Vidaara.orgClass 12 · Mathematics
CodeVID-M12-01-CMP-01
Composition & Invertible Functions — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
If $f(x)=x+2$ and $g(x)=3x$, then $(g\circ f)(1)=$
- A.$5$
- B.$9$
- C.$6$
- D.$3$
2.
A function is invertible if and only if it is:
- A.one-one only
- B.onto only
- C.bijective
- D.continuous
3.
If $f(x)=2x-1$, then $f^{-1}(x)=$
- A.$\dfrac{x+1}{2}$
- B.$\dfrac{x-1}{2}$
- C.$2x+1$
- D.$\dfrac{1}{2x-1}$
4.
For bijections $f,g$, $(g\circ f)^{-1}=$
- A.$g^{-1}\circ f^{-1}$
- B.$f^{-1}\circ g^{-1}$
- C.$f\circ g$
- D.$g\circ f$
5.
$(f^{-1})^{-1}=$
- A.$f^{-1}$
- B.$\dfrac1f$
- C.$f$
- D.$I$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
If $f(x)=x^2$ and $g(x)=x+1$, find $(f\circ g)(2)$.
7.
If $f(x)=3x+2$, find $f^{-1}(x)$.
8.
Show by example that composition of functions is not commutative.
9.
If $f$ and $g$ are inverses of each other, find $(f\circ g)(x)$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
If $f(x)=2x+1,\ g(x)=x^2$, find $(g\circ f)(x)$ and $(f\circ g)(x)$.
11.
Find the inverse of $f:\mathbb{R}\to\mathbb{R},\ f(x)=\dfrac{2x+3}{4}$.
12.
If $f(x)=x+7$ and $g(x)=x-7$, show that $g=f^{-1}$.
13.
Verify the reversal law $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$ for $f(x)=x+1,\ g(x)=2x$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Show that $f:\mathbb{R}\to\mathbb{R},\ f(x)=4x+3$ is invertible and find $f^{-1}$.
15.
Let $f:\mathbb{R}\to\mathbb{R},\ f(x)=x^3$. Show $f$ is invertible and find its inverse.
Answer Key
Section A — Multiple Choice Questions
- (B) $9$
- (C) bijective
- (A) $\dfrac{x+1}{2}$
- (B) $f^{-1}\circ g^{-1}$
- (C) $f$
Section B — Short Answer (2 marks)
- $9$.
- $f^{-1}(x)=\dfrac{x-2}{3}$.
- e.g. with $f(x)=x+1,\ g(x)=x^2$: $(g\circ f)(1)=4\ne(f\circ g)(1)=2$.
- $(f\circ g)(x)=x$.
Section C — Short Answer (3 marks)
- $(g\circ f)(x)=(2x+1)^2$; $(f\circ g)(x)=2x^2+1$.
- $f^{-1}(x)=\dfrac{4x-3}{2}$.
- $g\circ f=f\circ g=I$, so $g=f^{-1}$.
- Both equal $\dfrac{x-2}{2}$.
Section D — Long Answer (5 marks)
- Invertible; $f^{-1}(x)=\dfrac{x-3}{4}$.
- Bijective; $f^{-1}(x)=x^{1/3}$.
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