)
We can represent the solution on the number line as
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a) 1

(ii) \(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)
Solution:
\(\frac{x}{5}<\frac{2 x+1}{3}+\frac{1-3 x}{6}\)
⇒ \(\frac{x}{5}<\frac{4 x+2+1-3 x}{6}\)
⇒ \(\frac{x}{5}<\frac{x+3}{6}\)
⇒ 6x < 5x + 15
⇒ x < 15
If x ∈ R, the solution set is S = (-∞, 5)
We can represent the solution on the number line as
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a) 2

(iii) 2x + 1 ≥ 0
Solution:
2x + 1 ≥ 0
⇒ 2x ≥ -1
⇒ x ≥ -1/2
If x ∈ R, then the solution set is S = [\(-\frac{1}{2}\), ∞]
We can represent the solution on the number line as
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a) 3

(iv) \(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)
Solution:
\(\frac{x-1}{2} \leq \frac{x+1}{3}<\frac{3 x-1}{6}\)
⇒ 3x – 3 ≤ 2x + 2 < 3x – 1
⇒ 3x – 3 ≤ 2x + 2 and 2x + 2 < 3x – 1
⇒ x ≤ 5 and – x < – 3
⇒ x ≤ 5 and x > 3
⇒ 3 < x ≤ 5
If x ∈ R, the solution set is S = {3, 5}
We can represent the solution on the number line as
CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a) 4

Question 6.
In a triangle, ABC; AB, BC, and CA are x, 3x + 2, and x + 4 units respectively where x ∈ N. Find the length of its sides. (Hint: Apply triangle-inequality).
Solution:
Given AB = x
BC = 3x + 2
and CA = x + 4
Now AB + AC > BC (Triangle inequality)
⇒ x + x + 4 > 3x + 2
⇒ 2x + 4 > 3x + 2
⇒ – x > – 2
⇒ x < 2
As x ∈ N we have x = 1
The sides of triangle ABC are
AB = 1 unit
BC = 5 units
and CA = 5 units

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Ex 7(a)

Question 7.
The length of one side of a parallelogram is 1 cm. shorter than that of its adjacent side. If its perimeter is at least 26 c.m., find the minimum possible lengths of its sides.
Solution:
Let the longer side = x cm
∴ The smaller side = (x – 1) cm
Perimeter = 2(x + (x – 1)) = 4x – 2 cm
According to the question
4x – 2 ≥ 26
⇒ 4x ≥ 28
⇒ x > 7
The minimum value of x = 7.
∴ The minimum length of the sides is 7cm and 6 cm.

Question 8.
The length of the largest side of a quadrilateral is three times that of its smallest side. Out of the other two sides, the length of one is twice that of the smallest and the other is 1 cm. longer than the smallest. If the perimeter of the quadrilateral is at most 36 c.m., then find the maximum possible lengths of its sides.
Solution:
Let the smallest side = x cm.
Largest side = 3 times x = 3x cm.
The other two sides are 2x cm and x + 1 cm.
⇒ The perimeter = x + 3x + 2x + x + 1
= 7x + 1 cm
According to the question:
7x + 1 ≤ 36
⇒ 7x ≤ 35
⇒ x ≤ 5
Maximum value of x = 5
∴ The maximum possible length of sides are x = 5 cm, 3x = 15 cm, 2x = 10 cm, and x + 1 = 6 cm.

Question 9.
Find all pairs of consecutive odd numbers each greater than 20, such that their sum is less than 60.
Solution:
Let two consecutive odd numbers are
2n – 1 and 2n + 1
Now 2n – 1 > 20 and 2n + 1 > 20
But their sum = 2n – 1 + 2n + 1
= 4n < 60
⇒ n < 15
for n = 14 two numbers are 27, 29
for n = 13 two numbers are 25, 27
for n = 12 two numbers are 23, 25
for n = 11 two numbers are 21, 23
∴ All pairs are 21, 23; 23, 25; 25, 27 and 27, 29

Question 10.
Find all pairs of even numbers each less than 35, such that their sum is at least 50.
Solution:
Let two even numbers be x and y.
According to the question
x < 35, y < 35 and x + y ≥ 50
⇒ x ≤ 34, y ≤ 34 and x + y ≥ 50
⇒ x + y ≤ 70, x + y ≥ 50
⇒ 50 ≤ x + y ≤ 70
If x + y = 50 the numbers are {34, 16}, {32, 18}, {30, 20}, {28, 22}, {26, 24}
If x + y = 52 the numbers are {34, 18}, {32, 20}, {30, 22}, {28, 24}, {26, 26}
If x + y = 34 the numbers are {34, 20}, {32, 22}, {30, 24}
If x + y = 56 the numbers are {34, 22}, {32, 24}, {30, 26}, {28, 28}
If x + y = 58 the numbers are {34, 24}, {32, 26}, {30, 28}
If x + y = 60 the numbers are {34, 26}, {32, 28}, {30, 30}
If x + y = 62 the numbers are {34, 28}, {32, 30}
If x + y = 64 the numbers are {34, 30}, {32, 32}
If x + y = 68 the numbers are {34, 34}

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(a)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(a)

Question 1.
Compute the product A × B when
(i) A = {0} = B
(ii) A = {a, b}, B = {a, b, c}
(iii) A = Z, B = Φ
Solution:
(i) A = {0} = B
∴ A × B = {(0, 0)}

(ii) A = {a, b}, B = {a, b, c}
∴ A × B = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c)}

(iii) A = Z, B = Φ ∴ AxB = Φ

Question 2.
If |A| = m, |B| = n, what can you say about
(i) |A × B| (ii) |P(A) × P(B)|
Solution:
If |A| = m. |B| = n then

(i) lA × B| = mn.

(ii) |P(A)| = 2m . |P(B)| = 2n
∴  |P(A) × P(B)| =2m × 2n = 2m+n

Question 3.
Find x, y if
(i) (x, y) = (-3, 2)
(ii) {x + y, 1) = (1, x – y)
(iii) (2x + y, 1) = (x, 2x + 3y)
Solution:

(i) ∴ x = – 3, y = 2

(ii) ∴ x + y = 1, x – y = 1
∴ 2x = 2 or, x = 1
∴ y=0

(iii) ∴ 2x + y = x, 1 = 2x + 3y
∴ {x + y = 0} × 2
2x + 3y = 1
–      –      –
∴ – y = – 1 or, y = 1
∴ x = – 1

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(a)

Question 4.
If, A × B = B × A then what can you
Solution:
If A × B = B × A then A = B

Question 5.
|A × B| = 6. If ( -1, y ), (1, x), (0, y) are in A × B. Write other elements in A × B, where x ≠ y.
Solution:
Let |A × B| = 6 and (-1, y) (1, .x) (0, y) ∈ A × B
⇒ -1, 1, 0 ∈ A and x, y ∈ B
As |A × B| = 6 and 3 × 2 = 6
We have A = {-1, 1, 0} and B = {x, y}
Thus other elements of A × B is (-1, x) , (1, y), (0, x)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(d)

Question 1.
Fill in the blanks choosing the correct answer from the brackets.

(i) In Δ ABC, b =____________. (b cos B + c cos C, a cos A + c cos C, c cos A + a cos C)
Solution:
c cos A + a cos C

(ii) If a cot A = b cot B then Δ ABC is__________. (isosceles, right-angled, equilateral)
Solution:
isosceles

(iii) In Δ ABC if b sin C = c sin B = 2 then b sin C = ___________. (0, 1, 2)
Solution:
1

(iv) In Δ ABC if \(\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}=\frac{\cos \mathrm{C}}{c}\) then the tringle is_________ (equilateral, isosceles, scalene)
Solution:
equilateral

(v) If sin A = sin B and b = 1/2, then a = _______________. (2, 1/2, 1)
Solution:
a = 1/2

(vi) In Δ ABC if A = 60°, B = 45° a : b = __________. ( √2 : √3, √6 : 2, √3 : 2)
Solution:
√6 : 2

(vii) In Δ ABC if b2 + c2 < a2 , then _________ angle is obtuse. (A, B, C)
Solution:
A

(viii) If a cos B = b cos A. then cos B = _____________. \(\left(\frac{c}{a}, \frac{a}{2 c}, \frac{c}{2 a}\right)\)
Solution:
cos B = \(\frac{c}{2 a}\)

(ix) If a – b cos C, then __________ angle is a right angle. (A, B, C)
Solution:
∠B is a right angle

(x) If a = 12, b = 7, C = 30°, then Δ = ______________. (42, 84, 21)
Solution:
Δ = 21

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Question 2.
Prove that
(i) a sin A – b sin B = c sin (A – B)
Solution:
R.H.S. = c sin (A – B)
= 2R sin C sin (A – B)
= 2R sin (A + B) sin (A – B)
[∴ A + B + C = π or, A + B = π – C
or sin (A + B) = sin (π – C) sin C]
= 2R (sin2 A – sin2  B)
= 2R sin A sin A – 2R sin B sin B
= a sin A – b sin B = L.H.S.

(ii) b cos B + c cos C = a cos (B – C)
Solution:
R.H.S. = a cos (B – C)
= 2R sin A cos (B – C)
= 2R sin (B + C) cos (B – C)
= R sin (B + C + B – C) + sin (B + C – B + C)}
= R [(sin 2B + sin 2C)
= R(2sin B cos B + 2 sin C cos C)
= 2R sin B cos B + 2R sin C cos C
= b cos B + c cos C = L.H.S.

(iii) If (a + b + c)(b + c – a) = 3bc, then A = 60°
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

(iv) If \(\frac{b+c}{5}=\frac{c+a}{6}=\frac{a+b}{7}\) then sin A : sin B : sin C = 4 : 3 : 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 1

(v) If A: B: C = 1 : 2 : 3 then sin A: sin B: sin C = 1 : 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 2

(vi) If b2 + c2 – a2 = bc, then A = 60°
Solution:
If b2 + c2 – a2 = bc, then A = 60°
or, \(\frac{b^2+c^2-a^2}{2 b c}\) = 1/2 or, cos A = 1/2
or, A = 60°

(vii) If A : B: C = 1 : 2 : 7, then c: a = (√5 + 1) : (√5 – 1)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 3
But we know that \(\frac{\sin C}{\sin A}=\frac{c}{a}\)
∴ \(\frac{c}{a}=\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Question 3.
(i) If cos A = \(\frac{12}{13}\), cos B = \(\frac{5}{13}\) then find a : b.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 4

(ii) If a = 7, b = 3, c = 5 then find A.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 5

(iii) If a = 8, b = 6, c = 4 find tan \(\frac{B}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 6

(iv) If \(\frac{a}{\sec A}=\frac{b}{\sec B}\) and a ≠ b then find C.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 7

(v) If a = 48, b = 35, ∠C = 60° then find c.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 8

In Δ ABC prove that (Q. 4 to Q. 26)

Question 4.
a sin (B – C) + b sin (C – A) + c sin (A – B) = 0
Solution:
a sin (B – C) + b sin (C – A) + c sin (A – B)
= 2R sin A sin (B – C) + 2R sin B sin (C – A) + 2R sin C sin (A – B)
= 2R [sin (B + C) sin (B – C) + sin (C + A) sin (C – A) + sin (A + B) sin (A – B)]
= 2R[sin2 B – sin2 C + sin2 C – sin2 A + sin2 A – sin2 A]
= 2R x 0 = 0

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Question 5.
\(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b \cos C-c \cos B}{b \cos C+c \cos B}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 9

Question 6.
\(\sum \frac{a^2 \sin (B-C)}{\sin (B+C)}=0\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 10
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 11

Question 7.
a2(cos2 B – cos2 C) + b2(cos2 C – cos2 A) + c2(cos2 A – cos2 B) = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 12

Question 8.
\(\frac{b^2-c^2}{a^2} \sin 2 A+\frac{c^2-a^2}{b^2} \sin 2 B\) \(+\frac{a^2-b^2}{c^2} \sin 2 \mathrm{C}=0\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 13
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 14

Question 9.
\(\frac{a^2\left(b^2+c^2-a^2\right)}{\sin 2 \mathrm{~A}}=\frac{b^2\left(c^2+a^2-b^2\right)}{\sin 2 \mathrm{~B}}\) \(=\frac{c^2\left(a^2+b^2-c^2\right)}{\sin 2 C}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 15

Question 10.
\(\Sigma \frac{\cos A}{\sin B \cdot \sin C}=2\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 16
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 17

Question 11.
(a2 – b2 + c2) tan B = (a2 + b2 – c2) tan C
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 18

Question 12.
(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 19
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 20

Question 13.
\(\frac{b+c}{a}=\frac{\cos \mathbf{B}+\cos C}{1-\cos A}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 21
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 22

Question 14.
\(\sum a^3 \sin (B-C)=0\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 23

Question 15.
(b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c
Solution:
L.H.S. = (b + c) cos A + (c + a) cos B + (a + b) cos C
= b cos A + c cos A + a cos B + c cos B + a cos C + b cos C
= (b cos C+ c cos B) +(c cos A + a cos C) + (a cos B + b cos A)
= a + b + c = R.H.S.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Question 16.
2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2
Solution:
2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2
\(=2\left(b c \times \frac{\left(b^2+c^2-a^2\right)}{2 b c}+c a \times \frac{c^2+a^2-b^2}{2 c a}\right.\) \(\left.+a b \times \frac{a^2+b^2-c^2}{2 a b}\right)\)
= b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2– c2
= a2 + b2 + c2

Question 17.
a (b2 + c2) cos A + b (c2 + a2) cos B + c(a2 + b2) cos C = 3 abc.
Solution:
a (b2 + c2) cos A + b (c2 + a2) cos B + c(a2 + b2) cos C
= ab2 cos A + ac2 cos A + bc2 cos B + ba2 cos B + ca2 cos C + cb2 cos C
= ab2 cos A + ba2 cos B + ac2 cos A + ca2 cos C + bc2 cos B + cb2 cos C
= ab (b cos A + a cos B) + ac (c cos A + a cos C) bc (c cos B + b cos C)
= abc = abc + abc = 3abc

Question 18.
a3 cos (B – C) + b3 cos (C – A) + c3 cos (A – B) = 3 abc
Solution:
1st term of L.H.S. = a3 cos (B – C)
= a2 a cos (B – C)
= a2 . 2R sin A cos (B – C)
= 2a2R sin (B + C) cos (B- C)
= a2R [sin (B + C + B – C) + sin (B + C – B + C)]
= a2 R (sin 2B + sin 2C)
= a2R [2 sin B cos B + 2 sin C cos C]
= a2 [2R sin B cos B + 2R sin C cos C]
= a2 (b cos B + c cos C)
Similarly, 2nd term
= b2 (c cos C + a cos A) and
3rd term = c2 (a cos A + b cos B)
∴ L.H.S.= a2b cos B+a2c cos C+b2c cos C + b2a cos A + c2a cos A + c2b cos B
= ab (a cos B + b cos A) + bc (b cos C + c cos B) + ca (c cos A + a cos C)
= abc + bca + cab = 3abc = R.H.S.

Question 19.
a (cos B + cos C) = 2(b + c) sin2 \(\frac{A}{2}\)
Solution:
Refer to Q. N. 13.

Question 20.
(b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan \(\frac{B}{2}\) = (a + b – c) tan \(\frac{C}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 24
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 25

Question 21.
\((b+c-a)\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)\) \(=2 a \cot \frac{A}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 26

Question 22.
(a – b)2 cos2 \(\frac{C}{2}\) + (a + b)2 sin2 \(\frac{C}{2}\) = c2
Solution:
L.H.S = (a – b)2 cos2 \(\frac{C}{2}\) + (a + b)2 sin2 \(\frac{C}{2}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 27

Question 23.
1 – tan \(\frac{A}{2}\) tan \(\frac{B}{2}\) = \(\frac{c}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 28

Question 24.
(b – c) cot \(\frac{A}{2}\) + (c – a) cot \(\frac{B}{2}\) + (a – b) cot \(\frac{C}{2}\) = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 29
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 30

Question 25.
cot A + cot B + cot C = \(\frac{a^2+b^2+c^2}{4 \Delta}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 31

Question 26.
a2 cot A + b2 cot B + c2 cot C = 4Δ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 32
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 33

Question 27.
If \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\) then prove C = 60°.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 34
or, cos C = 1/2 or, ∠C = 60°

Question 28.
If a = 2b and A = 3B, find the measures of the angles of the triangle.
Solution:
If a = 2b and A = 3B, we have \(\frac{a}{b}\) = 2
or, \(\frac{\sin A}{\sin B}\) = 2
or, sin A = 2 sin B         …..(1)
Also  sin A = sin 3B (as a = 3B)    …..(2)
∴ From (1) and (2), we have
2 sin B = sin 3B = 3 sin B – 4 sin3 B
or, 4 sin3 B – sin B = 0
or, sin B(4 sin2 B – 1) = 0
or, sin B = 0, 4 sin2 B = 1
Now sin B = 0 ⇒ B = 0 (Impossible)
∴ sin2 B = \(\frac{1}{2}\) or, sin B = ± \(\frac{1}{2}\)
If sin B = \(\frac{1}{2}\) then ∠B = 30°
∴ A = 3B = 3 x 30° = 90°
∠C = 60°

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Question 29.
If a4 + b4 + c4 = 2c2 (A2 + b2), prove that m∠ACB = 45° or 135°.
Solution:
a4 + b4 + c4 = 2c2 (a2 + b2)
or, a4 + b4 + c4 + 2a2b2 – 2b2c2 – 2c2a2 = 2a2b2
or, (a2 + b2 – c2)2 = 2a2b2
or, a2 + b2 – c2 = ± √2 ab
or, \(\frac{a^2+b^2-c^2}{2 a b}=\pm \frac{1}{\sqrt{2}}\)
or, cos C = ± \(\frac{1}{\sqrt{2}}\)
∴ ∠C = 45° or 135°.

Question 30.
If x2 + x + 1, 2x + 1, and x2 – 1 are lengths of sides of a triangle, then prove that the measure of the greatest angle is 120°.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 35

Question 31.
if cos B = \(\frac{\sin A}{2 \sin C}\), prove that the triangle is isosceles.
Solution:
cos B = \(\frac{\sin A}{2 \sin C}\)
⇒ \(\frac{c^2+a^2-b^2}{2 c a}=\frac{a}{2 c}\) ⇒ c2 + a2 – b2 = a2
or, c2 = b2 or, c = b
∴ The triangle is isosceles.

Question 32.
If a tan A + b tan B = (a + b)tan \(\frac{1}{2}\) (A + B) prove that the triangle is isosceles.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 36
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 37
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 38

Question 33.
If (cos A + 2 cos C) : (cos A + 2 cos B) = sin B : sin C prove that the triangles are either isosceles or right-angled.
Solution:
\(\frac{\cos A+2 \cos C}{\cos A+2 \cos B}=\frac{\sin B}{\sin C}\)
⇒ cos A sin C = cos A sin B + 2 cos B sin B
⇒ cos A (sin B – sin C) + (sin 2B – sin 2c) = 0
⇒ cos A (sin B – sin C) + 2 cos (B + C) sin (B – C) = 0
⇒ cos A (sin B – sin C) – 2 cos A sin (B – C) = 0
(∴ cos (B + C) = cos (π – A) = – cos A)
⇒ cos A = 0 or sin B – sin C – 2 sin (B – C) = 0
cos A = 0 ⇒ A = 90°
i.e. the triangle is right-angled. sin B sin C – 2 sin (B – C) = 0
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 39

Question 34.
If cos A = sin B – cos C, prove that the triangle is right-angled.
Solution:
cos A = sin B – cos C
or, cos C + cos A = sin B
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 40
or, 2C = π
or, C = \(\frac{\pi}{2}\)

Question 35.
If a2, b2, c be in A.P., prove that cot A, cot B, cot C are also in A.P.
Solution:
If a2, b2, c be in A.P.
then b2 – a2 = c2 – b2
or, 2b2 = c2 + a2
or, \(b^2=\frac{c^2+a^2}{2}\)
or, 2b2 = c2 + a2 …..(1)
We have to prove that cot A, cot B, cot C are in A.P.
i.e. to prove cot B – cot A = cot C – cot B
i.e. 2 Cot B = cot C + cot A
∴ R.H.S. = cot C + cot A
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 41

Question 36.
If sin A: sin C = sin (A – B) : sin (B – C) prove that a2, b2, c2 are in A.P.
Solution:
\(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
or, sin A sin (B – C) = sin C sin (A – B)
or, sin (B + C) sin (B – C) = sin (A + B) sin (A – B)
or, sin2 B – sin2 C = sin2 A – sin2 B
or, 2 sin2 B = sin2 C + sin2 A
or, \(2 \frac{b^2}{4 \mathrm{R}^2}=\frac{c^2}{4 \mathrm{R}^2}+\frac{a^2}{4 \mathrm{R}^2}\)
or, 2b2 = c2 + a2
or, b2 – a2 = c2 – b2
∴ a2, b2, c2 are in A.P

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d)

Question 37.
If the side lengths a, b, and c are in A.P., then prove that cos \(\frac{1}{2}\) (A – C) = 2 sin \(\frac{1}{2}\) B.
Solution:
If a,b, and c are in A.P. then b – a – c – b or, 2b = c + a
We have to prove that
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 42
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 43

Question 38.
If the side lengths a, b, and c are in A.P., prove that cot \(\frac{1}{2}\) A, cot \(\frac{1}{2}\) B, cot \(\frac{1}{2}\) C are in A.P.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(d) 44

CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Ex 2(b)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 2 Sets Ex 2(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Exercise 2(b)

Question 1.
An examination was conducted in physics, chemistry, and mathematics. If P.C.M. denotes respectively the sets of students who passed in Physics, Chemistry, and Mathematics, express the following sets using union, intersection, and different symbols.
(a) Set of candidates who passed in Mathematics and Chemistry, but not in Physics.
(b) Set of candidates who passed in all three subjects.
(c) Set of candidates who passed in Mathematics only.
(d) Set of candidates who failed in Mathematics, but passed in at least one subject.
(e) Set of candidates who passed in at least two subjects.
(f) Set of candidates who failed in one subject only.
Solution:
An examination was conducted in Physics, Chemistry, and Mathematics. P, C, and M denoted the set of students who passed Physics, Chemistry, and Mathematics, respectively. Then.
(a) Set of candidates who passed in Mathematics and Chemistry, but not in Physics (M ∩ C) – P.
(b) Set of candidates who passed in all three subjects M ∩ C ∩ P.
(c) Set of candidates who passed in Mathematics only M – C – P.
(d) Set of candidates who failed in Mathematics, but passed in at least one subject (P ∪ C) – M.
(e) Set of candidates who passed in at least two subjects.
(f) Set of candidates who failed in one subject only.
(P ∩ C – M) ∪ (P ∩ M – C) ∪ (M ∩ C – P)

Question 2.
What can you say about sets A and B if
(i) A ∪ B= Φ
(ii) A Δ B = Φ
(iii) A \ B = Φ
(iv) A \ B = A
(v) A ∩ B= U, Where U is the Universal set, A \ B = U?
Solution:
(i) if A ∪ B = Φ then A = Φ =B
(ii) A Δ B = Φ ⇒ A = B
(iii) A – B = Φ ⇒ A ⊆ B
(iv) A – B = A ⇒ B = Φ
(v) A ∩ B = U ⇒ A = B = U
(vi) A – B = U ⇒ A = U and B = Φ

Question 3.
Are differences and symmetric commutative? Give reason.
Solution:
The difference of the two sets is not commutative but the symmetric of the two sets is commutative.
Reason:
Let x ∈ A – B ⇔ x ∈ A ∧ x ∉ B
≠ x ∈ B ∧ x ∉ A ⇔ x ∈ b – A
A- B ≠ B – A
But if y ∈ A Δ ⇔ y ∈ (A-B) ∪ (B – A)
⇒ y ∈ (B – A) ∪ (A – B) ⇔ y ∈ B Δ A
∴ A Δ B = B Δ A.

CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Ex 2(b)

Question 4.
If B ⊂ C, prove that A/B = A/C. Is this result true when a difference is replaced by a symmetric difference? Give reason.
Solution:
If B ⊂ C, then x ∈ A ⇒ x ∈ C
Now x ∈ A – C ⇔ x: x ∈ A ∧ x ∉ C
⇔ {x: x ∈ A ∧ x ∉ B}
⇔ {x: x ∈ A – C}
∴ A – C = A – B
but, A Δ B ≠ A Δ C.

Question 5.
Prove the following :
(i) (A\B)\C = (A\C)\B = A\(B ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∪ B)
(iii) A Δ (B Δ C) = (A Δ B) Δ C
(iv) A ⊂ B ⇔ B’ ⊂ A’ A ⇔ A’ ∪ B = U
⇔ B’ ∩ A = Φ, where U is the universal set.
(v) A ∪ B = U and A ∩ B = Φ
⇒ B = A’
(iv) A ∪ B = A for all A ⇒ B = Φ
Solution:

(i) Let x ∈ (A – B) – C    ……(1)
⇔ x ∈ A- B ∧ x ∉ C
⇔ (x ∈ A ∧ x ∉ B) ∧ x ∉ C   ……(2)
⇔ (x ∈ A ∧ x ∉ C) ∧ x ∉ B
⇔ x ∈ A – C ∧ x ∉ B
⇔ x ∈ (A – C) – B   ……(3)
∴ from (2), we have
x ∈ A ∧ ∉ B ∧ x ∉ C
⇔ x ∈ A ∧ (x ∉ B ∧ x ∉ C)
⇔ x ∈ A ∧ x ∉ B ∪ C
[∴ ~ (p ∨ q) = ~ p ∧ ~ q]
⇔ x ∈ A – (B ∪ C)   …….(4)
∴ From (1), (3), and (4), we have
(A – B) – C = (A – C)-B = A – (B ∪ C)

(ii) Let x ∈ A ∩ (B ∪ C)
⇔ x ∈ A ∧ x ∈ B Δ C
⇔ x ∈ A ∧ (x ∈ B – C ∨ x ∉ C – B)
⇔ x ∈ A ∧ (x ∈ B ∧ x ∉ C ∨ x ∈ C ∧ x ∉ B)
⇔ x ∈ A ∧ (x ∈ B ∧ x ∈ A ∧ C) ∨ x ∈ A ∧ (x ∈ B ∧ x ∈ B ∧ x ∈ A ∧ x ∉ B)
⇔ (x ∈ A ∩ B ∧ x ∉ A ∩ C) ∨ (x ∈ A ∩ C  ∨ x ∉ A ∩ B)
⇔ x ∈ (A ∩ B) – (A ∩ C) ∨ x ∈ (A ∩ C) – (A ∩ B)
⇔ x ∈ (A ∩ B) Δ (A ∩ C)   ……(2)
∴ From (1) and (2), we have
A ∩ (B Δ C) = (A ∩ B) Δ (A ∩ C)

(iii) A Δ (B Δ C)
= A ∪ (B Δ C)- A ∩ (B Δ C)
= A Δ (B ∪ C)- A Δ (B ∩ C)
[ ∴ A ∪ (B Δ C) = A Δ (B ∪ C)]
and A n (B Δ C) = A Δ (B ∩ C)
= (A Δ B) ∪ C- (A Δ B) ∩ C
[∴ A Δ (B ∪ C) = (A Δ B) ∪ C and A Δ (B ∩ C) = (A Δ B) ∩ C
= (A Δ B) Δ C
∴ A Δ (B Δ C)= (A Δ B) Δ C
(Proved)

(iv) If A ⊂ B then x ∈ B’ or ⇒ x ∉ B
⇒ x ∉ B ⇒ x ∈ A’ (∴ A ⊂ B)
∴ B’ ⊂ A’
Again, let y ∈ A ⇒ y ∉ A’ ⇒ y ∈ B’
( B’ ⊂ A’)
⇒ y ∈ B ∴ A ⊂ B
∴ A ⊂ B ⇔ B’ ⊂ A’
∴ Again as A ⊂ B, we have
U = A ∪ B = B = U, where U is the universal set of A and B.
∴ A’= B – A ⇒ A’ ∪ B
= (B – A) ∪ B = B = U
∴ A ⊂ B ⇒ A’ ∪ B = U
Again A’ ∪ B = U
⇒ A ∩ (A’ ∪ B) = A ∩ U = A
⇒ (A ∩ A’) ∪ (A ∩ B) = A
⇒ Φ ∪ (A ∩ B) = A
⇒ A ∩ B = A ⇒ A ⊂ B
Lastly, B’ = U’ = Φ
∴ B’ ∩ A = Φ

(v) Let A ∪ B = U and A ∩ B = Φ
∴ Let x ∈ B ⇔ x ∉ B’ ⇔ x ∉ U – B
⇔ x ∉ A ⇔ x ∉ A’

(vi) As A ∪ B = A for all A
we have B ⊂ A for all A
∴ B ⊂ A even for A = Φ Thus B = Φ

Question 6.
Prove all the results of sections 1.13 and 1.14 that are started without proof.
Solution:
(i) A ∪ B = B ∪ A
Let x ∈ A ∪ B ⇔ x ∈ A ∨ x ∈ B
⇔ x ∈ B ∨ x ∈ B ⇔ x ∈ B ∪ A

(ii) A ∩ B = B ∩ A
Let x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B
∴ A ∩ B = B ∩ A

(iii) A ∩ (B ∪ C)
= (A ∩ B) ∪ (A ∩ C)
Let x ∈ A ∩ (B ∪ C)
⇔ x ∈ A ∧ x ∈ A ∪ C
⇔ x ∈ A ∧ (x ∈ A ∨ x ∈ C)
⇔ (x ∈ A ∩ B ∨ x ∈ A ∩ C)
⇔ (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈C)
⇔ x ∈ (A ∩ B) ∪ (A ∩ C)
∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Question 7.
Prove that
(i) \(\mathbf{A}-\bigcup_{i=1}^n \mathbf{B}_i=\bigcap_{i=1}^n\left(\mathbf{A}-\mathbf{B}_i\right)\)
Solution:
Let x ∈ \(A-\bigcup_{i=1}^n B_i \Rightarrow x \in A \wedge x \notin \bigcup_{i=1}^n B_i\)
⇔ x ∈ A  ∧ x ∉(B1 ∪ B2 ∪….∪ Bn )
⇔ x ∈ A  ∧ (x ∉ B1 ∧ x ∉ B2 ∧…..∧ x ∉ Bn )
⇔ (x ∈ A  ∧ x ∉ B1 ) ∧ (x ∈ A ∧ x ∉ B2 ) ∧….∧ (x ∈ A  ∧ x ∉ Bn )
⇔ x ∈ A – B1 ∧ x ∈ A – B2 ∧……..∧ x ∈ A – Bn
⇔ x ∈ (A – B1 ) ∩ (A – Bi ) ∩…..∩ (A – Bn )
⇔ \(x \in \bigcap_{i=1}^n\left(\mathrm{~A}-\mathrm{B}_{\mathrm{i}}\right)\)
∴ \(\mathrm{A}-\cup_{i=1}^n \mathrm{~B}_{\mathrm{i}}=\bigcap_{i=1}^n\left(\mathrm{~A}-\mathrm{B}_{\mathrm{i}}\right)\)

(ii) ∴ \(\mathbf{A}-\bigcap_{i=1}^n \mathbf{B}_i=\bigcup_{i=1}^n\left(\mathbf{A}-\mathbf{B}_i\right)\)
Solution:
Let x ∈ \(A-\bigcap_{i=1}^n B_i\)
⇔ x ∈ A ∧ x ∈ \(\bigcap_{i=1}^n \mathrm{~B}_{\mathrm{i}}\)
⇔ x ∈ A ∧ x ∉ (B1 ∩ B2 ∩….∩ Bn )
⇔ x ∈ A ∧ (x ∉ B1 ∨ x ∉ B2 ∨….∨ x ∉ Bn )
⇔ (x ∈ A ∧ x ∉ B1 ) ∨ (x ∈ A ∧ x ∉ B2 ) ∨….∨ (x ∈ A ∧ x ∉ Bn )
⇔ x ∈ A – B1 ∨ x ∈ A – B2 ∨……..∨ x ∈ A – Bn
⇔ x ∈ (A – B1 ) ∪ (A – B2 )……(A – Bn )
⇔ x ∈ \(\cup_{i=1}^n\left(\mathrm{~A}-\mathrm{B}_{\mathrm{i}}\right)\)
∴ \(\mathrm{A} \bigcap_{i=1}^n \mathrm{~B}_1=\bigcup_{u=1}^n\left(\mathrm{~A}-\mathrm{B}_i\right)\)

CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Ex 2(b)

Question 8.
Prove that |A ∪ B ∪ C|
Solution:
= |A| + |B| + |C| + |A ∩ B ∩ C| – |A ∩ B| – |B ∩ C| – |C ∩ A|
L.H.S. =|A ∪ B ∪ C| = |A ∪ D|
where D = B ∪ C
= |A| + |D| – |A ∩ D|
|A ∪ B| =|A| + |B| – |A ∩ B|)
= |A| + |B ∪ C| – | A ∩ (B ∪ C)|
= |A| + |B| + |C| – |B ∩ C| – |(A ∩ B) ∪ (A ∩ C)|
= |A| + |B| + | C |- |B ∩ C| – [|A ∩ B| + |A ∩ C| – |(A ∩ B) ∩ (A ∩ C)|]
= |A| + |B| + |C| – |B ∩ C| – |A∩ B| – |A ∩ C| + |A ∩ B ∩ Cl
= |A| + |B| + |C| – | A ∩ Bl – |B ∩ C| – |C ∩ A| + |A ∩ B  ∩ Cl = R.H.S.

Question 9.
If X and Y are two sets such that X ∪ Y has 20 objects, X has 10 objects and Y has 15 objects; how many objects does X ∩ Y have?
Solution:
Given |X ∪ Y| = 20
|X| = 10
|Y| = 15
We know that |X ∪ Y|
= |X| + |Y| – |X ∩ Y|
⇒ 20 = 10 + 15 – |X ∩ Y|
⇒ |X ∩ Y| = 25 -20 = 5
∴ X ∩ Y has 5 elements.

Question 10.
In a group of 450 people, 300 can speak Hindi and 250 can speak English. How many people can speak both Hindi and English?
Solution:
Let H = The set of people who can speak Hindi
E = The set of people who can speak English.
According to the question we have
|H ∪ E| = 450, |H| = 300,
|E| = 250
We want to find 1 H ∩ E
|H ∩ E| = |H| + |E|-  |H ∪ E|
= 300 + 250 – 450 = 100
∴ 100 people can speak both Hindi and English.

Question 11.
In a group of people,37 like coffee, 52 like tea and each person in the group likes at least one of the two drinks. 19 people like both tea and
coffee, how many people are in the group?
Solution:
Let T = The set of persons who like Tea. ,
C = The set of persons who like coffee According to the question
|C| = 37, |T| = 52 and |T ∩ C| = 19
Total number of persons in the group
= |T ∩ C| = |T| + |C| – |T ∩ C|
= 37 + 52 – 19 = 70

Question 12.
In a class of 35 students, each student likes to play either cricket or hockey. 24 students like to play cricket and 5 students like to play both games; how many students play hockey?
Solution:
Let C = Set of students like to play cricket
H = The set of students like to play Hockey.
According to the question
|C ∪ H| = 35
|C| = 24, |C ∩ H| = 5
Now |C ∪ H| = |C| + |H| – |C ∩ H|
⇒ 35 = 24 + |H| – 5
⇒ |H| = 16
16 students like to play Hockey.

Question 13.
In a class of 400 students, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice or orange juice.
Solution:
Let A = The set of students take apple juice
O = The set of students take orange juice
According to the question
|A| = 100, |O| = 150 and
|A ∩ O| = 75
∴ Number of students take at least one of the juice = |A ∪ O|
= |A| + |O| – |A ∩ O|
= 100 + 150 – 75 = 175
Total number of students
= |U| = 400
Number of students taking neither of these juice
= |U| – |A ∪ O|
= 400 – 175 = 225

CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Ex 2(b)

Question 14.
In a group of 65 people, 40 like cricket, and 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Let C = The set of persons like cricket
T = The set of people who like tennis.
According to the question
CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Exercise 2(b)
|C| = 40 , |C ∩ T| = 10 and
|C ∪ T| = 65
A number of people like tennis only but not cricket = |C ∪ T| – |C|
= 65 – 40 = 15
Number of persons like tennis
= |C ∩ T| – |C| + |C ∩ T|
= 65 – 40 + 10 = 25

Question 15.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12
people liked products C and A, 14 people liked products B and C and 8 liked all the three products, find how many liked products C only.
Solution:
Let E = Set of persons like product A
F = Set of persons like product B
G = Set of persons like product C
CHSE Odisha Class 11 Math Solutions Chapter 2 Sets Exercise 2(b) 1
According to the question
a + b + d + e = 21
c + b + f + e = 26
g + f + d + e = 29
b + e = 14
f + e = 14
d + e = 12
e = 8
⇒ e = 8    g = 11
d = 4        c = 6
f = 6          a = 3
b = 6
Number of persons like product C only
O = g = 11

CHSE Odisha Class 11 Education Book Solutions (+2 1st Year)

CHSE Odisha 11th Class Education Book Solutions (+ 2 1st Year)

Unit 1 Fundamental of Education

Unit 2 Fundamentals of Educational Psychology

Unit 3 Education and Society

Unit 4 Method of Teaching

CHSE Odisha Class 11 Education Syllabus (+2 1st Year)

EDUCATION ELECTIVE (First Year)
Theory – 70 marks & Practical – 30 marks.
Theory Paper – I
FOUNDATIONS OF EDUCATION – I

Unit I Fundamental of Education (20 periods)
Meaning of Education, Aims of Education – Individual, Social, Democratic and Vocational, Function of education, Agencies of education, Formal, Informal, Non-formal, Active & Passive, Role Family, School, Community & Mass media as agencies of education.

Unit II Fundamentals of Educational Psychology (20 periods)
Meaning, Nature & Scope of educational psychology, Importance of educational psychology for the teacher, Growth & Development – Meaning, General Principles & factors affecting, growth & development, Stages of growth and development – Physical, Intellectual, Social & Emotional growth & development during infancy, Childhood and Adolescence.

Unit III Education and Society (20 periods)
Relationship between education & society Education for social change & social Control Education for social mobility, Education for citizenship & socialization, Gender disparity and the role of education Globalization and its impact on education

Unit IV Method of Teaching (20 periods)
(Any one of the following method subjects English, Odia, Mathematics, History, Geography & General Science)
Aims and Objectives, Methods of teaching applicable for elementary level, Teaching learning materials (TLM) purpose & use, General principles and Maxims of teaching, Objective based objective type test items, meaning & principles of construction.

PRACTICAL (60 periods)
A – Preparation of five lesson plans in the selected method subject. (30 periods)
B – Preparation of fifteen objective type test items, 5 each pertaining to knowledge, comprehension & skill objectives on a particular topic of the selected method subject. (30 periods)

BOOKS RECOMMENDED:
1. Bureau Uchcha Madhyamik Siksha (in Odia)
2. Bureau’s Higher Secondary Education I.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(b)

Question 1.
In the following questions, write ‘T’ for true and ‘F’ for false statements.
(i) If tan x + tan y = 5 and tan x, tan y = 1/2  then cot (x + y) = 10
Solution:
False

(ii) √3 (1 + tan 15°) = 1 – tan 15°
Solution:
False

(iii) If θ lies in 3rd quadrant, then cos \(\frac{\theta}{2}\) + sin \(\frac{\theta}{2}\) is positive.
Solution:
True

(iv) 2 sin 105°. sin 15° = 1/2.
Solution:
True

(v) If cos A = cos B = 1 then tan\(\frac{A+B}{2}\). tan\(\frac{A+B}{2}\) = 1
Solution:
False

(vi) cos 15° cos\(7 \frac{1}{2}^{\circ}\). sin \(7 \frac{1}{2}^{\circ}\) = 1
Solution:
False

(vii) sin 20° (3 – 4 cos2 70°) = \(\frac{\sqrt{3}}{2}\)
Solution:
True

(viii) √3 (3 tan 10° – tan3 10°) = 1 – 3tan2 10°
Solution:
True

(ix) \(\frac{2 \tan 7 \frac{1^{\circ}}{2}\left(1-\tan ^2 7 \frac{1^{\circ}}{2}\right)}{\left(1+\tan ^2 7 \frac{1^{\circ}}{2}\right)^2}\) = 1
Solution:
False

(x) The minimum value of sin θ. cos θ is (-1)2.
Solution:
False

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 2.
In the following questions, fill in the gaps with correct answers choice from the brackets.
(i) If α and β lie in 1st and 2nd quadrants respectively, and if sin α = 1/2, sin β = 1/3, then sin (α + β) = _______. \(\left(\frac{1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}, \frac{1}{2 \sqrt{3}}-\frac{\sqrt{2}}{3}, \frac{-1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}\right)\)
Solution:
\(\frac{1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}\)

(ii) if tan α = 1/2, tan β = 1/3, then α + β = ______      \(\left(\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{3}\right)\)
Solution:
\(\frac{\pi}{4}\)

(iii) The value of \(\frac{\cos 15^{\circ}+\sin 15^{\circ}}{\cos 15^{\circ}-\sin 15^{\circ}}\) = ______ \(\left(\frac{\sqrt{3}}{2}, \sqrt{3}, \frac{1}{\sqrt{3}}\right)\)
Solution:
√3

(iv) if \(\frac{1+\sin A}{\cos A}\) = √2 + 1, then the value of \(\frac{1-\sin A}{\cos A}\) is_________, \(\left(\frac{1}{\sqrt{2}-1}, \sqrt{2}-1, \sqrt{2}+1\right)\)
Solution:
√2 – 1

(v) sin 105°. cos 105° = \(\left(\frac{1}{2},-\frac{1}{4},-\frac{1}{2}\right)\)
Solution:
– 1/4

(vi) 2 sin\(67 \frac{1}{2}^{\circ}\) cos\(22 \frac{1}{2}^{\circ}\) = ___ \(\left(1-\frac{1}{\sqrt{3}}, 1+\frac{1}{\sqrt{2}},-1+\frac{1}{\sqrt{2}}\right)\)
Solution:
1 + \(\frac{1}{\sqrt{2}}\)

(vii) sin 35° + cos 5° =____ (2 cos 25°, √3 cos 25°, √3 sin 25°)
Solution:
√3 cos 25°

(viii) sin2 24° – sin2 26° =_____ \(\left(\frac{\sqrt{5}+1}{8}, \frac{\sqrt{5}-1}{8}, \frac{\sqrt{5}-1}{4}\right)\)
Solution:
\(\frac{\sqrt{5}-1}{8}\)

(ix) sin 70° (4 cos2 20° – 3) =_____ \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, \sqrt{3}\right)\)
Solution:
1/2

(x) cos 3θ + sin 3θ is maximum if θ =_____ (60°, 15°, 45°)
Solution:
15°

(xi) sin 15° – cos 15° = _____ (1/2, 0, positive, negative)
Solution:
Negative

(xii) If θ lies in the third quadrant and tan θ = 2, then the value of sin θ is ____. \(\left(\frac{1}{\sqrt{5}},-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\right)\)
Solution:
\(\frac{-2}{\sqrt{5}}\)

(xiii) The correct expression is. (sin 1° > sin 1, sin 1° < sin 1, sin 1° = sin 1, sin 1° = \(\frac{\pi}{180^{\circ}}\) sin 1)
Solution:
sin 1° < sin 1

(xiv) The correct expression is —. (tan 1 > tan 2, tan 1 < tan 2, tan 1 =  1/2 tan 2, tan 1 < 0)
Solution:
tan 1 > tan 2

Question 3.
Prove the following
(i) sin A. sin (B – C) + sin B sin (C – A) + sin C. sin(A – B) = 0
Solution:
L. H. S
= sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)
= sin A (sin B cos C – cos B sin C) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)
= sin A sin B cos C – sin A cos B sin C + cos A sin B sin C – sin A sin B cos C + sin A cos B sin C – cos A sin B sin C
= 0 = R. H. S

(ii) cos A. sin (B – C) + sin B sin (C – A) + cos C. sin(A – B) = 0
Solution:
L.H.S.
= cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B)
= cos A (sin B cos C – cos B sin C) + cos B (sin C cos A – cos C sin A) + cos C (sin A cos B – cos A sin B)
= cos A sin B cos C – cos A cos B sin C + cos A cos B sin C- sin A cos B cos C + sin A cos B cos C – cos A sin B cos C = 0 = R. H. S.

(iii) \(\frac{\sin (B-C)}{\sin B \cdot \sin C}\) + \(\frac{\sin (C-A)}{\sin C \cdot \sin A}\) + \(\frac{\sin (A-B)}{\sin A \cdot \sin B}\) = 0
Solution:
L. H. S.
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

(iv) tan2 A – tan2 B = \(\frac{\sin (\mathbf{A}+\mathbf{B}) \cdot \sin (\mathbf{A}-\mathbf{B})}{\cos ^2 \mathbf{A} \cdot \cos ^2 \mathbf{B}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 1
= tan2 A sec2 B – tan2 B sec2 A
= tan2 A (1 + tan2 B) – tan2 B (1 + tan2 A)
= tan2 A + tan2 A tan2 B – tan2 B tan2A tan2 B
= tan2 A – tan2 B = L. H. S

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 4.
Prove the following :
Solution:
(i) tan 75° + cot 75° = 4
Solution:
L.H.S = tan 75° + cot 75° = tan 75° + \(\frac{1}{\tan 75^{\circ}}\)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 2
(ii) sin2 18° + cos2 36° = 3/4
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 3

(iii) sin 18°. cos 36° = 1/4
Solution:
sin 18° cos 36°
\(=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)=\frac{5-1}{16}=\frac{1}{4}\)

(iv) sin 15° = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
Solution:
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
\(=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

(v) cot \(\frac{\pi}{8}\) – tan \(\frac{\pi}{8}\) = 2
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 4

(vi) \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = tan 54°
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 5

(vii) tan 10° + tan 35° + tan 10°. tan 35 = 1
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 6

Question 5.
Prove the following:
(i) cot 2A = \(\frac{\cot ^2 A-1}{2 \cot A}\)
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 7

(ii) \(\frac{\sin B}{\sin A}=\frac{\sin (2 A+B)}{\sin A}\) – 2cos(A + B)
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 89

= \(\frac{\sin B}{\sin A}\) = L.H.S

(iii) \(\frac{\sin 2 A+\sin 2 B}{\sin 2 A-\sin 2 B}=\frac{\tan (A+B)}{\tan (A-B)}\)
Solution:

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 9

(iv) \(\frac{\cot A-\tan A}{\cot A+\tan A}\) = cos2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 10

(v) \(\frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A}\) = tan 2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 11

(vi) cot A – tan A = 2 cot 2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 12

(vii) cot A – cosec 2A = cot 2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 13

(viii) \(\frac{\cos A-\sin A}{\cos A+\sin A}\) = sec 2A – tan 2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 14

(ix) tan θ (1 + sec 2θ) = tan 2θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 15

(x) \(\frac{\sin A+\sin B}{\sin A-\sin B}\) = tan \(\frac{A+B}{2}\) . cot \(\frac{A-B}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 16

(xi) sin 50° –  sin 70° + sin 10° = 0
Solution:
L. H. S. = sin 50° – sin 70° + sin 10°
= sin (60° – 10°) – sin (60° + 10°) + sin 10°
= – 2 cos 60° sin 10° + sin 10°
= – 2 × 1/2 sin 10° + sin 10°
= – sin 10° + sin 10° = 0 = R. H. S.

(xii) cos 80° + cos 40° – cos 20° = 0
Solution:
L. H. S. = cos 80° + cos 40° – cos 20°
= cos(60° + 20°) + cos (60° – 20°) – cos 20°
= 2 cos 60° cos 20° – cos 20° = 0

(xiii) 8 sin 10°. sin 50°. sin 70° = 1
Solution:
L. H. S. = 8 sin 10° sin 50° sin 70°
= 8 sin 10° sin (60° – 10°) sin (60° + 10°)
= 8 sin 10° (sin2 60° – sin2 10°)
= 8 sin 10°(3/4 – sin2 10°)
= 6 sin 10° – 8 sin3 10°)
= 2 (3 sin 10° – 4 sin3 10°)
= 2 sin (3 × 10°)
= 2 sin 30° = 2 × 1/2 = 1 = R.H.S

(xiv) 4 sin A sin (60° – A) sin (60° + A) – sin 3A = 0
Solution:
L. H. S. = 4 sin A sin (60° – A)
sin (60° + A) – sin 3A
= 4 sin A (sin2 60°- sin2 A) – sin 3A
= 4 sin A. 3/4 – 4 sin3 A – sin 3A
= (3 sin A – 4 sin3A) – sin 3A
= sin 3A – sin 3A = 0

(xv) tan 3A – tan 2A – tan A = tan 3A tan 2A tan A
Solution:
We have tan 3A = tan (2A + A)
or, tan 3A = \(\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}\)
or, tan 3A (1 – tan 2A tan A) = tan 2A + tan A
or, tan 3A – tan 3A tan 2A tan A = tan 2A + tan A
or, tan 3A – tan 2A – tan A = tan 3A tan 2A tan A (Proved)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 6.
Prove the following
(i) tan\(\frac{A}{2}\) = \(\sqrt{\frac{1-\cos A}{1+\cos A}}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 17

(ii) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = tan\(\left(\frac{\pi}{4}+\frac{A}{2}\right)\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 18

(iii) \(\frac{1+\tan \frac{A}{2}}{1-\tan \frac{A}{2}}\) = sec A + tan A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 19

(iv) sec θ + tan θ = tan\(\left(\frac{\pi}{4}+\frac{θ}{2}\right)\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 20

(v) cot\(\frac{A}{2}\) = \(\frac{\sin A}{1-\cos A}\)
Solution:
R.H.S = \(\frac{\sin A}{1-\cos A}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \sin ^2 \frac{A}{2}}\)
= cot\(\frac{A}{2}\) R.H.S

Question 7.
Find the maximum value of the following.
(i) 5 sin x + 12 cos x
Solution:
5 sin x + 12 cos x
Let 5 = r cos θ, 12 = r sin θ 52
∴ 52 = r2 cos2 θ, 122 = r2 sin2 θ
∴ 52 + 122 = r2 (cos2 θ + sin2 θ) = r2
∴ r = \(\sqrt{25+144}\) = 13
∴ The maximum value of 5 sin x + 12 cos x is 13.

(ii) 24 sin x – 7 cos x
Solution:
The maximum value of 24 sin x – 7 cos x is
\(\sqrt{(24)^2+(-7)^2}\)
= \(\sqrt{576+49}=\sqrt{625}\) = 25

(iii) 2 + 3 sin x + 4 cos x
Solution:
The maximum value of 3 sin x + 4 cos x is
\(\sqrt{3^2+4^2}\) = 5
∴ Maximum value of 2 + 3 sin x + 4 cos x is 2 + 5 = 7

(iv) 8 cos x – 15 sin x – 2
Solution:
Maximum value of 8 cos x- 15 sin x is \(\sqrt{(8)^2+(-15)^2}\)
= \(\sqrt{64+225}=\sqrt{289}\) = 17
∴ Maximum value of 8 cos x- 15 sin x – 2 is 17 – 2 = 15

Question 8.
Answer the following:
(i) If tan A = \(\frac{13}{27}\), tan B = \(\frac{7}{20}\) and A, B are acute, show that A + B = 45°.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 21

(ii) If tan θ = \(\frac{b}{a}\), find the value of a cos 2θ + b sin 2θ.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 22

(iii) If sec A – tan A = \(\frac{1}{2}\) and 0< A < 90° then show that sec A = \(\frac{5}{4}\)
Solution:
If sec A – tan A = \(\frac{1}{2}\),       ….(1)
0< A < 90°
⇒ sec A – tan A = \(\frac{\sec ^2 A-\tan ^2 A}{2}\)
= \(\frac{(\sec \mathrm{A}+\tan \mathrm{A})(\sec \mathrm{A}-\tan \mathrm{A})}{2}\)
or, sec A + tan A = 2    ……(2)
Now adding eqn. (1) and (2), We have
2 sec A = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)
or, sec A = \(\frac{5}{4}\)

(iv) If sin θ + sin Φ = a and cos θ + cos Φ = b the show that tan \(\frac{1}{2}\) (θ + Φ) \(\frac{a}{b}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 23

(v) If tan θ = \(\frac{a \sin x+b \sin y}{a \cos x+b \cos y}\) then show that a sin (θ – x) + b sin (θ – y) = 0
Solution:
tan θ = \(\frac{a \sin x+b \sin y}{a \cos x+b \cos y}\)
or, \(\frac{\sin \theta}{\cos \theta}=\frac{a \sin x+b \sin y}{a \cos x+b \cos y}\)
or, a sin θ cos x + b sin θ cos y = a cos θ sin x + b cos θ sin y
or, a (sin θ cos x – cos θ sin x) + b (sin θ cos y – cos θ sin y) = 0
or, a sin (θ – x) + b sin (θ – y) = 0

(vi) If A + C = B, show that tan A. tan B. tan C = tan B – tan A – tan C.
Solution:
A + C = B
or, tan (A + C) = tan B
or, \(\frac{\tan A+\tan C}{1-\tan A \tan C}\)
or, tan A + tan C = tan B – tan A tan B tan C
or, tan A tan B tan C = tan B – tan A – tan C

(vii) If tan A = \(\frac{1}{5}\), tan B = \(\frac{2}{3}\) then show that cos 2A = sin 2B.
Solution:
tan A = \(\frac{1}{5}\), tan B = \(\frac{2}{3}\)
∴ cos 2A = \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 24

(viii) If cos 2A = tan2 B, then show that cos 2B = tan2 A In Δ ABC, prove that.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 25

(ix) tan\(\frac{B+C}{2}\) = cot\(\frac{A}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 26

(x) cos (A + B) + sin C = sin (A + B) – cos C
If A + B + C = π and cos A = cos B. cos C show that (xi  and xii)
Solution:
We have A + B = π – C or, cos (A + B)
or, cos (A + B)
= cos (π – C) = – cos C
and sin (A + B) = sin (π – C) = sin C
∴ cos (A + B) + sin C
= – cos C + sin (A + B)
= sin (A + B) – cos C.

(xi) tan B + tan C = tan A
Solution:
[∴ A + B + C = π ⇒ B + C = π – A
⇒ sin (B + C) = sin (π – A) = sin A]
L. H. S. = tan B + tan C
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 27

(xii) 2 cot B. cot C = 1
Solution:
We have A + B + C = π
⇒ B + C = π – A
⇒ cos (B + C) = cos (π – A) = – cosA
⇒ cos B cos C – sin B sin C = – cos B cos C
(cos B cos C = cos A)
⇒ 2 cos B cos C = sin B sin C
⇒ \(\frac{2 \cos B \cos C}{\sin B \sin C}\) = 1
⇒ 2 cot B cot C = 1

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 9.
Prove the following:
(i) cos (A – D) sin (B – C) + cos (B – D) sin (C – A) + cos (C- D) sin (A – B) = 0
Solution:
L. H. S. = cos (A – D) sin (B – C) + cos (B – D) sin (C – A) + cos (C – D) sin (A- B)
= \(\frac{1}{2}\) [2 cos (A- D) sin (B – C) + 2 cos (B – D) sin (C – A) + 2 cos (C – D) sin (A – B)]
= \(\frac{1}{2}\) [ sin (A- D + B – C) – sin (A – D – B + C) + sin (B – D + C – A)- sin (B- D- C + A) + sin ( C – D + A – B)- sin (C – D- A + B)]
= \(\frac{1}{2}\) × 0 = 0 = R.H.S

(ii) sin 2A + sin 2B + sin 2 (A – B) = 4 sin A. cos B. cos (A – B)
Solution:
L. H. S. = sin 2A + sin 2B + sin 2 (A – B)
= 2 sin \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) + 2 sin (A – B) cos (A – B)
= 2 sin (A + B) cos (A- B) + 2 sin (A – B) cos (A – B)
= 2 cos (A – B)[ sin (A-+ B) + sin(A – B)]
= 2 cos (A – B) × 2 sin A cos B
= 4 sin A cos B cos (A – B) = R. H. S.

(iii) cos 2A + cos 2B + cos 2 (A – B) + 1 = 4 cos A. cos B. cos (A – B)
Solution:
cos 2A + cos 2B + cos 2 (A- B) + 1
= 2 cos \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) + 2 cos2 (A – B)
= 2 cos (A + B) cos (A- B) + 2 cos2 (A – B)
= 2 cos (A – B) [cos (A + B) + cos (A – B)]
= 2 cos (A – B) x 2 cos A cos B
= 4 cos A cos B cos (A – B)
= R. H. S

(iv) sin 2A + sin 2B + sin 2C- sin 2 (A + B + C)
Solution:
L. H. S. = sin 2A + sin 2B + sin 2C- sin 2 (A + B + C)
= 2 sin \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) = 2 cos \(\frac{2 \mathrm{C}+2(\mathrm{~A}+\mathrm{B}+\mathrm{C})}{2}\) × \(\frac{2 \mathrm{C}-2(\mathrm{~A}+\mathrm{B}+\mathrm{C})}{2}\)
= 2 sin (A + B) cos (A – B) + 2 cos (A + B + 2C) sin (A – B)
= 2 sin (A + B) [cos (A- B) – cos (A + B + 2C)]
= 2 sin (A + B) × 2 sin \(\frac{A-B+A+B+2 C}{2}\) sin \(\frac{A+B+2 C-A+B}{2}\)
= 4 sin (A + B) sin (C + A) sin (B + C) = R. H. S.

(v) sin A + sin 3A + sin 5A = sin 3A (1 + 2 cos 2A)
Solution:
L. H. S. = sin A + sin 3A + sin 5A
= sin 3A + sin 5A + sin A
= sin 3A + 2 sin \(\frac{5 \mathrm{~A}+\mathrm{A}}{2} cos \frac{5 \mathrm{~A}-\mathrm{A}}{2}\)
= sin 3A + 2 sin 3A cos 2A
= sin 3A (1 + 2 cos 2A) = R. H. S

(vi) sin A – sin 3A + sin 5A = sin 3A (2 cos 2A – 1)
Solution:
L. H. S. = sin A – sin 3A + sin 5A
= sin A + sin 5A – sin 3A
= sin 5A + sin A – sin 3A
= 2 sin \(\frac{5 \mathrm{~A}+\mathrm{A}}{2} cos \frac{5 \mathrm{~A}-\mathrm{A}}{2}\) – sin 3A
= 2 sin 3A cos 2A – sin 3A
= sin 3A (2 cos 2A – 1) = R. H. S.

(vii) cos (A + B) + sin (A – B) = 2 sin (45° + A) cos (45° + B)
Solution:
R. H. S. = 2 sin (45° + A) cos (45° + B)
= sin (45° + A + 45° + B) + sin (45° + A – 45° – B)
= sin (90° + A + B) sin (A – B)
= cos (A + B) sin (A – B) = L. H. S.

(viii) cos (120° + A) cos (120° – A) + cos (120° + A) cos A + cos A cos (120° – A) + \(\frac{3}{4}\) = 0
Solution:
L. H. S.
= cos (120° + A) cos (120° – A) + cos (120° + A) cos A + cos A cos (120° – A) + \(\frac{3}{4}\)
= cos2 A – sin2 120° + cos A [ cos (120° + A) + cos (120° – A)] + \(\frac{3}{4}\)
= cos2 A – \(\frac{3}{4}\) + cos A ( 2 cos 120°. cos A) + \(\frac{3}{4}\)
= cos2 A + 2 cos 120°. cos2 A
= cos2 A + 2 \(\left(-\frac{1}{2}\right)\) . cos2 A
= cos2 A – cos2 A = 0

(ix) cos 4A – cos 4B = 8 (cos A – cos B) (cos A + cos B) (cos A – sin B) (cos A + sin B)
Solution:
R. H. S. = 8 (cos A- cos B) (cos A + cos B)
(cos A – sin B) (cos A + sin B)
=  8 (cos2 A – cos2 B) (cos2 A – sin2 B)
= – 8(cos2 B – cos2 A)(cos2 A – sin2 B)
= – 8 sin (A +B) sin (A – B) cos (A + B) cos (A – B)
= – 2 x [ 2 sin (A + B) cos (A + B)] [2 sin (A – B) cos (A – B)]
= – 2 sin (2A + 2B) sin (2A – 2B)
= – [cos (2A + 2B – 2A + 2B) – cos (2A + 2B + 2A – 2B)]
= – cos 4B + cos 4A
= cos 4A – cos 4B = L. H. S.

Question 10.
Prove the following:
(i) \(\frac{1-\tan ^2\left(45^{\circ}-A\right)}{1+\tan ^2\left(45^{\circ}-A\right)}\) = sin 2A
Solution:
L.H.S \(\frac{1-\tan ^2\left(45^{\circ}-A\right)}{1+\tan ^2\left(45^{\circ}-A\right)}\) = sin 2A = cos 2(45° – A)
= cos (90° – 2A) = sin 2A = R. H. S.

(ii) \(\frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\) = 2 tan 2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 28

(iii) \(\frac{1-\cos 2 A+\sin 2 A}{1+\cos 2 A+\sin 2 A}\) = tan A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 29

(iv) \(\frac{\sin (\mathbf{A}+\mathbf{B})+\cos (\mathbf{A}-\mathbf{B})}{\sin (\mathbf{A}-\mathbf{B})+\cos (\mathbf{A}+\mathbf{B})}\) = sec 2B + tan 2B
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 30

(v) \(\frac{\cos 7 \alpha+\cos 3 \alpha-\cos 5 \alpha-\cos \alpha}{\sin 7 \alpha-\sin 3 \alpha-\sin 5 \alpha+\sin \alpha}\) = cot 2α
Solution:
L.H.S = \(\begin{array}{r}
\cos 7 \alpha+\cos 3 \alpha \\
-\cos 5 \alpha-\cos 3 \alpha \\
\hline \sin 7 \alpha-\sin 3 \alpha \\
-\sin 5 \alpha+\sin \alpha
\end{array}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 31

(vi) \(\frac{\sin \theta+\sin 3 \theta+\sin 5 \theta+\sin 7 \theta}{\cos \theta+\cos 3 \theta+\cos 5 \theta+\cos 7 \theta}\) = tan 4θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 32

Question 11.
Prove the following:
(i) Express 4 cos A. cos B. cos C as the sum of four cosines.
Solution:
4 cos A cos B cos C
= 2 (2cos A cos B) cos C
= 2 [ cos (A + B) + cos (A – B)] cos C
= 2 cos (A + B) cos C + 2 cos (A – B) cos C
= cos (A + B + C) + cos (A + B – C) + cos (A – B + C) + cos (A – B – C)

(ii) Express cos 2A + cos 2B + cos 2C + cos 2 (A + B + C) as the product of three cosines.
Solution:
cos 2A + cos 2B + cos 2C + cos 2(A + B + C)
= 2 cos \(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\) cos \(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\) = 2 cos \(\frac{2 C+2(A+B+C)}{2}\) × cos \(\frac{2 C-2(A+B+C)}{2}\)
= 2 cos (A + B) cos (A- B) + 2 cos (A + B + 2C) cos (A + B)
= 2 cos (A + B) [cos (A- B) + cos (A + B + 2C)]
= 2 cos (A + B) × 2 cos \(\frac{(A-B+A+B+2 C)}{2}\) cos \(\frac{(A-B-A-B-2 C)}{2}\)
= 4 cos (A + B) cos (C + A) cos (B + C)

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 12.
Prove the following:
(i) cos6 A – sin6 A = cos 2A(1 – \(\frac{1}{4}\) sin2 2A)
Solution:
L.H.S = cos6 A – sin6 A
= (cos2 A)3– (sin2 A)3 = (cos2 A – sin2 A)3 + 3 cos2 A sin2 A (cos2 A – sin2 A)
= cos3 2A + \(\frac{3}{4}\) sin2 2A cos 2A
= cos2 A (cos2 2A + \(\frac{3}{4}\) sin2 2A)
= cos 2A (1 – sin2 2A + \(\frac{3}{4}\) sin2 2A)
= cos 2A (1 – \(\frac{1}{4}\) sin2 2A) = R.H.S

(ii) cos6A + sin6 A = \(\frac{1}{4}\) (1 + 3 cos22A)
Solution:
L.H.S = cos6 A + sin6 A
= (cos2 A)3 + (sin2 A)3
= (cos2 A)3 + (sin2 A)3 – 3 cos2 A. sin2 A (cos2 A + sin2 A)
= 1 – \(\frac{3}{4}\) sin2 2A = 1 – \(\frac{3}{4}\) (1 – cos2 2A)
= 1 – \(\frac{3}{4}\) + \(\frac{3}{4}\) cos2 2A
= \(\frac{1}{4}\) + \(\frac{3}{4}\) cos2 2A = \(\frac{1}{4}\) (1 + 3cos2 2A) = R.H.S.

(iii) cos3 A. cos 3A + sin3 A sin3 A = cos3 2A
Solution:
L.H.S = cos3 A cos 3A + sin3 A sin 3A
= cos3 A (4 cos3 A- 3 cos A) + sin3 A (3 sin A- 4 sin3 A)
= 4 cos6A- 3 cos4A + 3 sin4A- 4 sin6A
= 4 (cos6A- sin6A) – 3(cos4A- sin4A)
= 4 {(cos2A)3– (sin2A)3} – 3 {(cos2A)2– (sin2A)2}
= 4 (cos2A- sin2A) {(cos2A)2 + cos2A sin2A + (sin2A)2} – 3 (cos2A- sin2A) (cos2A + sin2A)
= (cos2A- sin2A) [4 {(cos2A + sin2A)2-2 cos2A sin2A + cos2A sin2A} -3×1]
= cos 2A ( 4- 4 sin2A cos2A- 3)
= cos 2A ( 1 – 4 sin2A cos2A)
= cos 2A (1 – sin22A)
= cos 2A cos22A = cos32A = R. H. S.

(iv) sin4 θ = \(\frac{3}{8}\) – \(\frac{1}{2}\) cos 2θ + \(\frac{1}{8}\) cos 4θ
Solution:
R.H.S = \(\frac{3}{8}\) – \(\frac{1}{2}\) cos 2θ + \(\frac{1}{8}\) cos 4θ
= \(\frac{3}{8}\) – \(\frac{1}{2}\) (1 – 2 sin2 θ) + \(\frac{1}{8}\) (1 – 2 sin2 θ)
= \(\frac{3}{8}\) – \(\frac{1}{2}\) + sin2 θ + \(\frac{1}{8}\) – \(\frac{1}{4}\)  sin2
= sin2 θ – \(\frac{1}{4}\) × 4 sin2 θ cos2 θ
= sin2 θ (1 – cos2 θ)
= sin2 θ sin2 θ = sin4 = L.H.S

(v) cot 3A = \(\frac{\cot ^3 A-3 \cot A}{3 \cot ^2 A-1}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 33

(vi) tan 4θ = \(\frac{4 \tan \theta-4 \tan ^3 \theta}{1-6 \tan ^2 \theta+\tan ^4 \theta}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 34

(vii) \(\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}\) = cot 2A
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 35

(viii) \(\frac{\cot A}{\cot A-\cot 3 A}-\frac{\tan A}{\tan 3 A-\tan A}\) = 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 36
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 37

Question 13.
Find the value of
sin 3°,cos 3°, 2 sin \(\frac{\pi}{32}\)
Solution:
sin 3° = sin (18° – 15°)
= sin 18° cos 15° – cos 18° sin 15°
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 38
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 39
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 40

Question 14.
If sin A + sin B = a and cos A + cos B = b, then show that
(i) tan(A + B) = \(\frac{2 a b}{b^2-a^2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 41

(ii) sin (A + B) = \(\frac{2 a b}{b^2-a^2}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 42

(iii) cos (A + B) = \(\frac{b^2-a^2}{b^2+a^2}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 43

Question 15.
Prove the following:
(i) \(\frac{1+\sin A-\cos A}{1+\sin A+\cos A}\) = tan \(\frac{A}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 44

(ii) \(8 \sin ^4 \frac{1}{2} \theta-8 \sin ^2 \frac{1}{2} \theta\) + 1 = cos 2θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 45

(iii) \(\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}\) \(+\cos ^4 \frac{7 \pi}{8}=\frac{3}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 46
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 47

(iv) cos2 \(\frac{\alpha}{2}\) (1- 2cos α)2 + sin2 α(1+ 2 cos α)2 =1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 48

Question 16.
Prove the following:
(i) sin 20°. sin 40°. sin 60°. sin 80° = \(\frac{3}{16}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 49

(ii) cos 36°. cos 72°. cos 108°. cos144° = \(\frac{7}{16}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 50
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 51

(iii) cos 10°. cos 30°. cos 50°. cos 70° = \(\frac{3}{16}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 52

(iv) cos 20°. cos 40°. cos 60°. cos 80° = \(\frac{1}{16}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 53

(v) tan 6°. tan 42°.tan 66°. tan 78° = 1 [Hints: Use the identity tan 3A = tan A tan (60° – A) tan (60° + A)]
Solution:
We have tan 3A
= tan A tan (60° – A) tan (60° + A)
Now putting A = 6° and 18°.
in (1) we have
tan 18° = tan 6° tan 54° tan 66°     …(2)
and tan 54°
= tan 18° tan 42° tan 78°            ……(3)
Multiplying (2) and (3) we have
tan 18° tan 54°
= tan 6° tan 54° tan 66° tan 18° . tan 42°. tan 78°
or, 1 = tan 6° tan 42° tan 66° tan 78°.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 17.
Prove the following:
(i) cos 7 \(\frac{1}{2}\)° = √6 + √3 + √2 + 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 54

(ii) cot 22 \(\frac{1}{2}\)° = √2 + 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 55

(iii) cot 37 \(\frac{1}{2}\)° = √6 – √3 – √2 + 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 56

(iv) tan 37 \(\frac{1}{2}\)° = √6 + √3 – √2 + 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 57

(v) cos \(\frac{\pi}{16}\) = 1/2 \(\sqrt{2+\sqrt{2+\sqrt{2}}}\)
∴ 2 cos \(\frac{\pi}{16}\)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 58

Question 18.
(i) If sin A = K sin B, prove that tan 1/2(A – B) = \(\frac{K-1}{K+1}\) tan 1/2(A – B)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 59

(ii) Ifa cos (x + α) = b cos (x – α) show that (a + b) tan x = (a – b) cot α.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 60
or, cot x cot α = \(\frac{a+b}{a-b}\)
or, (a – b) cot α = \(\frac{a+b}{\cot x}\) = (a + b) tan x
or, (a + b) tan x =(a-b) cot α

(iii) An angle 0 is divided into two parts α, β such that tan α: tan β = x : y.
Prove that sin (α – β) = \(\frac{x-y}{x+y}\) sin θ.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 61

(iv) If sin θ + sin Φ = a, cos θ + cos Φ = b, show that
\(\frac{\sin \frac{\theta+\phi}{2}}{a}=\frac{\cos \frac{\theta+\phi}{2}}{b}=2 \frac{\cos \frac{\theta-\phi}{2}}{a^2-b^2}\)
Solution:
sin θ + sin Φ = a
cos θ + cos Φ = b
We have
a2 + b2 = (sin θ + sin Φ)2 + (sin θ + sin Φ)2
= sin2 θ + sin2 Φ + 2 sin θ. sin Φ + cos2 θ +cos2 Φ + 2 cos θ. cos Φ
= 2 + 2 (cos θ cos Φ + sin θ sin Φ)
= 2 + 2 cos (θ – Φ)
= 2 [1+ cos (θ – Φ)]
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 62
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 63

(v) If a cos α + b sin α = c = a cos β + b sin α then prove that
\(\frac{a}{\cos \frac{1}{2}(\alpha+\beta)}=\frac{b}{\sin \frac{1}{2}(\alpha+\beta)}\) \(=\frac{c}{\cos \frac{1}{2}(\alpha-\beta)}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 64
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 65

(vi) Prove that \(\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n\) = 2 cosn \(\frac{A-B}{2}\) or zero according as n is even or odd.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 66

Question 19.
(i) If (1 – e) tan2 \(\frac{\boldsymbol{\beta}}{2}\) = (1 + e) tan2 \(\frac{\boldsymbol{\alpha}}{2}\), prove that cos β = \(\frac{\cos \alpha-e}{1-e \cos \alpha}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 67
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 68

(ii) If cos θ = \(\frac{\cos \mathbf{A}-\cos \mathbf{B}}{1-\cos \mathbf{A} \cdot \cos \mathbf{B}}\) prove that one of the values of
tan \(\frac{θ}{2}\) is tan \(\frac{A}{2}\) . tan \(\frac{B}{2}\).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 69
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 70

(iii) If tan θ = \(\frac{\sin x \cdot \sin y}{\cos x+\cos y}\) then prove that one of the values of tan \(\frac{1}{2}\) θ tan \(\frac{1}{2}\) x and tan \(\frac{1}{2}\) y.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 71
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 72
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 73

(iv) If sec (Φ + a) + sec (Φ – α) = 2 sec Φ. show that cos Φ = ± √2 cos \(\frac{α}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 74
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 75

(v) If tan A + tan B = a and cot A + cot B = b. then show that cot (A + B) = \(\frac{1}{a}\) – \(\frac{1}{b}\).
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 76

(vi) If cot θ = cos (x + y) and cot Φ = cos (x – y) show that tan (θ = Φ) = \(\frac{2 \sin x \cdot \sin y}{\cos ^2 x+\cos ^2 y}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 77

(vii) If tan β = \(\frac{n^2 \sin \alpha \cdot \cos \alpha}{1-n^2 \sin ^2 \alpha}\), then show that \(\frac{\tan (\alpha-\beta)}{\tan \alpha}\) = 1 – n2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 78
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 79

(viii)If 2 tan α = 3 tan β, then prove that tan(α – β) = \(\frac{\sin 2 \beta}{5-\cos 2 \beta}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 80
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 81

(ix) If α, β are acute angles and cos 2α = \(\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}\) then prove that tan α = √2 tan β.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 82

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 20.
If A + B + C = π, then prove the following.
(i) cos 2A + cos 2B + cos 2C + 1 + 4 cos A. cos B. cos C = 0
Solution:
A + B + C = π  or, A + B = π – C
or, cos (A + B) = cos (π – C) = – cos C
∴ cos 2A + cos 2B + cos 2C
= 2 cos \(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\) cos \(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\)
= 2 cos (A + B) cos (A- B) + 2 cos2 C – 1
= – 2 cos C cos (A – B) + 2 cos2 C –  1
= – 1 – 2 cos C cos (A – B) – cos C
= – 1 – 2 cos C [cos (A- B) + cos (A + B)]
= – 1 – 2 cos C × 2 cos A cos B
= – 1 – 4 cos A cos B cos C
∴ L. H. S. = cos 2A + cos 2B + cos 2C + 1 + 4 cos A cos B cos C
= – 1 – 4 cos A cos B cos C + 1 + 4 cos A cos B cos C = 0 = R. H. S.

(ii) sin 2A + sin 2B – sin 2C = 4 cos A. cos B. sin C
Solution:
L. H. S. = sin 2A + sin 2B – sin 2C
= 2 sin \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) – sin 2C
= 2 sin(A + B) cos (A – B) -2 sinC cos C
= 2 sin C cos (A- B) – 2 sin C cos C [ A + B= π – C]
or, sin (A + B) = sin (n- C) = sin C]
= 2 sin C [cos (A – B)- cos C]
= 2 sin C [cos (A – B) – cos (A + B)]
= 2 sin C. 2 cos A. cos B
= 4 cos A cos B sin C = R. H. S.

(iii) cos A + cos B + cos C = 1 + 4 sin \(\frac{1}{2}\) A. sin \(\frac{1}{2}\) B. sin \(\frac{1}{2}\) C.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 83

(iv) sin A + cos B- sin C = 4 sin \(\frac{1}{2}\) A. sin \(\frac{1}{2}\) B. cos \(\frac{1}{2}\) C.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 84
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 85

(v) cos2 A + cos2 B + 2cos A. cos B. cos C = sin2 C
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 86

(vi) sin2 \(\frac{A}{2}\) + sin2 \(\frac{B}{2}\) + sin2 \(\frac{C}{2}\) = 1 – 2 sin \(\frac{A}{2}\). sin \(\frac{B}{2}\). sin \(\frac{C}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 87

(vii) sin \(\frac{A}{2}\) + sin \(\frac{B}{2}\) + sin \(\frac{c}{2}\) = 4 sin \(\frac{\pi-A}{4}\) sin \(\frac{\pi-B}{4}\). sin \(\frac{\pi-C}{4}\) + 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 88
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 89
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 90

(viii) cos2 \(\frac{A}{2}\) + cos2 \(\frac{B}{2}\) – cos2 \(\frac{C}{2}\) = 2 cos \(\frac{A}{2}\). cos \(\frac{B}{2}\). sin \(\frac{C}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 91
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 92

(ix) sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin \(\frac{B-C}{2}\). sin \(\frac{C-A}{2}\). sin \(\frac{A-B}{2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 93
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 94
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 95

Question 21.
(i) Show that (2 cos θ – 1) (2 cos 2θ – 1) (2 cos2 2θ – 1) ….. (2 cos 2n-1 θ – 1) \(=\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}\)
Solution:
We have
(2 cos θ + 1) (2 cos θ – 1)
= 4 cos2 θ – 1=4 cos2 θ – 2+1
= 1 + 2(2 cos2 θ – 1)
= 1+2 cos 2θ = 2 cos 2θ+1
And, (2 cos 2θ + 1) (2 cos 2θ – 1)
= 4 cos2 2θ – 1 = 2 ( 2 cos2 2θ – 1) + 1
= 2 cos 4 θ + 1 = 2 cos 22 θ + 1
Similarly, we can prove,
( 2 cos 22 θ + 1) ( 2 cos 22 θ- 1)
= 2 cos 23 θ + 1
Proceeding in this way, we can prove,
(2 cos 2θ + 1) (2 cos 2θ – 1)
(2 cos 2θ- 1) (2 cos 22 θ – 1) ……  (2 cos 2n-1 θ-1)
= 2 cos 2n θ + 1
or, (2 cos θ-1) (2 cos 2θ-1) (2 cos 22 θ- 1) (2 cos 2n-1 θ-1)
\(=\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}\)

(ii) Show that 2n cos θ. cos 2θ. cos22 θ ……… 2n-1 θ = 1 If θ = \(\frac{\pi}{2^n+1}\)
Solution:
we have, θ = \(\frac{\pi}{2^n+1}\)
or, 2n θ + θ = π
or, 2n θ  = π – θ
or, 2n θ = sin (π – θ) = sin θ
or, \(\frac{\sin 2^n \theta}{\sin \theta}\) = 1   …….(1)
Again, sin 2n θ = 2 sin 2n-1 θ cos 2n-1 θ
= 2 × 2 sin 2n-2 θ cos 2n-2 θ cos 2n-1 θ
= 22 × 2 sin 2n-3 θ cos 2n-3 θ  cos 2n-2 θ cos 2n-v θ
= 23 sin 2n-3 θ cos 2n-3 θ cos 2n-2 θ cos 2n-1 θ

………………………..

………………………..

………………………..

= 2n sin θ cos θ cos 2θ cos 22 θ …….. cos 2n-2 θ cos 2n-1 θ

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 96
or,  2n cos θ cos 2θ cos 22 θ …… cos 2n-1 θ = 1

(iii) Prove that \(\frac{\tan 2^n \theta}{\tan \theta}\) = = (1 + sec 2θ) (1 + sec22 θ) …. (1 + sec2n θ)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 97
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 98

Question 22.
If x + y + z = xyz, prove that
(i) \(\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\) = =\(\frac{4 x y z}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\)
Solution:
x + y + z = xyz   (Given)
Let x = tan α, y = tan β, z = tan γ
∴ tan α + tan β + tan γ = tan α .tan β .tan γ or; tan α + tan β
= – tan γ + tan α tan β tan γ
= – tan γ(1 – tan α tan β)
or, \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) = – tan γ
or, tan (α + β) = tan (π- γ)
or, α + β = π-γ
or, 2α + 2β = 2π- 2γ
or, tan (2α + 2β = tan (2π- 2γ)
or, tan (2α + 2β) = tan (2π- 2γ)
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 99

(ii) \(\frac{3 x-x^3}{1-3 x^2}+\frac{3 y-y^3}{1-3 y^2}+\frac{3 z-z^3}{1-3 z^2}\) = \(\frac{3 x-x^3}{1-3 x^2} \cdot \frac{3 y-y^3}{1-3 y^2} \cdot \frac{3 z-z^3}{1-3 z^2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 100
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 101

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b)

Question 23.
If \(\text { If } \frac{\sin ^4 \alpha}{a}+\frac{\cos ^4 \alpha}{b}=\frac{1}{a+b}\) show that \(\frac{\sin ^8 \alpha}{a^3}+\frac{\cos ^8 \alpha}{b^3}=\frac{1}{(a+b)^3}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 102
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Ex 4(b) 103

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Odisha State Board CHSE Odisha Class 11 Political Science Solutions Unit 3 Indian Constitution Short Answer Questions.

CHSE Odisha 11th Class Political Science Unit 3 Understanding Political Theory Short Answer Questions

Short Question With Answers

Question 1.
What were the functions of the constituent assembly of India?
Answer:
The constitution assembly was formed to prepare the new constitution of India and To act as the union legislature till the new parliament was formed.

Question 2.
What was the importance of the Drafting committee?
Answer:
The drafting committee played an important role in drafting the new constitution of India. It was the future ruling pattern of India.

Question 3.
What do you mean by objective resolution?
Answer:
Objective resolution refers to the aims and objectives of the constitutional frames expressed in the form of a resolution. It was approved by the constituent assembly on 22nd January 1947.

Question 4.
Why Indian constitution has been so large?
Answer:
The constitution of India is so large because it contains provisions both for the union government and states. The length of the constitution increased in view of the detailed analysis of every provision to make it clear to the people.

Question 5.
What is an ideal constitution?
Answer:
An ideal constitution is a document that has the ability to develop and V change in accordance with changes in time and circumstances. Its language must be simple clear andunamigous

Question 6.
What is the preamble?
Answer:
A preamble is a brief introduction about the constitution. It contains ideals and objectives of the constitution in a nutshell.

Question 7.
What does the term we the people of India imply?
Answer:
The term we the people of India implies the source of the constitution and its democratic character. It was framed by a group of representatives of Indian people.

Question 8.
What are the main sources of the constitution?
Answer:
The Government of India Act. 1935. The parliamentary statutes and debates. The constitution of UK, Canada, Australia, Ireland, Germany, Spain and South Africa etc.

Question 9.
Why India is called a secular state?
Answer:
India is called a secular state because
There is no state religion in India and All religious communities are treated equally by the state. People enjoy the freedom of religion and worship.

Question 10
Why India is called a sovereign state?
Answer:
India is called a sovereign state because it is independent and free from the influence of any foreign country. The decision of the government of India is found delating internal administration and foreign policy.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 11.
Why India is called a socialist state?
Answer:
India is called a socialistic state because, it seeks to provide social, economic, and political justice to the people. The government gives priority to the public sector and seeks to abolish economic exploitation and provide basic minimum needs to all.

Question 12.
Why India is called a democracy?
Answer:
India is called a democracy because the constitution has set up parliamentary representative democracy in India. The people enjoy fundamental rights and the media and judiciary remain free from Govt, control.

Question 13.
Why India is called a Republic?
Answer:
India is called a republic because the head of state, the president is elected by the people indirectly Again any ordinary citizen can contest and assume the office of the president of India.

Question 14.
Why Indian constitution is called an enacted constitution?
Answer:
Indian constitution is called an enacted constitution because it was ‘ framed by a constituent assembly.’ It was enacted on 26th November 1949 by the constituent assembly.

Question 15.
India is a union of states?
Answer:
Art. 1 declares India as a union of states. That means the federal structure of ” India can’t be changed by the decision of states.

Question 16.
Why India is called a quasi-federal state?
Answer:
Prof. K.C. where describes India as a quasi-federal state. It means that India is organized on federal lines but it works as a unitary state.

Question 17.
Is the Indian constitution rigid?
Answer:
Indian constitution can’t be considered exclusively either as a rigid or flexible constitution. The procedure of amendment is a mixture of both rigidity and flexibility.

Question 18.
What is single citizenship?
Answer:
Single citizenship means only the union government has the power to grant citizenship. It is found in a unitary state, but in India single citizenship is preferred to maintain national unity and integrity.

Question 19.
What is judicial review?
Answer:
Judicial review refers to the power of the supreme court to examine the constitutional validity of the law. If it finds any law or decision violating the constitution the court declares such law unconstitutional.

Question 20.
What do you mean by fundamental rights?
Answer:
The fundamental rights are the most valuable rights which are mentioned in part – III of the constitution. These rights are justiciable in nature, therefore in case of violation, the court takes immediate redressal measures.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 21.
What is Right to Equality?
Answer:
The right to equality is a fundamental right which is guaranteed under Art 14 to 18 of the constitution. This right is available both for citizens.

Question 22.
What is equality before law?
Answer:
Equality before law means all are equal in the eyes of law. The law courts the judges and the legal system gives equal protection to all citizens.

Question 23.
Importance of Art. 19?
Answer:
Art. 19 guarantees six fundamental freedoms to Indian citizens. These include freedom of speech and expression freedom to assemble peacefully without arms, freedom of association, freedom of movement, freedom of settlement & freedom of profession trade or business.

Question 24.
Traffic in human beings?
Answer:
Traffic in human beings means exploitation and unlawful treatment of weaker sections of society. It prohibits men and upper classes from exploiting women, children, and backward classes.

Question 25.
How the fundamental rights differ from legal rights?
Answer:
The fundamental rights are guaranteed and protected by the institution, but the legal rights are protected by law. The legal rights can be amended by ordinary legislation, but fundamental rights can be amended only by constitutional amendment.

Question 26.
Which fundamental rights are available both for citizens and aliens?
Answer:
Right to equality under Art. 14 and Right to life liberty and property under Art. 21 are available both for citizens and aliens

Question 27.
Right to education?
Answer:
Right to education is now a fundamental right under Art. 21 (A) It is inserted into the fundamental rights by the 86th amendment act. 2002. It provides for free and compulsory education to all children below 14 years of age.

Question 28.
Right against exploitation?
Answer:
This right is guaranteed under Art. 23 and 24 of the constitution. It seeks to give protection to the weaker sections of society, including women, children and backward classes from the exploitation of rich and male-dominated society.

Question 29.
What is protective discrimination?
Answer:
The constitution of India under Art. 15 (3) and provide for protective discrimination. It enables the scheduled castes, scheduled tribes, women- and backward classes to avail special reservations in admission to educational institutions and in public service.

Question 30.
Habeas Corpus?
Answer:
Habeas Corpus is a writ issued by a superior court against any public authority. It is issued against unlawful arrest and detention.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 31.
Mandamus?
Answer:
The writ of Mandamus is issued by a superior court to any person or ‘ public authority to perform a legal duty. It is a direction to perform ministerial acts.

Question 32.
What is Prohibition?
Answer:
The writ of prohibition is issued either by Supreme Court or High Court to an inferior court against excessive jurisdiction. It prevents a lower court from handling cases beyond the jurisdiction

Question 33.
Certiorari?
Answer:
A certiorari is a writ issued by the superior court to any lower court against excessive jurisdiction. It is issued to quash the order and decision of the court or tribunal.

Question 34.
Quo-Warranto?
Answer:
The writ of quo warm to is issued by a superior court to any public authority against the illegal assumption of public office. It prohibits illegal appointments to public service.

Question 35.
Importance of Art. 32?
Answer:
The Indian Constitution Under Art. 32 grants right to constitutional remedies. As per the article, in the case of. violation of the fundamental rights of a citizen, he can move to the Supreme Court for immediate redressal.

Question 36.
Objectives of Directive Principles?
Answer:
The objectives of directive principles of state policy are To make India a welfare state and To create conditions for socio-economic democracy in India.

Question 37.
Fundamental-Duties?
Answer:
The fundamental duties are the moral responsibilities of the citizens towards the state. These are non-justiciable in nature.

Question 38.
Right to the property?
Answer:
The right to property was a fundamental right up to 1978, but it was deleted in 1978 by 44th Amendment Act. Now it is a legal duty under Art. 300 A.

Question 39.
Right to Education?
Answer:
Right to Education is a fundamental right under Art. 21 (A). It provides for free and compulsory education to all children belonging to the age group of 6 to 14 years.

Question 40.
Right to life?
Answer:
Right to life is a fundamental right under Art. 21. It states that no one shall be deprived of his life and liberty except, according to procedure established by law.

Question 41.
Right to Information Act. 2005?
Answer:
The Right to Information Act, 2005 seeks to promote transparency and accountability in the working of the government. It helps in containing corruption and makes the citizens aware about the activities of the government.

Question 42.
What is preventive detention?
Answer:
The constitution of India under Art 22 provides for preventive detention of criminal and anti-socials beyond 24 hours. Under this provision any person whose presence is apprehensive of creating social disorder and violence, he may be kept under legal custody for three months even without violating law.

Question 43.
What constitutes the basic structure of the constitution?
Answer:
Basic structure comprises of some values ideals and principles of the constitution which lay its foundation, According to eminent justice, it includes parliamentary democracy, fundamental rights, preamble, secularism, republican model, independent judiciary and centralized federation, etc.

Question 44.
What is free legal aid?
Answer:
The constitution of India under Art. 39A under the Directive Principle makes provision for free legal aid to the poor and underprivileged sections of society; Under the provision, the state government bears the cost of fighting legal battles against those exploiting women, poor and distressed people.

Question 45.
Constituent Assembly of India?
Answer:
The new constitution of India was framed by die constituent Assembly, (b) It comprises of all the eminent lawyers. Intellectuals, social activists, and political leaders of India of that time. It was a representative body that prepared the constitution within 2 years 11 months and 18 days. It played the role of parliament till the new parliament was elected. Its members were elected indirectly by the provincial legislatures on the basis of population.

Question 46.
Cabinet mission plan?
Answer:
Cabinet mission. came to India in the year 1946 to discuss about the grant of independence to India. Prime minister clement allee sent three of his cabinet minister, Sir Stafford Cripps, A’.V. Alexander and Lord Pathik Lawrence as members of the mission. The mission held discussions with Indian leaders, leaders of the Muslim League, and administrators about the ways and means to concede independence. The mission proposed for a ‘Constituent assembly to be formed to prepare the new constitution of India. It recommended for a federal setup in India with Indian provinces and princely states.

Question 47.
Drafting committee?
Answer:
The drafting committee was instrumental for framing the new. constitution of India. It comprises of Dr. B .R.Ambedkar as the chairman and N. Gopalswamy Ayengarm, Alladi Krishnaswamy, Nyer, K.M. Munsi, Mahammad Saddik, N. Madhab Rao and T.T. Krishiiamachari as members. This committee prepared the draft of the whole constitution and got it approved. It had 55 sessions continuing for 114 days and on 4th November 1948 submitted the draft constitution for approved. Dr. Ambedkar played a leading role as its chairman for which he is regarded as the father of Indian constitution.

Question 48.
Objective Resolution?
Answer:
Objective Resolution of the constituent Assembly was initiated by Pandit Nehru on 13th December 1946. It was approved on. 22nd January 1947 by the constituent assembly. This resolution contained the aims and aspirations of the founding fathers relating to the future of India. It proposed to make India, a sovereign Independent Republic. Besides, it also resolved to ensure social political and economic justice to the people of India along with liberty and equality.

Question 49.
Write at least five important features of Indian Constitution?
Answer:

  • Longest written constitution.
  • A balance between rigidity and flexibility
  • Federal in form but unitary in spirit
  • Parliamentary democracy; and
  • Secular State

Question 50.
What does the preamble to the constitution imply?
Answer:
The preamble implies an introduction to the Constitution. It embodies the objective, ideals, sentiments and aspirations of the nation. It is a key to understanding the nature and spirit of the Indian Policy. It speaks about the source of the Constitution and the date of its adoption. The preamble indicates the general purpose for which people have ordained and established Constitution.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 51.
What are the source of the Indian Constitution?
Answer:
The contents of the Indian Constitution have been derived from different sources out of which the important ones are given below.
The Government of India Act, of 1935 has been the most important source of the Constitution. The British Constitution has influenced it the most. The fundamental rights have come from the American constitution. The Irish constitution, the Weimar Constitution of Germany the Canadian Constitution and the Constitution of the South African Republic have also influenced the Indian Constitution.

Question 52.
How does the Preamble depict the characteristics of Indian Constitution?
Answer:
The preamble outlines the following characteristics of Indian Constitution. India is a sovereign independent community. India is a republic. India has a democratic form of government. The constitution ensures justice, liberty, equality and fraternity. It seeks to harmonize dignity of the individual with unity of the nation. It establishes popular sovereignty and seeks to follow socialistic and secular principles.

Question 53.
India is a Sovereign Socialist Secular Democratic Republic Discuss?
Answer:
The Preamble reads India as a Sovereign, Socialist Secular Democratic Republic. India is sovereign both internally and externally and it believes in socialistic principles. India has no state religion and the Constitution grants equal status to all religions. Indian policy is organized on democratic principles and the President being the head of state is elected hence it is a republic.

Question 54.
Why is Indian Constitution so long?
Answer:
Indian constitution is the longest written Constitution in the world having 395 Articles and 12 Schedules. The factors responsible for the extra length of the Constitution are as follows. The constitution is meant both for the Union and the States. Everything in the Constitution has been discussed in detail.

The diversity of the Indian culture necessitated special provisions for which the Constitution has been so long. Besides the above factors, detailed chapters on fundamental rights, Directive Principles, emergency provisions, center-state relations and fundamental duties have added to the size of the constitution.

Question 55.
Why is Indian Constitution supposed to be a bag of borrowings?
Answer:
The Indian Constitution was framed following the major democratic constitution of the world. The British Constitution America, Irish, Australian and Canadian Constitutions, etc. have significantly influenced the framers, but that does not mean that the principles and ideals of the Constitution are borrowed: The framers have borrowed most of the provisions from Western countries and modified them to suit to Indian conditions and the people. It will be wrong to brand as a borrowed constitution.

Question 56.
Indian Constitution is federal in form but unitary in spirit. Discuss?
Answer:
Indian constitution has designed the country on the principles of a federation but it works more or less as a unitary state. The framers adopted a federal form of government on account of its vastness but declared it as a union of states. Despite all federal features, the union government is given an upper hand in every sphere. The union Govt enjoys more powers and it can at times change the federal structure into a unitary one. The framers have made a happy blend between unitarism and federalism for which it is supposed to be federal in form but unitary in spirit.

Question 57.
Why is India called a Quasi Federal State?
Answer:
A quasi-federal state is one which is neither completely unitary nor federal in character. Prof. K.C. where has criticized India’s federal character as was quasi-federal. India is organized on the basis of the federation but works on unitary lines. The union government enjoys more powers than the states and the framer can suspend the state government on various grounds. The Union Govt, through the office of the Governor, All India Services, Planning Commission, National Development Council and Finance Commission, etc. can influence the Govt, at the state. Therefore, India is called a quasi-federal state.

Question 58.
Is Indian Constitution a rigid one?
Answer:
No, Indian Constitution does not come along the lines of a rigid Constitution as that of the U.S.A. Most parts of the Constitution can be amended by simple procedure of legislation. Only the federal features of the Constitution are rigid. Those matters can be amended if a majority of both houses of Parliament coupled with the 2/3rd majority of those present and voting support the motion.

Therefore, the Constitutional amendment is sent to the state legislatures for ratification. If at least half of the states approve the proposal the amendment is effected. This rigid procedure applies only to a few items.

Question 59.
Indian Constitution is flexible. Discuss?
Answer:
A flexible Constitution is one which can be amended easily, by the simple procedure of legislation. Most parts of the Constitution can be amended Iggy this process but there are still important matters which need to be amended by the rigid process. As the constitutional framers have made a compromise between rigidity and flexibility it can not be described as i flexible Constitution.

Question 60.
Briefly state the Preamble of the Indian Constitution?
Answer:
The Preamble of the Indian Constitution reads as follows: “WE ‘THE PEOPLE OF INDIA” laving solemnly resolved to constitute India intoa SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC AND TO SECURE TO ALL ITS CITIZENS. JUSTICE, Social, Economic and Political LIBERTY of thought, expression, faith, belief and worship.

EQUALITY of status and opportunity, land to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the nation. IN OUR CONSTITUENT ASSEMBLY, this twenty-sixth day of November 1949, do hereby ADOPT ENACT AND GIVE TO OURSELVES THIS CONSTITUTION.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 61.
What is the importance of the term ‘We the people of India’?
Answer:
The Preamble starts with the phrase ‘We the people of India which illustrates that the people of India are the maker of the Constitution. It speaks about the representative character of the Constitution and accepts people as the ultimate sovereign. The constitution is a popular document which is not imposed on the people but deliberately prepared by a representative Constituent Assembly.

Question 62.
Is India is a sovereign state?
Answer:
Yes, India is a sovereign state and the preamble of the constitution declares about its sovereign character. Internally and externally. India is supreme. Externally, the country is free from any outside authority and it has accepted the membership of international organizations voluntarily internally, it is competent to adopt its domestic policies and regulate the behavior of its nationals and organizations.

Question 63.
Is India is a Secular State?
Answer:
Yes, India is a secular state. A secular state is one which is neither anti-religious nor irreligious. It grants equal status and freedom to, all religions. There is no state religion in India and no religious discrimination is practiced. The Constitution has guaranteed right to freedom of religion to the citizens which symbolizes the secular character of the constitution.

Question 64.
Why is lndia called a democratic republic?
Answer:
The Preamble declares India as a democratic republic. The Indian political system is organized on a comprehensive democratic basis. The parliament and the state legislature are elective bodies and the franchise is extended to all adult citizens universally. The Govt, of India, is elected by the people periodically and remains. responsible.to them. It is a republic because the President being the head of State is elected indirectly. Thus India is called a democratic republic.

Question 65.
Is India a Welfare State?
Answer:
Yes, India is a secular state ordained under Art. 38 of the Constitution. The Directive Principles also embody the principles of a welfare state. The constitution states that the ownership and control of material resources shall be distributed equally to subserve common good. The Govt provides special assistance to the educationally, and economically backward sections of the community. The five-year plans and the socialistic principles are implemented in the country to make India a welfare state.

Question 66.
What is single citizenship?
Answer:
Single citizenship means that in India the power to grant citizenship has been conferred upon the union government only. The state has no authority in this regard. All citizens residing anywhere within India irrespective of their residence or place of birth, are the citizens of India. This idea is supposed to promote a sense of unity among Indians.

Question 67.
What are the five pillars of the Constitution?
Answer:
The five pillars of the constitution are

  • the election commission,
  • the Finance Commission
  • the Union Public Service Commission
  • the Attorney General and
  • the Auditor-General of India.

These constitute the heart of the democratic structure.

Question 68.
What are the basic philosophy of the Constitution?
Answer:
The basic principles embodying the philosophy of the constitution are as follows:

  • Popular sovereignty,
  • Federal system
  • Fundamental rights of the citizens,
  • Directive principles of state policy,
  • Judicial independence
  • Parliamentary system
  • Supremacy of the Union Government.

Question 69.
How the Indian Constitution can be amended?
Answer:
The Indian constitution can be amended by the Parliament under Art. 368. The parliament by required majority can make addition, variation or repeal any provision of the constitution. Such a proposal can be initiated in either House of Parliament and after approval, it is submitted to the President for assent. The President, can’t reject such a proposal and after President’s assent, necessary amendments may be effected.

Question 70.
Why India is called a socialist state?
Answer:
India is called a socialist state, because of the following reasons. The Govt, of India, acts on socialistic lines. The Govt is committed to securing socio-economic and political justice to the people by ending all forms of exploitation. It undertakes economic planning to ensure equitable distribution of wealth and income.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 71.
What is the importance of the Preamble to the constitution?
Answer:
A preamble is an introduction to the constitution. It underlines the philosophical foundations of the constitution and its objectives. It outlines in a nutshell the ideals and objectives of our constitution. It lays down the pattern of our political setup, that is to make India a sovereign, socialist, secular, democratic republic. It speaks about the source of constitution And the date of its, adoption.

Question 72.
Is, the Preamble a part of the constitution?
Answer:
This is a controversial question. As per the decision of the Supreme Court on the transfer of the Berubari Union and exchange of Enclaves, the Preamble is not a part of the constitution. But the apex court modified its decision: in the Keshavananda Bharati case and regarded the preamble as a part of the constitution. In fact, it forms a part of the basic structure as it defines the. structure and philosophy of the constitution.

Question 73.
How the Constitution of India has safeguarded the interests of S.Cs and STs?
Answer:
The Indian constitution under Part XVI has specified special provisions for safeguarding the interests of Scheduled Castes Under Art – 330 seats have been reserved for them in the Lok Sabha and under Art. 331 and 332 in the State Legislatures. They are given special facilities to join public service, to get promotions, and to get admission into colleges, universities, and- professional institutions.

Question 74.
Which four new languages have been added to the VIIIth Schedule by the 100th Amendment Act?
Answer:
The four languages which have got an entry into the VIIIth Schedule by the 100th Amendment Act are, Bodo, Dogri, Maithili and Santhali.

Question 75.
Name five major sources of Indian Constitution?
Answer:
The five major sources of Indian Constitution are:

  • The Govt, of India Act, 1935
  • Provisions of foreign constitutions.
  • Records of debates and discussions in the constituent Assembly.
  • Parliamentary statutes and judicial decisions.
  • The ideals and values of freedom struggle.

Question 76.
86th Constitutional Amendment, 2002?
Answer:
The Parliament passed the 86th Amendment Act, in 2002. It provides for Inclusion of Art. 21 (A) Whereby all children under the age group of 6 to 14 years of age are given the fundamental right of free and compulsory education. It added a new duty in Art. 51(A) increasing the number to 11. By this amendment, it has become the duty of the parents to send their children from 6 to 14 years of age to school.

Question 77.
What do you mean by fundamental Rights?
Answer:

  • Fundamental Rights are those rights guaranteed in the constitution.
  • It ensures the development of individual personality.
  • These rights are guaranteed and protected by the state.
  • The fundamental rights are superior to ordinary laws.
  • These rights are the bedrock of Indian democracy.
  • The Government cannot make laws violating any of these rights.

Question 78.
Name the six Fundamental Rights?
Answer:

  • Right to Equality
  • Right to Freedom
  • Right against Exploitation
  • Right of Freedom of Religion
  • Educational and Cultural Rights
  • Right to Constitutional Remedies

Question 79.
What do you mean by Right to equality?
Answer:
Right to equality is the first fundamental right guaranteed to the citizens. It refers to equality before law and equal protection of laws. If eliminates the possibility of all discrimination against a citizen on grounds of religion, race, sex, caste or place of birth. It guarantees equality of opportunity to all in matters of public employment. It seeks to abolish untouchability to ensure social equality. The constitution prohibits all titles and honors for the Indian except that of military or academic character to maintain proper equality.

Question 80.
What are the six freedom guaranteed under Art19?
Answer:
The six freedoms guaranteed under Art. 19 are:

  • Freedom of speech and expression.
  • Freedom to assemble peacefully without arms.
  • Freedom to form associations or unions.
  • Freedom to move freely throughout the territory of India.
  • Freedom to reside and settle in any part of India.
  • Freedom to practice any profession, occupation, trade, or business.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 81.
What does Right against Exploitation imply?
Answer
The right against exploitation is a fundamental right which seeks to prohibit all forms of exploitation of the children and women. It prohibits forced labor and traffic in human beings. It prevents child below 14 years to be employed in any factory or mine where there is danger to life. It also opposes the illegal use of women for immoral purposes. This right is meant for the weaker sections of the community.

Question 82.
What does the Right to Freedom of Religion imply?
Answer:
The Right to Freedom of Religion implies secular character of the state. It has been explained under four articles:

  • Art. 25 states that all persons are equally entitled to freedom of conscience and the right to profess, propagate or practice any religion.
  • Art. 26 grants the freedom to establish and maintain religious institutions for religious or charitable purposes.
  • Art. 27 provides that, no person shall be ” compelled to pay any tax for the promotion of any particular religion.
  • Art. 28 states that no religious instruction shall be provided in any educational institution wholly maintained by the state funds.

Question 83.
What is the meaning of Right to constitutional remedies?
Answer:
This is the sixth fundamental right which provides remedial measures for the enforcement of fundamental rights. This right is conferred under Art. 32 of the Constitution and it empowers the Supreme Court to uphold the fundamental rights of the citizens. If the fundamental rights of a citizen are violated he can move to the Supreme Court under Art. 32 or High Court under Art. 226 for redressal. The court can issue appropriate writs to provide- relief to the affected persons.

Question 84.
What are the different kinds of writs issued by the courts for the redressal of fundamental rights?
Answer:
There are five kinds of writs issued by the courts for the redressal of fundamental rights. These writs are

  • Habeas Corpus
  • Mandamus
  • Prohibition
  • Certiorari, and
  • Quo-Warranto

Question 85.
What is the purpose of the writ of Habeas Corpus?
Answer:
The writ of Habeas Corpus is issued to maintain and enforce the fundamental rights of citizens. It is a Latin word which means to have one’s own body. The purpose of Habeas Corpus is to protect a citizen from unlawful arrest and detention. If a person is arrested or detained without any valid reason the court orders for his release by issuing this writ. It preserves the liberty of a person who is confined without legal justification.

Question 86.
What for the writ of Mandamus is issued?
Answer:
The writ of mandamus is issued by the court commanding a person or authority to do his duty. This writ is used for public purposes to enforce the performance of public duties. It also enforces some privacy rights when they are withheld by the public.

Question 87.
What is prohibition?
Answer:
It is a judicial writ issued by a superior court to an inferior court to prevent the lower court from usurping jurisdiction. The writ of prohibition is meant to check the possibility of abuse in a jurisdiction. If any case pending before the court is beyond its jurisdiction, the higher court by iásue of the writ prohibits the lower cÓurt not to hear such case.

Question 88.
What is Certiorari?
Answer:
A certiorari is a writ issued by a higher court to a lower count for the withdrawal of a case from the latter to the former. Such a writ may be issued even before the trial of the court comes to know that anything beyond the jurisdiction of the court is pending for decision before it. Both the writs of prohibition and certiorari are intended to deal with complaints of excess of jurisdiction.

Question 89.
What is the purpose of Quo warranto?
Answer:
Quo-warranto is a writ which is issued to prevent the illegal assumption of public office. Quo-warrantà’means by what authority. The court issuing the writ has to consider whether there has b&n usurpation of public office or not. It has been there even before the framing of the constitution.

Question 90.
Can the Parliament of India amend the fundamental rights?
Answer:
Yes, the Parliament of India can amend any portion of the fundamental rights excluding the basis structure of the Keshavananda Bharati and others Vs the State of Kerì1a held that the f parliament has right to an end all párts of the Constitution including Part Ill without destroying the basic structure of the constitution.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 91.
What do you mean by Directive Principles of State Policy?
Answer:
Directive Principles of State Policy is an important feature of the Indian constitution. These principles are discussed under part IV of the constitution and they are in the nature of constitutional directions o the state and central government to implement in administration. These directives are nonjustifiable but they are considered fundamental in the governance of the country These principles intend to play a positive role in the establishment of a socialist welfare state.

Question 92.
What Is the importance of Directive Principles?
Answer:
The Directive Principles of State Policy have immense utility and importance in making every modem state welfare state. These principles set the foundation of socio-economic democracy in India. These are not obligatory, but they are positive guidelines for the states to observe and implement them in the day-to-day administration of the state. They are not enforceable in any court but they are considered fundamental in the governance of the country.

Question 93.
Name of the socialistic principles of the Directives?
Answer:
The socialistic principles of the Directives contain the following.

  • The ownership and control of the material resources of the community must be distributed to subserve common good.
  • There must be equal pay for equal work for both men and women.
  • The operation of the economic system should not lead to concentration of wealth and means of production to common detriment.
  • The health and strength of the workers and tender age of children should not be abused.

Question 94.
Mention at least three Gandhian Principles of the Directives?
Answer:

  • Organization of village Panchayats and to provide them with the power of self-government.
  • Promotion of educational and economic interests of the S.C. and S.T.
  • Promotion of cottage industry on individual and cooperative lines in the rural sector.

Question 95.
What constitutes the liberal principles of the directives?
Answer:
The liberal principles of the directive principles include:

  • Separation of executive from the judiciary in public services.
  • Provision for uniform civil code throughout the country.
  • Protection of monuments and objects of historical and artistic importance.
  • Promotion of international peace and security and maintaining adjustable and honorable relations with nations.
  • To foster respect for international law and treaty obligations and settlement of international disputes through arbitration.

Question 96.
Fundamental Duties?
Answer:
The fundamental duties are mentioned in Part – IV A, under Art – 51 (A) of the constitution. These are inserted into the text of the constitution by the 42nd Amendment Act of 1976 on the recommendations of Dr. Swaran Singh’s Committee.These are moral in character. These duties are ten in number and in 2002, the 86th Amendment has inserted another duty to the list.

CHSE Odisha Class 11 Political Science Unit 3 Understanding Political Theory Short Answer Questions

Question 97.
Features of Indian Fundamental Rights?
Answer:
The basic features of Indian fundamental rights can be discussed below:

  • These rights are elaborate and comprehensive.
  • Both negative and positive rights.
  • These rights are not absolute.
  • There are certain rights for minorities.
  • They are binding upon the union, states, and other state authorities.

Question 98.
Right to Education?
Answer:
The Indian Constitution by the 86 Amendment Act of 2002 has added a new fundamental right under Art.- 21 A, called right to education. Under this article, the govt provides free and compulsory education to all children from the age group of 6 to 14 years in such as manner as the state may by law determine.

Question 99.
The Fundamental Rights in India are defective in many respects.
Answer:
The constitution under part – III has prescribed a comprehensive list of fundamental rights for Indian citizens. But there are certain weaknesses inherent in these rights as are given below:

  • The constitution has imposed several limitations on fundamental rights.
  • These rights are coded in difficult languages.
  • There are no socioeconomic rights.
  • The Parliament can amend fundamental rights.
  • No mention about natural rights.