Class 12

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
If f is an odd function, then write the value of \(\int_{-a}^a \frac{f(\sin x)}{f(\cos x)+f\left(\sin ^2 x\right)}\) dx
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(b) 0

Question 2.
If p and q are respectively degree and order of the differential equation y = e^dy/dx then write the relation between p and q.
(a) p ≠ q
(c) p ≡ q
(b) p = q
(d) None of these
Solution:
(b) p = q

Question 3.
Write the value of \(\int_0^1\){x} dx where {x} stands for fractional part of x.
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{2}{3}\)
Solution:
(a) \(\frac{1}{2}\)

Question 4.
Write the value of:
\(\int_0^{\pi / 2} \frac{\sin x}{\sin x+\cos x}\) dx – \(\int_0^{\pi / 2} \frac{\cos x}{\sin x+\cos x}\) dx
(a) 1
(b) 2
(c) 0
(d) π
Solution:
(c) 0

Question 5.
Write the value of \(\int_{\frac{\pi}{4}}^{\frac{\pi}{4}}\)sin⁵ x cos x dx
(a) 0
(b) 1
(c) cos x
(d) sin x
Solution:
(a) 0

Question 6.
Write the particular solution of the equation \(\frac{d y}{d x}\) = sin x given that y(π) = 2
(a) y = cos x + 1
(b) y = -cos x + 1
(c) y = -cos x – 1
(d) y = -sin x + 1
Solution:
(b) y = -cos x + 1

Question 7.
Write the degree of the following differential equation:
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{2 y^3+\left(\frac{d y}{d x}\right)^4}{\sqrt{\frac{d^2 y}{d x^2}}}\)
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

Question 8.
Write the order ofthe following differential equation:
\(\frac{d^2 y}{d x^2}\) = \(\frac{2 y^3+\left(\frac{d y}{d x}\right)^4}{\sqrt{\frac{d^2 y}{d x^2}}}\)
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2

Question 9.
What is F(x) if F(x) = \(\int_0^x\)e^2t sin 3t dt?
(a) e^2x sin 3x
(b) e^2x cos 3x
(c) e^x sin 3x
(d) e^2x sin x
Solution:
(a) e^2x sin 3x

Question 10.
\(\int \frac{d x}{\cos ^2 x \sin ^2 x}\) = ?
(a) -2 cos 2x + C
(b) -2 cot 2x + C
(c) -2 sin 2x + C
(d) 2 cot 2x + C
Solution:
(b) -2 cot 2x + C

Question 11.
If \(\int_1^2\)f(x) dx= λ, then what is the value of \(\)f(3 – x) dx?
(a) λ
(b) λ²
(c) 1λ
(d) 2λ
Solution:
(a) λ

Question 12.
What is the value of \(\int_{-1}^1 \frac{d x}{1+x^2}\)?
(a) \(\frac{2 \pi}{2}\)
(b) 2π
(c) π
(d) \(\frac{\pi}{2}\)
Solution:
(d) \(\frac{\pi}{2}\)

Question 13.
Write the order of the following differential equation:
\(\frac{d^3 y}{d x^3}\) = \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^4\) + y
(a) 1
(b) 3
(c) 2
(d) 0
Solution:
(b) 3

Question 14.
Write the degree of the following differential equation:
\(\frac{d^3 y}{d x^3}\) = \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^4\) + y
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(a) 1

Question 15.
Write the particular solution of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (1 + x)⁴, y = 0 when x = -1.
(a) y = \(\frac{(1+x)^2}{5}\)
(b) y = \(\frac{(2+x)^5}{5}\)
(c) y = \(\frac{(1-x)^5}{5}\)
(d) y = \(\frac{(1+x)^5}{5}\)
Solution:
(d) y = \(\frac{(1+x)^5}{5}\)

Question 16.
Evaluate the integral ∫2x cosec² x² dx?
(a) cot x² + C
(b) -cot x² + C
(c) -cot 2x² + C
(d) cot 2x² + C
Solution:
(b) -cot x² + C

Question 17.
What is the value of \(\frac{d}{d x} \int_{250}^{300}\left(x^4+5 x^3\right)^2\) dx
(a) 0
(b) 1
(c) -1
(d) 2
Solution:
(a) 0

Question 18.
Write down the integral of ∫\(e^{x^2}\) 2x dx.
(a) \(e^{2 x^2}\)
(b) 2\(e^{2 x^2}\)
(c) \(e^{x^2}\)
(d) None of the above
Solution:
(c) \(e^{x^2}\)

Question 19.
What is the integral of ∫log e^x dx?
(a) \(\frac{2 x^2}{2}\) + C
(b) \(\frac{2 x^2}{3}\) + C
(c) \(\frac{x^2}{2}\) + C
(d) None of the above
Solution:
(c) \(\frac{x^2}{2}\) + C

Question 20.
What is the value of \(\int_{-2}^2\)|x| dx?
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0

Question 21.
\(\int_{-1}^1\)|1 – x| dx = ______.
(a) 0
(b) 1
(c) 2
(d) -1
Solution:
(c) 2

Question 22.
If ∫x³\(e^{c x^4}\)dx = \(\frac{1}{20} \mathrm{e}^{\mathrm{cx}}\) then C = ______.
(a) 0
(b) 2
(c) 4
(d) 5
Solution:
(d) 5

Question 23.
\(\int_a^b\)f(x) dx = 1 ⇒ \(\int_a^b\)k f(t)dt ______.
(a) k
(b) -k
(c) 2k
(d) None of the above
Solution:
(b) -k

Question 24.
\(\int_{-1}^1\)f(x) dx = k and f is an even function then \(\int_{-1}^1\)f(x) = ______.
(a) k
(b) -k
(c) 2k
(d) None of the above
Solution:
(c) 2k

Question 25.
If ∫\(\int_0^1\)f(x) dx = 4, \(\int_0^2\)f(t) dt and \(\int_4^2\)f(u) du = 1 then \(\int_1^4\)f(x) dx = ______.
(a) 0
(b) 1
(c) 3
(d) -3
Solution:
(d) -3

Question 26.
I(f) = \(\int_a^x\)f(t) dt and Df = f'(x) then (ID – DI) f = ______.
(a) -f(a)
(b) 2f(a)
(c) f(a)
(d) None of the above
Solution:
(a) -f(a)

Question 27.
\(\int_0^\pi\)cos¹⁰¹ x dx = ______.
(a) 0
(b) 1
(c) -1
(d) 101
Solution:
(a) 0

Question 28.
Let f satisfies all the conditions of Rolle’s theorem in [1, 6] then \(\int_1^6\)f'(x) dx = ______.
(a) 0
(b) 1
(c) -1
(d) 6
Solution:
(a) 0

Question 29.
\(\int_{-2}^2\)|x| dx = ______.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4

Question 30.
Integrate ∫log x dx
(a) x. log x + x + C
(b) x. log x – x + C
(c) log x – x + C
(d) None of these
Solution:
(b) x. log x – x + C

Question 31.
Evaluate \(\int_0^2\)[x – 1] dx
(a) 0
(b) 1
(c) -1
(d) 2
Solution:
(b) 1

Question 32.
What is the value of: ∫\(\frac{f^{\prime}(x)-f(x)}{e^x}\) dx?
(a) e^x f(x) + C.
(b) e^2x f(x) + C.
(c) e^-x f(x) + C.
(d) None of the above
Solution:
(c) e^-x f(x) + C.

Question 33.
What is the value of \(\int_0^1\)x(1 – x)⁹⁹ dx?
(a) \(\frac{1}{100}\)
(b) \(\frac{1}{10}\)
(c) \(\frac{1}{1010}\)
(d) \(\frac{1}{10100}\)
Solution:
(d) \(\frac{1}{10100}\)

Question 34.
Solution of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy + x + y + 1 is ______.
(a) 2x + \(\frac{x^2}{2}\) + C
(b) x + \(\frac{x}{2}\) + C
(c) x + \(\frac{2 x^2}{2}\) + C
(d) x + \(\frac{x^2}{2}\) + C
Solution:
(d) x + \(\frac{x^2}{2}\) + C

Question 35.
f(x) = \(\int_0^x\)t sin t dt then f ‘(x) = ______.
(a) x cos x
(b) x sin t
(c) x sin x
(d) x tan x
Solution:
(c) x sin x

Question 36.
What is the value of the integral \(\int_a^b \frac{|x|}{x}\)dx?
(a) |b| – |a|
(b) |a| – |b|
(c) |b| + |a|
(d) |a| + |b|
Solution:
(a) |b| – |a|

Question 37.
What is the value of ∫x^x (1 + ln x) dx?
(a) x^2x + C
(b) x^x + C
(c) 2x^x + C
(d) x² + C
Solution:
(b) x^x + C

Question 38.
Evaluate: \(\int_0^{\mathrm{p} / 2}\)ln(cot x) dx.
(a) 0
(b) 1
(c) cot x
(d) sin x
Solution:
(a) 0

Question 39.
Evaluate: \(\int_{-3}^4\)|x| dx
(a) \(\frac{2}{25}\)
(b) \(\frac{25}{2}\)
(c) \(\frac{25}{4}\)
(d) \(\frac{25}{-3}\)
Solution:
(b) \(\frac{25}{2}\)

Question 40.
Evaluate: \(\int_0^{\frac{\pi}{2}}\)(cos x – sin x) dx
(a) 0
(b) 1
(c) -1
(d) π
Solution:
(a) 0

Question 41.
Evaluate: \(\int_0^{\frac{\pi}{2}}\)log tan x dx.
(a) 1
(b) -1
(c) 0
(d) π
Solution:
(c) 0

Question 42.
Integrate: \(\frac{d x}{3 e^x-1}\)
(a) \(\ln \left(\frac{e^{3 x}-1}{e^x}\right)\) + C
(b) \(\ln \left(\frac{3 e^x+1}{e^x}\right)\) + C
(c) \(\ln \left(\frac{3 e^x-1}{e^x}\right)\) + C
(d) \(\ln \left(\frac{3 e^x+1}{e^{3 x}}\right)\) + C
Solution:
(c) \(\ln \left(\frac{3 e^x-1}{e^x}\right)\) + C

Question 43.
Evaluate: \(\int_0^1 \ln \left(\frac{1}{x}-1\right)\)dx
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(c) 0

Question 44.
Evaluate: ∫e^x\(\left(\frac{1-\sin x}{1-\cos x}\right)\)dx
(a) -e^x cot\(\frac{x}{2}\) + C
(b) e^x tan\(\frac{x}{2}\) + C
(c) e^x cot\(\frac{x}{2}\) + C
(d) -e^x sin\(\frac{x}{2}\) + C
Solution:
(a) -e^x cot\(\frac{x}{2}\) + C

Question 45.
Evaluate: \(\int_0^1\)x log(1 + x) dx
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{2}{3}\)
Solution:
(b) \(\frac{1}{4}\)

Question 46.
What is the integrating factor of the equation y’ + y cot x = cosec x?
(a) cot x
(b) sin x
(c) cos x
(d) cosec x
Solution:
(b) sin x

(B) Very Short Type Questions With Answers

Question 1.
Write the order of the differential equation whose solution is given by
y = (c₁ + c₂) cos (x + c₃) + c₄\(e^{x+c_5}\) where c₁, c₂, c₄ and c₅ are arbitrary constants.
Solution:
y = (c₁ + c₂) cos (x + c₃) + c₄\(e^{x+c_5}\)
y = (c₁ + c₂) cos (x + c₃) + c₄\(e^{c_5}\).e^x
= A cos(x + c₃) + Be^x
Where c₁ + c₂ = A, c₄\(e^{c_5}\) = B
As there are 3 independent constants the order of the differential equation is 3.

Question 2.
If p and q are respectively degree and order of the differential equation y = e^dy/dx, then write the relation between p and q.
Solution:
Given differential equation is
y = \(e^{\frac{d y}{d x}}\) ⇒ \(\frac{d y}{d x}\) = ln y
Whose order = 1 = p
Degree = 1 = q
∴ p = q

Question 3.
Write the value of \(\int_0^1\){x} dx where {x} stands for fractional part of x.
Solution:

Question 4.
Write the order of the differential equation of the family of circles
ar² + ay² + 2gx + 2fy + c = 0
ax² + ay² + 2gx + 2fy + c = 0
Solution:
As there are 3 independent constants, the order of the differential equation is 3.

Question 5.
If p and q are the order and degree of the differential equation
y\(\left(\frac{d y}{d x}\right)^2\) + x² \(\frac{d^2 y}{d x^2}\) + xy = sin x, then choose the correct statement out of (i) p > q, (ii) p = q, (iii) p < q.
Solution:
Order of the given differential = p = 2
Degree of the given differential equation = q = 1
∴ p > q

Question 6.
Write the order of the differential equation of the system of ellipses:
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
Solution:
As there are two unknown constants in the system of ellipses \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 the order of the differential equation is 2.

Question 7.
What do you mean by integration? Write your answer in one sentence.
Solution:
Integration is the antiderivative of a function.

Question 8.
Write the differential equation of the family of straight lines parallel to the y-axis.
Solution:
\(\frac{d x}{d y}\) = 0 is the differential equation of family of lines parallel to y-axis.

Question 9.
Write the value of ∫\(\int_{-\pi / 4}^{\pi / 4}\)sin⁵ x cos x dx.
Solution:
Let f(x) = sin⁵ x cos x
f(-x) = sin⁵ (-x) cos (-x)
= -sin⁵ x cos x = -f(x)
i.e. f is an odd function.
Thus \(\int_{-\pi / 4}^{\pi / 4}\)sin⁵ x cos x dx = 0

Question 10.
Write the degree of the differential equation ln\(\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)\) = y
Solution:
The degree of the differential equation ln\(\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)\) = y is 1.

Question 11.
What is F'(t) if F(t) = \(\int_a^t\)e^3x .cos 2x dx ?
Solution:
F(t) = \(\int_a^t\)e^3x .cos 2x dx
⇒ F'(t) = e^3x cos 2t

Question 12.
Write the order and degree of the following differential equation:
\(\frac{d^2 y}{d x^2}\) = \(\frac{2 y^3+\left(\frac{d y}{d x}\right)^4}{\sqrt{\frac{d^2 y}{d x^2}}}\)
Solution:
Order = 2, Degree = 3

Question 13.
∫\(\frac{\cot x d x}{\ln \sin x}\) = ?
Solution:
∫\(\frac{\cot x d x}{\ln \sin x}\) = ln(ln sin x) + C

Question 14.
What is F'(x) if F(x) = \(\int_0^{\mathbf{x}}\)e^2t sin 3t dt?
Solution:
If F(x) = \(\int_0^{\mathbf{x}}\)e^2t sin 3t dt then F'(x) = e^2x sin 3x

Question 15.
∫\(\frac{d x}{\cos ^2 x \sin ^2 x}\) = ?
Solution:
∫\(\frac{d x}{\cos ^2 x \sin ^2 x}\) = 4∫\(\frac{d x}{\sin ^2 2 x}\)
= 4∫cosec² 2x dx = -2 cot 2x + C

Question 16.
What is the value of ∫\(\frac{d}{d x}\)f(x) dx – \(\frac{d}{d x}\)(∫f(x) dx)?
Solution:
∫\(\frac{d}{d x}\)f(x) dx – \(\frac{d}{d x}\)(∫f(x) dx)
= f(x) + C – f(x) = C (constant)

Question 17.
If \(\int_1^2\)f(x) dx = λ, then what is the value \(\int_1^2\)f(3 – x) dx?
Solution:
If \(\int_1^2\)f(x) dx = λ, then \(\int_1^2\)f(3 – x) dx = λ

Question 18.
What is the value of \(\int_{-1}^1 \frac{d x}{1+x^2}\)?
Solution:
\(\int_{-1}^1 \frac{d x}{1+x^2}\) = \(\left[\tan ^{-1} x\right]_{-1}^1\)
= tan^-1 1 – tan^-1 (-1)
= tan^-1 1 + tan^-1 1
= 2tan^-1 (1) = 2 . \(\frac{\pi}{4}\) = \(\frac{\pi}{2}\)

Question 19.
Write the order and the degree of the following differential equation:
\(\frac{d^3 y}{d x^3}\) = \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{d y}{d x}\right)^4\) + y
Solution:
Order = 3
Degree = 1

Question 20.
Write the particular solution of \(\frac{d y}{d x}\) = (1 + x)⁴, y = 0 when x = -1.
Solution:
\(\frac{d y}{d x}\) = (1 + x)⁴ ⇒ \(\frac{(1+x)^5}{5}\) + C
Given y = 0 for x = -1
⇒ o = o + c ⇒ c = o
∴ The particular solution is y = \(\frac{(1+x)^5}{5}\)

(C) Short Type Questions With Answers

Question 1.
Evaluate: ∫\(\frac{2 x+1}{\sqrt{x^2+10 x+29}}\)dx
Solution:

Question 2.
Evaluate: \(\int_0^{\pi / 2} \frac{\cos x d x}{(2-\sin x)(3+\sin x)}\)
Solution:

Question 3.
Evaluate: ∫\(\frac{d x}{(1+x) \sqrt{1-x^2}}\)
Solution:

Question 4.
Solve: cosec x \(\frac{d^2 y}{d x^2}\) = x.
Solution:
cosec x \(\frac{d^2 y}{d x^2}\) = x => \(\frac{d^2 y}{d x^2}\) = x sin x
⇒ \(\frac{d y}{d x}\) = ∫x sin x dx + A
= x (-cos x) – ∫(-cos x) dx + A
= -x cos x + sin x + A
⇒ y = -∫x cos x dx + ∫sin x dx + A∫dx + B
= [x sin x – ∫sin x dx] – cos x + Ax = B
⇒ y = -x sin x – 2 cos x + Ax + B is the solution.

Question 5.
Find the particular solution of the following differential equation:
\(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\) given that y = √3 when x = 1
Solution:

Question 6.
Evaluate: \(\int_0^a x^2\left(a^2-x^2\right)^{5 / 2}\) dx
Solution:

Question 7.
Evaluate: \(\int_0^a \frac{d x}{e^{4 x}-5}\)
Solution:

Question 8.
Evaluate: ∫x² tan^-1 x dx.
Solution:

Question 9.
If f(x) = e^x + \(\frac{1}{1+x^2}\) and f(0) = 1, then find f(x).
Solution:
f(x) = e^x + \(\frac{1}{1+x^2}\)
⇒ f(x) = ∫\(\left(e^x+\frac{1}{1+x^2}\right)\)dx + C
= e^x + tan^-1 x + C
f(0) = 1
⇒ 1 = 1 + 0 + C => C = 0
Thus f(x) = e^x + tan^-1 x

Question 10.
Evaluate: ∫(log x)² dx
Solution:
I = ∫(log x)² dx
= (log x)². x – 2∫(log x) . \(\frac{1}{x}\) . x . dx
= x (log x)² – 2 ∫log x. dx
= x (log x)² – 2 {(log x) x – ∫dx}
= x (log x)² – 2x log x + 2x + C

Question 11.
Evaluate: ∫\(\frac{2 x+9}{(x+3)^2}\)dx
Solution:

Question 12.
Solve: ydy + e^-y x sin x dx = 0
Solution:
ydy = e^-y x sin x dx = 0
⇒ y e^y dy + x sin x dx = 0
⇒ ∫y e^y dy + ∫x sin x dx =C
⇒ y e^y – e^y + (-x cos x) + sin x = C
⇒ e^y (y – 1) – x cos x + sin x = C is the general solution.

Question 13.
Evaluate: ∫\(\frac{d x}{x \ln x \sqrt{(\ln x)^2-4}}\)
Solution:

Question 14.
Find the particular solution of the differential equation \(\frac{d^2 y}{d x^2}\) = 6x given that y = 1 and \(\frac{d y}{d x}\) = 2 when x = 0.
Solution:
\(\frac{d^2 y}{d x^2}\) = 6x ⇒ \(\frac{d y}{d x}\) = 6 . \(\frac{x^2}{2}\) + A
\(\frac{d y}{d x}\) = 3x² + A ⇒ y = x³ + Ax + B
Using the givne conditions x = 0, \(\frac{d y}{d x}\) = 2, y = 1, we get
2 = 0 + A ⇒ A = 2
and 1 = 0 + 0 + B ⇒ B = 1
The particular solution is y = x³ + 2x + 1

Question 15.
Evaluate: \(\int_0^{\frac{3}{2}}\)[x²] dx
Solution:

Question 16.
Find the differential equation whose general solution is ax² + by = 1, where a and b are arbitrary constants.
Solution:

Question 17.
Integrate: ∫\(\frac{\sin 6 x+\sin 4 x}{\cos 6 x+\cos 4 x}\) dx.
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(a)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) \(\vec{a}\) = î + 2ĵ + k̂, \(\vec{b}\) = 2î – 2ĵ + 2k̂ and \(\vec{c}\) = -î + 2 ĵ + k̂ then
(a) \(\vec{a}\) and \(\vec{b}\) have the same direction
(b) \(\vec{a}\) and \(\vec{c}\) have opposite directions.
(c) \(\vec{b}\) and \(\vec{c}\) have opposite directions
(d) no pair of vectors have same direction
Solution:
(d) no pair of vectors have same direction

(ii) If the vectors \(\vec{a}\) = 2î + 3ĵ – 6k̂ and \(\vec{b}\) = -α î – ĵ + 2k̂ are parallel, then α = ______.
(a) 2
(b) \(\frac{2}{3}\)
(c) –\(\frac{2}{3}\)
(d) \(\frac{1}{3}\)
Solution:
(c) –\(\frac{2}{3}\)

(iii) If the position vectors of two points A and B are 3î + k̂, and 2î + ĵ – k̂, then the vector \(\overrightarrow{BA}\) is
(a) -î + ĵ – 2k̂
(b) î + ĵ
(c) î – ĵ + 2k̂
(d) î – ĵ – 2k̂
Solution:
(c) î – ĵ + 2k̂

(iv) If \(|k \vec{a}|\) = 1, then
(a) \(\vec{a}=\frac{1}{k}\)
(b) \(\vec{a}=\frac{1}{|k|}\)
(c) \(k=\frac{1}{|\vec{a}|}\)
(d) \(k=\frac{+1}{|\vec{a}|}\)
Solution:
(d) \(k=\frac{+1}{|\vec{a}|}\)

(v) The direction cosines of the vectors \(\overrightarrow{PQ}\) where \(\overrightarrow{OP}\) = (1, 0, -2) and \(\overrightarrow{OQ}\) = (3, -2, 0) are
(a) 2, -2, 2
(b) 4, -2, -2
(c) \(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
(d) \(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}\)
Solution:
(c) \(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)

Question 2.
Rectify the mistakes, if any
(i) \(\vec{a}-\vec{a}\) = 0
Solution:
\(\overrightarrow{0}\)

(ii) The vector \(\overrightarrow{0}\) has unique direction.
Solution:
indefinite direction

(iii) All unit vectors are equal.
Solution:
equal magnitude

(iv) \(|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\vec{b}\)
Solution:
\(\vec{a}=\vec{b} \Rightarrow|\vec{a}|=|\vec{b}|\)

(v) Subtraction of vectors is not commutative.
Solution:
true

Question 3.
(i) If \(\vec{a}\) = (2, 1), \(\vec{b}\) = (-1, 0), find \(3 \vec{a}+2 \vec{b}\).
Solution:
\(3 \vec{a}+2 \vec{b}\) = 3 (2, 1) + 2 (-1, 0)
= (6 – 2, 3 + 0)
= (4, 3 )

(ii) If \(\vec{a}\) = (1, 1, 1) , \(\vec{b}\) = (-1, 3, 0) and \(\vec{c}\) =(2, 0, 2), find \(\vec{a}+2 \vec{b}-\frac{1}{2} \vec{c}\).
Solution:
\(\vec{a}+2 \vec{b}-\frac{1}{2} \vec{c}\)
= (1, 1, 1) + 2 (-1, 3, 0) – \(\frac{1}{2}\)(2, 0, 2)
= (1 – 2 – 1, 1 + 6 – 0, 1 + 0 – 1)
= (-2, 7, 0)

Question 4.
If A, B, C and D are the vertices of a square, find \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}\).
Solution:
Let ABCD be a square.

Question 5.
The given points A, B, C are the vertices of a triangle. Determine the vectors \(\overrightarrow{A B}, \overrightarrow{B C} \text { and } \overrightarrow{C A}\) and the lengths of these vectors in the following cases.
(i) A (4, 5, 5), B (3, 3, 3), C (1, 2, 5)
Solution:

(ii) A (8, 6, 1), B (2, 0, 1), C (-4, 0, -5)
Solution:

Question 6.
Find the vector from origin to the midpoint of the vector \(\overrightarrow{{P}_1 {P}_2}\) joining the points P₁(4, 3) and P₂(8, -5).
Solution:
P₁ = (4, 3) and P₂ = (8, -5)
If P is the mid-point of P₁P₂ then P = (6, -1).
Position vector of P = \(\overrightarrow{{OP}}\) = 6î – ĵ

Question 7.
Find the vectors from the origin to the points of trisection the vector \(\overrightarrow{{P}_1 {P}_2}\) joining P₁ (-4, 3) and P₂ (5, -12).
Solution:

Question 8.
Find the vector from the origin to the intersection of the medians of the triangle whose vertices are A (5, 2, 1), B(-4, 7, 0) and C (5, -3, 5).
Solution:

Question 9.
Prove that the sum of all the vectors drawn from the centre of a regular octagon to its vertices is the null vector.
Solution:

Question 10.
Prove that the sum of the vectors represented by the sides of a closed polygon taken in order is a zero vector.
Solution:
Consider a closed polygon ABCDEFA.

Question 11.
(a) Prove that:
(i) \(|\overrightarrow{a}+\overrightarrow{{b}}| \leq|\overrightarrow{a}|+|\overrightarrow{b}|\)
State when the equality will hold;
Solution:

(ii) \(|\overrightarrow{a}-\overrightarrow{b}| \geq|\overrightarrow{a}|-|\overrightarrow{b}|\)
Solution:

(b) What is the geometrical significance of the relation \(|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|\)?
Solution:

Question 12.
Find the magnitude of the vector \(\overrightarrow{PQ}\), its scalar components and the component vectors along the coordinate axes, if P and Q have the coordinates.
(i) P (-1, 3), Q (1, 2)
Solution:

(ii) P (-1, -2), Q (-5, -6)
Solution:

(iii) P (1, 4, -3), Q (2, -2, -1).
Solution:

Question 13.
In each of the following find the vector \(\overrightarrow{PQ}\), its magnitude and direction cosines, if P and Q have co-ordinates.
(i) P (2, -1, -1), Q (-1, -3, 2);
Solution:

(ii) P (3, -1, 7), Q (4, -3, -1).
Solution:

Question 14.
If \(\vec{a}\) = (2, -2, 1), \(\vec{b}\) = (2, 3, 6) and \(\vec{c}\) = (-1, 0, 2), find the magnitude and direction of
\(\vec{a}-\vec{b}+2 \vec{c}\).
Solution:

Question 15.
Determine the unit vector having the direction of the given vector in each of the following problems:
(i) 5î – 12ĵ
Solution:

(ii) 2î + ĵ
Solution:

(iii) 3î + 6ĵ – k̂
Solution:

(iv) 3î + ĵ – 2k̂
Solution:

Question 16.
Find the unit vector in the direction of the vector \(\overrightarrow{r_1}-\overrightarrow{r_2}\), where \(\vec{r}_1\) = î + 2ĵ + k̂ and \(\vec{r}_2\) = 3î + ĵ – 5k̂.
Solution:

Question 17.
Find the unit vector parallel to the sum of the vectors \(\vec{a}\) = 2î + 4ĵ – 5k̂ and \(\vec{b}\) = î + 2ĵ + 3k̂. Also find its direction cosines.
Solution:

Question 18.
If the sum of two unit vectors is a unit vector, show that the magnitude of their difference is √3.
Solution:

Question 19.
The position vectors of the points A, B, C and D are 4î + 3ĵ – k̂, 5î + 2ĵ + 2k̂, 2î – 2ĵ – 3k̂ and 4î – 4ĵ + 3k̂ respectively. Show that AB and CD are parallel.
Solution:
Given that the
position vector of A = 4î + 3ĵ – k̂
position vector of B = 5î + 2ĵ + 2k̂
position vector of C = 2î – 2ĵ – 3k̂
position vector of D = 4î – 4ĵ + 3k̂

Question 20.
In each of the following problems, show by vector method that the given points are collinear.
(i) A (2, 6, 3), B (1, 2, 7) and C (3, 10, -1)
Solution:
Given that A = (2, 6, 3), B = (1, 2, 7) and C = (3, 10, -1)
Then

(ii) P (2, -1, 3), Q (3, -5, 1) and R (-1, 11, 9).
Solution:
Given that P = (2, -1, 3) Q = (3, -5, 1) and R = (-1, 11, 9)

Hence the points P, Q, R are collinear. (Proved)

Question 21.
Prove that the vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂, 3î – 4ĵ – 4k̂ are the sides of a right angled triangle.
Solution:
Let A, B and C be the points whose position vectors are 2î – ĵ – k̂, î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂ respectively.

Question 22.
Prove by vector method that:
(a) the medians of a triangle are concurrent;
Solution:

The symmetry of the result shows that the point G also lies on the other two medians.
Hence the medians are concurrent. (Proved)

(b) the diagonals of a parallelogram bisect each other;
Solution:

(c) the line segment joining the midpoints of two sides of a triangle is parallel to the third and half of it;
Solution:

(d) the lines joining the midpoints of consecutive sides of a quadrilateral is a parallelogram;
Solution:

⇒ SR = PQ and SR || PQ
Hence PQRS is a parallelogram.
(Proved)

(e) in any triangle ABC, the point P being on the side \(\overrightarrow{B C} \text {; if } \overrightarrow{P Q}\) is the resultant of the vectors \(\overrightarrow{A P}, \overrightarrow{P B}\) and \(\overrightarrow{P C}\) then ABQC is a parallelogram;
Solution:

Hence ABQC is parallelogram. (Proved)

(f) In a parallelogram, the line joining a vertex to the midpoint of an opposite side trisects the other diagonal.
Solution:

⇒ P divides BD into the ratio 1 : 2.
Similarly we can show that Q divides BD into the ratio 2 : 1.
Hence P, Q are the points of trisection of the diagonal BD. (Proved)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(b)

Solve the following differential equations.
Question 1.
\(\frac{d y}{d x}\) + y = e^-x
Solution:
Given equation is \(\frac{d y}{d x}\) + y = e^-x … (1)
This is a linear differential equation.
Here P = 1, Q = e^-x
So the integrating factor
I.F. = e^{∫P dx} = e^∫dx = e^x
The solution of (1) is given by
ye^x = ∫e^-x . e^x dx = ∫dx = x + C
⇒ y – xe^-x + Ce^-x

Question 2.
(x² – 1)\(\frac{d y}{d x}\) + 2xy = 1
Solution:
Given equation is (x² – 1)\(\frac{d y}{d x}\) + 2xy = 1

Question 3.
(1 – x²)\(\frac{d y}{d x}\) + 2xy = x \(\sqrt{1-x^2}\)
Solution:
Given equation is

Question 4.
x log x \(\frac{d y}{d x}\) + y = 2 log x
Solution:
Given equation is

Question 5.
(1 + x²)\(\frac{d y}{d x}\) + 2xy = cos x
Solution:

Question 6.
\(\frac{d y}{d x}\) + y sec x = tan x
Solution:
Given equation is
\(\frac{d y}{d x}\) + y sec x = tan x
This is a linear equation where
P = sec x, Q = tan x
I.F. = e^{∫sec dx}
= e^{(sec x + tan x)} = sec x + tan x
The solution is y . (sec x + tan x)
= ∫(sec x + tan x) tan x dx
= ∫(sec x tan x + tan² x) dx
= ∫(sec x . tan x + sec² x – 1) dx
= ∫(sec x + tan x) – x + C
⇒ (y – 1) (sec x + tan x) + x = C

Question 7.
(x + tan y) dy = sin 2y dx
Given equation can be written as
Solution:

Question 8.
(x + 2y³)\(\frac{d y}{d x}\) = y
Solution:
Given equation can be written as

Question 9.
sin x\(\frac{d y}{d x}\)+ 3y = cos x
Solution:

Question 10.
(x + y + 1)\(\frac{d y}{d x}\) = 1
Solution:

Question 11.
(1 + y²) dx + (x – \(e^{-\tan ^{-1} y}\)) dy = 0
Solution:
Given equation can be written as

Question 12.
x\(\frac{d y}{d x}\) + y = xy²
Solution:
Given equation can be written as

⇒ z = -x ln x + Cx
⇒ \(\frac{1}{y}\) = -x ln x + Cx
⇒ 1 = -xy ln x + Cxy
∴ The solution is (C – ln x) xy = 1

Question 13.
\(\frac{d y}{d x}\) + y = y² log x
Solution:

Question 14.
(1 + x²)\(\frac{d y}{d x}\) = xy – y²
Solution:
The given equation can be written as

Question 15.
\(\frac{d y}{d x}\) + \(\frac{y}{x-1}\) = \(x y^{\frac{1}{2}}\)
Solution:
The given equation can be written as

Question 16.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x², y(1) = 1
Solution:
The given equation can be written as
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x², y(1) = 1 … (1)
This is a linear equation.

Question 17.
\(\frac{d y}{d x}\) + 2y tan x = sin x, y\(\left(\frac{\pi}{3}\right)\) = 0.
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(a)

Question 1.
Determine the order and degree of each of the following differential equations.
(i) y sec² x dx + tan x dy = 0
Solution:
Order: 1, Degree: 1

(ii) \(\left(\frac{d y}{d x}\right)^4\) + y⁵ = \(\frac{d^3 y}{d x^3}\)
Solution:
Order: 3, Degree: 1

(iii) a\(\frac{d^2 y}{d x^2}\) = \(\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{\frac{3}{2}}\)
Solution:
Order: 2, Degree: 2

(iv) tan^-1\(\sqrt{\frac{d y}{d x}}\) = x
Solution:
Order: 1, Degree: 1

(v) ln\(\left(\frac{d^2 y}{d x^2}\right)\) = y
Solution:
Order: 2, Degree: 1

(vi) \(\frac{\frac{d y}{d t}}{y+\frac{d y}{d t}}\) = \(\frac{y t}{d y}\)
Solution:
Order: 1, Degree: 2

(vii) \(\frac{d^2 y}{d u^2}\) = \(\frac{3 y+\frac{d y}{d u}}{\sqrt{\frac{d^2 y}{d u^2}}}\)
Solution:
Order: 2, Degree: 3

(viii) \(e^{\frac{d z}{d x}}\) = x²
Solution:
Order: 1, Degree: 1

Question 2.
Form the differential equation by eliminating the arbitrary constants in each of the following cases.
(i) y = A sec x
Solution:
y = A sec x
Then \(\frac{d y}{d x}\) = A sec x tan x = y tan x

(ii) y = C tan^-1 x
Solution:

(iii) y = Ae^t + Be^2t
Solution:

(iv) y = Ax² + Bx
Solution:

(v) y = -acos x + b sin x
Solution:

(vi) y = a sin^-1 x + b cos^-1 x
Solution:

(vii) y = at + be^t
Solution:

(viii) y = a sin t + be^t
Solution:

(ix) ax² + by = 1
Solution:

Question 3.
Find the general solution ofthe following differential equations.
(i) \(\frac{d y}{d x}\) = \(\frac{e^{2 x}+1}{e^x}\)
Solution:
\(\frac{d y}{d x}\) = \(\frac{e^{2 x}+1}{e^x}\)
⇒ y = ∫(e^x + e^-x) dx = e^x – e^-x + C

(ii) \(\frac{d y}{d x}\) = x cos x
Solution:
\(\frac{d y}{d x}\) = x cos x
⇒ y = ∫x cos x dx
= x . sin x – ∫sin x dx – x sin x + cos x + C

(iii) \(\frac{d y}{d x}\) = t⁵ log t
Solution:

(iv) \(\frac{d y}{d x}\) = 3t² + 4t + sec² t
Solution:
\(\frac{d y}{d x}\) = 3t² + 4t + sec² t
⇒ y = t³ + 2t² + tan t + C

(v) \(\frac{d y}{d x}\) = \(\frac{1}{x^2-7 x+12}\)
Solution:

(vi) \(\frac{d y}{d u}\) = \(\frac{u+1}{\sqrt{3 u^2+6 u+5}}\)
Solution:

(vii) (x² + 3x + 2) dy – dx = 0
Solution:

(viii) \(\frac{d y}{d t}\) = \(\frac{\sin ^{-1} t e^{\sin ^{-1} t}}{\sqrt{1-t^2}}\)
Solution:

Question 4.
Solve the following differential equations.
(i) \(\frac{d y}{d x}\) = y + 2
Solution:

(ii) \(\frac{d y}{d t}\) = \(\sqrt{1-y^2}\)
Solution:
\(\frac{d y}{d t}\) = \(\sqrt{1-y^2}\)
⇒ \(\frac{d y}{\sqrt{1-y^2}}\) = dt
⇒ sin^-1 y = t + C

(iii) \(\frac{d y}{d z}\) = sec y
Solution:
\(\frac{d y}{d z}\) = sec y
⇒ cos y dy = dz
⇒ sin y = z + C

(iv) \(\frac{d y}{d x}\) = e^y
Solution:

(v) \(\frac{d y}{d x}\) = y² + 2y
Solution:

(vi) dy + (y² + 1) dx = 0
Solution:

(vii) \(\frac{d y}{d x}\) + \(\frac{e^y}{y}\) = 0
Solution:

(viii) dx + cot x dt = 0
Solution:
dx + cot x dt = 0
⇒ tan x dx + dt = 0
⇒ ∫tan x dx + ∫dt = C₁
⇒ In sec x + t = C₁
⇒ In sec x = C₁ – t
⇒ sec x = \(e^{C_1}\) . e^-t
⇒ cos x = \(e^{-C_1}\) . e^t
⇒ cos x = Ce^t where C = \(e^{-C_1}\)

Question 5.
Obtain the general solution of the following differential equations.
(i) \(\frac{d y}{d x}\) = (x² + 1) (y² + 1)
Solution:

(ii) \(\frac{d y}{d t}\) = e^2t+3y
Solution:

⇒ 2e^-3y + 3e^2t + 6C₁ = 0
⇒ 2e^-3y + 3e^2t = C
where C = -6C₁

(iii) \(\frac{d y}{d z}\) = \(\frac{\sqrt{1-y^2}}{\sqrt{1-z^2}}\)
Solution:

(iv) \(\frac{d y}{d z}\) = \(\frac{x \log x}{3 y^2+4 y}\)
Solution:

(v) x²\(\sqrt{y^2+3}\) dx + y\(\sqrt{x^3+1}\) dy = 0
Solution:

(vi) tan y dx + cot x dy = 0
Solution:
tan y dx + cot x dy = 0
⇒ tan x . dx + cot y dy = 0
⇒ ∫tan x dx + ∫cot y dy = 0
⇒ -ln cos x + ln siny = ln C
⇒ ln\(\frac{\sin y}{\cos x}\) = ln C
⇒ \(\frac{\sin y}{\cos x}\) = C
⇒ sin y = C cos x

(vii) (x² + 7x + 12) dy + (y² – 6y + 5) dx = 0
Solution:

(viii) y dy + e^-y x sin x dx = 0
Solution:
y dy + e^-y x sin x dx = 0
⇒ ye^y dy + x sin x dx = 0
⇒ ∫ye^y dy + ∫x sin dx = C
[Integrating by parts.
⇒ ye^y – ∫e^y dy + x(-cos x) – ∫(-cos x) dx = C
⇒ ye^y – e^y – x cos x + sin x = C
⇒ (y – 1) e^y – x cos x + sin x = C

Question 6.
Solve the following second order equations.
(i) \(\frac{d^2 y}{d x^2}\) = 12x² + 2x
Solution:

(ii) \(\frac{d^2 y}{d t^2}\) =e^2t +e^-t
Solution:

(iii) \(\frac{d^2 y}{d \vartheta^2}\) = -sin υ + cos υ + sec² υ
Solution:
\(\frac{d^2 y}{d \vartheta^2}\) = -sin υ + cos υ + sec² υ
Integrating we get
\(\frac{d y}{d υ}\) = ∫sin υ dυ + ∫cos υ dυ + ∫sec² υ dυ
= cos υ + sin υ + tan υ + A
Again integratingwe get
y = ∫(cos υ + sin υ + tan υ + A)dυ + B
where A, B are arbritrary constants.
⇒ y = sin υ – cos υ + ln |sec υ| + A.υ. + B

(iv) cosec x \(\frac{d^2 y}{d x^2}\) = x
Solution:
cosec x \(\frac{d^2 y}{d x^2}\) = x
\(\frac{d^2 y}{d x^2}\) = x sin x
Integrating we get
\(\frac{d y}{d x}\) = ∫x sin x dx + A
= x . (-cos x) – ∫(-cos x) dx + A
= -x cos x + ∫cos x dx + A
= -x cos x + sin x + A
Again integrating we get
y = -∫x cos x dx + ∫sin x + ∫A dx + B
= -{x sin x -∫1 . sin x dx} – cos x + Ax + B
= -x sin x – 2cos x + Ax + B

(v) x²\(\frac{d^2 y}{d x^2}\) + 2 = 0
Solution:

(vi) sec x \(\frac{d^2 y}{d x^2}\) = sec 3x
Solution:

(vii) \(\frac{d^2 y}{d x^2}\) = sec² x + cos² x
Solution:

(viii) e^-x\(\frac{d^2 y}{d x^2}\) = x
Solution:
e^–x\(\frac{d^2 y}{d x^2}\) = x
⇒ \(\frac{d^2 y}{d x^2}\) = xe^x
Integrating we get
\(\frac{d y}{d x}\) = ∫xe^x dx = ∫e^x dx + Ax + B
= xe^x – e^x – e^x + Ax + B
= (x – 2)e^x + Ax + B

Question 7.
Find the particular solutions of the following equations subject to the given conditions.
(i) \(\frac{d y}{d x}\) = cos x, given that y = 2 when x = 0.
Solution:
\(\frac{d y}{d x}\) = cos x
Integrating we get
y = ∫cos x dx = sin x + C
Given that when x = 0, y = 2
So 2 = C
∴ The particular solution is y = sin x + 2

(ii) \(\frac{d y}{d t}\) = cos² y subject to y = \(\frac{\pi}{4}\) when t = 0.
Solution:
\(\frac{d y}{d t}\) = cos² y
⇒ sec² y dy = dt
∫sec² dy = ∫dt
⇒ tan y = t + C
When t = 0, y = \(\frac{\pi}{4}\)
So tan \(\frac{\pi}{4}\) = C ⇒ C = 1
∴ The particular solution is tan y = t + 1

(iii) \(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\) given that y = √3 when x = 1.
Solution:

(iv) \(\frac{d^2 y}{d x^2}\) = 6x given that y = 1 and \(\frac{d y}{d x}\) = 2 when x = 0.
Solution:
\(\frac{d^2 y}{d x^2}\) = 6x ⇒ \(\frac{d y}{d x}\) = 3x² + 2
When x = 0, \(\frac{d y}{d x}\) = 2
So 2 = A
∴ \(\frac{d y}{d x}\) = 3x² + 2
Again integrating we get
y = x³ + 2x + B
When x = 0, y = 1
So B = 1.
∴ The particular solution is y = x³ + 2x + 1

Question 8.
(i) Solve : \(\frac{d y}{d x}\) = sec (x + y)
Solution:

(ii) Solve : \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)
Solution:

(iii) Solve : \(\frac{d y}{d x}\) = cos (x + y)
Solution:

(iv) Solve : \(\frac{d y}{d x}\) + 1 = e^x+y
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(d)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) \(\vec{a} \cdot \vec{b} \times \vec{a}\) = _______.
(a) \(\overrightarrow{0}\)
(b) 0
(c) 1
(d) \(\vec{a}^2 \vec{b}\)
Solution:
\(\vec{a} \cdot(\vec{b} \times \vec{a})\) = \((\vec{b} \times \vec{a}) \cdot \vec{a}\)
= \(\vec{b} \cdot(\vec{a} \times \vec{a})\) = \(\vec{b} \cdot \overrightarrow{0}\)
= 0 [∴ Dot product is commutative and in the scalar triple product the dot and cross can be interchanged.]

(ii) \((-\vec{a}) \cdot \vec{b} \times(-\vec{c}))\) = _______.
(a) \(\vec{a} \times \vec{b} \cdot \vec{c}\)
(b) \(-\vec{a} \cdot(\vec{b} \times \vec{c})\)
(c) \(\vec{a} \times \vec{c} \cdot \vec{b}\)
(d) \(\vec{a} \cdot(\vec{c} \times \vec{b})\)
Solution:
\((-\vec{a}) \cdot \vec{b} \times(-\vec{c})\) = \(\vec{a} \cdot(\vec{b} \times \vec{c})\)

(iii) For the non-zero vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{a} \cdot(\vec{b} \times \vec{c})\) = 0 if
(a) \(\vec{b} \perp \vec{c}\)
(b) \(\vec{a} \perp \vec{b}\)
(c) \(\vec{a} \| \vec{c}\)
(d) \(\vec{a} \perp \vec{c}\)
Solution:
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = \((\vec{a} \times \vec{b}) \cdot \vec{c}\)
\(\vec{c} \perp(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})\)
but \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\)
∴ \(\vec{a} \| \vec{b}\)

Question 2.
Find the scalar triple product \(\vec{b} \cdot(\vec{c} \times \vec{a})\) where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are respectively.
(i) î + ĵ, î – ĵ, 5î + 2ĵ + 3k̂
Solution:

= 1 (0 – 3) + 1 (0 – 3) + 0 (5 – 2)
= 3 – 3 = -6

(ii) 5î – ĵ + 4k̂, 2î + 3ĵ + 5k̂, 5î – 2ĵ
Solution:

= 5 (18 + 10) + 1 (12 – 25) + 4 (- 4 – 15)
= 140 – 13 – 76 = 140 – 89 = 51

Question 3.
Find the volume of the parallelopiped whose sides are given by the vectors.
(i) î + ĵ + k̂, k̂, 3î – ĵ + 2k̂
Solution:

= 1 (0 + 1) – 1 (0 – 3) + 1 (0 – 0)
= 1 + 3 = 4 cube units.

(ii) (1, 0, 0), (0, 1, 0), (0, 0, 1).
Solution:

Question 4.
Show that the following vector are co-planar
(i) î – 2ĵ + 2k̂, 3î + 4ĵ + 5k̂, -2î + 4ĵ – 4k̂
Solution:

(ii) î + 2ĵ + 3k̂, -2î – 4ĵ + 5k̂, 3î + 6
Solution:

Question 5.
Find the value of λ so that the three vectors are co-planar.
(i) î + 2ĵ + 3k̂, 4î + ĵ + λk̂ and λî – 4ĵ + k̂
Solution:

(ii) (2, -1, 1), (1, 2, -3) and (3, λ, 5)
Solution:

⇒ 2 (10 + 3λ) + 1 (5 + 9) + 1 (λ – 6) = 0
⇒ 20 + 6λ +14 + λ – 6 = 0
⇒ 7λ + 28 = 0 ⇒ λ = -4

Question 6.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) mutually perpendiculars, show that \([\vec{a} .(\vec{b} \times \vec{c})]^2\) = a²b²c²
Solution:

Question 7.
Show that \([\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]\) = 2\([\vec{a} \vec{b} \vec{c}]\)
Solution:

Question 8.
Prove that \([\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]\) = \([\vec{a} \vec{b} \vec{c}]^2\)
Solution:

Question 9.
For \(\vec{a}\) = î + ĵ, \(\vec{b}\) = -î + 2k̂, \(\vec{c}\) = ĵ + k̂ obtain \(\vec{a} \times(\vec{b} \times \vec{c})\) and also verify the formula \(\vec{a} \times(\vec{b} \times \vec{c})\) = \((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}\).
Solution:

Question 10.
Prove that \(\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})\) and hence prove that \(\vec{a} \times(\vec{b} \times \vec{c}), \vec{b} \times(\vec{c} \times \vec{a}), \vec{c} \times(\vec{a} \times \vec{b})\) are coplanar.
Solution:

Question 11.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) unit vectors and \(\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}\) find the angles that â makes with b̂ and ĉ, where b̂, ĉ are not parallel.
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Additional Exercise Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Additional Exercise

Question 1.
∫\(\sqrt{1-\sin 2 x}\) dx
Solution:
I = ∫\(\sqrt{1-\sin 2 x}\) dx
= ∫\(\sqrt{(\cos x-\sin x)^2}\) dx
= ∫(cos x – sin x) dx
= sin x + cos x + c

Question 2.
∫\(\frac{d x}{1+\sin x}\)
Solution:
I = ∫\(\frac{d x}{1+\sin x}\)
= ∫\(\frac{1-\sin x}{\cos ^2 x}\)
= ∫sec² x – sec x tan x dx
= tan x – sec x + c

Question 3.
∫\(\frac{\sin x}{1+\sin x}\) dx
Solution:

Question 4.
∫\(\frac{\sec x}{\sec x+\tan x}\) dx
Solution:
I = ∫\(\frac{\sec x}{\sec x+\tan x}\) dx
= ∫\(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) dx
= ∫sec² x – sec x tan x dx
= tan x – sec x + c

Question 5.
∫\(\frac{1+\sin x}{1-\sin x}\) dx
Solution:
I = ∫\(\frac{1+\sin x}{1-\sin x}\) dx
= ∫\(\frac{(1+\sin x)^2}{\cos ^2 x}\) dx
= ∫[sec² x+ tan² x+ 2sec x tan x) dx
= ∫[2sec² x – 1 + 2sec x tan x) dx
= 2tan x – x + 2sec x + c

Question 6.
∫tan^-1 (sec x + tan x) dx
Solution:

Question 7.
∫\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
Solution:
I = ∫\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
= ∫\(\frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}\) dx
= 2 ∫(cos x + cos α) dx
= 2 sin x + 2x cos α + c

Question 8.
∫tan-1\(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\) dx
Solution:

Question 9.
∫\(\frac{d x}{\sqrt{x+1+} \sqrt{x+2}}\)
Solution:

Question 10.
∫\(\frac{2+3 x}{3-2 x}\) dx
Solution:

Question 11.
∫\(\frac{d x}{\sqrt{x}+x}\)
Solution:

Question 12.
∫\(\frac{d x}{1+\tan x}\)
Solution:

Question 13.
∫\(\frac{x+\sqrt{x+1}}{x+2}\) dx (Hints put : \(\sqrt{x+1}\) = t)
Solution:

Question 14.
∫sin^-1\(\sqrt{\frac{x}{a+x}}\) dx (Hints put : x = a tan² t)
Solution:

Question 15.
∫e^x\(\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right)\) dx
Solution:

Question 16.
∫\(\frac{\left(x^2+1\right) e^x}{(x+1)^2}\) dx
Solution:

Question 17.
∫\(\frac{x^2-1}{x^4+x^2+1}\) dx
Solution:

Question 18.
∫\(\frac{x^2 d x}{x^4+x^2+1}\)
Solution:

Question 19.
∫\(\sqrt{\cot x}\) dx
Solution:

Question 20.
∫\((\sqrt{\tan x}+\sqrt{\cot x})\) dx
Solution:

Question 21.
∫\(\frac{\mathrm{dx}}{x\left(x^4+1\right)}\)
Solution:

Question 22.
∫\(\frac{\mathrm{dx}}{e^x-1}\)
Solution:

Question 23.
∫\(\frac{(x-1)(x-2)(x-3)}{(x+4)(x-5)(x-6)}\) dx
Solution:

Question 24.
∫\(\frac{d x}{\left(e^x-1\right)^2}\)
Solution:

Question 25.
∫\(\frac{d x}{\sin x \cos ^2 x}\)
Solution:

Question 26.
\(\int_2^4 \frac{\left(x^2+x\right) d x}{\sqrt{2 x+1}}\)
Solution:

Question 27.
\(\int_{-a}^a \sqrt{\frac{a-x}{a+x}}\) dx
Solution:

Let a² – x² = t²
-2x dx = 2t dt
x = -a ⇒ 0 t = 0
x = a ⇒ t = 0
= 0
I = aI₁ – I₂ = aπ

Question 28.
\(\int_0^{\pi / 2}(\sqrt{\tan x}+\sqrt{\cot x})\) dx
Solution:

Question 29.
\(\int_0^{\pi / 2} \frac{\cos x d x}{1+\cos x+\sin x}\)
Solution:

Question 30.
\(\int_0^1\)x (1 – x)ⁿ dx
Solution:

Question 31.
\(\int_0^{\pi / 2}\)sin 2x log (tan x) dx
Solution:

Question 32.
\(\int_0^{\pi / 2} \frac{\sin ^2 x d x}{\sin x+\cos x}\)
Solution:

Question 33.
\(\int_0^{\pi / 2} \frac{\sin ^2 x d x}{1+\sin x \cos x}\)
Solution:

Question 34.
\(\int_0^{\pi / 2} \frac{x d x}{\sin x+\cos x}\)
Solution:

Question 35.
Prove that \(\int_0^\pi\) x sin³ x dx = \(\frac{2 \pi}{3}\)
Solution:

Question 36.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x d x}{\sin x+\cos x}\)
Solution:

Question 37.
\(\int_0^\pi\)|cos x| dx
Solution:

Question 38.
\(\int_1^4\)(|x – 1| + |x – 2| + |x – 3|) dx
Solution:

Question 39.
\(\int_{-\pi / 2}^{\pi / 2}\)(sin |x| + cos |x|) dx
Solution:

Question 40.
\(\int_0^\pi\)log (1 + cos x) dx
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(l) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(l)

Question 1.
\(\int_0^{\frac{\pi}{2}}\)sin¹⁰ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)sin¹⁰ θ dθ = \(\frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\) = \(\frac{405 \pi}{7680}\)

Question 2.
\(\int_0^{\frac{\pi}{2}}\)cos¹² θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)cos¹² θ dθ = \(\frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\) = \(\frac{4455 \pi}{92160}\)

Question 3.
\(\int_0^{\frac{\pi}{2}}\)sin¹¹ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)sin¹¹ θ dθ = \(\frac{10}{11} \cdot \frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\) = \(\frac{3840}{4455}\)

Question 4.
\(\int_0^{\frac{\pi}{2}}\)cos⁹ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)cos⁹ θ dθ = \(\frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\) = \(\frac{384}{405}\)

Question 5.
\(\int_0^1 \frac{x^7}{\sqrt{1-x^2}}\) dx
Solution:

Question 6.
\(\int_0^1 \frac{x^5\left(4-x^2\right)}{\sqrt{1-x^2}}\) dx
Solution:

Question 7.
\(\int_0^a x^3\left(a^2-x^2\right)^{\frac{5}{2}}\) dx
Solution:

Question 8.
\(\int_0^1 x^5 \sqrt{\frac{1+x^2}{1-x^2}}\) dx
Solution:

Question 9.
\(\int_0^{\infty} \frac{x^2}{\left(1+x^6\right)^n}\) dx
Solution:

Question 10.
\(\int_0^\pi\)sin⁸ θ dθ
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(k) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(k)

Evaluate the following Integrals:
Question 1.
(i) \(\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan x}\)dx
Solution:

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\)dx
Solution:

(iii) \(\int_0^1 \frac{\ln (1+x)}{2+x^2}\)dx (x = tan θ)
Solution:

(iv) \(\int_0^\pi \frac{x d x}{1+\sin x}\)
Solution:

Question 2.
(i) \(\int_{-a}^a\)x⁴ dx
Solution:

(ii) \(\int_{-a}^a\)(x⁵ + 2x² + x) dx
Solution:

(iii) \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)cos² x dx
Solution:

(iv) \(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin⁵ x dx
Solution:
Let f(x) = sin⁵ x
Then f(-x) = sin⁵ (-x)
= -sin⁵ x = -f(x)
So f(x) is an odd function.
Thus \(\int_{-a}^a\)f(x) dx = 0
\(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin⁵ x dx = 0

Question 3.
(i) \(\int_0^\pi\)cos³ x dx
Solution:

(ii) \(\int_0^\pi\)cos² x dx
Solution:

(iii) \(\int_0^\pi\)sin³ x cos x dx
Solution:
\(\int_0^\pi\)sin³ x cos x dx
[Put sin x = t, then cos x dx = dt
When x = 0, t = 0, when x = π, t = 0
\(\int_0^\pi\)t³ dt = 0

(iv) \(\int_0^\pi\)sin x cos² x dx
Solution:

Question 4.
Show that
(i) \(\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}\) dx = \(\frac{\pi}{2} \ln \frac{1}{2}\)
Solution:

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x}\) dx = 0
Solution:

(iii) \(\int_0^\pi\)x ln sin x dx = \(\frac{\pi^2}{2} \ln \frac{1}{2}\)
Solution:

Question 5.
(i) \(\int_0^{\pi / 2}\)ln (tan x + cot x) dx
Solution:

(ii) \(\int_0^\pi \frac{x \tan x-\sin x}{1+\sin x \cos x}\) dx
Solution:

(iii) \(\int_1^3 \frac{\sqrt{x} d x}{\sqrt{4-x}+\sqrt{x}}\)
Solution:

(iv) \(\int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x}\)
Solution:

(v) \(\int_0^1\)x (1 – x)¹⁰⁰ dx
Solution:

(vi) \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}\)
Solution:

(vii) \(\int_0^{50}\)e^x-[x] dx
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(c)

Find the solutions of the following differential equations:
Question 1.
(x + y) dy + (x – y) dx = 0
Solution:
Given equation can be written as

Question 2.
\(\frac{d y}{d x}\) = \(\frac{1}{2}\left(\frac{y}{x}+\frac{y^2}{x^2}\right)\)
Solution:

Question 3.
(x² – y²) dx + 2xy dy = 0
Solution:
Given equation can be written as

Question 4.
x\(\frac{d y}{d x}\) + \(\sqrt{x^2+y^2}\) = y
Solution:
Given equation can be written as

Question 5.
x (x + y) dy = (x² + y²) dx
Solution:


This is the required solution.

Question 6.
y² + x² \(\frac{d y}{d x}\) = xy \(\frac{d y}{d x}\)
Solution:
Given equation can be written as

This is the required solution.

Question 7.
x sin\(\frac{y}{x}\) dy = \(\left(y \sin \frac{y}{x}-x\right)\)dx
Solution:
Given equation can be written as

Question 8.
x dy – y dx= \(\sqrt{x^2+y^2}\) dx
Solution:

This is the required solution.

Question 9.
\(\frac{d y}{d x}\) = \(\frac{y-x+1}{y+x+5}\)
Solution:
Given equation is


This is the required solution.

Question 10.
(x – y) dy = (x + y + 1) dx
Solution:
Given equation can be written as


This is the required solution.

Question 11.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:
Given equation can be written as


This is the required solution.

Question 12.
\(\frac{d y}{d x}\) = \(\frac{3 x-7 y+7}{3 y-7 x-3}\)
Solution:
Given equation can be written as


This is the required solution.

Question 13.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution:
Given equation can be written as

⇒ 2z + ln (z – 1) = 3x + C
⇒ 2 (2x + y) + ln (2x + y – 1 ) = 3x + C
⇒ (x + 2y) + ln (2x + y – 1 ) = C
This is the required solution.

Question 14.
(2x + 3y – 5)\(\frac{d y}{d x}\) + 3x + 2y – 5 = 0
Solution:
Given equation can be written as

Question 15.
(4x + 6y + 5) dx – (2x + 3y + 4) dy = 0
Solution:
Given equation can be written as


This is the required solution.

CHSE Odisha 12th Class Political Science Book Solutions (+ 2 2nd Year)

CHSE Odisha Class 12 Political Science Book Solutions in English Medium

Unit 1 Democracy in India

• Democracy in India Objective Questions
• Democracy in India Short Answer Questions
• Democracy in India Long Answer Questions

Unit 2 Democratic Process in India-I

• Democratic Process in India-I Objective Questions
• Democratic Process in India-I Short Answer Questions
• Democratic Process in India-I Long Answer Questions

Unit 3 Democratic Process in India-II

• Democratic Process in India-II Objective Questions
• Democratic Process in India-II Short Answer Questions
• Democratic Process in India-II Long Answer Questions

Unit 4 India in World Politics

• India in World Politics Objective Questions
• India in World Politics Short Answer Questions
• India in World Politics Long Answer Questions

Unit 5 Issues in International Politics

• Issues in International Politics Objective Questions
• Issues in International Politics Short Answer Questions
• Issues in International Politics Long Answer Questions

CHSE Odisha Class 12 Political Science Book Solutions in Odia Medium

Chapter 1 ଗଣତନ୍ତ୍ର

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 2 ଭାରତରେ ରାଜନୈତିକ ଦଳୀୟ ବ୍ୟବସ୍ଥା

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 3 ଭାରତରେ ସଂଘୀୟବାଦ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 4 ଭାରତରେ ଗ୍ରାମାଞ୍ଚଳ ଓ ସହରାଞ୍ଚଳ ସ୍ୱାୟତ୍ତ ଶାସନ ବ୍ୟବସ୍ଥା

• Objective Questions
• Short Answer Questions
• Long Answer Questions Part-1
• Long Answer Questions Part-2

Chapter 5 ଜାତି ଗଠନରେ ପ୍ରତିବନ୍ଧକ

• Objective & Short Answer Questions
• Long Answer Questions

Chapter 6 ଭାରତୀୟ ରାଜନୀତିରେ ସାଂପ୍ରତିକ ପ୍ରସଙ୍ଗ-ଜନପ୍ରିୟ ଆନ୍ଦୋଳନ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 7 ଭାରତର ବୈଦେଶିକ ନୀତି

• Objective Questions
• Short & Long Answer Questions

Chapter 8 ଆନ୍ତର୍ଜାତୀୟ ସଙ୍ଗଠନ

• Objective & Short Answer Questions
• Long Answer Questions

Chapter 9 ସାଂପ୍ରତିକ ବିଶ୍ଵରେ ନିରାପତ୍ତା ପ୍ରସଙ୍ଗର ପରିବର୍ତ୍ତିତ ପୃଷ୍ଠଭୂମି

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 10 ପରିବେଶ ଓ ପ୍ରାକୃତିକ ସମ୍ପଦ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

CHSE Odisha Class 12 Political Science Syllabus (+2 2nd Year)

There shall be two papers in Political Science modeled on the Syllabi of CBSE. Paper-I: TitleFoundation of Political Theory and Indian Government at work (For First Year). Paper-II: Title-Democracy and Nation Building in India and International affairs (For Second Year).

The subject of Political Science modeled on the Syllabi of CBSE consists of two papers as mentioned above. Paper-I is to be covered in the +2 First Year class and Paper-ll is to be covered in the +2 Second Year class. Each paper is divided into two sections and each section is further subdivided into two/three units. Thus there are five units in both Paper-I and Paper II. Periods have been allocated for the respective units approximately. Teachers are advised to take at least those numbers of periods to cover the particular unit. The major concepts and principles should be taught in such a manner as to stimulate higher mental abilities among students like application, logical thinking, analysis, etc., and not factual information. Paper-setters and Examiners are requested to keep the above in mind while setting questions and examining, respectively. Questions should of short (one word/ multiple-choice/ one sentence), medium (50/100 words/ five sentences), and long (500 words or thereabout). Also, Questions of the final/AHS Examination shall cover all five units of Paper -II.

Objectives of the course/syllabus are, as briefly mentioned above are:

• To enable the students to acquire knowledge about the important concepts, theories, principles, provisions, processes and Institutions of the Indian constitution, and some rudimentary knowledge about International affairs;
• To acquaint the students with the changing dimension of politics and political theory both in the national and international knowledge domain;
• To develop an interest among the students regarding problems of the political domain and to find out the possible solution to those problems.

Suggested Reading:

1. Political Theory- For Class-XI (Published by NCERT, New Delhi)
2. Indian Constitution at Work- For Class-XI (Published by NCERT, New Delhi)
3. Contemporary World Politics, For Class-XII (Published by NCERT, New Delhi)
4. Politics in India, For Class-XII (Published by NCERT, New Delhi)

Second Year CHSE (2022-2023)
Political Science Paper-II
(DEMOCRACY IN INDIA AND INTERNATIONAL POLITICS)

Part A: Politics in India

Unit I Democracy in India

1. Democracy: Meaning, Types, and Features; Challenge to Democratic Process in India – Inequality, Illiteracy, Regionalism, Naxalite Problem, Gender Inequality. (8 Periods)
2. Party System in India: Meaning, Types; One Party Dominance, Coalition Politics; Regional Parties. (8 Periods)

Unit II Democratic Process in India-I

1. Federalism in India: Features, Centre-State relation, Recent Trends in Indian Federalism. (8 Periods)
2. Local Government in India – Rural and Urban Local Bodies, Composition and Functions. (8 Periods)

Unit III Democratic Process in India-II

1. Challenges to Nation-Building: Meaning, Communalism, Casteism, Regionalism, Terrorism; Remedies. (8 Periods)
2. Contemporary issues in Indian Politics: Popular Movements – Women Movement, Environment protection Movement, Development – Displacement Movements. (8 Periods)

Part B: Contemporary World Politics

UNIT-IV (India in World Politics)

1. Indian Foreign Policy: Basic Features; India and its neighbours-China, Pakistan. (8 Periods)
2. International Organizations: UN: Major Organs-General Assembly; Security Council; International Court of Justice; Reforms of the UN. India’s position in the UN. International Economic Organizations- World Bank and the IMF. (8 Periods)

UNIT-IV (Issues in International Politics)

1. Changing Dimension of Security in Contemporary World: Traditional Security Concerns: Arms Race and Disarmament. Non-Traditional Security Concerns: Human security: Global Poverty, Inequality, Health, and Education. (8 Periods)
2. Environment and Natural Resources: Global Environmental Concerns; Development and Environment; Global Warming and Climate Change. (8 Periods)

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

CHSE Odisha Class 12 Text Book Solutions

CHSE Odisha 12th Class Psychology Book Solutions (+ 2 2nd Year)

CHSE Odisha 12th Class Psychology Book Solutions in English Medium

Unit 1 Life Span Development & Self and Personality

• Unit 1 Objective & Short Answer Type Questions
• Unit 1 Long Answer Questions Part-1
• Unit 1 Long Answer Questions Part-2
• Unit 1 Long Answer Questions Part-3
• Unit 1 Long Answer Questions Part-4

Unit 2 Stress: Meeting Life Challenges & Physical Environment and Behaviour

• Unit 2 Objective & Short Answer Type Questions
• Unit 2 Long Answer Questions Part-1
• Unit 2 Long Answer Questions Part-2

Unit 3 Group Processess and Leadership & Counselling Process

• Unit 3 Objective & Short Answer Type Questions
• Unit 3 Long Answer Questions Part-1
• Unit 3 Long Answer Questions Part-2

Unit 4 Psychological Disorder & Therapeutic Approaches

• Unit 4 Objective & Short Answer Type Questions
• Unit 4 Long Answer Questions Part-1
• Unit 4 Long Answer Questions Part-2
• Unit 4 Long Answer Questions Part-3
• Unit 4 Long Answer Questions Part-4

Unit 5 Statistics in Psychology

• Unit 5 Objective Questions and Answers

CHSE Odisha 12th Class Psychology Book Solutions in Odia Medium

Unit 1 ଜୀବନବ୍ୟାପୀ ବିକାଶ & ସ୍ଵୟଂ ତଥା ବ୍ୟକ୍ତିତ୍ଵ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Unit 2 ମନୋ-ସାମାଜିକ ଚାପ & ଭୌତିକ ପରିବେଶ ଓ ବ୍ୟବହାର

Unit 3 ସମୂହ ପ୍ରକ୍ରିୟା ଓ ନେତୃତ୍ଵ & ପରାମର୍ଶ ପ୍ରକ୍ରିୟା

• Objective Questions
• Short & Long Answer Questions

Unit 4 ମାନସିକ ବିକାର & ମାନସିକ ବିକାର ଚିକିତ୍ସା

• Objective Questions
• Short & Long Answer Questions

Unit 5 ମନୋବିଜ୍ଞାନରେ ପରିସଂଖ୍ୟାନର ବ୍ୟବହାର

• Objective Questions
• Short & Long Answer Questions

CHSE Odisha Class 12 Psychology Syllabus (+2 2nd Year)

Psychology is introduced as an elective subject at the higher secondary stage of education. The course deals with Psychological knowledge and practices which are contextually rooted, It emphasizes the complexity of cognitive and behavioural processes of human beings within a socio-cultural context. It encourages critical reasoning, allowing students to appreciate the role of social factors in behaviour and illustrates how biology and experience shape behaviour.

In the second year, there will be a council examination and the theory paper carries 70 marks and the practical paper carries 30 marks. The question paper pattern for both papers is same.

QUESTION PAPER PATTERN

Theory Paper:
Group-A: Objective type (Compulsory)
Q.No.1: Multiple choice (Fill up the blanks from all units) 1 mark each × 10 = 10 marks
Q.No.2: Statements “True” or “False” 1 mark each × 10 = 10 marks

Group-B: Short Type
Q.No.3: Short type answer (Answer within two /three sentences) and one has to answer 10 bits out of 12 bits.
2 marks each × 10 = 20 marks
Q.No.4: Short type answer (Answer within six sentences) and one has to answer 3 bits out of 5 bits.
3 mark each × 3 = 09 marks

Group-C: Long Type
Q.No.5: Question No-5 to Q.No-10 (Questions will be from all the units and one has to answer any three questions).
7 marks each × 3 = 21 marks

PRACTICAL
There will be 4 number of questions and the examinee is to choose/draw 2 number of questions through the lottery and is to conduct any one question out of the two questions.
Distribution of marks :
Record – 03
Viva -Voce – 07
Conduction & Report writing – 20
Total marks – 30

PSYCHOLOGY IN APPLICATION
SECOND YEAR
Total Marks – 100
Theory – 70 marks
Practical – 30 marks

Theory
UNIT-I
1. Life span development [10 Periods]
This chapter deals with variations in development and developmental tasks across the life span.
a) Meaning of development – Life span perspective
b) Principles of development
c) Stages of development: Pre-natal stage, Infancy, childhood stage, Adolescence, Adulthood and old age.

2. Self and Personality [9 Periods]
This chapter focuses on the study of self and personality in the context of different approaches in an effort to appraise the person. The assessment of personality will also be discussed.
a) Concept of self and personality
b) Personality types and traits
c) Assessment of Personality
This chapter deals with the nature of stress and strategies to cope with stress.
a) Meaning, Nature and causes of stress.
b) Coping strategies to deal with stress.

UNIT-II
3. Stress: Meeting life challenges [6 Periods]
This chapter deals with the nature of stress and strategies to cope with stress.
a) Meaning, Nature and causes of stress.
b) Coping strategies to deal with stress.

4. Physical environment and behaviour. [6 Periods]
This chapter focuses on the application of psychological understanding of human-environment relationship.
a) Human impact on the environment: Noise Pollution, Crowding, Natural disasters.
b) Impact of environment on human behaviour.

UNIT-III
5. Group Processes and leadership. [7 Periods]
This chapter deals with the concept of a group and the role of the leader in a group.
a) Groups: Nature, types and formation.
b) Leadership: Nature, functions and styles of leadership.

6. Counselling Processes [6 Periods]
This chapter focuses on helping the client in living a meaningful and fulfilling life.
a) Meaning and concept of counselling; Goals of counselling.
b) Characteristics of an effective counsellor.

UNIT-IV
7. Psychological disorder [10 Periods]
This chapter discusses the concept of normality and abnormality and the major psychological disorders.
a) Concept of normality and abnormality, criteria of studying abnormal behaviour
b) Causal factors associated with abnormal behaviour.
c) Major Psychological disorders: Anxiety disorders, somatoform disorders and mood disorders.

8. Therapeutic Approaches [6 Periods]
This chapter discusses the purpose and processes to treat Psychological disorders:
a) Nature and Processes of therapy
b) Types of therapy: Psychotherapy, Behaviour therapy, Cognitive therapy and Biomedical therapy.

UNIT-V
9. Statistics in Psychology [10 Periods]
This chapter deals with some basic statistical methods to be used in psychological studies.
a) Frequency distribution
b) Measures of Central Tendency: Computation and uses of mean, median and mode.

PRACTICALS

1. RCPM (Children) / RPM (Adults)
2. Case History Method (Preparation of at least one case profile)
3. Personality Test (Type A/B)
4. Piagetian Task (Conservation of Liquid Quantity)

Books Recommended:
1. Bureau’s Higher Secondary +2 Psychology Part-II Published by Odisha State Bureau of Test Book Preparation and Production, Bhubaneswar.
2. Psychology Part-II, NCERT.

CHSE Odisha Class 12 Text Book Solutions