Mathematics Questions and Answers – Composition of Functions and Invertible Function

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

1. The composition of functions is both commutative and associative.
a) True
b) False
View Answer

Answer: b
Explanation: 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.
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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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: d
Explanation: 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.
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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) \(cos(\frac{x^3+6}{7})\)
View Answer

Answer: a
Explanation: 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
View Answer

Answer: b
Explanation: 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.
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6. The function f:R→R defined by f(x)=5x+9 is invertible.
a) True
b) False
View Answer

Answer: a
Explanation: 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
View Answer

Answer: b
Explanation: 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)
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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)}
View Answer

Answer: a
Explanation: 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
View Answer

Answer: a
Explanation: 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)=\(\sqrt{x-9}\).
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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) \(\frac{2x-3}{7x+9}\)
View Answer

Answer: b
Explanation: 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)=\(\frac{7y-9}{2x+3}\).

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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