JEE Mains Maths · Relations and Functions

Composition, Inverse and Iterates

Functions built from other functions: composing two, recovering one from a composite, inverting a function, recognising a self-inverse map, and applying the same function many times.

All Relations and Functions notes JEE Mains Maths strategy

Why this matters

Eighteen PYQs. Some compose two given functions or recover one from the composite; some find an inverse or the value that makes a function its own inverse; a few apply a function ten or a hundred times, which is only possible once a pattern shows. Three ideas cover the page.

Concept 1 of 3: Composing functions, and recovering one from the other

(f\circ g)(x)=f(g(x))

: apply

g

first, then

f

. To find

f

when

f(g(x))

is given, put

t=g(x)

, write

x

in terms of

t

, and substitute. When

f

and

g

are polynomials, the degree of

f\circ g

is the product of the degrees, and matching leading coefficients fixes the unknowns.

Definition

$(f\circ g)(x)=f(g(x))$ ; usually $f\circ g\ne g\circ f$ .
Recover $f$ : $t=g(x)$ , $x=g^{-1}(t)$ , then $f(t)=(f\circ g)(g^{-1}(t))$ .
$\deg(f\circ g)=\deg f\cdot\deg g$ .

Composition

(f\circ g)(x)=f\big(g(x)\big)

Worked example

f(x)=2x+1

g(x)=x^2

. Find

f(g(2))

and

g(f(2))

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q74Moderate

Example 1 · Relations and Functions · Composition, Inverse and Iterates

For

x\in R

, two real valued functions

f(x)

and

g(x)

are such that,

g(x) =\sqrt{x}+ 1

and

f \circ g(x) = x + 3 -\sqrt{x}

. Then

f(0)

is equal to

The inner function acts first

f\circ g

means

g

first. Reading it left to right gives

g\circ f

, a different function.

Drill 7 more on composing functions, and recovering one from the other

Concept 2 of 3: Inverse functions and self-inverse maps

Write

y=f(x)

and solve for

x

: that expression in

y

f^{-1}(y)

. Only a one-one onto function has an inverse. A map

\frac{ax+b}{cx+d}

is its own inverse, so

f(f(x))=x

, exactly when

a+d=0

. For an increasing function, the graph of

f^{-1}

is the mirror image in

y=x

, so

f(x)=f^{-1}(x)

happens only on that line: solve

f(x)=x

Definition

$y=f(x)\iff x=f^{-1}(y)$ .
$\frac{ax+b}{cx+d}$ has inverse $\frac{dx-b}{-cx+a}$ ; it is self-inverse iff $a+d=0$ .
$f$ increasing: $f(x)=f^{-1}(x)\iff f(x)=x$ .

Self-inverse Möbius map

f(x)=\frac{ax+b}{cx+d},\quad f\circ f=\mathrm{id}\iff a+d=0

Worked example

Find the inverse of

f(x)=\frac{2x+3}{x-5}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 12 · Q46Moderate

Example 2 · Relations and Functions · Composition, Inverse and Iterates

Let

f:\mathbb{R}-\left\{ \frac{\alpha}{6} \right\}\rightarrow R

be defined by

f(x) =\frac{5x + 3}{6x -\alpha}

. Then the value of

\alpha

for which (fof)

(x) =x

, for all

x \in R -\left\{ \frac{\alpha}{6} \right\}

, is:

Only for increasing functions

A decreasing function can meet its inverse off the line

y=x

. The shortcut

f(x)=x

is safe only when

f

is increasing.

Drill 5 more on inverse functions and self-inverse maps

Concept 3 of 3: Applying a function many times

Compute

f\circ f

, then

f\circ f\circ f

, until a pattern appears: either a cycle, where some

f^k

is the identity so only

n

modulo

k

matters, or a formula in

n

. A map

\frac{ax+b}{cx+d}

composes like the matrix

\begin{pmatrix}a&b\\c&d\end{pmatrix}

, so

f^n

comes from the

n

th power of that matrix.

Definition

$f^1=f$ , $f^{n+1}=f\circ f^n$ .
Cycle: $f^k=\mathrm{id}\Rightarrow f^n=f^{\,n\bmod k}$ .
$\frac{ax+b}{cx+d}\leftrightarrow\begin{pmatrix}a&b\\c&d\end{pmatrix}$ ; composition $\leftrightarrow$ matrix product.
$f(x)=\frac{x}{\sqrt{1+kx^2}}\Rightarrow f^n(x)=\frac{x}{\sqrt{1+nkx^2}}$ .

A cycle

f^k=\mathrm{id}\ \Rightarrow\ f^{n}=f^{\,n \bmod k}

Worked example

f(x)=\frac1{1-x}

. Find

f^{10}(2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q61Moderate

Example 3 · Relations and Functions · Composition, Inverse and Iterates

Let

f(x) =\frac{x - 1}{x + 1}\mathbf{,x \in R - \{ 0, - 1,1)}

. If

f^{n + 1}(x) = f\left( f^{n}(x) \right)

for all

n\in N

, then

f^{6}(6) +f^{7}(7)

is equal to:

Find the cycle on a general x

A value that repeats at one particular

x

proves nothing about the others. Show

f^k(x)=x

as an identity before reducing

n

Drill 3 more on applying a function many times

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Composing functions, and recovering one from the other
Composition
$(f\circ g)(x)=f\big(g(x)\big)$
Inverse functions and self-inverse maps
Self-inverse Möbius map
$f(x)=\frac{ax+b}{cx+d},\quad f\circ f=\mathrm{id}\iff a+d=0$
Applying a function many times
A cycle
$f^k=\mathrm{id}\ \Rightarrow\ f^{n}=f^{\,n \bmod k}$

Watch out for (3)

The inner function acts first
→ Composing functions, and recovering one from the other
Only for increasing functions
→ Inverse functions and self-inverse maps
Find the cycle on a general x
→ Applying a function many times

Test yourself on Relations and Functions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Relations and Functions chapter test Drill all 18 questions