250+ TOP MCQs on Composition of Functions and Invertible Function | Class 12 Maths

Mathematics written test Questions & Answers on “Composition of Functions and Invertible Function”.

1. The composition of functions is both commutative and associative.
a) True
b) False
Answer: b
Clarification: The given statement is false. The composition of functions is associative i.e. fο(g ο h)=(f ο g)οh. The composition of functions is not commutative i.e. g ο f ≠ f ο g.

2. If f:R→R, g(x)=3x²+7 and f(x)=√x, then gοf(x) is equal to _______
a) 3x-7
b) 3x-9
c) 3x+7
d) 3x-8
Answer: c
Clarification: Given that, g(x)=3x²+7 and f(x)=√x
∴ gοf(x)=g(f(x))=g(√x)=3(√x)²+7=3x+7.
Hence, gοf(x)=3x+7.

3. If f:R→R is given by f(x)=(5+x⁴)^1/4, then fοf(x) is _______
a) x
b) 10+x⁴
c) 5+x⁴
d) (10+x⁴)^1/4
Answer: d
Clarification: Given that f(x)=(5+x⁴)^1/4
∴ fοf(x)=f(f(x))=(5+{(5+x⁴)^1/4}⁴)^1/4
=(5+(5+x⁴))^1/4=(10+x⁴)^1/4.

4. If f:R→R f(x)=cos⁡x and g(x)=7x³+6, then fοg(x) is ______
a) cos⁡(7x³+6)
b) cos⁡x
c) cos⁡(x³)
d) (cos(frac{x^3+6}{7}))
Answer: a
Clarification: Given that, f:R→R, f(x)=cos⁡x and g(x)=7x³+6
Then, fοg(x) = f(g(x))=cos⁡(g(x))=cos⁡(7x³+6).

5. A function is invertible if it is ____________
a) surjective
b) bijective
c) injective
d) neither surjective nor injective
Answer: b
Clarification: A function is invertible if and only if it is bijective i.e. the function is both injective and surjective. If a function f:A→B is bijective, then there exists a function g:B→A such that f(x)=y⇔g(y)=x, then g is called the inverse of the function.

6. The function f:R→R defined by f(x)=5x+9 is invertible.
a) True
b) False
Answer: a
Clarification: The given statement is true. A function is invertible if it is bijective.
For one – one: Consider f(x₁)=f(x₂)
∴ 5x₁+9=5x₂+9
⇒x₁=x₂. Hence, the function is one – one.
For onto: For any real number y in the co-domain R, there exists an element x=(frac{y-9}{5}) such that f(x)=(f(frac{y-9}{5})=5(frac{y-9}{5}))+9=y.
Therefore, the function is onto.

7. If f:N→N, g:N→N and h:N→R is defined f(x)=3x-5, g(y)=6y² and h(z)=tan⁡z, find ho(gof).
a) tan⁡(6(3x-5))
b) tan⁡(6(3x-5)²)
c) tan⁡(3x-5)
d) 6 tan⁡(3x-5)²
Answer: b
Clarification: Given that, f(x)=3x-5, g(y)=6y² and h(z)=tan⁡z,
Then, ho(gof)=hο(g(f(x))=h(6(3x-5)²)=tan⁡(6(3x-5)²)
∴ ho(gof)=tan⁡(6(3x-5)²)

8. Let M={7,8,9}. Determine which of the following functions is invertible for f:M→M.
a) f = {(7,7),(8,8),(9,9)}
b) f = {(7,8),(7,9),(8,9)}
c) f = {(8,8),(8,7),(9,8)}
d) f = {(9,7),(9,8),(9,9)}
Answer: a
Clarification: The function f = {(7,7),(8,8),(9,9)} is invertible as it is both one – one and onto. The function is one – one as every element in the domain has a distinct image in the co – domain. The function is onto because every element in the codomain M = {7,8,9} has a pre – image in the domain.

9. Let f:R₊→[9,∞) given by f(x)=x²+9. Find the inverse of f.
a) (sqrt{x-9})
b) (sqrt{9-x})
c) (sqrt{x^2-9})
d) x²+9
Answer: a
Clarification: The function f(x)=x²+9 is bijective.
Therefore, f(x)=x²+9
i.e.y=x²+9
x=(sqrt{y-9})
⇒f^-1 (x)=(sqrt{x-9}).

10. Let the function f be defined by f(x)=(frac{9+3x}{7-2x}), then f^-1(x) is ______
a) (frac{9-3x}{7+2x})
b) (frac{7x-9}{2x+3})
c) (frac{2x-7}{3x+9})
d) (frac{2x-3}{7x+9})
Answer: b
Clarification: The function f(x)=(frac{9+3x}{7-2x}) is bijective.
∴ f(x)=(frac{9+3x}{7-2x})
i.e.y=(frac{9+3x}{7-2x})
7y-2xy=9+3x
7y-9=x(2y+3)
x=(frac{7y-9}{2y+3})
⇒f^-1 (x)=(frac{7y-9}{2x+3}).