Class 12

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(b)

Question 1.
Let X={x, y} and Y={u, v}. Write down all the functions that can be defined from X to Y. How many of these are (i) one-one (ii) onto and (ii) one-one and onto?
Solution:
The functions from X = {x, y} to y = {u, v} are:
f1 = {(x, u), (y, v)}
f2 = {(x, v), (y, u)}
f3 = {(x, u), (y, u)}
f4 = {(x, v), (y, v)}

Out of these 4 functions there are:
(i) 2 one-one functions
(ii) 2 onto functions
(iii) 2 one-one and onto function.

Question 2.
Let X and Y be sets containing m and n elements respectively.
(i) What is the total number of functions from X to Y.
(ii) How many functions from X to Y are one-one according as men, m > n and m = n?
Solution:
If | x | = m and | y | = n then
(i) Number of functions =n^m
(ii) If m < n then number of one-one functions = ⁿP_m.
If m > n then number of one-one functions = 0
If m = n then number of one-one functions = m!

Question 3.
Examine each of the following functions if it is
(i) injective (ii) surjective, (iii) bijective and (iv) none of the three
(a) f : R → R, f(x) = x²
(b) f : R → [-1, 1], f(x) = sin x
(c) f : R₊ → R + , f(x) = x + 1/x
where R₊ = {x ∈ R : x > 0}
(d) f : R → R, f(x) = x³ + 1
(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
(f) f : R → R, f(x) = [x] = the greatest integer ≤ x.
(g) f : R → R, f(x) = | x |
(h) f : R → R, f(x) = sgn x
(i) f : R → R, f = id_R = the identity function on R.
Solution:
(a) f : R → R, f(x) = x²
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂)
⇒ x₁² = x₂²
⇒ x₁  = ± x₂
∴ f is not one-one.
Hence f is not injective or bijective.
Rng f = [0, ∞) ≠ R
∴ f is not surjective.

(b) f : R → [-1, 1], f(x) = sin x
For x₁ , x₂ ∈ R
let f(x₁) = f(x₂) ⇒ sin x₁ = sin x₂ 
⇒ x₁ = nπ + (- 1)ⁿ  x₂
⇒ x₁ = x₂ (not always)
∴ f is not injective and also not bijective.
But f is onto, as ∀ y ∈ [-1, 1]
there is a x ∈ R such that f(x) = sin x.
i.e., f is surjective.

(c) f : R₊ → R₊ , f(x) = x + \(\frac{1}{x}\)
f(2) = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)
f(\(\frac{1}{2}\)) = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)
f(2) = f(\(\frac{1}{2}\))
but 2 ≠ \(\frac{1}{2}\)
∴ f is not injective (one-one).
Again, for 1 ∈ R₊ (domain)
⇒ there is no x ∈ R₊^(Dom)
such that x + \(\frac{1}{2}\) = 1
∴ f is not onto.

(d) f : R → R, f(x) = x³ + 1
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂
∴ f is injective.
Let f(x) = y ⇒ y = x³ + 1
⇒ x³ = y – 1
⇒ x = (y – 1)^1/3 which exists ∀ y ∈ R
∴ f is onto.
∴ f is bijective.

(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
for x₁, x₂ ∈ (-1, 1)
Let f(x₁) = f(x₂) ⇒ \(\frac{x_1}{1-x_1^2}\) = \(\frac{x_2}{1-x_2^2}\)
⇒x₁ – x₁ x₂² = x₂ – x₁² x₂
⇒ x₁ – x₂ + x₁² x₂ – x₁ x₂² = 0
⇒ (x₁ – x₂) (1 + x₁ x₂) = 0
⇒ x₁ = x₂ (for x₁ x₂ ∈ (-1, 1) x₁ x₂ ≠ -1)
∴ f is injective.
Again let y = \(\frac{x}{1-x^2}\) ⇒ y – x²y = x
⇒ x²y + x – y = 0
⇒ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{1+4 y^2}}{2 y}\) ∉ (-1, 1) ∀ y ∈¸.
∴f is surjective.
∴f is bijective.

(f) f : R → R
f(x) = [x]
f(1.2) = f(1.5) =1
∴ f is not injective.
Rng f = Z ⊂ R
∴ f is not surjective.
∴ Hence it is not also bijective.

(g) f : R → R
f(x) = | x |
As f(-1) = f(1) = 1
∴ f is not injective.
Again Rng f = [0, ∞) ⊂ R
⇒ f is not surjective.
Thus f is not bijective.

(h) f : R → R
f(x) = Sgn (x) = \(\left\{\begin{array}{cc}
1, & x>0 \\
0, & x=0 \\
-1, & x<0
\end{array}\right.\)
As f(1) = f(2) = 1
We have f is not injective.
Again Rng f = {- 1, 0, 1} ≠ R
∴ R is not surjective.
⇒ R is not bijective.

(i) f : R → R
f = idx
∴ f(x) = x
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂) where x₂ x₂ ∈ R
⇒ x₁ = x₂
∴ f is one-one.
Again Rng f = R (codomain)
∴ f is onto.
Thus f is a bijective function.

Question 4.
Show that the following functions are injective.
(i) f(x) = sin x on \(\left[0, \frac{\pi}{2}\right]\)
(ii) f(x) = cos x [0, π]
(iii) fix) = log_a x on (0, ∞), (a > 0 and a ≠ 1)
(iv) f(x) = a^x on R. (a > 0 and a ≠ 1)
Solution:
(i) f(x) = sin x . \(\left[0, \frac{\pi}{2}\right]\)
for α, β ∈ \(\left[0, \frac{\pi}{2}\right]\)
Let f(α) = f(β) ⇒ sin α = sin β
⇒ α = β, as α, β ∈ \(\left(0, \frac{\pi}{2}\right)\) and no other values of α is possible.
∴ f is one-one.

(ii) f(x) = cos x, [0, π]
for α, β ∈ [0, π]
Let f(α) = f(β) ⇒ cos α = cos β
⇒ α = β, since α, β ∈ [0, π] and cos x is +ve in 1st quadrant and -ve in 2nd quadrant.
∴ f is one-one.

(iii) f(x) = log_a x [1, ∞]
for α, β ∈ [1, ∞]
Let f(α) = f(β)
⇒ log_a α = log_a β ⇒ α = β
∴ f is one-one.

(iv) f(x) = a^x, (a > 0), x ∈ R
for x₁, x₂ ∈ R
Let f(x₁) = f(x₂) ⇒ a^x₁ = a^x₂
⇒ x₁ = x₂
∴ f is one-one.

Question 5.
Show that functions f and g defined by f(x) = 2 log x and g(x) = log x² are not equal even though log x² = 2 log x.
Solution:
f(x) = 2 log x
g(x) = log x²
Dom f(x) = (0, ∞)
Dom g(x) = R – {0}
As Dom f(x) ≠ Dom g(x) we have f(x) ≠ g(x), though log x² = 2 log x

Question 6.
Give an example of a function which is
(i) Surjective but not injective.
(ii) injective but not surjective.
(iii) neither injective nor surjective.
(iv) bijective
Solution:
(i) f(x) = sin x
from R → [-1, 1]
which is surjective but not injective.

(ii) f(x) = \(\frac{x}{2}\) from Z → R
is injective but not surjective.

(iii) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
is neither surjective nor injective.
[Refer Q. No. 3(e)].

(iv) f(x) = x³ + 1, f : R → R
is bijective.
[Refer No. 3 (d)].

Question 7.
Prove that the following sets are equivalent:
{1, 2, 3, 4, 5, 6,…}
{2, 4, 6, 8, 10,…}
{1, 7, 5, 7, 9,…}
{1, 4, 9, 16, 25,…}
Solution:
Let A = {1, 2, 3, 4, 5, 6, ……..}
B = {2, 4, 6, 8, 10, …….}
C = {1, 3, 5, 7, 9, ……}
D = {1, 4, 9, 16, 25, ……}
Let f : A → B defined as f(x) = 2x
Clearly f is bijective.
There is a one-to-one correspondence between A and B.
⇒ A and B are equivalent.
Let g : A → C defined as g(x) = 2x – 1
Clearly f is bijective.
⇒ There is a one-to-one correspondence between A to C
∴ A and C are equivalent.
Let h : A → D defined as h(x) = x².
Clearly h is bijective.
⇒ There is a one-to-one correspondence between A to D.
⇒ A and D are equivalent.
Thus A, B, C and D are equivalent.

Question 8.
Let f = {(1, a), (2, b), (3, c), (4, d)} and g = {{a, x), (b, x), (c, y), (d, x)}. Determine gof and fog if possible. Test whether fog = gof.
Solution:
f = {(1, a), (2, b), (3, c), (4, d)}
g = {{a, x), (b, x), (c, y), (d, x)}
gof (1) = g(a) = x
gof (2) = g(b) = x
gof (3) = g(c) = y
gof (4) = g(a) = x
∴ gof = {(1, x), (2, x), (3, y), (4, x)} Here fog is not defined.

Question 9.
Let f = {(1, 3), (2, 4), (3, 7)} and g = {(3, 2), (4, 3), (7, 1)}. Determine gof and fog if possible. Test whether fog = gof.
Solution:
f = {(1, 3), (2, 4), (3, 7)}
g = {(3, 2), (4, 3), (7, 1)}
We have fog (3) = f (g(3)) = f(2) = 4
fog (4) = f (g(4)) = f(3) = 7
fog (7) = f (g(7)) = f(1) = 3
∴  fog = {(3, 4), (4, 7), (7, 3)}
Again gof (1) = g (f(1)) = g(3) = 2,
gof (2) = g (f(2)) = g(4) = 3,
gof (3) = g (f(3)) = g(7) = 1
∴ gof = {(1, 2), (2, 3), (3, 1)}
∴ gof ≠ fog
So the composition of functions is not necessarily commutative.

Question 10.
Let f(x) =√x and g(x) = 1 – x².
(i) Find natural domains of f and g.
(ii) Compute fog and gof and find their natural domains.
(iii) Find natural domain of h(x) = 1 – x.
(iv) Show that h = gof only on R₀ = {x ∈ R : x ≥ 0} and not on R.
Solution:
Let f(x) = √x, g(x) = 1 – x²
(i) ∴ Dom f = R₊ U_{0}, Dom g = R

(ii) fog (x) = f (g(x))
= f (1 – x²) = \(\sqrt{1-x^2}\)
∴ fog (x) exists when 1 – x² ≥ 0
⇒ x² ≤ 1 ⇒ -1 ≤ x ≤ 1 i.e., x ∈ [-1, 1]
∴ Dom fog = [-1, 1]
Again gof (x) = g (f(x))
= g( √x ) = 1 – ( √x )² = 1 – x
∴ Dom gof = R₀ = (0, ∞)

(iii) Domain of h(x) = 1 – x is R.

(iv) We have proved in (ii) that gof (x) = 1 – x.
∴ h(x) = gof (x) ⇒ h = gof only when x ∈ R₀ as dom f is R₀ = [0, ∞]

Question 11.
Find the composition fog and gof and test whether fog = gof when f and g are functions on R given by the following:
(i) f(x) = x³ + 1, g(x) = x² – 2
(ii) f(x) = sin x, g(x) = x⁵
(iii) f(x) = cos x, g(x) = sin x²
(iv) f(x) = g(x) = (1 – x³)^1/3
Solution:
(i) f(x) = x³ + 1, g(x) = x² – 2
∴ fog (x) = f (g(x)) = f(x² – 2)
= (x² – 2)³ + 1
gof (x) = g (f(x)) = g(x³ + 1)
= (x³ + 1)² – 1
fog ≠ gof

(ii) f(x) = sin x, g(x) = x⁵
∴ fog (x) = f (g(x)) = f(x⁵) = sin x⁵
∴ gof (x) = g (f(x))
= g(sin x) = (sin x)⁵ = sin⁵ x
fog ≠ gof

(iii) f(x) = cos x, g(x) = sin x²
∴ fog (x) = f (g(x))
= f( sin x²) = cos (sin x²)
and gof (x) = g (f(x)) = g(cos x)
= sin (cos x)² = sin (cos² x)
fog ≠ gof

(iv) f(x) = g(x) = (1 – x³)^1/3
fog (x) = f (g(x))
= (1 – (g(x))³)^1/3
= [1 – (1 – x³)]^1/3 = x
gof (x) = g (f(x))
= [1 – (f(x))³]^1/3 = x
⇒ fog = gof

Question 12.
(a) Let f be a real function. Show that h(x) = f(x) + f(-x) is always an even function and g(x) = f(x) – f(-x) is always an odd function.
(b) Express each of the following function as the sum of an even function and an odd function:
(i) 1 + x + x² , (ii) x², (iii) e^x, (iv) e^x + sin x
Solution:
(a) We have h(x) = f(x) + f(-x)
∴ h(-x) = f(-x) + f(x) = h(x)
∴ h is always an even function.
Further, g(x) = f(x) – f(-x)
∴ g(-x) = f(-x) – f(x)
= – [f(x) – f(-x)] = – g(x).
∴ g is always an odd function.

(b) (i) Let f(x) = 1 + x + x²
∴ f(-x) = 1 – x + x²
∴ g(x) = \(\frac{f(x)+f(-x)}{2}\)
= \(\frac{1+x+x^2+1-x+x^2}{2}\)
= x² + 1 and
g(-x) = (-x)² + 1 = x² + 1
∴ g is an even function.
h(x) = \(\frac{f(x)-f(-x)}{2}\)
= \(\frac{\left(1+x+x^2\right)-\left(1-x+x^2\right)}{2}\) = x
h(-x) = -x = -h(x)
⇒ h is an odd function.
∴ f(x) = g(x) + f(x)
where g is even and h is odd.

(ii) Let f(x) = x²
So that f(-x) = (-x)² = x²
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\) = \( \frac{x^2+x^2}{2}\) = x²
g(-x) = g(x)
∴ g is an even function.
and h(x) = \( \frac{f(x)-f(-x)}{2}\) = \(\frac{x^2-x^2}{2}\) = 0
h(x) = 0 is both even and odd.
∴ f(x) = g(x) = f(x),
where g is even and h is odd.

(iii) Let f(x) = e^x
f(-x) = e^-x
g(x) = \( \frac{e^x+e^{-x}}{2}\)
g(-x) = g(x)
g is an even function.
and h(x) = \( \frac{e^x-e^{-x}}{2}\)
h(-x) = \( \frac{e^{-x}-e^x}{2}\) = -h²(x)
⇒ h is an odd function.
∴ f(x) = g(x) + h(x),
where g is even and h is odd.

(iv) Let f(x) = e^x + sin x
f(-x) = e^-x + sin (-x) = e^-x –  sin x
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\)
= \( \frac{e^x+\sin x+e^{-x}-\sin x}{2}\)
= \( \frac{e^x+e^{-x}}{2}\) and h(x) = \( \frac{f(x)-f(-x)}{2}\)
= \( \frac{e^x+\sin x-e^{-x}+\sin x}{2}\)
= \( \frac{e^x-e^{-x}+2 \sin x}{2}\)
∴ f(x) = g(x) + h(x)
where g is even and g is odd.

Question 13.
Let X = {1, 2, 3, 4} Determine whether f : X → X defined as given below have inverses.
Find f^-1 if it exists:
(i) f = {(1, 4), (2, 3), (3, 2), (4, 1)}
(ii) f = {(1, 3), (2, 1), (3, 1), (4, 2)}
(iii) f = {(1, 2), (2, 3), (3, 4), (4, 1)}
(iv) f = {(1, 1), (2, 2), (2, 3), (4, 4)}
(v) f = {(1, 2), (2, 2), (3, 2), (4, 2)}
Solution.
(i) x = {1, 2, 3, 4}
f is bijective. Hence f^-1 exists.
f^-1 = {(4, 1), (3, 2), (2, 3), (1, 4)}

(ii) f(2) = f(3) = 1
⇒ f is not injective
∴ f is not invertible.

(iii) f is bijective. Hence f^-1 exists.
f^-1 = {(2,1 ), (3, 2), (4, 3), (1, 4)}

(iv) f is not a function as
f(2) = 2 and f(2) = 3

(v) f is not injective hence not invertible.

Question 14.
Let f : X → Y.
If there exists a map g : Y → X such that gof = id_x and fog = id_y, then show that
(i) f is bijective and (ii) g = f^-1
[Hint: Since id_x is a bijective function, gof = id_x is bijective. By Theorem 2(iv) f is injective. Similarly fog is bijective ⇒ f is surjective by Theorem 2(iii)]
Solution:
Let f : x → y and g : y – x
where gof = id_x and fog = id_y
we know that id_x and id_y are bijective functions.
⇒ gof and fog are both bijective functions.
⇒ f is a bijective function.

(ii) As f is bijective (by (i)) we have f^-1 exists.
and f^-1 : y → x where f^-1of = id_x and fof^-1 = id_y
But g : y → x with gof = id_x and fog = id_y
∴ g = f^-1

Question 15.
Construct an example to show that f(A ∩ B) ≠ f(A) ∩ f(B) where A ∩ B ≠ Ø
Solution:
Let f(x) = cos x.
Let A = \(\left\{0, \frac{\pi}{2}\right\}\), B = \(\left\{\frac{\pi}{2}, 2 \pi\right\}\).
∴ f(A) = \(\left\{\cos 0, \cos \frac{\pi}{2}\right\}\)
= {1, 0} = {0, 1}
∴ f(B) = \(\left\{\cos \frac{\pi}{2}, \cos 2 \pi\right\}\) = {0, 1}
∴ f(A) ∩ f(B) = {0, 1}
Again,
A ∩ B = \(\left\{\frac{\pi}{2}\right\}\) and f(A ∩ B) = cos \(\frac{\pi}{2}\) = {0}
∴ f(A ∩ B) ≠ f(A) ∩ f(B)

Question 16.
Prove that for any f : X → Y, foid_x = f = id_yof.
Solution:
Let f : X → Y, so that y = f(x), x ∈ X.
∴ foid_x = fof^-1 of (x) = fof^-1 (f(x))
= f(x) = y (∵ id_x = fof^-1‍)
Again, (id_yof)(x) = (fof^-1) of (x)
= (fof^-1)(y) = f (f^-1(y)) = f(x) = y  ….(2)
∴ From (1) and (2)
we have foid_x = f = id_yof

Question 17.
Prove that f : X → Y is surjective iff for all B ⊆ Y, f (f^-1(B)) = B.
Solution:
Let f : X → Y is surjective.
i.e. for all y ∈ Y, ∃ a x ∈ X such that
y = f(x).
∴ x = f^-1(y) ⇔ f(x)
= f (f^-1(y)) ∈ f (f^-1(B)).
for y = B ⊂ Y ⇔ y ∈ f (f^-1(B)).
∴ y ∈ f (f^-1(B)) ⇔ y ∈ B
∴ f (f^-1(B)) = B

Question 18.
Prove that f : X → Y is injective iff f (f^-1(A)) = A for all A ⊆ X.
Solution:
f : X → Y is injective.
Let x ∈ A ⇔ f(x) ∈ f(A) (∵ f is injective)
⇔ x ∈ f (f^-1(A))
∴ A = f (f^-1(A)) for all A ⊆ X.

Question 19.
Prove that f : X → Y is injective iff for all subsets A, B of X, f(A ∩ B) = f(A) ∩ f(B).
Solution:
f : X → Y is injective.
Let A and B are subsets of X.
Let f(x) ∈ f(A ∩ B)
⇔ x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B
⇔ f(x) ∈ f(A) ∧ f(x) ∈ f(B) (∵ f is injective)
⇔ f(x) = f(A) ∩ f(B)
∴ f(A ∩ B) = f(A) ∩ f(B)

Conversely, suppose that
f(A ∩ B) = f(A) ∩ f(B)
Let f is not injective.
The if f(x) ∈ f(A ∩ B) ⇔ x ∈ A ∩ B
⇔ x ∈ A ∧ ⇔ x ∈ B
≠ f(x) ∈ f(A) ∧ ⇔ f(x) ∈ f(B)
⇔ f(x) ∈ f(A) ∩ f(B)
∴ f(A ∩ B) = f(A) ∩ f(B) is false.
so f must be injective.

Question 20.
Prove that f : X → Y is surjective iff for all A ⊆ X, (f(A))‘ ⊆ f(A‘), where A‘ denotes the complement of A in X.
Solution:
f : X → Y is surjective.
Then for all y ∈ Y ∃ x ∈ X
such that f(x) = y.
Let y ∈ [f(A)]‘ ⇒ y ∉ f(A)
⇒ f(x) ∉ f(A) ⇒ x ∉ A ⇒ x ∈ A‘
⇒ f(x) ∈ f(A‘) ⇒ y ∈ f(A‘)
∴ [f(A)]‘ ⊂ f(A‘)

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(a)

Question 1.
A merchant sells two models X and Y of TV with cost price ₹25000 and ₹50000 per set respectively. He gets a profit of ₹1500 on model X and ₹2000 on model Y. The sales cannot exceed 20 sets in a month. If he cannot invest more than 6 lakh rupees, formulate the problem of determining the number of sets of each type he must keep in stock for maximum profit.
Solution:
To get maximum profit let x TVs of model X and Y TVs of model Y must be kept in stock.
∴ Total profit = Z = 1500x + 2000y which is to be maximum.
According to the question, the sales cannot exceed 20 sets i.e. x + y ≤ 20.
Again total investment does not exceed 6 lakh.
25000x + 50000y ≤ 600000
⇒ x + 2y ≤ 24
∴ The LPP is:
maximise: Z = 1500x + 2000y
subject to: x + y ≤ 20
x + 2y ≤ 24
x, y ≥ 0

Question 2.
A company manufactures and sells two models of lamps L₁ and L₂, the profit being ₹15 and ₹10 respectively. The process involves two workers W, and W2 who are available for this kind of work 100 hours and 80 hours per month respectively, W1 assembles L₁ in 20 and L₂ in 30 minutes. W2 paints L₁ in 20 and L₂ in 10 minutes. Assuming that all lamps made can be sold, formulate the LPP for determining the production figures for maximum profit.
Solution:
Let x units of L₁ and y units of L₂: are to be produced to get maximum profit.
Total profit = Z = 15x + 10y
According to the question
20x + 30y ≤ 600
and 20x + 10y ≤ 480
=> 2x + 3y ≤ 600
2x + y ≤ 480
.-. The LPP is maximize: Z = 15x + 10y
subject to: 2x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0.

Question 3.
A factory uses three different resources for the manufacture of two different products, 20 units of the resource A, 12 units of B and 16 units of C being available. One unit of the first product requires 2, 2 and 4 units of the resources and one unit of the second product requires 4, 2 and 0 units of the resources taken in order. It is known that the first product gives a profit of ₹20 per unit and the second ₹30 per unit. Formulate the LPP so as to earn maximum profit.
Solution:
Let to earn maximum profit the factory produces x units of first product and y units of the second product.
The given data can be summarised as below:

	Resource A	Resource B	Resource C	Profit per unit in ₹
Product – I	2	2	4	20
Product – II	4	2	0	30
Availability	20	12	16

Total profit = 20x + 30y
which is to be maximum.
According to the question
2x + 4y ≤ 20
2x + 2y ≤ 12
4x + 0y ≤ 16
.-. The LPP is
maximize: Z = 20x + 30y
subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.

Question 4.
A man plans to start a poultry farm by investing at most ₹3000. He can buy old hens for ₹80 each and young ones for ₹140 each, but he cannot house more than 30 hens. Old hens lay 4 eggs per week and young ones lay 5 eggs per week, each egg being sold at ₹5. It costs ₹5 to feed an old hen and ₹8 to feed a young hen per week. Formulate his problem determining the number of hens of each type he should buy so as to earn a profit of more than ₹300 per week.
Solution:
Let to get maximum profit he has to purchase x old hens and y young hens.
Total cost = 80x + 140y ≤ 3000
⇒ 4x + 7y ≤ 150
Total number of hens x + y ≤ 30
Number of eggs per week = 4x + 5y
Total income per week = 20x + 25y
Total cost to feed per week = 5x + 8y
∴ Weekly profit = 15x + 17y
∴ 15x + 17y > 300
and also total profit = z = 15x + 17y is to be maximum.
∴ The LPP is
maximize: Z = 15x + 17y
subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y ≥ 300
x, y ≥ 0.

Question 5.
An agro-based company produces tomato sauce and tomato jelly. The quantity of material, machine hour, labour (man-hour) required to produce one unit of each product and the availability of raw material ore given in the following table:

	Sauce	Jelly	availability
Man-hour	3	2	10
Machine hour	1	2.5	7.5
Raw material	1	1.2	4.2

Assume that one unit of sauce and one unit of jelly each yield a profit of ₹2 and ₹4 respectively. Formulate the LPP so as to yield maximum profit.
Solution:
Let the company produces x units of sauce and y units of jelly.
Total profit = 2x + 4y to be maximum.
Man hour = 3x + 2y ≤ 10
Machine hour = x + 2.5y ≤ 7.5
⇒ 2x + 5y ≤ 15
Raw material = x + 1.2y ≤ 4.2
⇒ 5x + 6y ≤ 21
∴ The LPP is
maximize: Z = 2x + 4y
subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0.

Question 6.
(Allocation Problem.) A farmer has 5 acres of land on which he wishes to grow two crops X and Y. He has to use 4 cart loads and 2 cartloads of manure per acre for crops X and Y respectively. But not more than 18 cartloads of manure is available. Other expenses are ₹200 and ₹500 per acre for the crops X and Y respectively. He estimates profit from crops X and Y at the rates ₹1000 and ₹800 per acre respectively. Formulate the LPP as to how much land he should allocate to each crop for maximum profit.
Solution:
Let x acres are allocated for crop X and y acres for crop Y.
Total profit = 100x + 800y to be maximum.
According to the question
x + y ≤ 5
Manure = 4x + 2y ≤ 1 8 ⇒ 2x + y ≤ 9
∴ The LPP is
maximize: Z = 1000x + 800y
subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0.

Question 7.
(Transportation Problem) A company has two factories at locations X and Y. He has to deliver the products from these factories to depots located at three places A, B and C. The production capacities at X and Y are respectively 12 and 10 units and the requirements at the depots are 8, 8 and 6 units respectively. The cost of transportation from the factories to the depots per unit of the product is given below.

(Cost in ₹)
	To →	A	B	C
From	X	210	160	250
	Y	170	180	140

The company has to determine how many units of product should be transported from each factory to each depot so that the cost of transportation is minimum. Formulate this LPP.
Solution:
Let x units are transported from X to A and y units from X to B. The transportation matrix is

Total cost of transportation
= 210x + 160y + 250 (12 – x – y) +170 (8 – x) +180 (8 – y) + 140 (x + y – 6)
= 4960 – 70x – 130y to be minimum.
Now all costs of transportation are ≥ 0.
∴ x ≥ 0, y ≥ 0.
12 – x – y ≥ 0 ⇒ x + y ≤ 12
8 – x ≥ 0 ⇒ x ≤ 8
8 – y ≥ 0 ⇒ y ≤ 8
x + y – 6 ≥ 0 x + y ≥ 6
∴ The LPP is
Minimize: Z = 4960 – 70x – 130y
subject to: x + y ≤ 12
x ≤ 8
y ≤ 8
x + y ≥ 6
x, y ≥ 0

Question 8.
(Diet Problem) Two types of food X and Y are mixed to prepare a mixture in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. These vitamins are available in one kg of food as per the table given below.

	Vitamins
food	A	B	C
X	1	2	3
Y	2	2	1

One kg. of food X cost ₹16 and one kg. of food Y costs ₹20. Formulate the LPP so as to determine the least cost of the mixture containing the required amount of vitamins.
Solution:
Let x units of food X and y units of food Y are to be mixed to prepare the mixture.
Cost of the mixture = 16x + 20y to be minimum.
According to the question
Vitamin A content = x + 2y ≥ 10
Vitamin B content = 2x + 2y ≥ 12
Vitamin C content = 3x + y ≥ 8
∴ The LPP is
minimize: Z = 16x + 20y
subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0.

Question 9.
Special purpose coins each weighing 10gms are to be manufactured using two basic metals M₁ and M₂ and a mix of other metals M₃. M₁, M₂ and M₃ cost ₹500, ₹800 and ₹50 per gram respectively. The strength of a coin demands that not more than 7gm. of M₁ and a minimum of 3 gm of M₁ should be used. The amount of M₃ in each coin is maintained at 25% of that of M₁. Since the demand for the coin is related to its price, formulate the LPP to find the minimum cost of a coin.
Solution:
Let x gm of M₁ and y g of M₂ are used to make the coin. According to the demand of the coin \(\frac{x}{4}\) g of M₃ is to be mixed.
Cost of the coin = 500x + 800y + \(\frac{50 x}{4}\)
= (512.5)x + 800y
which is to be minimum.
Again weight of the coin = 10g
⇒ x + y + \(\frac{x}{4}\) = 10
⇒ 5x + 4y = 40
According to the question x ≤ 7, y ≤ 3.
Thus the L.P.P is
minimize: Z = (512.5)x + 800y
subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0.

Question 10.
A company produces three types of cloth A, B and C. Three kinds of wool, say red, green and blue are required for the cloth. One unit length of type A cloth needs 2 metres of red and 3 metres of blue wool; one unit length of type B cloth needs 3 metres of red, 2 metres of green and 2 metres of blue wool and one unit length of type C cloth needs 5 metres of green and 4 metres of blue wool. The firm has a stock of only 80 metres of red, 100 metres of green and 150 metres of blue wool. Assuming that income obtained from one unit length of cloth is ₹30, ₹50 and ₹40 of types A, B and C respectively, formulate the LPP so as to maximize income.
Solution:
Let x units of cloth A, y units of cloth B and z units of cloth C are to be produced from the available materials to get the maximum income. The given data can be summarised as:

	Red wool	Green wool	Blue wool	Income
Cloth A	2	–	3	30
Cloth B	3	2	2	50
Cloth C	–	5	4	40
Availability	80	100	150

Total Income = 30x + 50y + 40z
which is to be maximum.
According to the question
2x + 3y ≤ 80
2y + 5z ≤ 100
3x + 2y + 4z ≤ 150
Thus the L.P.P. is
minimize: Z = 30x + 50y + 40z
subject to: 2x + 3y ≤ 80
2y + 5z ≤ 100
3x + 2y + 4z ≤ 150
x, y, z ≥ 0

Question 11.
A person wants to decide the constituents of a diet which will fulfil his daily requirements of proteins, fats and carbohydrates at minimum cost. The choice is to be made from three different types of food. The yields per unit of these foods are given in the following table.

food	yield/unit			cost/unit
	Protein	Fat	Carbonate
f1	3	2	6	45
f2	4	2	3	40
f3	8	7	7	85
Minimum Requirement	100	200	800

Formulate the LPP.
Solution:
Let the diet constitues x units of f1, y units of f2 and z units of f3.
Total cost = 45x + 40y + 85z, which is to be minimum.
According to the question
3x + 4y + 8z ≥ 1000
2x + 2y + 7z ≥ 200
6x + 3y + 7z ≥ 800
Thus the LPP is
Minimize: Z = 45x + 40y + 85z
Subject to: 3x + 4y + 8z ≥ 1000
2x-+2y + 7z ≥ 200
6x + 3y + 7z ≥ 800
x, y, z ≥ 0

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(c)

Question 1.
Show that the operation ∗ given by x ∗ y = x + y – xy is a binary operation on Z, Q and R but not on N.
Solution:
The operation ∗ given by
x ∗ y = x + y – xy
Clearly for all x, y ∈ Z
x + y – xy ∈ Z
⇒ x ∗ y ∈ Z
∴ ∗ is a binary operation on Z.
For all x, y ∈ Q
x + y – xy ∈ Q
⇒ x ∗ y ∈ Q
⇒ ∗ is a binary operation on Q for all x, y, ∈ R.
x + y – xy ∈ R
⇒ x ∗ y ∈ R
⇒ ∗ is a binary operation on R
Again 3, 4 ∈ N.
3 + 4 – 3 x 4 = 7 – 12 = – 5 ∉ N
i.e., x, y ∈ N
≠ x ∗ y ∈ N
∴ ∗ is not a binary operation on N.

Question 2.
Determine whether the following operations as defined by ∗ are binary operations on the sets specified in each case. Give reasons if it is not a binary operation.
(i) a ∗ b = 2a + 3b on Z.
(ii) a ∗ b = ma – nb on Q⁺ where m and n ∈ N.
(iii) a ∗ b = a + b (mod 7) on {0, 1, 2, 3, 4, 5, 6}
(iv) a ∗ b = min {a, b} on N.
(v) a ∗ b = GCD {a, b} on N.
(vi) a ∗ b = LCM {a, b} on N.
(vii) a ∗ b = LCM {a, b} on {0, 1, 2, 3, 4……, 10}
(viii) a ∗ b = \(\sqrt{a^2+b^2}\) on Q⁺
(ix) a ∗ b =a × b (mod 5) on {0, 1, 2, 3, 4}.
(x) a ∗ b = a² + b² on N.
(xi) a ∗ b = a + b – ab on R – {1}.
Solution:
(i) For all a, b ∈ Z
2a + 3b ∈ Z
⇒ a ∗ b ∈ Z
∗ is a binary operation on Z.

(ii) Let a = 1, b = 2
m = 1, n = 3
ma – nb = 1 – 6 = – 5 ∉ Q⁺
∴ a, b ∈ Q⁺ ≠ a ∗ b ∈ Q⁺
⇒ ∗ is not a binary operation on Q⁺

(iii) a ∗ b = a + b (mod 7) ∈ {0, 1, 2, 3, 4, 5, 6}
for a, b ∈ 7
∗ is a binary operation on the given set.

(iv) a, b ∈ N ⇒ min {a, b} ∈ N
∴ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(v) for all a, b ∈ N, GCD [a, b] ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(vi) for all a, b ∈ N, LCM {a, b} ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(vii) Let A = {0, 1, 2, ….. 10}
4, 5 ∈ A but 4 ∗ 5 = LCM {4, 5}
= 20 ∉ A
⇒ ∗ is not a binary operation on A.

(viii) for all a, b ∈ Q⁺
a ∗ b = \(\sqrt{a^2+b^2}\) ∉ Q⁺
⇒ ∗ is not a binary operation on Q⁺.

(ix) For all a, b ∈ {0, 1, 2, 3, 4}
a ∗ b = a × b (mod 5) ∈ {0, 1, 2, 3, 4}
∴ ∗ is a binary operation on the given set.

(x) for all a, b ∈ N, a * b = a² + b² ∈ N
∴ ∗ is a binary operation on N.

(xi) For all a, b ∈ R – {1}
a ∗ b = a + b – ab ∈ R – {1}
∴ ∗ is a binary operation on R – {1}

Question 3.
In case ∗ is a binary operation in Q2 above, test whether it is (i) associative
(ii) commutative, Test further if the identity element exists and the inverse element for any element of the respective set exists.
Solution:
(i) On Z the binary operation is
a ∗ b = 2a + 3b
Commutative:
b ∗ a = 2b + 3a ≠ a ∗ b
∴ ∗ is not commutative.

Associative:
(a ∗ b) ∗ c = (2a + 3b) ∗ c
= 2 (2a + 3b) + 3c
= 4a + 6b + 3c
a ∗ (b ∗ c) = a ∗ (2b + 3c)
= 2a + 3 (2b + 3c)
= 2a + 6b + 9c
As (a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∗ is not associative.

Existance of identity:
Let e is the identity
∴ e ∗ a = a
⇒ 2e + 3a = a
⇒ e = -2a / 2 = -a
which depends on a.
∴ Identity element does not exist.

(iii) A = (0, 1, 2, 3, 4, 5, 6}
Commutative:
a ∗ b = a + b (mod 7)
= The remainder obtained when a + b is divided by 7.
b ∗ a = b + a (mod 7) = a + b (mod 7)
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c = {a + b (mod 7)} ∗ c
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
a ∗ (b ∗ c) = a ∗ {b + c (mod 7)}
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
∴ (a ∗ b) ∗ c = a + (b ∗ c)
∴ ∗ is associative.

Existance of identity:
Let e is the identity
⇒ e ∗ a = a ∗ e = a
⇒ e + a mod 7 = a
⇒ e = 0
∴ 0 is the identity.

Existance of inverse:
Let a^-1 = the inverse of a
⇒ a ∗ a^-1 = a^-1 ∗ a = e = 0
⇒ a + a^-1 (mod 7) = 0
⇒ a + a^-1 is divisible by 7.
1^-1 = 6, 6^-1 = 1
2^-1 = 5, 5^-1 = 2
3^-1 = 4, 4^-1 = 3

(iv) a ∗ b = min {a, b} on N.
Commutative:
a ∗ b = min {a, b}
b ∗ a – min {b, a} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = min {a, b} ∗ c
= min {a, b, c}
a ∗ (b ∗ c)= a ∗ min {b, c}
= min {a, b, c}
⇒ a ∗ (b ∗ c) = (a ∗ b) ∗ c
∴ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ For all a ∈ N
e ∗ a = a ∗ e = a
⇒ min {e, a} = a
No such element exists in N.
∴ ∗ has no identity element on N.

(v) a ∗ b = GCD {a, b} on N.
b ∗ a = GCD {b, a} = GCD {a, b} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = GCD {a, b} ∗ c
= GCD {a, b, c}
a ∗ (b ∗ c) = a ∗ GCD {b, c}
= GCD {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ a ∗ e = e ∗ a = a
⇒ GCD {e, a} = a
No such element exists in N
⇒ ∗ has no indentity element.

(vi) a ∗ b = LCM {a, b} on N
Commutative:
a ∗ b = LCM {a, b}
= LCM {b, a}
= b ∗ a
∴ ∗ is commutative.

Associative
(a ∗ b) ∗ c = LCM {a, b} ∗ c
= LCM {a, b, c}
a ∗ (b ∗ c) =» a ∗ LCM {b, c}
= LCM {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
∴ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ LCM {e, a} = a
⇒ e – 1
∴ 1 is the identity element.

Existance of inverse:
Let a^-1 is the inverse of a
⇒ a * a^-1 = e = 1
⇒ LCM [a, a^-1} = 1
a = a^-1 = 1
Only 1 is invertible with 1^-1 = 1.

(ix) a ∗ b = a × (mod 5) on {0, 1, 2, 3, 4}

Commutative:
a ∗ b = a x b (mod 5)
= Remainder on dividing a x b by 5
= Remainder on dividing b x a by 5
= b x a (mod 5)
= b x a
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c=a x b (mod 5) ∗ c
= a x b x c (mod 5)
a and a ∗ {b ∗ c} = a ∗ {b x c (mod 5)}
= a x b x c (mod 5)
∴ (a ∗ b) ∗ c -=a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of identity:
Let e is the identity
∴ For all a ∈ {0, 1, 2, 3, 4}
a ∗ a = e ∗ a = a
a × e (mod 5) = a
⇒ e = 1
∴ 1 is the identity element.

Existance of inverse:
Let a^-1 is the inverse of a
∴ a ∗ a^-1= a^-1 ∗ a = e = 1
⇒ a x a^-1 (mod 5) = 1
⇒ 1^-1 = 1
2^-1 = 3, 3^-1 = 2, 4^-1 = 4
0 has no inverse.

(x) a ∗ b = a² + b² on N.
Commutative:
a ∗ b = a² + b²
b ∗ a = b² + a² = a² + b² = a ∗ b
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c = (a² + b²) ∗ c
= (a² + b²)² + c²
a ∗ (b ∗ c) = a ∗ (b² + c²)
= a² + (a² + b²)²
(a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∴ ∗ is not associative.

Existance of Identity:
Let e is the identity
a ∗ e = e ∗ a = a
⇒ a² + e² = a
⇒ e = \( \sqrt{a-a^2}\) which depends on a
∴ Identity does not exist.

(xi) a ∗ b = a + b – ab on R – {1}
Commutative:
a ∗ b = a + b – ab
b ∗ a = b + a – ba
a ∗ b = b ∗ a
∴ ∗ is commutative.

Associative:
a ∗ (b ∗ c) = a ∗ (b + c – bc)
= a + (b + c – bc) – a (b + c – bc)
= a + b + c – bc – ab – ac + abc
(a ∗ b) ∗ c = (a + b – ab) ∗ c
= a + b – ab + c – (a + b – ab) c
= a + b + c – ab – bc – ca + abc
∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ a + e – ae = a
⇒ e (1 – a) = 0
⇒ e = 0 (∵ a ≠ 1)
∴ 0 is the identity.

Existance of inverse:
Let a^-1 is the inverse of a
⇒ a ∗ a^-1 = a^-1 ∗ a = e
⇒ a + a^-1 – aa^-1 = 0
⇒ a^-1 (1 – a) = – a
⇒ a^-1 = \(\frac{a}{a-1}\) for a ∈ R – {1}

Question 4.
Construct the composition table/multiplication table for the binary operation ∗ defined on {0, 1, 2, 3, 4} by a ∗ b = a × b {mod 5). Find the identity element if any. Also find the inverse elements of 2 and 4.
[This operation is called multiplication moduls 5 and denoted by x⁵. In general, on a finite subset of N, x_m denotes the operation of multiplication modulo m where m is a fixed positive integer].
Solution:
A = {0. 1, 2, 3, 4}
a ∗ b = a × b mod 5

∗	0	1	2	3	4
0	0	0	0	0	0
1	0	1	2	3	4
2	0	2	4	1	3
3	0	3	1	4	2
4	0	4	3	2	1

As 3rd row is identical to the first row we have 1 is the identity clearly 2^-1 = 3 and 4^-1 = 4.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(b)

Question 1.
Maximize Z = 5x₁+ 6x₂
Subject to: 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation, we get 2x₁ + 3x₂ = 6
Step – 2 Let us draw the graph

x1	3	0
x2	0	0

Step – 3 Clearly (0,0) statisfies 2x₁ + 3x₂ ≤ 6
The shaded region is the feasible region with vertices 0(0,0), A(3,0), B(0,2).
Step – 4

Corner point	Z = 5x1+ 6x2
0(0.0)	0
A(3,0)	15 → maximum
B(0,2)	12

Z is maximum at A (3,0)
∴ The solution of LPP is x₁ = 3, x₂ = 0
Z_max = 15

Question 2.
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation we get x₁ + 2x₂ = 0
Step – 2 Let us draw the graph of x₁ + 2x₂ = 4

x1	0	4
x2	2	0

Step – 3 Clearly 0(0,0) does not satisfy
x₁ + 2x₂ > 4, x₁ > 0, x₂ > 0 is the first quadrant.
The feasible region is the shaded region with vertices A(4, 0), B(0, 2).
Step – 4 Z (4, 0) = 24
Z (0, 2) = 14 → minimum
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B (0, 2).
Let us draw the half-plane 6x₁ + 7x₂ < 14

x1	0	3.5
x2	2	-1

As this half-plane has no point common with the feasible region, we have Z is minimum for x₁= 0, x₂ = 2 and the minimum value of Z = 14.

Question 3.
Maximize Z = 20x₁+ 40x₂
Subject to: x₁ + x₂ ≤ 1
6x₁ + 2x₂ ≤ 3
x₁, x₂ ≥ 0.
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ = 1    …. (1)
6x₁ + 2x₂ = 3   …. (2)
x₁, x₂ ≥ 0
Step – 2 Let us draw the graph:
Table – 1

x1	0	1
x2	1	0

Table – 2

x1	0	0.5
x2	1.5	0

Step – 3 As (0, 0) satisfies both the inequations the shaded region is the feasible region.
Step – 4 Solving
x₁ + x₂ = 1
6x₁ + 2x₂ = 3
we have x₁ = ¼ x₂ = ¾
The vertices are O(0, 0), A(0.5, 0), B(0,1) and C(¼, ¾)
Now Z(O) = 0
Z(A) = 10
Z(B) = 40
Z(C) = 20 × ¼ + 40 × ¾ = 35
∴ Z attains maximum at B for x₁= 0, x₂ = 1
Z_max = 40

Question 4.
Minimize: Z = 30x₁ + 45x₂
Subject to: 2x₁ + 6x₂ ≥ 4
5x₁ + 2x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Consider the constraints as equations
2x₁ + 6x₂ = 4
5x₁ + 2x₂ = 5
Step – 2
Table – 1

x1	2	-1
x2	0	1

Table – 2

x1	1	0
x2	0	2.5

Step – 3 Clearly 0(0,0) does not satisfy 2x₁ + 6x₂ ≥ 4 and 5x₁ + 2x₂ ≥ 5.
Thus the shaded region is the feasible region.
Solving the equations we get
x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\).
∴ The vertices are A(2, 0)
B(\(\frac{11}{13}\), \(\frac{5}{13}\)) and C(0, \(\frac{5}{2}\)).
Step – 4 Z(A) = 60
Z(B) = \(\frac{555}{13}\) → minimum
Z(C) = \(\frac{225}{2}\)
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B(\(\frac{11}{13}\), \(\frac{5}{13}\))
Let us draw the half plane
30x₁ + 45x₂ < \(\frac{555}{13}\)

x1	\(\frac{11}{13}\)	0
x2	\(\frac{5}{13}\)	\(\frac{27}{39}\)

As this half plane and the feasible region has no point in common we have Z is minimum for x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\), and Z_min = \(\frac{555}{13}\)

Question 5.
Maximize: Z = 3x₁+ 2x₂
Subject to: -2x₁ + x₂ ≤ 1
x₁ ≤ 2
x₁+ x₂ ≤ 3
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-2x₁ + x₂ = 1        …..(1)
x₁ = 2                   …..(2)
x₁+ x₂ = 3            …..(3)
Step – 2 Let us draw the lines.
Table – 1

x1	0	-1
x2	1	-1

Table – 2

x1	2	2
x2	0	1

Table – 3

x1	0	3
x2	3	0

Step – 3 (0, 0) satisfies all the constraints and x₁, x₂ > 0 is the 1st quadrant the shaded region is the feasible region.

Step – 4 Solving -2x₁ + x₂ = 1
x₁+ x₂ = 3
we have 3x₁ = 2
⇒ x₁ = \(\frac{2}{3}\), x₂ = 3 – \(\frac{2}{3}\) = \(\frac{7}{3}\)
From x₁+ x₂ = 3 and x₁ = 2 we have x₁ = 2, x₂ = 1
∴ The vertices are 0(0, 0), A(2, 0), B(2, 1), C(\(\frac{2}{3}\), \(\frac{7}{3}\)), D(0, 1)
Z(0) = 0, Z(A) = 6, Z(B) = 8, Z(C) = 3.\(\frac{2}{3}\) + 2.\(\frac{7}{3}\) = \(\frac{20}{3}\), Z(D) = 2
Z is maximum at B.
∴ The solution of given LPP is x₁ = 2, x₂ = 1, Z_(max) = 8.

Question 6.
Maximize: Z = 50x₁+ 60x₂
Subject to: x₁ + x₂ ≤ 5
x₁+ 2x₂ ≤ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + x₂ = 5     ….(1)
x₁+ 2x₂ = 4    ….(2)
Step – 2 Let us draw the graph
Table – 1

x1	5	5
x2	0	0

Table – 2

x1	4	0
x2	0	2

Step – 3 0(0,0) satisfies x₁ + x₂ ≤ 5 and does not satisfy x₁+ 2x₂ ≤ 4
Thus the shaded region is the feasible region.
Step – 4 The corner points are A(4,0), B(5,0), C(0,5) , D(0,2)
∴

Corner point	z = 50x1+ 60x2
A(4,0)	200
B (5,0)	250 → maximum
C(0,5)	300
D(0,2)	120

Z is maximum for x₁ = 0, x₂ = 5, Z_(max) = 300.

Question 7.
Maximize: Z = 5x₁+ 7x₂
Subject to: x₁ + x₂ ≤ 4
5x₁+ 8x₂ ≤ 30
10x₁+ 7x₂ ≤ 35
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get,
x₁ + x₂ = 4           …. (1)
5x₁+ 8x₂ = 30      …. (2)
10x₁+ 7x₂ = 35    …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	4	0
x2	0	4

Table – 2

x1	6	2
x2	0	2.5

Table – 3

x1	0	3.5
x2	5	0

Step – 3 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get (\(\frac{2}{3}\), \(\frac{10}{3}\))
From (1) and (3) we get
x₁ = \(\frac{7}{3}\), x₁ = \(\frac{5}{3}\)
∴ The corner points are 0(0,0), A(\(\frac{7}{2}\), 0), B(\(\frac{7}{3}\), \(\frac{5}{3}\)), C(\(\frac{2}{3}\), \(\frac{10}{3}\)), D(0, \(\frac{15}{4}\))
Step – 4

Corner point	z = 5x1+ 7x2
0(0,0)	0
A(\(\frac{7}{2}\), 0)	\(\frac{35}{2}\)
B(\(\frac{7}{3}\), \(\frac{5}{3}\))	\(\frac{70}{3}\)
C(\(\frac{2}{3}\), \(\frac{10}{3}\))	\(\frac{80}{3}\)
D(0, \(\frac{15}{4}\))	\(\frac{105}{4}\)

Z attains its maximum value \(\frac{80}{3}\) for x₁ = \(\frac{2}{3}\) and x₂ = \(\frac{10}{3}\).

Question 8.
Maximize: Z = 14x₁ – 4x₂
Subject to: x₁ + 12x₂ ≤ 65
7x₁ – 2x₂ ≤ 25
2x₁+ 3x₂ ≤ 10
x₁, x₂ ≥ 0
Also find two other points which maximize Z.
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + 12x₂ = 65   …. (1)
7x₁ – 2x₂ = 25    …. (2)
2x₁ + 3x₂ = 10   …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	65	5
x2	0	5

Table – 2

x1	5	10
x2	5	22.5

Table – 3

x1	5	2
x2	0	2

Step – 3 Clearly 0(0,0) satisfies x₁ + 12x₂ ≤ 65 and 7x₁ – 2x₂ ≤ 25 but does not satisfy 2x₁+ 3x₂ ≤ 10. Thus shaded region is the feasible region.
Equation (1) and (2) meet at (5, 5).
From (2) and (3)

∴ The corner points of the feasible region are A(0, \(\frac{10}{3}\)), B(\(\frac{19}{5}\), \(\frac{4}{5}\)), C(5, 5), D(0, \(\frac{65}{12}\)).
Step – 4

Corner point	z = 14x1 – 4x2
A(0, \(\frac{10}{3}\))	\(\frac{-40}{3}\)
B(\(\frac{19}{5}\), \(\frac{4}{5}\))	50 → maximum
C(5, 5)	50 → maximum
D(0, \(\frac{65}{12}\))	\(\frac{65}{3}\)

Z is maximum for x₁ = \(\frac{19}{5}\), x₂ = \(\frac{4}{5}\) or x₁ = 5, x₂ = 5 and Z_max = 50
There is no other point that maximizes Z.

Question 9.
Maximize: Z = 10x₁ + 12x₂ + 8x₃
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁ + x₂ + x₃ = 20
x₁, x₂ ≥ 0
[Hints: Eliminate x₃ from all expressions using the given equation in the set of constraints, so that it becomes an LPP in two variables]
Solution:
Eliminating x₃ this LPP can be written as Maximize Z = 2x₁ + 4x₂ + 160
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁, x₂ ≥ 0
Step – 1 Treating the consraints as equations we get
x₁ + 2x₂ = 30    …..(1)
5x₁ – 7x₃ = 12   …..(2)
Step – 2 Let us draw the graph
Table – 1

x1	30	0
x2	0	15

Table – 2

x1	8	1
x2	8	20

Step – 3 Clearly 0(0,0) satisfies x₁ + 2x₂ ≤ 30 and does not satisfy 12x₁ + 7x₂ ≤ 152
∴ The shaded region is the feasible region.

Step – 4

Z is maximum for x₁ = 30, x₂ = 0 and Z_max = 220

Question 10.
Maximize: Z = 20x₁ + 10x₂
Subject to: x₁ + 2x₂ ≤ 40
3x₁ + x₂ ≥ 30
4x₁+ 3x₂ ≥ 60
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equalities we have:
x₁ + 2x₂ = 40   ….(1)
3x₁ + x₂ = 30   ….(2)
4x₁+ 3x₂ = 60  ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + 2x₂ ≤ 40 and does not satisfy 3x₁ + x₂ ≥ 30 and 4x₁+ 3x₂ ≥ 60, x₁, x₂ ≥ 0 is the first quadrant.
∴ The shaded region is the feasible region.
Step – 4 x₁ + 2x₂ = 40 and 3x₁ + x₂ = 30

∴ The vetices are A(15, 0), B(10, 0), C(4, 18) and D(6, 12)
Z(A) = 300, Z(B) = 800
Z (C) = 20 x 4 + 10 x 18 = 260
Z (D) = 120 + 120 = 240
Z attains minimum at D(6 ,12).
∴ The required solution x₁ = 6, x₂ =12 and Z_min = 240

Question 11.
Maximize: Z = 4x₁ + 3x₂
Subject to: x₁ + x₂ ≤ 50
x₁ + 2x₂ ≥ 80
2x₁+ x₂ ≥ 20
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ ≤ 50    ….(1)
x₁ + 2x₂ ≥ 80  ….(2)
2x₁+ x₂ ≥ 20   ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + x₂ < 50, x₁ + 2x₂ < 80 but does not satisfy
2x₁ + x₂ > 20, x₁ > 0, x₂ > 0 is the 1st quadrant.
Hence the shaded region is the feasible region.
Step – 4 x₁ + x₂ = 50
x₁ + 2x₂ = 80
=> x₂ = 30, x₁ = 20
The vertices of feasible region are
A(10, 0), B(50, 0), C(20, 30), D (0, 40) and E (0, 20)

Point	Z = 4x1 + 3x2
A(10,0)	40
5(50,0)	200
C(20,30)	170
D(0,40)	120
E(0,120)	60

Question 12.
Optimize: Z = 5x₁ + 25x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 (0, 0) satisfies -0.5x₁ + x₂ ≤ 2, but does not satisfy x₁ + x₂ ≥ 2 and -x₁+ 5x₂ ≥ 5, x₁ > 0, x₂ > 0 is the 1st quadrant.
The shaded region is the feasible region with vertices A(\(\frac{5}{6}\), \(\frac{7}{6}\)) and B(0, 2).
Step – 4 Z can be made arbitrarily large.
∴ Problem has no maximum.
But Z(A) = \(\frac{100}{3}\), Z(B) = 50
Z is minimum at A(\(\frac{5}{6}\), \(\frac{7}{6}\)).
But the feasible region is unbounded.
Hence we cannot immediately decide, Z is minimum at A.
Let us draw the half plane
5x₁ + 25x₂ < \(\frac{100}{3}\)
⇒ 3x₁ + 15x₂ < 20
As there is no point common to this half plane and the feasible region.
we have Z is minimum for x₁ = \(\frac{5}{6}\), x₂ = \(\frac{7}{6}\) and the minimum value = \(\frac{100}{3}\)

Question 13.
Optimize: Z = 5x₁ + 2x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.

As Z can be made arbitrarily large, the given LPP has no maximum.
Z is minimum at B (0, 2). But we cannot immediately decide, Z is minimum at B.
Let us draw the half plane 5x₁ + 2x₂ < 4

x1	0	4/5
x2	2	0

As there is no point common to this half plane and the feasible region,
we have Z is minimum for x₁ = 0, x₂ = 2 and the minimum value of Z = 4.

Question 14.
Optimize: Z = -10x₁ + 2x₂
Subject to: -x₁ + x₂ ≥ -1
x₁ + x₂ ≤ 6
x₂ ≤ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-x₁ + x₂ = -1     ….(1)
x₁ + x₂ = 6        ….(2)
x₂ = 5                ….(3)
Step – 2 Let us draw the graph
Table – 1

x1	1	0
x2	0	-1

Table – 2

x1	6	0
x2	0	1

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
The vertices are 0(0,0) , A(1,0), B(\(\frac{7}{2}\), \(\frac{5}{2}\)) ,C(1, 5) and D (0, 5)
Step – 4 Z(O) = 0
Z(A) = -10
Z(B) = – 30
Z(C) = 0
Z(D) = 10
∴ Z is maximum for x₁= 0, x, = 5 and Z_(max) = 10
Z is minimum for x₁ = \(\frac{7}{2}\)  x₂ = \(\frac{5}{2}\) and Z_(min) = -30

Question 15.
Solve the L.P.P.s obtained in Exercise 3(a) Q.1 to Q. 9 by graphical method.
(1) Maximise: Z = 1500x + 2000y
Subject to: x + y < 20
x + 2y < 24
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 20
x + 2y = 24
Step – 2 Let us draw of graph.

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get
y = 14
x = 16
With vertices 0(0, 0), A(20, 0), B(16, 4), C(0, 12).
Step – 4 Z(0) = 0
Z(A) = 30,000
Z(B) = 32,000 → Maximum
Z(C) = 24000
Z is maximum for x = 16, y = 4 with Z = 32000
To get maximum profit he must keep 16 sets of model X and 4 sets of model Y.
Maximum profit = 1500 × 16 + 2000 × 4 = ₹32,000

(2) Maximize: 15x + 10y
Subject: x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
2x +3y = 600
2a + y = 480
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
The corner point are 0(0, 0), A (240, 0) B(210, 60),C(0, 200)
Step – 4 Z(0) = 6
Z(A) = 3600
Z(B) = 3150 + 600
= 3750 → maximum
Z(C) = 2000
Thus Z is maximum for x = 210 and y = 60
and Z_(max) = 3750

(3) Maximize: Z = 20x + 30y
Subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10       …(1)
x + y = 6           …(2)
x = 4
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
Solving (1) and (2) we get x = 2, y = 4.
The vertices and 0(0, 0) , A(4, 0), B(4, 2), C(2, 4), D (0, 5).
Step – 4 Z(0) =0
Z(A) = 80
Z (B) =140
Z(C) = 1 60 → maximum
Z (D) = 150
∴ Z is Maximum when x = 2, y = 4 and Z_(max) = 160

(4) Maximize: Z = 15x + 17y
Subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y > 300
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
4x + 7y = 150      ….(1)
x + y = 30            ….(2)
15x + 17y = 300  ….(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
4x + 7y ≤ 150, x + y ≤ 30, but does not satisfy 15x + 17y ≥ 300.
∴ The shaded region is the feasible region.
From (1) and (2) we get

∴ Z is maximum for x = 20. y = 10 and Z_(max) = 470

(5) Maximize: Z = 2x + 4y
Subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
3x + 2y = 10  …(1)
2x + 5y = 15  …(2)
5x + 6y = 21  …(3)
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
From (1) and (3) we get

From (2) and (3) we get

Step-4 Z(O) = 0

(6) Maximize: Z = 1000x + 800y
Subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 5    ….(1)
2x + y = 9  ….(2)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
∴ Thus the shaded region is the feasible region.
From (1) and (2) we get x = 4, y = 1.
∴ The vertices are A(0, 0), A(4.5, 0), B(4, 1) and C(0, 5).
Step – 4 Z(0) =0
Z (A) = 4500
Z (B) = 4800 → Maximum
Z (C) = 4000
Z is maximum for x = 4 and y = 1, Z_(max) = 4800

(7) Minimize: Z = 4960 – 70x – 130y
Subject to: x + y ≤ 12
x + y ≥ 6
x ≤ 8
y ≤ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 12   ….(1)
x + y = 6     ….(2)
x = 8           ….(3)
y = 4           ….(4)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints except x + y > 6.
The shaded region is the feasible region.
The vertices are A(6, 0), B(8, 0), C(8, 4), D(4, 8), E(0, 8) and F(0, 6).
Step – 4 Z (A) = 4540
Z (B) = 4400
Z (C) = 3880
Z (D) = 3640 → Minimum
Z (E) = 3920
Z (F) = 4180
∴ Z is maximum for x = 4 and y = 8 and Z_(min) = 3640.

(8) Minimize: Z = 16x + 20y
Subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10  ….(1)
x + y = 6      …(2)
3x + y = 8    …(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints. Thus the shaded region is the feasible region.
From (1) and (2) we get y = 4, x = 2.
From (2) and (3) we get x = 1, y = 5.
The vertices are A(10, 0), B(2, 4), C(1, 5), D(0, 8).
Step – 4 Z (A) = 160
Z (B) = 112 → Minimum
Z (C) =116
Z (D) = 160
As the region is unbounded, let us draw the half plane Z < Z_(min)
⇒ 16x + 20y < 112
⇒ 4x + 5y < 28

x1	7	0
x2	0	5.6

There is no point common to the shaded region and the half plane 4x + 5y ≤ 28 other than B(2, 4).
∴ Z is minimum for x = 2, y = 4 and Z_(min) = 112.

(9) Minimize: Z = (512.5)x + 800y
Subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0
Solution:
Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3

x1	8	0
x2	0	10

Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3
Step – 2 The line segment AB is the feasible region.
Step – 3 Z (A) = 3587.5 + 1000 = 4587.5
Z (B) = 2870 + 2400 = 5270
Clearly Z is minimum for
x = 7, y = \(\frac{5}{4}\) and Z_(min) = 4587.5

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 2 Inverse Trigonometric Functions Ex 2 Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

Question 1.
Fill in the blanks choosing correct answer from the brackets:
(i) If A = tan^-1 x, then the value of sin 2A = ________. (\(\frac{2 x}{1-x^2}\), \(\frac{2 x}{\sqrt{1-x^2}}\), \(\frac{2 x}{1+x^2}\))
Solution:
\(\frac{2 x}{1+x^2}\)

(ii) If the value of sin^-1 x = \(\frac{\pi}{5}\) for some x ∈ (-1, 1) then the value of cos^-1 x is ________. (\(\frac{3 \pi}{10}\), \(\frac{5 \pi}{10}\),\(\frac{3 \pi}{10}\))
Solution:
\(\frac{3 \pi}{10}\)

(iii) The value of tan^-1 x (2cos\(\frac{\pi}{3}\)) is ________. (1, \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(iv) If x + y = 4, xy = 1, then tan^-1 x + tan^-1 y = ________. (\(\frac{3 \pi}{4}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{2}\)

(v) The value of cot^-1 2 + tan^-1 \(\frac{1}{3}\) = ________. (\(\frac{\pi}{4}\), 1, \(\frac{\pi}{2}\))
Solution:
\(\frac{\pi}{4}\)

(vi) The principal value of sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________. (\(\frac{2 \pi}{3}\), \(\frac{\pi}{3}\), \(\frac{4 \pi}{3}\))
Solution:
\(\frac{\pi}{3}\)

(vii) If sin^-1 \(\frac{x}{5}\) + cosec^-1 \(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x = ________. (2, 3, 4)
Solution:
x = 3

(viii) The value of sin (tan^-1 x + tan^-1 \(\frac{1}{x}\)), x > 0 = ________. (0, 1, 1/2)
Solution:
1

(ix) cot^-1 \(\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\) = ________. (2π – \(\frac{x}{2}\), \(\frac{x}{2}\), π – \(\frac{x}{2}\))
Solution:
π – \(\frac{x}{2}\)

(x) 2sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{24}{25}\) = ________. (π, -π, 0)
Solution:
π

(xi) if Θ = cos^-1 x + sin^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which Θ lies is ________. [(\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), [0, \(\frac{\pi}{2}\)), (0, \(\frac{\pi}{2}\)])
Solution:
(0, \(\frac{\pi}{2}\)]

(xii) sec² (tan^-1 2) + cosec² (cot^-1 3) = ________. (16, 14, 15)
Solution:
15

Question 2.
Write whether the following statements are true or false.
(i) sin^-1 \(\frac{1}{x}\) cosec^-1 x = 1
Solution:
False

(ii) cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{2}{3}\) = tan^-1 \(\frac{17}{6}\)
Solution:
True

(iii) tan^-1 \(\frac{4}{3}\) + cot^-1 (\(\frac{-3}{4}\)) = π
Solution:
True

(iv) sec^-1 \(\frac{1}{2}\) + cosec^-1 \(\frac{1}{2}\) = \(\frac{\pi}{2}\)
Solution:
False

(v) sec^-1 (-\(\frac{7}{5}\)) = π – cos^-1 \(\frac{5}{7}\)
Solution:
True

(vi) tan^-1 (tan 3) = 3
Solution:
False

(vii) The principal value of tan^-1 (tan \(\frac{3 \pi}{4}\)) is \(\frac{3 \pi}{4}\)
Solution:
False

(viii) cot^-1 (-√3) is in the second quadrant.
Solution:
True

(ix) 3 tan^-1 3 = tan^-1 \(\frac{9}{13}\)
Solution:
False

(x) tan^-1 2 + tan^-1 3 = – \(\frac{\pi}{4}\)
Solution:
False

(xi) 2 sin^-1 \(\frac{4}{5}\) = sin^-1 \(\frac{24}{25}\)
Solution:
False

(xii) The equation tan^-1 (cotx) = 2x has exactly two real solutions.
Solution:
True

Question 3.
Express the value of the foilowing in simplest form.
(i) sin (2 sin^-1 0.6)
Solution:
sin (2 sin^-1 0.6)

(ii) tan (\(\frac{\pi}{4}\) + 2 cot^-1 3)
Solution:

(iii) cos (2 sin^-1 x)
Solution:

(iv) tan (cos^-1 x)
Solution:

(v) tan^-1 (\(\frac{x}{y}\)) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:

(vi) cosec (cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\))
Solution:

(vii) sin^-1 \(\frac{1}{\sqrt{5}}\) + cos^-1 \(\frac{3}{\sqrt{10}}\)
Solution:

(viii) sin cos^-1 tan sec √2
Solution:
sin cos^-1 tan sec √2
= sin cos^-1 tan sec \(\frac{\pi}{4}\)
= sin cos^-1 1 = sin 0 = 0

(ix) sin (2 tan^-1 \(\sqrt{\frac{1-x}{1+x}}\))
Solution:

(x) tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
Solution:

(xi) sin cot^-1 cos tan^-1 x.
Solution:

(xii) tan^-1 \(\left(x+\sqrt{1+x^2}\right)\)
Solution:

Question 4.
Prove the following statements:
(i) sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\frac{36}{85}\)
Solution:

(ii) sin^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)
Solution:

(iii) tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
Solution:
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

(iv) tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) tan ( 2tan^-1 \(\frac{1}{5}\) – \(\frac{\pi}{4}\) ) + \(\frac{7}{17}\) = 0
Solution:

Question 5.
Prove the following statements:
(i) cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:

(ii) sin^-1 \(\frac{4}{5}\) + 2 tan^-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:

(iii) 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{70}\) + tan^-1 \(\frac{1}{99}\) = \(\frac{\pi}{4}\)
Solution:

(iv) 2 tan^-1 \(\frac{1}{5}\) + sec^-1 \(\frac{5 \sqrt{2}}{7}\) + 2 tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) cos^-1 \(\frac{12}{13}\) + 2 cos^-1 \(\sqrt{\frac{64}{65}}\) + cos^-1 \(\sqrt{\frac{49}{50}}\) = cos^-1 \(\frac{1}{\sqrt{2}}\)
Solution:


(vi) tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\) = 6
Solution:
tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\)
= tan² tan^-1 √2 + cot² cot^-1 (2)
= 2 + 4 = 6

(vii) cos tan^-1 cot sin^-1 x = x.
Solution.

Question 6.
Prove the following statements:
(i) cot^-1 (tan 2x) + cot^-1 (- tan 2x) = π
Solution:

(ii) tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)
Solution:

(iii) tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\)) = tan^-1 a – tan^-1 c.
Solution:
tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\))
= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c
= tan^-1 a – tan^-1 c.

(iv) cot^-1 \(\frac{p q+1}{p-q}\) + cot^-1 \(\frac{q r+1}{q-r}\) + cot^-1 \(\frac{r p+1}{r-p}\) = 0
Solution:

(v)

Solution:

Question 7.
Prove the following statements:
(i) tan^-1 \(\frac{2 a-b}{b \sqrt{3}}\) + tan^-1 \(\frac{2 b-a}{a \sqrt{3}}\) = \(\frac{\pi}{3}\)
Solution:

(ii) tan^-1 \(\frac{1}{x+y}\) + tan^-1 \(\frac{y}{x^2+x y+1}\) = tan^-1 \(\frac{1}{x}\)
Solution:

(iii) sin^-1 \(\sqrt{\frac{x-q}{p-q}}\) = cos^-1 \(\sqrt{\frac{p-x}{p-q}}\) = cot^-1 \(\sqrt{\frac{p-x}{x-q}}\)
Solution:

(iv) sin² (sin^-1 x + sin^-1 y + sin^-1 z) = cos² (cos^-1 x + cos^-1 y + cos^-1 z)
Solution:

(v) tan (tan^-1 x + tan^-1 y + tan^-1 z) = cot (cot^-1 x + cot^-1 y + cot^-1 z)
Solution:

Question 8.
(i) If sin^-1 x + sin^-1 y + sin^-1 z = π, show that x\(\sqrt{1-x^2}\) + x\(\sqrt{1-y^2}\) + x\(\sqrt{1-z^2}\) = 2xyz
Solution:
Let sin^-1 x = α, sin^-1 y = β, sin^-1 z = γ
∴ α + β + γ = π
∴ x = sin α, y = sin β, z = sin γ
or, α + β = π – γ
or, sin(α + β) = sin(π – γ) = sin γ
and cos(α + β) = cos(π – γ) = – cos γ

(ii) tan^-1 x + tan^-1 y + tan^-1 z = π show that x + y + z = xyz.
Solution:

(iii) tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\). Show that xy + yz + zx = 1
Solution:

or, 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 \(\frac{y z}{x r}\) + tan^-1 \(\frac{z x}{y r}\) + tan^-1 \(\frac{x y}{z r}\) = \(\frac{\pi}{2}\)
Solution:

(v) In a triangle ABC if m∠A = 90°, prove that tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\) = \(\frac{\pi}{4}\). where a, b, and c are sides of the triangle.
Solution:
L.H.S. tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\)

Question 9.
Solve
(i) cos (2 sin^-1 x) = 1/9
Solution:

(ii) sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
Solution:
sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
or, sin^-1 (1 – x) = \(\frac{\pi}{2}\) – sin^-1 x = cos^-1 x
or, sin^-1 (1 – x) = sin^-1 \(\sqrt{1-x^2}\)
or, 1 – x = \(\sqrt{1-x^2}\)
or, 1 + x² – 2x = 1 – x²
or, 2x² – 2x  = 0
or, 2x (x – 1) = 0
∴ x = 0 or, 1

(iii) sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ – 2 sin^-1 x = \(\frac{\pi}{2}\) – sin^-1 (1 – x)
⇒ cos^-1 (1 – x)
⇒ cos (– 2 sin^-1 x) = 1 – x      ….. (1)
Let sin^-1 Θ ⇒ sin Θ
Now cos (– 2 sin^-1 x) = cos (-2Θ)
= cos 2Θ = 1 – 2 sin² Θ = 1 – 2x²
Using in (1) we get
1 – 2x² = 1 – x
⇒ 2x² – x = 0 ⇒ x (2x – 1) = 0
⇒ x = 0, ½, But x = ½ does not
Satisfy the given equation, Thus x = 0.

(iv) cos^-1 x + sin^-1 \(\frac{x}{2}\) = \(\frac{\pi}{6}\)
Solution:

(v) tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}\) = \(\frac{\pi}{4}\)
Solution:

(vi) tan^-1 \(\frac{1}{2 x+1}\) + tan^-1 \(\frac{1}{4 x+1}\) = tan^-1 \(\frac{2}{x^2}\)
Solution:

(vii) 3 sin^-1 \(\frac{2 x}{1+x^2}\) – 4 cos^-1 \(\frac{1-x^2}{1+x^2}\) + 2 tan^-1 \(\frac{2 x}{1-x^2}\) = \(\frac{\pi}{3}\)
Solution:

(viii) cot^-1 \(\frac{1}{x-1}\) + cot^-1 \(\frac{1}{x}\) + cot^-1 \(\frac{1}{x+1}\) = cot^-1 \(\frac{1}{3x}\)
Solution:

(ix) cot^-1 \(\frac{1-x^2}{2 x}\) =  cosec^-1 \(\frac{1+a^2}{2 a}\) – sec^-1 \(\frac{1+b^2}{1-b^2}\)
Solution:

(x) sin^-1 \(\left(\frac{2 a}{1+a^2}\right)\) + sin^-1 \(\left(\frac{2 b}{1+b^2}\right)\) = 2 tan^-1 x
Solution:

(xi) sin^-1 y – cos^-1 x = cos^-1 \(\frac{\sqrt{3}}{2}\)
Solution:

(xii) sin^-1 2x + sin^-1 x = \(\frac{\pi}{3}\)
Solution:

Question 10.
Rectify the error ifany in the following:
sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{33}{65}\)
Solution:

Question 11.
Prove that:
(i) cos^-1 \(\left(\frac{b+a \cos x}{a+b \cos x}\right)\) = 2 tan^-1 \(\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)\)
Solution:

(ii) tan \(\left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\) + tan \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\)
Solution:

(iii) tan^-1 \(\sqrt{\frac{x r}{y z}}\) + tan^-1 \(\sqrt{\frac{y r}{y x}}\) + tan^-1 \(\sqrt{\frac{z r}{x y}}\) = π where r = x + y +z.
Solution:

Question 12.
(i) If cos^-1 (\(\frac{x}{a}\)) + cos^-1 (\(\frac{y}{b}\)) = Θ, prove that \(\frac{x^2}{a^2}\) – \(\frac{2 x}{a b}\) cos Θ + \(\frac{y^2}{b^2}\) = sin² Θ.
Solution:

(ii) If cos^-1 (\(\frac{x}{y}\)) + cos^-1 (\(\frac{y}{3}\)) = Θ, prove that 9x² – 12xy cos Θ + 4y² = 36 sin² Θ.
Solution:

(iii) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = sin^-1 (\(\frac{c^2}{a b}\)) prove that b²x² + 2xy \(\sqrt{a^2 b^2-c^4}\) a²y² = c²
Solution:

(iv) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = α prove that \(\frac{x^2}{a^2}\) + \(\frac{2 x y}{a b}\) cos α + \(\frac{y^2}{b^2}\) = sin² α
Solution:

(v) If sin^-1 x + sin^-1 y + sin^-1 z = π prove that x² + y² + z² + 4x²y²z² = 2 ( x²y² + y²z² + z²x² )
Solution:

Question 13.
Solve the following equations:
(i) tan^-1 \(\frac{x-1}{x+1}\) + tan^-1 \(\frac{2 x-1}{2 x+1}\) = tan^-1 \(\frac{23}{36}\)
Solution:

(ii) tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 x = \(\frac{\pi}{4}\)
Solution:

(iii) cos^-1 \(\left(x+\frac{1}{2}\right)\) + cos^-1 x+ cos^-1 \(\left(x-\frac{1}{2}\right)\) = \(\frac{3 \pi}{2}\)
Solution:

(iv) 3tan^-1 \(\frac{1}{2+\sqrt{3}}\) – tan^-1 \(\frac{1}{x}\) = tan^-1 \(\frac{1}{3}\)
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(b)

Question 1.
State which of the following matrices are symmetric, skew-symmetric, both or not either:

Solution:
(i) Symmetric
(ii) Neither Symmetric nor skew-symmetric
(iii) Symmetric
(iv) Skew symmetric
(v) Both
(vi) Neither symmetric nor skew-symmetric
(vii) Skew symmetric

Question 2.
State ‘True’ or ‘False’:
(i) If A and B are symmetric matrices of the same order and AB – BA ≠ 0, then AB is not symmetric.
Solution:
True

(ii) For any square matrix A, AA’ is symmetric.
Solution:
True

(iii) If A is any skew-symmetric matrix, then A² is also skew-symmetric.
Solution:
False

(iv) If A is symmetric, then A², A³, …, Aⁿ are all symmetric.
Solution:
True

(v) If A is symmetric then A – A¹ is both symmetric and skew-symmetric.
Solution:
False

(vi) For any square matrix (A – A¹)² is skew-symmetric.
Solution:
True

(vii) A matrix which is not symmetric is skew-symmetric.
Solution:
False

Question 3.
(i) If A and B are symmetric matrices of the same order with AB ≠ BA, final whether AB – BA is symmetric or skew symmetric.
Solution:
A and B are symmetric matrices;
Thus A’ = A and B’ = B
Now (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB = – (AB – BA)
∴ AB – BA is skew symmetric.

(ii) If a symmetric/skew-symmetric matrix is expressed as a sum of a symmetric and a skew-symmetric matrix then prove that one of the matrices in the sum must be zero matrix.
Solution:
We know that zero matrix is both symmetric as well as skew-symmetric.
Let A is symmetric.
∴ A = A + O where A is symmetric and O is treated as skew-symmetric. If B is skew-symmetric then we can write B = O + B where O is symmetric and B is skew-symmetric.

Question 4.
A and B are square matrices of the same order, prove that
(i) If A, B and AB are all symmetric, then AB – BA = 0
Solution:
Let A, B and AB are all symmetric.
∴A’ = A, B’ = B and (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB
⇒ AB – BA = 0

(ii) If A, B and AB are all skew symmetric then AB + BA = 0
Solution:
Let A, B and AB are all skew symmetric matrices
∴ A’ = -A, B’ = -B and (AB)’ = -AB
Now (AB)’ = -AB
⇒ B’A’ = -AB
⇒ (-B) (-A) = -AB
⇒ BA = -AB
⇒ AB + BA = 0

Question 5.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A’ = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
2 & 1 & 5 \\
0 & 3 & 3
\end{array}\right]\)

(i) A+A’ is symmetric

(ii) A-A’ is skew-symmetric

Question 6.
Prove that a unit matrix is its own inverse. Is the converse true?
IfA = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\) show that A² = I and hence A= A^-1.
Solution:
No the converse is not true for example:

Question 7.
Here A is an involuntary matrix, recall the definition given earlier.
Solution:

Question 8.
Show that \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\) is its own inverse.
Solution:

Question 9.
Express as a sum of a symmetric and a skew symmetric matrix.

Solutions:

Question 10.
What is the inverse of

Question 11.
Find inverse of the following matrices by elementary row/column operation (transformations):
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find the inverse of the following matrices using elementary transformation:
(i) \(\left[\begin{array}{lll}
\mathbf{0} & \mathbf{0} & 2 \\
\mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{2} & \mathbf{0} & \mathbf{0}
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]\)
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 5 Determinants Ex 5(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 5 Determinants Exercise 5(b)

Question 1.
Write the number of solutions of the following system of equations.
(i) x – 2y = 0
Solution:
No solution

(ii) x – y = 0 and 2x – 2y = 1
Solution:
Infinite

(iii) 2x + y = 2 and -x – 1/2y = 3
Solution:
No solution

(iv) 3x + 2y = 1 and x + 5y = 6
Solution:
One

(v) 2x + 3y + 1 = 0 and x – 3y – 4 = 0
Solution:
One

(vi) x + y + z = 1
x + y + z = 2
2x + 3y + z = 0
Solution:
No solution

(vii) x + 4y – z = 0
3x – 4y – z = 0
x – 3y + z = 0
Solution:
One

(viii) x + y – z = 0
3x – y + z = 0
x – 3y + z = 0
Solution:
One

(ix) a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0
and \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
Infinite solutions as Δ = Δ₁ = Δ₂ = Δ₃ = 0

Question 2.
Show that the following system is inconsistent.
(a – b)x + (b – c)y + (c – a)z = 0
(b – c)x + (c – a)y + (a – b)z = 0
(c – a)x + (a – b)y + (b – c)z =1
Solution:

Question 3.
(i) The system of equations
x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6 has
(a) infinitely many solutions
(b) no solution
(c) a unique solution
(d) none of the three
Solution:
(a) infinitely many solutions

(ii) If the system of equations
2x + 5y + 8z = 0
x + 4y + 7z = 0
6x + 9y – z = 0
has a nontrivial solution, then is equal to
(a) 12
(b) -12
(c) 0
(d) none of the three
Solution:
(b) -12

(iii) The system of linear equations
x + y + z = 2
2x + y – z = 3
3x +2y + kz = 4
has a unique solution if
(a) k ≠ 0
(b) -1 < k < 1
(c) -2 < k < 2
(d) k = 0
Solution:
(a) k ≠ 0

(iv) The equations
x + y + z = 6
x + 2y + 3z = 10
x + 2y + mz = n
give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) none of the three
Solution:
(c) m = 3, n = 10

(v) The system of equations
2x – y + z = 0
x – 2y + z = 0
x – y + 2z = 0
has infinite number of nontrivial solutions for
(a) = 1
(b) = 5
(c) = -5
(d) no real value of
Solution:
(c) = -5

(vi) The system of equations
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z =0
with has
(a) more than two solutions
(b) one trivial and one nontrivial solutions
(c) No solution
(d) only trivial solutions
Solution:
(a) more than two solutions

Question 4.
Can the inverses of the following matrices be found?
(i) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
|A| = 0
∴ A^-1 can not be found.

(ii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
∴ |A| = 4 – 6 = -2 ≠ 0
∴ A^-1 exists.

(iii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) = 1 – 1 = 0
∴ A^-1 does not exist.

(iv) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) = 4 – 4 = 0
∴ A^-1 does not exist.

(v) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1 ≠ 0
∴ A^-1 exists.

Question 5.
Find the inverse of the following:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{cc}
i & -i \\
i & i
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
x & -x \\
x & x^2
\end{array}\right]\), x ≠ 0, x ≠ -1
Solution:

Question 6.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & -1 & 2 \\
1 & 3 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
-2 & 2 & 3 \\
1 & 4 & 2 \\
-2 & -3 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
2 & 2 & 1 \\
1 & 2 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 3 & 0 \\
2 & -1 & 6 \\
5 & -3 & 1
\end{array}\right]\)
Solution:

Question 7.
Which of the following matrices are invertible?
(i) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
2 & 1 & -2 \\
1 & 2 & 1 \\
3 & 6 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
2 & 1 & -4 \\
-1 & 0 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -2 & 1 \\
3 & 2 & 4
\end{array}\right]\)
Solution:

Question 8.
Examining consistency and solvability, solve the following equations by matrix method.
(i) x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:

(ii) x + 2y – 3z = 4
2x + 4y – 5z = 12
3x – y + z = 3
Solution:
Let

(iii) 2x – y + z = 4
x + 3y + 2z = 12
3x + 2y + 3z = 16
Solution:

(iv) x + y + z = 4
2x + 5y – 2x = 3
x + 7y – 7z = 5
Solution:

(v) x + y + z = 4
2x – y + 3z = 1
3x + 2y – z = 1
Solution:

(vi) x + y – z = 6
2x – 3y + z = 1
2x – 4y + 2z = 1
Solution:

(vii) x – 2y = 3
3x + 4y – z = -2
5x – 3z = -1
Solution:

(viii) x + 2y + 3z = 14
2x – y + 5z = 15
2y + 4z – 3x = 13
Solution:

(ix) 2x + 3y +z = 11
x + y + z = 6
5x – y + 10z = 34
Solution:

Question 9.
Given the matrices
A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
write down the linear equations given by AX = C and solve it for x, y, z by matrix method.
Solution:

Question 10.
Find X, if \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & -1 \\
2 & 1 & -1
\end{array}\right]\) X = \(\left[\begin{array}{l}
6 \\
0 \\
1
\end{array}\right]\) where X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\)
Solution:

Question 11.
Answer the following:
(i) If every element of a third order matrix is multiplied by 5, then how many times its determinant value becomes?
Solution:
125

(ii) What is the value of x if \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=,\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\) ?
Solution:
x = 6

(iii) What are the values of x and y if \(\left|\begin{array}{ll}
x & y \\
1 & 1
\end{array}\right|=2,\left|\begin{array}{ll}
x & 3 \\
y & 2
\end{array}\right|=1\) ?
Solution:
x = 5, y = 3

(iv) What is the value of x if \(\left|\begin{array}{ccc}
x+1 & 1 & 1 \\
1 & 1 & -1 \\
-1 & 1 & 1
\end{array}\right|\) = 4?
Solution:
x = 0

(v) What is the value of \(\left|\begin{array}{ccc}
\mathbf{o} & -\mathbf{h} & -\mathbf{g} \\
\mathbf{h} & \mathbf{0} & -\mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{0}
\end{array}\right|\)?
Solution:
0

(vi) What is the value of \(\left|\begin{array}{l}
\frac{1}{a} 1 \mathrm{bc} \\
\frac{1}{b} 1 c a \\
\frac{1}{c} 1 a b
\end{array}\right|\)
Solution:
0

(vii) What is the co-factor of 4 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
4 & 5 & 0 \\
2 & 0 & 1
\end{array}\right|\)
Solution:
-2

(viii)In which interval does the determinant \(\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\) lie?
Solution:
[2, 4]

(ix) Ifx + y + z = n, what is the value of Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & -\tan A & 0
\end{array}\right|\) Where A, B, C are the angles of triangle.
Solution:
0

Question 12.
Evaluate the following determinants:
(i) \(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
= 2\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\) = 0
(∵ C₁ = C₃)

(ii) \(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\) = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
14 & 17 & 11
\end{array}\right|\)
( R₁ = R₁ – R₂, R₂ = R₂ – R₃)
= 0 (∵ R₁ = R₂)

(iii) \(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
= 7\(\left|\begin{array}{ccc}
32 & 777 & 32 \\
105 & 888 & 105 \\
116 & 999 & 116
\end{array}\right|\) = 0
(∵ C₁ = C₂)

(iv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 4 \\
3 & 4 & 6
\end{array}\right|\)
Solution:

(v) \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 7 \\
8 & 14 & 20
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:

= 225 – 256 – 4(100 – 144) + 9(64 – 81)
= -31 – 4(-44) + 9(-17)
= -31 + 176 – 153 = -184 + 176
= -8

(vii) \(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
= 2\(\left|\begin{array}{cc}
1 & -5863 \\
1 & 4137
\end{array}\right|\)
(expanding along 2nd column)
= 2(4137 + 5863)
= 2 × 10000 = 20000

(viii) \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
0 & a^2 & b \\
b^2 & 0 & a^2 \\
a & b^2 & 0
\end{array}\right|\)
Solution:

= -a² (0 –  a²) + b (b⁴ –  0) = a⁵ + b⁵

(x) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
\boldsymbol{x}-\boldsymbol{y} & \boldsymbol{y}-\boldsymbol{z} & z-\boldsymbol{x} \\
\boldsymbol{p}-\boldsymbol{q} & \boldsymbol{q}-\boldsymbol{r} & \boldsymbol{r}-\boldsymbol{p}
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-r & r-p
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0 ( ∵ C₁ = 0)

(xi) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0

(xii) \(\left|\begin{array}{ccc}
-\cos ^2 \theta & \sec ^2 \theta & -0.2 \\
\cot ^2 \theta & -\tan ^2 \theta & 1.2 \\
-1 & 1 & 1
\end{array}\right|\)
Solution:

(Expanding along 3rd row)
= (-cos² θ + sec² θ) (-tan² θ – 1.2) – (sec² θ + 0.2) (cot² θ – tan² θ)
= sin² θ – 1.2 cos² θ – sec² θ tan² θ – 1.2 sec² θ – cosec² θ +  sec² θ tan² θ – 0.2 cot² θ + 0.2 tan² θ
= sin² θ – cosec² θ + 1.2 (cos² θ – sec² θ) + 0.2 (tan² θ – cot² θ) ≠ 0
The question seems to be wrong.

Question 13.
If \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\) = 0 what are x and y?
Solution:

or, xy – 0 = 0 ⇒ xy = 0, ⇒ x = 0, or y = 0

Question 14.
For what value of x \(\left|\begin{array}{ccc}
2 x & 0 & 0 \\
0 & 1 & 2 \\
-1 & 2 & 0
\end{array}\right|\) = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 4 \\
0 & 3 & 5
\end{array}\right|\)?
Solution:

Question 15.
Solve \(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
or, (x – a) \(\left|\begin{array}{cc}
x+b & 0 \\
0 & x+c
\end{array}\right|\) = 0
or, (x + a) (x + b) (x + c) = 0
x = -a, x = -b, x = -c

Question 16.
Solve \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Solution:

Question 17.
Solve \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|\) = 0
Solution:

Question 18.
Show that x = 2 is a root of \(\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|\) = 0 Solve this completely.
Solution:
Putting x = 2,

= (x – 1) (-15x + 30 – 5x² + 10x)
= (x – 1) (-5x² – 5x + 30)
= -5(x – 1) (x² + x – 6)
= -5(x – 1) (x + 3) (x – 2) = 0
⇒ x = 1 or, -3 or 2.

Question 19.
Evaluate \(\left|\begin{array}{ccc}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\) – \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

= (a – b) (b – c) [(-a + c) – (b + c – a – b)]
= (a – b) (b – c) (-a + c – c + a) = 0

Question 20.
\(\left|\begin{array}{lll}
a & a^2-b c & 1 \\
b & b^2-a c & 1 \\
c & c^2-a b & 1
\end{array}\right|\)
Solution:

Question21.
For what value of X the system of equations
x + y + z = 6, 4x + λy – λz = 0, 3x + 2y – 4z = -5 does not possess a solution?
Solution:

= 24 – 6λ – 2λ = 24 – 8λ
when Δ = 0
We have 24 – 8λ, = 0 or, λ = 3
The system of equations does not posses solution for λ = 3.

Question 22.
If A is a 3 × 3 matrix and |A| = 2, then which matrix is represented by A × adj A?
Solution:

Question 23.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\)
show that (I + A) (I – A)^-1 = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) where I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

Question 24.
Prove the following:
(i) \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:

(ii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) = (b – c) (c – a) (a – b) (a + b + c)
Solution:

= (a – b) (b – c) (b² + bc + c² – a² – ab – b²)
= (a – b) (b- c) (c² – a² + bc – ab)
= (a – b) (b – c) {(c – a) (c + a) + b(c – a)}
= (a – b) (b – c) (c – a) (a + b + c) = R.H.S.
(Proved)

(iii) \(\left|\begin{array}{lll}
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{b}
\end{array}\right|\) = 3abc – a³ – b³ – c³
Solution:

= (a + b + c) {(b – c) (a – b) – (c – a)²}
= (a + b + c) (a + b + c) (ab – b² – ca + bc – c² – a² + 2ca)
= (a + b + c) (-a² – b² – c² + ab + bc + ca)
= -(a + b + c) (a² + b² + c² – ab – bc – ca)
=- (a³ + b³ + c³ – 3abc)
= 3abc – a³ – b³ – c³

(iv) \(\left|\begin{array}{lll}
b^2-a b & b-c & b c-a c \\
a b-a^2 & a-b & b^2-a b \\
b c-a c & c-a & a b-a^2
\end{array}\right|\) = 0
Solution:

= (b² – a² + bc – ac) (a – b) {(-a + b) (c – a) – (bc – ac – ab + a²)}
= (b² – a² + bc – ac) (a – b) (- ca + a² + bc – ab – bc + ac + ab – a²)
= (b² – a² + bc – ac) (a – b) × 0 = 0
= R.H.S.
(Proved)

(v) \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\) = 4a²b²c²
Solution:

(vi) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c a \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a² + b² + c² ) (a + b + c) (b – c) (c – a) (a – b)
Solution:

= (a – b) (b – c) (a² + b² + c²) (-a² – ab + bc + c²)
= (a – b) (b – c) (a² + b² + c²) {(c² – a²) + b(c – a)}
= (a² + b² + c²) (a – b) (b – c) (c – a) (c + a + b)

(vii) \(\left|\begin{array}{lll}
b+c & a+b & a \\
c+a & b+c & b \\
a+b & c+a & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b +c) {(a – b) (a – c) – (c – b) (b – c)}
= (a + b + c) (a² – ac – ab + bc – bc + c² + b² – bc)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= (a³ + b³ + c³ – 3abc)

(viii) \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2(b + c) (c + a) (a + b)
Solution:

= -2(a + b) (b + c) (-a – b – c + b)
= 2(a + b) (b + c) (c + a)

(ix) \(\left|\begin{array}{ccc}
a x-b y-c z & a y+b x & a z+c x \\
b x+a y & b y-c z-a x & b z+c y \\
c x+a z & a y+b z & c z-a x-b y
\end{array}\right|\) = (a² + b² + c²) (ax + by + cz) (x² + y² + z²)
Solution:

Question 25.
If 2s = a + b + c show that \(\left|\begin{array}{ccc}
a^2 & (s-a)^2 & (s-a)^2 \\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2
\end{array}\right|\) = 2s³ (s – a) (s – b) (s – c)
Solution:
Let s – a = A, s – b = B, s – c = C
A + B + C = 3s – (a + b + c)
= 3s – 2s = s
Also B + C = s – b + s – c = 2s – (b + c)
= (a + b + c) – b + c = a
Similarly C + A = b, A + B = c

= 2 ABC (A + B + C)²
[Refer Q.No.9 (xii) of Exercise 5(a)]
= 2(s – a) (s – b)(s – c) s³

Question 26.
if \(\left|\begin{array}{ccc}
x & x^2 & x^3-1 \\
y & y^2 & y^3-1 \\
z & z^2 & z^3-1
\end{array}\right|\) = 0 then prove that xyz =1 when x, y, z are non zero and unequal.
Solution:

= (x – y) (y – z) (z – x) (xyz – 1)
It is given that
(x – y) (y – z) (z – x) (xyz – 1) = 0
⇒ xyz – 1 (as x ≠ y ≠ z)

Question 27.
Without expanding show that the following determinant is equal to Ax + B where A and B are determinants of order 3 not involving x.
\(\left|\begin{array}{ccc}
x^2+x & x+1 & x-2 \\
2 x^2+3 x-1 & 3 x & 3 x-3 \\
x^2+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right|\)
Solution:

Question 28.
If x, y, z are positive and are the pth, qth and rth terms of a G.P. then prove that \(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|\) = 0
Solution:
Let the G.P. be
a, aR, aR², aR³ …..aR^n-1
p th term = aR^p-1
q th term = aR^q-1
r th term = aR^r-1
x = aR^p-1, y= aR^q-1, z = aR^r-1
log x = log a + (p – 1) log R,
log y = log a + (q – 1) log R,
log z = log a + (r – 1) log R

Question 29.
If D_j = \(\left|\begin{array}{ccc}
j & a & n(n+2) / 2 \\
j^2 & b & n(n+1)(2 n+1) / 6 \\
j^3 & c & n^2(n+1)^2 / 4
\end{array}\right|\) then prove that \(\sum_{j=1}^n\)D_j = 0.
Solution:

Question 30.
Ifa₁, a₂,……a_n are in G.P. and a_i > 0 for every i, then find the value of
\(\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\
\log a_{n+2} & \log a_{n+3} & \log a_{n+4}
\end{array}\right|\)
Solution:

Question 31.
If f(x)= \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin ^2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin ^2 x
\end{array}\right|\) what is the least value of f(x)?
Solution:

As minimum value of sin 2x is 0. So the minimum value of above function f(x) is 2.

Question 32.
If f_r(x), g_r(x), h_r(x), r = 1, 2, 3 are polynomials in x such that f_r(a) = g_r(a) = h_r(a) and
F(x) = \(\left[\begin{array}{lll}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{array}\right]\) find F'(x) at x = a.
Solution:
We have

[Since f₁a) = g₁(a) = h₁(a), f₂(a) = g₂(a) = h₂(a) and f₃(a) = g₃(a) = h₃(a) So that each determinant is zero due to presence of two identical rows.]

Question 33.
If f(x) = \(\left[\begin{array}{ccc}
\cos x & \sin x & \cos x \\
\cos 2 x & \sin 2 x & 2 \cos 2 x \\
\cos 3 x & \sin 3 x & 3 \cos 3 x
\end{array}\right]\) find f'(\(\frac{\pi}{2}\)).
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(a)

Question 1.
State the order of the following matrices.
(i) [abc]
(ii) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
x & y \\
y & z \\
z & x
\end{array}\right]\)
(iv) \(\left[\begin{array}{cccc}
1 & 0 & 1 & 4 \\
2 & 1 & 3 & 0 \\
-3 & 2 & 1 & 3
\end{array}\right]\)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)

Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a sqare matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p²

Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i) \(\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)\)
(ii) \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\)
(iii) \(\left(\begin{array}{ll}
a & b \\
c & d \\
e & f \\
g & h
\end{array}\right)\)
(iv) (1, 2, 3)

Question 4.
Let A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) What is the order of A?
(ii) Write down the entries a₃₁, a₂₅, a₂₃
(iii) Write down A^T.
(iv) What is the order of A^T?
Solution:
A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) Order of A is (3 x 5)
(ii) a₃₁ = 3, a₂₅= 2, a₂₃ = 6
(iii) A^T = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of A^T is (5 x 3).

Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\), B = \(\left[\begin{array}{c}
-6 \\
9
\end{array}\right]\)
Solution:

(ii) A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Solution:

(iii) A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{5}
\end{array}\right]\), B = \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{2} \\
\frac{1}{2} & \frac{4}{5}
\end{array}\right]\)
Solution:

(iv) A = \(\left[\begin{array}{cc}
1 & a-b \\
a+b & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & b \\
-a & 5
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
-1 & 4 & 3 \\
1 & 2 & -3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & -5 \\
1 & -3 & -3 \\
1 & 2 & 4
\end{array}\right]\)
Solution:

Question 6.
(i) Find the 2×2 matrix X
if X + \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
Solution:

(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] derermine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1

(iii) If \(\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]\) – \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) determine x₁, x₂, y₁, y₂.
Solution:

(iv) Find a matrix which when added to \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:

Question 7.
Calculate whenever possible, the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\) is impossible because number of columns of 1st ≠ number of rows of second.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
1 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\)
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\), C = \(\left[\begin{array}{ll}
2 & 2 \\
1 & 3
\end{array}\right]\)
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:

Question 9.
Find the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:

(vi) \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(x) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False

(ii) k[0] = 0, k ∈ R
Solution:
False

(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False

(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True

(v) \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) + \(\left[\begin{array}{cc}
-1 & 0 \\
2 & 0
\end{array}\right]\) = 0
Solution:
True

(vi) \(\left[\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right]\) = 3 \(\left[\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right]\)
Solution:
False

(vii) With five elements a matrix can not be constructed.
Solution:
False

(viii)The unit matrix is its own transpose.
Solution:
True

Question 11.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 13
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find A – α I, α ∈ R.
Solution:

Question 12.
Find x and y in the following.
(i) \(\left[\begin{array}{cc}
x & -2 y \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -8 \\
0 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{l}
x \\
y
\end{array}\right]+\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1
\end{array}\right]\)
Solution:

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.

Question 14.
Write down the matrix

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) a_ij = i + j
(ii) a_ij = i – j
(iii) a_ij = i × j
(iv) a_ij = i / j
Solution:

Question 16.
If \(\left[\begin{array}{cc}
2 x & y \\
1 & 3
\end{array}\right]+\left[\begin{array}{cc}
4 & 2 \\
0 & -1
\end{array}\right]=\left[\begin{array}{ll}
8 & 3 \\
1 & 2
\end{array}\right]\)
Solution:

Question 17.
Find A such that
\(\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 0 & -2 \\
3 & 1 & -1
\end{array}\right]+A=\left[\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 0 \\
1 & 3 & 2
\end{array}\right]\)
Solution:

Question 18.
If

Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)

Question 20.
Find A if \(\left[\begin{array}{l}
4 \\
1 \\
3
\end{array}\right]\) A = \(\left[\begin{array}{rrr}
-4 & 8 & 4 \\
-1 & 2 & 1 \\
-3 & 6 & 3
\end{array}\right]\)
Solution:

Question 21.
Find B if B² = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:

∴ a² + bc = 17, ab + bd= 8
ca + cd = 8, bc + d² = 17
∴ a² + bc = bc + d²
or, a² + d² or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a² + bc = 17
or, a² + b . b = 17 (∵ b = c)
or, a² + b² = 17
Also (a + b)² = a² + b² + 2ab
∴ (a + b)² = 17 + 8 = 25
or, a + b = 5
And (a – b)² = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

Question 22.
Find x and y when

Question 23.
Find AB and BA given that:

Question 24.
Evaluate

Question 25.
If

Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:

Question 26.
Find A and B where

Question 27.
If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) and I be the 2 × 2 unit matrix find (A – 2I) (A – 3I)
Solution:

Question 28.
Verify that [AB]^T = B^TA^T where

Question 29.
Verify that A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) satisfies the equation x² – (a + d)x + (ad – bc)I = 0 where I is the 2 x 2 matrix.
Solution:

Question 30.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), show that A³ – 23 A – 40 I = 0
Solution:

Question 31.

Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A² – B² = (A – B) (A + B)
(ii) A² + 2AB + B² = (A + B)²
(iii) A² – 2AB + B² = (A – B)²
Solution:
(i) (A – B) (A + B)
= A² + AB – BA – B²
= A² + AB – AB- B²(∵ AB = BA)
= A² – B²
(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² (∵ AB = BA)
= A² + 2AB + B²
(iii) (A – B)² = (A – B) (A – B)
= A² – AB – BA + B²
= A² – AB – AB + B² (∵ AB = BA)
= A² – 2AB + B²

Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A² – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A² – AβI – αIA + αβI²
= A² – βAI – αA + αβI
(∵ IA = A, I² = I)
= A² – βA – αA + αβI) (∵ AI = A)
= A² – (α + β) A + αβI

Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar mltiβlication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Question 42.

Question 43.

	Men	Women	Children
Family A →	4	6	2
Family B →	2	2	4

	Family B
	Calory	Proteins
Men	2400	45
Women	1900	55
Children	1800	33

Solution:
The given informations can be written in matrix form as

∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]

⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(b)

Question 1.
Differentiate from definition
(i) e^3x
Solution:

(ii) 2^x2
Solution:

(iii) In (3x + 1)
Solution:

(iv) logx⁵ (Hint : logx⁵ = \(\frac{\ln 5}{\ln x}\))
Solution:

(v) In sin x
Solution:

(vi) x² a^2x
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(f) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(f)

Find derivatives of the following functions.
Question 1.
x^x
Solution:
Let y = x^x
Then In y = x . In x
⇒ \(\frac{d}{d x}\)(In y) = \(\frac{d}{d x}\)(x . In x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = In x + x . \(\frac{1}{x}\) = In x + 1 = 1 + In x
⇒ \(\frac{d y}{d x}\) = y (1 + In x) = x^x (1 + In x) = x^x In (ex)

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

Question 3.
x^{sin x}
Solution:
y = x^{sin x}
⇒ In y = sin x . In x
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = cos x In x + sin x . \(\frac{1}{x}\)
⇒ \(\frac{d y}{d x}\) = x^{sin x} (cos x . In x + \(\frac{\sin x}{x}\))

Question 4.
(log x)^{tan x}
Solution:
y = (log x)^{tan x}
⇒ log y = tan x . log (log x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = sec² x log (log x) + tan x . \(\frac{1}{\log x}\) . \(\frac{1}{x}\)
⇒ \(\frac{d y}{d x}\) = (log x)^{tan x} {sec² x . log log x + \(\frac{\tan x}{x \log x}\)}

Question 5.
\(2^{\left(2^x\right)}\)
Solution:
y = \(2^{\left(2^x\right)}\)
⇒ In y = 2^x . In 2
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = 2^x . In 2 . In 2
⇒ \(\frac{d y}{d x}\) = \(2^{\left(2^x\right)}\) . 2^x . (In 2)²

Question 6.
\((1+\sqrt{x})^{x^2}\)
Solution:

Question 7.
\(\left(\sin ^{-1} x\right)^{\sqrt{1-x^2}}\)
Solution:

Question 8.
\((\tan x)^{\log x^3}\)
Solution:

Question 9.
x^1/x + (sin x)^x
Solution:

Question 10.
(cos x)^x + x^{cos x}
Solution:

Question 11.
(x² + 1)^2/3 (3x + 1)^1/4 √x
Solution:

Question 12.
\(\frac{(x+1)(x+2)^2(x+3)^3}{(x-1)(x-2)^2(x-3)^3}\)
Solution:

Question 13.
(sin x)^x \(\sqrt{\sin x}\left(1+x^2\right)^{\frac{1}{2}+x}\)
Solution:

Question 14.
(sec x + tan x)^{cot x}
Solution:
y = (sec x + tan x)^{cot x}
⇒ In y = cot x . In (sec x + tan x)
⇒ \(\frac{1}{y}\)\(\frac{d y}{d x}\) = -cosec² x . In (sec x + tan x) + cot x . \(\frac{1}{\sec x+\tan x}\) × (sec x . tan x + sec² x)
= -cosec² x . In (sec x + tan x) + cot x . sec x
∴ \(\frac{d y}{d x}\) = y {-cosec² x . In (sec x + tan x) + cosec x}
= (sec x + tan x)^{cot x} {cosec x – cosec² x . In (sec x + tan x)}

Question 15.
(2^√x)^1+√x
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(e) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(e)

Differentiate the following functions by proper substitution.
Question 1.
sin^-1 2x\( \sqrt{1-x^2} \)
Solution:
y = sin^-1 2x\( \sqrt{1-x^2} \)   [Put x = sin θ
= sin^-1 (2 sin θ . cos θ)
= sin^-1 sin 2θ = 2θ = 2 sin^-1 x.
\(\frac{d y}{d x}\) = \(\frac{2}{\sqrt{1-x^2}}\)

Question 2.
tan^-1 \(\frac{2 x}{1-x^2}\)
Solution:
y = tan^-1 \(\frac{2 x}{1-x^2}\)
= tan^-1 \(\frac{2 \tan \theta}{1-\tan ^2 \theta}\) = tan^-1 (tan 2θ)
= 2θ = 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1-x^2}\)

Question 3.
tan^-1 \(\sqrt{\frac{1-t}{1+t}}\)
Solution:

Question 4.
\(\left[\left(\frac{1+t^2}{1-t^2}\right)^2-1\right]^{\frac{1}{2}}\)
Solution:

Question 5.
tan^-1 \(\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{x a}}\right)\)
Solution:

Question 6.
sin^-1 (\(\frac{2 x}{1+x^2}\))
Solution:
y = sin^-1 \(\frac{2 x}{1+x^2}\)  [Put x = tan θ
= sin^-1 \(\frac{2 \tan \theta}{1+\tan ^2 \theta}\) = sin^-1 sin θ
= 2θ = 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2 x}{1+x^2}\)

Question 7.
sec^-1 \(\left(\frac{\sqrt{a^2+x^2}}{a}\right)\)
Solution:

Question 8.
sin^-1 \(\left(\frac{2 \sqrt{t^2-1}}{t^2}\right)\)
Solution:

Question 9.
cos^-1 \(\left(\frac{1-t^2}{1+t^2}\right)\)
Solution:
y = cos^-1 \(\left(\frac{1-t^2}{1+t^2}\right)\) [Put t = tan θ
= cos^-1 \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)
= cos^-1 cos 2θ = 2 tan^-1 t
∴ \(\frac{d y}{d x}\) = \(\frac{2}{1+t^2}\)

Question 10.
cos^-1 (2t² – 1)
Solution:
y = cos^-1 (2t² – 1) [Put t = tan θ
= cos^-1 (2 cos² θ – 1)
= cos^-1 cos 2θ = 2θ = 2 cos^-1 t
∴ \(\frac{d y}{d x}\) = – \(\frac{2}{\sqrt{1-t^2}}\)