Functions for JEE: Domain, Range, Inverse and Equations

A function is a rule that gives every input exactly one output. That single sentence is the whole chapter: the domain asks which inputs are allowed, the range asks which outputs actually occur, one-one and onto ask how inputs and outputs are paired, and the inverse, the composite, even and odd, periodic and functional equations all describe that pairing in different ways. JEE uses functions in almost every calculus question, so speed here pays everywhere.

This note follows the order in which JEE asks the chapter: the meaning of a function, domain, range, the standard functions and their graphs, transformations, one-one and onto, counting functions, composition, the inverse, even and odd functions, periodic functions and functional equations. Every worked example is a JEE Main 2025 or 2026 or JEE Advanced 2026 question unless it says otherwise, and every answer was re-derived from scratch.

What is a function

For sets $A$ and $B$, the cartesian product $A\times B$ is the set of all ordered pairs $(a,b)$ with $a\in A$, $b\in B$. Any subset of $A\times B$ is a relation from $A$ to $B$.

A relation $f$ from $A$ to $B$ is a function, written $f:A\to B$, when

1. every element of $A$ has an image, and
2. no element of $A$ has two images.

So in a function each input is used, and used once. Two inputs may share an output; that is allowed.

Always $f(A)\subseteq B$. The range can be smaller than the co-domain.

Vertical line test. A curve in the plane is the graph of a function of $x$ exactly when every vertical line meets it at most once.

Two graphs side by side: on the left a circle centred at the origin cut by a dashed vertical line that meets it at two marked points; on the right an upward parabola cut by a dashed vertical line that meets it at one marked point

Real valued functions. When the range lies in $\mathbb{R}$, the function is real valued. For two such functions on a common domain,

$$(f\pm g)(x)=f(x)\pm g(x)$$ $$(fg)(x)=f(x)\,g(x)$$ $$\left(\frac fg\right)(x)=\frac{f(x)}{g(x)},\quad g(x)\ne0$$

The domain of $f\pm g$ and $fg$ is $D_f\cap D_g$; for $\dfrac fg$, remove also the zeros of $g$.

Example 1 (IITianForum). Let $A=\{1,2,3\}$ and $B=\{4,5,6\}$. Which of these relations are functions from $A$ to $B$? For each function, write its range.

$$R_1=\{(1,4),(2,5),(3,4)\}$$ $$R_2=\{(1,4),(2,6)\}$$ $$R_3=\{(1,4),(1,5),(2,6),(3,6)\}$$

▸ SHOW SOLUTION

Step 1. Check each input

$R_1$: $1,2,3$ each appear once as a first entry. Function.

$R_2$: $3$ has no image. Not a function.

$R_3$: $1$ has two images, $4$ and $5$. Not a function.

Step 2. Range of R₁

Outputs that occur: $4$ and $5$.

Answer: only $R_1$ is a function; its range is $\{4,5\}$, a proper subset of the co-domain $B$.

Domain

The domain of a formula is the set of all real $x$ for which every part of the formula gives a real number. Write one condition for each part, solve each, and intersect.

Two facts settle most log conditions:

$$a>1:\ \log_a u>k\iff u>a^k$$ $$0<a<1:\ \log_a u>k\iff 0<u<a^k$$

The second line reverses the inequality because $\log_a$ is decreasing when the base is less than $1$.

Example 2 (IITianForum). Find the domain of

$$f(x)=\sqrt{x-2}+\frac{1}{\log_{10}(5-x)}$$

▸ SHOW SOLUTION

Step 1. Root

$$x-2\ge0\Rightarrow x\ge2$$

Step 2. Log in the denominator

$5-x>0$ gives $x<5$. The log must not be zero: $5-x\ne1$, so $x\ne4$.

Step 3. Intersect

Answer: $[2,4)\cup(4,5)$

Example 3 (JEE Main 2026, 24 January Shift 1). If the domain of the function

$$f(x)=\log_{(10x^2-17x+7)}(18x^2-11x+1)$$

is $(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}$, then $90(a+b+c+d+e)$ equals

▸ SHOW SOLUTION

Step 1. Argument positive

$$18x^2-11x+1=(2x-1)(9x-1)>0$$$$x<\tfrac19\ \text{or}\ x>\tfrac12$$

Step 2. Base positive

$$10x^2-17x+7=(x-1)(10x-7)>0$$$$x<\tfrac7{10}\ \text{or}\ x>1$$

Step 3. Base not equal to 1

$$10x^2-17x+6=(5x-6)(2x-1)\ne0$$

$x=\tfrac12$ is already excluded by Step 1, so only $x=\tfrac65$ is removed.

Step 4. Intersect

$$\left(-\infty,\tfrac19\right)\cup\left(\tfrac12,\tfrac7{10}\right)\cup(1,\infty)-\left\{\tfrac65\right\}$$

So $a=\tfrac19$, $b=\tfrac12$, $c=\tfrac7{10}$, $d=1$, $e=\tfrac65$.

$$90(a+b+c+d+e)$$$$=10+45+63+90+108$$

Answer: $316$, option (4)

Example 4 (JEE Main 2026, 23 January Shift 1). Let the domain of the function $f(x)=\log_3\log_5\log_7(9x-x^2-13)$ be the interval $(m,n)$. Let the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ have eccentricity $\dfrac n3$ and length of latus rectum $\dfrac{8m}{3}$. Then $b^2-a^2$ is equal to

▸ SHOW SOLUTION

Step 1. Outside in

$\log_3(\cdot)$ needs $\log_5\log_7 u>0$, so $\log_7 u>1$, so $u>7$:

$$9x-x^2-13>7$$$$x^2-9x+20<0\Rightarrow(x-4)(x-5)<0$$

So $m=4$, $n=5$.

Step 2. Hyperbola

$$e^2=\frac{25}{9}=1+\frac{b^2}{a^2}\Rightarrow\frac{b^2}{a^2}=\frac{16}{9}$$$$\frac{2b^2}{a}=\frac{32}{3}\Rightarrow\frac{b^2}{a}=\frac{16}{3}$$

Divide: $a=3$, then $b^2=16$.

Answer: $b^2-a^2=7$, option (4)

Example 5 (JEE Main 2026, 22 January Shift 2). Let the domain of the function

$$f(x)=\log_3\log_5\bigl(7-\log_2\lambda\bigr)$$ $$\qquad+\sin^{-1}\left\lvert\frac{3x-7}{17-x}\right\rvert,$$

where $\lambda=x^2-10x+85$, be $(\alpha,\beta]$. Then $\alpha+\beta$ is equal to

▸ SHOW SOLUTION

Step 1. The log part

$\lambda=(x-5)^2+60>0$ always. The outer log needs $\log_5(7-\log_2\lambda)>0$:

$$7-\log_2\lambda>1\Rightarrow\lambda<2^6=64$$$$x^2-10x+21<0\Rightarrow x\in(3,7)$$

Step 2. The inverse sine part

$$(3x-7)^2\le(17-x)^2$$$$8x^2-8x-240\le0$$$$(x-6)(x+5)\le0$$

So $x\in[-5,6]$.

Step 3. Intersect

$(3,7)\cap[-5,6]=(3,6]$, so $\alpha=3$, $\beta=6$.

Answer: $9$, option (3)

Example 6 (JEE Main 2026, 2 April Shift 1). If the domain of the function

$$f(x)=\sqrt{\log_{0.6}\left\lvert\frac{2x-5}{x^2-4}\right\rvert}$$

is $(-\infty,a]\cup\{b\}\cup[c,d)\cup(e,\infty)$, then the value of $a+b+c+d+e$ is ______.

▸ SHOW SOLUTION

Step 1. Reduce to one inequality

The root needs the log $\ge0$. The base $0.6$ is less than $1$, so the inequality reverses:

$$0<\left\lvert\frac{2x-5}{x^2-4}\right\rvert\le1$$

So $x\ne\pm2$ and $x\ne\tfrac52$.

Step 2. Lower half: ≥ −1

$$\frac{2x-5}{x^2-4}+1=\frac{(x+1)^2-10}{(x-2)(x+2)}\ge0$$$$x\le-1-\sqrt{10}\ \text{or}\ -2<x<2$$$$\text{or}\ x\ge-1+\sqrt{10}$$

Step 3. Upper half: ≤ 1

$$1-\frac{2x-5}{x^2-4}=\frac{(x-1)^2}{(x-2)(x+2)}\ge0$$$$x<-2,\ x=1,\ x>2$$

Step 4. Intersect, remove 5/2

$$\left(-\infty,-1-\sqrt{10}\,\right]\cup\{1\}$$$$\cup\left[-1+\sqrt{10},\tfrac52\right)$$$$\cup\left(\tfrac52,\infty\right)$$

So $a=-1-\sqrt{10}$, $b=1$, $c=-1+\sqrt{10}$, $d=e=\tfrac52$.

Answer: $a+b+c+d+e=4$

Example 7 (JEE Main 2026, 21 January Shift 1). If the domain of the function

$$f(x)=\cos^{-1}\frac{2x-5}{11-3x}$$ $$\qquad+\sin^{-1}(2x^2-3x+1)$$

is the interval $[\alpha,\beta]$, then $\alpha+2\beta$ is equal to

▸ SHOW SOLUTION

Step 1. Inverse sine

$2x^2-3x+1\ge-1$ means $2x^2-3x+2\ge0$, true for all $x$ (discriminant $-7$). And

$$2x^2-3x+1\le1\Rightarrow x(2x-3)\le0$$

So $x\in\left[0,\tfrac32\right]$.

Step 2. Inverse cosine

$$\frac{2x-5}{11-3x}+1=\frac{6-x}{11-3x}\ge0$$$$\frac{2x-5}{11-3x}-1=\frac{5x-16}{11-3x}\le0$$

Together: $x\le\tfrac{16}{5}$ or $x>\tfrac{11}{3}$.

Step 3. Intersect

$\left[0,\tfrac32\right]$ lies inside $\left(-\infty,\tfrac{16}5\right]$, so the domain is $\left[0,\tfrac32\right]$.

Answer: $\alpha+2\beta=0+3=3$, option (2)

Example 8 (JEE Main 2026, 22 January Shift 1). If the domain of

$$f(x)=\sin^{-1}\frac{5-x}{3+2x}+\frac{1}{\log(10-x)}$$

is $(-\infty,\alpha]\cup[\beta,\gamma)-\{\delta\}$, then $6(\alpha+\beta+\gamma+\delta)$ is

▸ SHOW SOLUTION

Step 1. Inverse sine

$\lvert5-x\rvert\le\lvert3+2x\rvert$ with $x\ne-\tfrac32$. Square:

$$25-10x+x^2\le4x^2+12x+9$$$$3x^2+22x-16\ge0$$$$(3x-2)(x+8)\ge0$$

So $x\le-8$ or $x\ge\tfrac23$.

Step 2. Log in the denominator

$10-x>0$ and $10-x\ne1$: $x<10$, $x\ne9$.

Step 3. Intersect

$$(-\infty,-8]\cup\left[\tfrac23,10\right)-\{9\}$$$$6\left(-8+\tfrac23+10+9\right)=6\cdot\frac{35}{3}$$

Answer: $70$, option (1)

Example 9 (JEE Main 2026, 24 January Shift 2). If the domain of the function

$$f(x)=\sin^{-1}\frac{1}{x^2-2x-2}$$

is $(-\infty,\alpha]\cup[\beta,\gamma]\cup[\delta,\infty)$, then $\alpha+\beta+\gamma+\delta$ is equal to

▸ SHOW SOLUTION

Step 1. Use t

Let $t=x^2-2x-2=(x-1)^2-3$, $t\ne0$. The condition $\left\lvert\dfrac1t\right\rvert\le1$ is simply

$$\lvert t\rvert\ge1$$

Step 2. Two cases

$t\ge1$: $(x-1)^2\ge4$, so $x\le-1$ or $x\ge3$.

$t\le-1$: $(x-1)^2\le2$, so $1-\sqrt2\le x\le1+\sqrt2$.

Step 3. Domain

$$(-\infty,-1]\cup\left[1-\sqrt2,1+\sqrt2\right]\cup[3,\infty)$$

Answer: $-1+(1-\sqrt2)+(1+\sqrt2)+3=4$, option (2)

The reciprocal step in Example 9 is worth remembering: $\left\lvert\dfrac1t\right\rvert\le1\iff\lvert t\rvert\ge1$. It turns a rational inequality into two quadratic ones.

Domain with the greatest integer function

$[x]$ is the greatest integer not exceeding $x$. On each interval $n\le x<n+1$ it is the constant $n$, so a condition containing $[x]$ is solved one interval at a time.

Example 10 (JEE Main 2026, 6 April Shift 1). Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x)=\sin^{-1}\left(\dfrac{x+[x]}{3}\right)$ is $[\alpha,\beta)$, then $\alpha^2+\beta^2$ is equal to

▸ SHOW SOLUTION

Step 1. Condition

$$-3\le x+[x]\le3$$

Step 2. Interval by interval

On $[n,n+1)$, $x+[x]=x+n$ runs over $[2n,2n+1)$.

$n$	$x+[x]$	inside $[-3,3]$?
$-2$	$[-4,-3)$	no
$-1$	$[-2,-1)$	yes
$0$	$[0,1)$	yes
$1$	$[2,3)$	yes
$2$	$[4,5)$	no

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Step 3. Domain

$[-1,0)\cup[0,1)\cup[1,2)=[-1,2)$, so $\alpha=-1$, $\beta=2$.

Answer: $\alpha^2+\beta^2=5$, option (2)

Example 11 (JEE Main 2026, 4 April Shift 1). Let $[\cdot]$ denote the greatest integer function. If the domain of $f(x)=\cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)$ is $[\alpha,\beta]$, then $12(\alpha+\beta)$ is equal to

▸ SHOW SOLUTION

Step 1. Condition

$$-3\le4x+2[x]\le3$$$$-3-4x\le2[x]\le3-4x$$

Step 2. Interval [0, 1)

$[x]=0$: $-3-4x\le0\le3-4x$ gives $0\le x\le\tfrac34$.

Step 3. Interval [−1, 0)

$[x]=-1$: $-3-4x\le-2$ gives $x\ge-\tfrac14$; the other side holds for all $x<0$. So $-\tfrac14\le x<0$.

Step 4. Other intervals

For $x\ge1$, $4x+2[x]\ge6>3$. For $x<-1$, $4x+2[x]<-8$. Nothing more.

Step 5. Domain

$\left[-\tfrac14,0\right)\cup\left[0,\tfrac34\right]=\left[-\tfrac14,\tfrac34\right]$.

Answer: $12\left(-\tfrac14+\tfrac34\right)=6$, option (1)

The step graph y equals 2 times the greatest integer of x, drawn as horizontal segments with filled left ends and hollow right ends, together with two parallel falling lines y equals 3 minus 4x and y equals minus 3 minus 4x; the part of the x axis from minus one quarter to three quarters, where the steps lie between the two lines, is highlighted

The picture shows Example 11 at a glance: $x$ is in the domain exactly when the step $2[x]$ lies between the two falling lines.

Range

The range is the set of values $f(x)$ actually takes. JEE uses six methods.

1. Solve for $x$. Put $y=f(x)$, make $x$ the subject, and see which $y$ give a valid $x$ in the domain.
2. Discriminant. If $y=f(x)$ becomes a quadratic in $x$, real $x$ needs discriminant $\ge0$.
3. Substitution. Put $t=$ a simpler piece ($\lvert x\rvert$, $\cos x$, $2^x$), find the range of $t$, then the range of $y$ in terms of $t$.
4. Monotonicity. A function increasing on $[a,b]$ has range $[f(a),f(b)]$.
5. Standard bounds. $-\sqrt{a^2+b^2}\le a\sin x+b\cos x\le\sqrt{a^2+b^2}$, and $t+\dfrac1t\ge2$ for $t>0$.
6. Graph. Read the lowest and highest heights the curve reaches.

Two graphs side by side: on the left the hyperbola y equals 2x plus 3 over x minus 1 with dashed asymptotes x equals 1 and y equals 2, the curve never touching the horizontal asymptote; on the right the curve y equals x plus 1 over x with its two branches, the lower branch peaking at height minus 2 and the upper branch bottoming at height 2, and the horizontal band between minus 2 and 2 shaded as a gap the curve never enters

Example 12 (IITianForum). Find the range of $f(x)=\dfrac{2x+3}{x-1}$.

▸ SHOW SOLUTION

Step 1. Solve for x

$$y(x-1)=2x+3\Rightarrow x(y-2)=y+3$$$$x=\frac{y+3}{y-2}$$

Step 2. Which y work

$y=2$ gives no $x$. For $y\ne2$ the value of $x$ is never $1$ (that would need $y+3=y-2$).

Answer: $\mathbb{R}-\{2\}$. The horizontal asymptote $y=2$ is the missing value (left panel of the figure).

Example 13 (JEE Main 2025, 7 April Shift 2). If the range of the function $f(x)=\dfrac{5-x}{x^2-3x+2}$, $x\ne1,2$, is $(-\infty,\alpha]\cup[\beta,\infty)$, then $\alpha^2+\beta^2$ is equal to

▸ SHOW SOLUTION

Step 1. Quadratic in x

$$yx^2+(1-3y)x+(2y-5)=0$$

$y=0$ gives $x=5$, allowed. For $y\ne0$, real $x$ needs

$$(1-3y)^2-4y(2y-5)\ge0$$$$y^2+14y+1\ge0$$

Step 2. Excluded points

$x=1$ would need $-4=0$ and $x=2$ would need $-3=0$, so no root is lost.

Step 3. Range

Roots of $y^2+14y+1=0$: $y=-7\pm4\sqrt3$. So $\alpha=-7-4\sqrt3$, $\beta=-7+4\sqrt3$.

$$\alpha^2+\beta^2=2(49+48)$$

Answer: $194$, option (4)

Example 14 (IITianForum). Find the range of

$$f(x)=\frac{3\lvert x\rvert-2}{2\lvert x\rvert+1}$$

▸ SHOW SOLUTION

Step 1. Substitute t = |x|

$t$ takes every value in $[0,\infty)$.

$$y=\frac{3t-2}{2t+1}=\frac32-\frac{7}{2(2t+1)}$$

Step 2. Monotone in t

As $t$ grows, $\dfrac{7}{2(2t+1)}$ falls, so $y$ rises. At $t=0$, $y=-2$. As $t$ grows without bound, $y$ gets as close to $\tfrac32$ as we like but never equals it.

Answer: $\left[-2,\tfrac32\right)$

Example 15 (JEE Main 2026, 5 April Shift 1). The sum of all the integral values of $p$ such that the equation $3\sin^2x+12\cos x-3=p$, $x\in\mathbb{R}$, has at least one solution, is

▸ SHOW SOLUTION

Step 1. Equation has a solution iff p is in the range

Put $c=\cos x\in[-1,1]$:

$$3(1-c^2)+12c-3=12-3(c-2)^2$$

Step 2. Range

$c-2\in[-3,-1]$, so $(c-2)^2\in[1,9]$ and the range is $[-15,9]$.

Step 3. Sum of integers from −15 to 9

The terms $-9$ to $9$ cancel, leaving $-(10+11+12+13+14+15)$.

Answer: $-75$, option (3)

Example 16 (JEE Main 2026, 28 January Shift 1). In a G.P., if the product of the first three terms is $27$ and the set of all possible values for the sum of its first three terms is $\mathbb{R}-(a,b)$, then $a^2+b^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Terms

Take the terms $\dfrac Ar,\ A,\ Ar$. Product $A^3=27$, so $A=3$.

$$S=3\left(1+r+\frac1r\right)$$

Step 2. Range of r + 1/r

For $r>0$, $r+\dfrac1r\ge2$ (AM ≥ GM). For $r<0$, $r+\dfrac1r\le-2$. So $r+\dfrac1r\in(-\infty,-2]\cup[2,\infty)$ (right panel of the figure).

Step 3. Range of S

$S\ge3(1+2)=9$ or $S\le3(1-2)=-3$. So $S\in\mathbb{R}-(-3,9)$.

Answer: $a^2+b^2=9+81=90$

Standard functions and their graphs

Modulus and signum

$$\lvert x\rvert=x\ (x\ge0),\qquad \lvert x\rvert=-x\ (x<0)$$

$\operatorname{sgn}x$ is $1$ for $x>0$, $0$ at $x=0$ and $-1$ for $x<0$. For $x\ne0$, $\operatorname{sgn}x=\dfrac{\lvert x\rvert}{x}$.

Two graphs side by side: on the left the V shaped graph of y equals modulus of x with its corner at the origin; on the right the signum function, a horizontal ray at height 1 for positive x starting with a hollow dot, a filled dot at the origin, and a horizontal ray at height minus 1 for negative x ending with a hollow dot

Greatest integer and fractional part

$[x]$ is the greatest integer $\le x$, and $\{x\}=x-[x]$. Examples: $[2.7]=2$, $[-2.7]=-3$, $\{-2.7\}=0.3$.

Two graphs side by side: on the left the staircase y equals greatest integer of x, each step a horizontal segment with a filled dot at its left end and a hollow dot at its right end; on the right the sawtooth y equals fractional part of x, rising lines of slope 1 from filled dots on the x axis to hollow dots at height 1, under a dashed line at height 1

Exponential and logarithm

$y=a^x$ and $y=\log_a x$ are inverses of each other, so their graphs are mirror images in the line $y=x$. For $a>1$ both increase; for $0<a<1$ both decrease.

The exponential curve y equals e to the x and the logarithm curve y equals log x drawn as mirror images in the dashed line y equals x; the exponential passes through the point 0 comma 1 and stays above the x axis, the logarithm passes through 1 comma 0 and stays to the right of the y axis

Example 17 (IITianForum). Let $[\cdot]$ denote the greatest integer function. All values of $x$ for which the product

$$\left[x-\tfrac13\right]\left[x+\tfrac23\right]$$

is a prime number form the set $[x_1,x_2)\cup[x_3,x_4)$. Find $x_1^2+x_2^2+x_3^2+x_4^2$.

▸ SHOW SOLUTION

Step 1. Consecutive integers

$x+\tfrac23=\left(x-\tfrac13\right)+1$, so with $n=\left[x-\tfrac13\right]$ the product is $n(n+1)$.

Step 2. When is n(n+1) prime

$n(n+1)$ is a product of two consecutive integers, hence even. The only even prime is $2$:

$$n(n+1)=2\Rightarrow n=1\ \text{or}\ n=-2$$

Step 3. Back to x

$n=1$: $1\le x-\tfrac13<2$, so $x\in\left[\tfrac43,\tfrac73\right)$.

$n=-2$: $-2\le x-\tfrac13<-1$, so $x\in\left[-\tfrac53,-\tfrac23\right)$.

$$\frac{25+4+16+49}{9}=\frac{94}{9}$$

Answer: $\dfrac{94}{9}$

Example 18 (JEE Main 2026, 28 January Shift 2). The sum of all the elements in the range of

$$f(x)=\operatorname{Sgn}(\sin x)+\operatorname{Sgn}(\cos x)$$ $$\qquad+\operatorname{Sgn}(\tan x)+\operatorname{Sgn}(\cot x),$$

$x\ne\dfrac{n\pi}{2}$, where $\operatorname{Sgn}(t)=1$ for $t>0$ and $-1$ for $t<0$, is

▸ SHOW SOLUTION

Step 1. Quadrant by quadrant

$\tan x$ and $\cot x$ always have the same sign, equal to the sign of $\sin x\cos x$.

Quadrant	$\sin$	$\cos$	$\tan,\cot$	$f$
I	$+$	$+$	$+$	$4$
II	$+$	$-$	$-$	$-2$
III	$-$	$-$	$+$	$0$
IV	$-$	$+$	$-$	$-2$

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Step 2. Range

$\{4,-2,0\}$, sum $2$.

Answer: option (2)

Example 19 (IITianForum). How many integers lie in the range of $f(x)=\operatorname{sgn}\bigl(\ln(x^2+x+1)\bigr)$?

▸ SHOW SOLUTION

Step 1. Range of the inside

$$x^2+x+1=\left(x+\tfrac12\right)^2+\tfrac34\ge\tfrac34$$

So the quadratic takes every value in $\left[\tfrac34,\infty\right)$.

Step 2. Sign of the log

$\ln u<0$ for $\tfrac34\le u<1$, $\ln u=0$ at $u=1$, $\ln u>0$ for $u>1$. All three occur.

Answer: the range is $\{-1,0,1\}$, so $3$ integers.

Transformations of graphs

Start from the graph of $y=f(x)$. Each change in the formula changes the graph in a fixed way ($k>0$, $a>0$).

Four small graphs in a row, each showing the dashed curve y equals root x and one transformed copy drawn solid: the first raised by one unit, the second shifted two units right, the third reflected in the x axis so it lies below, the fourth reflected in the y axis so it lies to the left

Example 20 (IITianForum). Find the number of real solutions of

$$\left\lvert x^2-4\lvert x\rvert+3\right\rvert=\tfrac12$$

▸ SHOW SOLUTION

Step 1. Use t = |x|

The left side is unchanged when $x$ is replaced by $-x$, so each positive solution $t$ gives two values $x=\pm t$. With $t\ge0$,

$$t^2-4t+3=\tfrac12\ \text{or}\ t^2-4t+3=-\tfrac12$$

Step 2. First equation

$$t^2-4t+\tfrac52=0\Rightarrow t=2\pm\sqrt{\tfrac32}$$

Both are positive ($\sqrt{1.5}<2$).

Step 3. Second equation

$$t^2-4t+\tfrac72=0\Rightarrow t=2\pm\sqrt{\tfrac12}$$

Both positive.

Step 4. Count

Four positive values of $t$, each giving two values of $x$.

Answer: $8$ solutions

Three graphs side by side built from the parabola y equals x squared minus 4x plus 3: the first is y equals the modulus of that parabola, with the dip between 1 and 3 flipped above the axis; the second is y equals f of modulus x, a W shape symmetric about the y axis; the third is the modulus of f of modulus x, an M and W shape crossed by the dashed horizontal line y equals one half at eight marked points

One-one, onto and bijective

For $f:A\to B$:

Every function is exactly one of one-one or many-one, and exactly one of onto or into.

Four arrow diagrams side by side, each with a set A on the left and a set B on the right drawn as ovals with dots: the first shows three dots going to three different dots of a four dot set with one dot left unused, marked one-one into; the second shows two arrows meeting at the same dot and every dot of B used, marked many-one onto; the third shows a perfect pairing, marked one-one onto; the fourth shows two arrows meeting and a dot of B unused, marked many-one into

Tests for one-one

1. Definition. Assume $f(a)=f(b)$ and show $a=b$; or find two inputs with the same output.
2. Horizontal line test. $f$ is one-one exactly when no horizontal line meets the graph twice.
3. Monotonicity. If $f$ is strictly increasing or strictly decreasing on its domain, it is one-one. For a differentiable $f$ on an interval, $f'(x)>0$ throughout (or $f'<0$ throughout) is enough; $f'\ge0$ with equality only at isolated points also works.
4. Quick rejections. An even function on a symmetric domain, a periodic function, and a continuous function with a turning point are all many-one.

Two graphs side by side: on the left the cubic y equals x cubed, cut by a dashed horizontal line that meets it once; on the right the parabola y equals x squared, cut by a dashed horizontal line that meets it at two marked points

Tests for onto

Find the range and compare it with the co-domain. A polynomial of odd degree from $\mathbb{R}$ to $\mathbb{R}$ is always onto; one of even degree never is.

Example 21 (JEE Main 2026, 28 January Shift 2). Given below are two statements.

Statement I: $f:\mathbb{R}\to\mathbb{R}$, $f(x)=\dfrac{x}{1+\lvert x\rvert}$ is one-one.

Statement II: $f:\mathbb{R}\to\mathbb{R}$, $f(x)=\dfrac{x^2+4x-30}{x^2-8x+18}$ is many-one.

▸ SHOW SOLUTION

Step 1. Statement I

For $x\ge0$, $f(x)=\dfrac{x}{1+x}=1-\dfrac1{1+x}$, increasing, with values in $[0,1)$. For $x<0$, $f(x)=\dfrac{x}{1-x}$, increasing, with values in $(-1,0)$. The two pieces do not overlap, so $f$ is strictly increasing on $\mathbb{R}$: one-one. True.

Step 2. Statement II

$f(0)=-\dfrac{30}{18}=-\dfrac53$. Solve $f(x)=-\dfrac53$ (the denominator is never zero):

$$3(x^2+4x-30)=-5(x^2-8x+18)$$$$8x^2-28x=0\Rightarrow x=0,\ \tfrac72$$

Two inputs, one output: many-one. True.

Answer: option (2)

Example 22 (JEE Main 2026, 6 April Shift 2). Let $f:\mathbb{R}\to\mathbb{R}$ be defined as

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

Then $f$ is

▸ SHOW SOLUTION

Step 1. Not one-one

Both quadratics read the same from either end ($2,-3,2$ and $3,1,3$). So multiplying the numerator and denominator of $f\left(\dfrac1x\right)$ by $x^2$ gives $f(x)$ back: $f\left(\dfrac1x\right)=f(x)$ for $x\ne0$. For example,

$$f(2)=f\left(\tfrac12\right)=\frac{4}{17}$$

Step 2. Range

$$(3y-2)x^2+(y+3)x+(3y-2)=0$$

Real $x$ needs $(y+3)^2-4(3y-2)^2\ge0$:

$$(7y-1)(5y-7)\le0\Rightarrow\tfrac17\le y\le\tfrac75$$

The range is bounded, not $\mathbb{R}$: not onto.

Answer: option (4)

Example 23 (JEE Main 2025, 24 January Shift 2). The function $f:(-\infty,\infty)\to(-\infty,1)$, defined by $f(x)=\dfrac{2^x-2^{-x}}{2^x+2^{-x}}$, is

▸ SHOW SOLUTION

Step 1. Simplify

Multiply above and below by $2^x$:

$$f(x)=\frac{4^x-1}{4^x+1}=1-\frac{2}{4^x+1}$$

Step 2. One-one

$4^x$ increases, so $\dfrac{2}{4^x+1}$ decreases and $f$ increases strictly. One-one.

Step 3. Range

$4^x$ takes all values in $(0,\infty)$, so $\dfrac{2}{4^x+1}$ takes all values in $(0,2)$ and $f$ takes all values in $(-1,1)$. The co-domain $(-\infty,1)$ is larger: not onto.

Answer: option (1)

Example 24 (JEE Main 2025, 28 January Shift 2). Let $f:[0,3]\to A$ be defined by $f(x)=2x^3-15x^2+36x+7$ and $g:[0,\infty)\to B$ be defined by $g(x)=\dfrac{x^{2025}}{x^{2025}+1}$. If both the functions are onto and $S=\{x\in\mathbb{Z}:x\in A\ \text{or}\ x\in B\}$, then $n(S)$ is equal to

▸ SHOW SOLUTION

Step 1. Onto means A is the range of f

$$f'(x)=6(x-2)(x-3)$$

$f$ increases on $[0,2]$ and decreases on $[2,3]$.

$$f(0)=7,\quad f(2)=35,\quad f(3)=34$$

So $A=[7,35]$.

Step 2. Range of g

$g(x)=1-\dfrac{1}{x^{2025}+1}$ increases from $0$ towards $1$, so $B=[0,1)$.

Step 3. Integers

From $B$: only $0$. From $A$: $7,8,\dots,35$, that is $29$ integers.

Answer: $n(S)=30$, option (1)

Example 25 (IITianForum). Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $x+2\lvert f(x)\rvert=3f(x)$ for every real $x$. Show that $f$ is one-one and onto.

▸ SHOW SOLUTION

Step 1. Case f(x) ≥ 0

$x+2f(x)=3f(x)$ gives $f(x)=x$. This needs $x\ge0$.

Step 2. Case f(x) < 0

$x-2f(x)=3f(x)$ gives $f(x)=\dfrac x5$. This needs $x<0$.

Step 3. The function

$$f(x)=x\ (x\ge0),\qquad f(x)=\frac x5\ (x<0)$$

For each $x$ exactly one case applies, so $f$ is well defined.

Step 4. Bijective

Both pieces increase, and the left piece takes all negative values while the right piece takes all values $\ge0$. So $f$ is strictly increasing with range $\mathbb{R}$.

Answer: $f$ is one-one and onto.

Example 26 (IITianForum). If $f:\mathbb{R}\to\mathbb{R}$,

$$f(x)=x^3-(k-1)x^2+(k+5)x-7$$

is a bijection, find the sum of all integral values of $k$.

▸ SHOW SOLUTION

Step 1. Onto is automatic

A cubic from $\mathbb{R}$ to $\mathbb{R}$ takes every real value. So bijective means one-one, which for a cubic means $f'(x)\ge0$ for all $x$.

Step 2. Derivative

$$f'(x)=3x^2-2(k-1)x+(k+5)$$

This is never negative when its discriminant is $\le0$:

$$4(k-1)^2-12(k+5)\le0$$$$k^2-5k-14\le0\Rightarrow(k-7)(k+2)\le0$$

Step 3. Integers

$k=-2,-1,\dots,7$. Sum $=25$.

Answer: $25$. At $k=-2$ and $k=7$, $f'$ is zero at a single point only, so $f$ is still strictly increasing.

Counting functions

Let $n(A)=m$ and $n(B)=n$.

For general $n$ the onto count is inclusion-exclusion over the elements of $B$ that are missed:

$$\sum_{r=0}^{n}(-1)^r\binom nr(n-r)^m$$

Example 27 (IITianForum). For sets with $n(A)=3$ and $n(B)=6$, the number of functions from $A$ to $B$ is $x$, of which $y$ are one-one. Find $x+y$.

▸ SHOW SOLUTION

Step 1. All functions

Each of the $3$ elements has $6$ choices: $x=6^3=216$.

Step 2. One-one

$y=6\cdot5\cdot4=120$.

Answer: $x+y=336$

Example 28 (JEE Main 2026, 4 April Shift 1). The number of functions $f:\{1,2,3,4\}\to\{a,b,c\}$ which are not onto is

▸ SHOW SOLUTION

Step 1. Total

$$3^4=81$$

Step 2. Onto

$$3^4-3\cdot2^4+3\cdot1^4=81-48+3=36$$

Answer: $81-36=45$, option (2)

Example 29 (JEE Main 2026, 8 April Shift 2). A person has three different bags and four different books. The number of ways in which he can put these books in the bags so that no bag is empty, is

▸ SHOW SOLUTION

Step 1. It is an onto count

Each book goes to one bag: a function from $4$ books to $3$ bags. No bag empty: onto.

$$3^4-3\cdot2^4+3=36$$

Step 2. Check by grouping

Sizes must be $2,1,1$: choose the pair, $\binom42=6$ ways, then arrange three groups in three bags, $3!=6$ ways.

Answer: $36$, option (2)

Example 30 (JEE Main 2026, 23 January Shift 2). Consider two sets

$$A=\{x\in\mathbb{Z}:\lvert\lvert x-3\rvert-3\rvert\le1\},$$ $$B=\{x\in\mathbb{R}-\{1,2\}:\ h(x)=0\},$$ $$h(x)=\frac{(x-2)(x-4)}{x-1}\ln\lvert x-2\rvert.$$

Then the number of onto functions $f:A\to B$ is

▸ SHOW SOLUTION

Step 1. A

$$-1\le\lvert x-3\rvert-3\le1\Rightarrow2\le\lvert x-3\rvert\le4$$

So $5\le x\le7$ or $-1\le x\le1$: $A=\{-1,0,1,5,6,7\}$, six elements.

Step 2. B

The factor $x-2$ is not zero ($x\ne2$). So $x=4$ or $\lvert x-2\rvert=1$, giving $x=3$ (and $x=1$, excluded). $B=\{3,4\}$.

Step 3. Onto count

$$2^6-2=62$$

Answer: option (1)

Example 31 (JEE Main 2026, 5 April Shift 1). Let $A=\{1,2,3,4,5,6\}$. The number of one-one functions $f:A\to A$ such that $f(1)\ge3$, $f(3)\le4$ and $f(2)+f(3)=5$, is ______.

▸ SHOW SOLUTION

Step 1. The pair (f(2), f(3))

Distinct values with sum $5$ and $f(3)\le4$: $(1,4),(4,1),(2,3),(3,2)$. Four pairs.

Step 2. f(1)

$f(1)\in\{3,4,5,6\}$, minus the values already used. Each pair above uses exactly one of $3,4$ and none of $5,6$, so $f(1)$ has $3$ choices every time.

Step 3. The rest

The remaining $3$ values go to $4,5,6$ in $3!=6$ ways.

Answer: $4\times3\times6=72$

Example 32 (JEE Advanced 2026, Paper 2). Consider the sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5,6,7\}$. Let $S$ be the set of all functions $f:A\to B$ such that $f(2)\ne2$ and $f(4)\ne4$. Consider the set

$$T=\{f\in S:\ g(f(x))=2^x\ \text{for some } g\}$$

where $g:B\to\mathbb{N}$ and the condition holds for all $x\in A$. Then the number of elements in $T$ is ______.

▸ SHOW SOLUTION

Step 1. Such a g exists iff f is one-one

If $f(x_1)=f(x_2)$ then $2^{x_1}=g(f(x_1))=g(f(x_2))=2^{x_2}$, so $x_1=x_2$. Conversely, if $f$ is one-one, set $g(f(x))=2^x$ on the range of $f$ and give $g$ any natural value elsewhere.

Step 2. One-one functions

$$7\cdot6\cdot5\cdot4\cdot3=2520$$

Step 3. Remove f(2) = 2 or f(4) = 4

$f(2)=2$: $6\cdot5\cdot4\cdot3=360$. Same for $f(4)=4$. Both: $5\cdot4\cdot3=60$.

$$2520-360-360+60$$

Answer: $1860$

Composite functions

For $f:A\to B$ and $g:B\to C$, the composite $g\circ f:A\to C$ is

$$(g\circ f)(x)=g\bigl(f(x)\bigr)$$

First apply $f$, then $g$. For formulas, the domain of $g\circ f$ is

$$\{x\in D_f:\ f(x)\in D_g\}$$

Example 33 (IITianForum). Let

$$f(x)=x+1\quad\text{for } x<1$$ $$f(x)=2x\quad\text{for } x\ge1$$

Find $f(f(x))$.

▸ SHOW SOLUTION

Step 1. Where f(x) crosses 1

For $x<1$, $f(x)=x+1$, which is $<1$ when $x<0$ and $\ge1$ when $0\le x<1$. For $x\ge1$, $f(x)=2x\ge2$.

Step 2. Piece x < 0

$f(x)=x+1<1$, so $f(f(x))=(x+1)+1=x+2$.

Step 3. Piece 0 ≤ x < 1

$f(x)=x+1\ge1$, so $f(f(x))=2(x+1)$.

Step 4. Piece x ≥ 1

$f(x)=2x\ge1$, so $f(f(x))=4x$.

Answer: $f(f(x))=x+2$ for $x<0$, $2x+2$ for $0\le x<1$, $4x$ for $x\ge1$.

Two graphs side by side: on the left the piecewise function made of the line y equals x plus 1 for x less than 1 and the line y equals 2x for x at least 1, joining at the point 1 comma 2; on the right its composite with itself, made of three line pieces y equals x plus 2, y equals 2x plus 2 and y equals 4x, with a dashed guide at x equals 1; the formula also changes at x equals 0, on the y axis

Example 34 (JEE Main 2026, 28 January Shift 1). If $g(x)=3x^2+2x-3$, $f(0)=-3$ and $4g(f(x))=3x^2-32x+72$, then $f(g(2))$ is equal to

▸ SHOW SOLUTION

Step 1. Quadratic in f

Write $f$ for $f(x)$:

$$12f^2+8f-12=3x^2-32x+72$$$$12f^2+8f-(3x^2-32x+84)=0$$

Step 2. Solve

The discriminant is a perfect square:

$$64+48(3x^2-32x+84)=(12x-64)^2$$$$f=\frac{-8\pm(12x-64)}{24}$$

Step 3. Use f(0) = −3

The $+$ sign gives $f(0)=-3$; the $-$ sign gives $\tfrac73$. So

$$f(x)=\frac{x-6}{2}$$

Step 4. Value

$g(2)=12+4-3=13$, and $f(13)=\tfrac72$.

Answer: option (3)

Example 35 (JEE Main 2025, 23 January Shift 1). Let $f(x)=\log_e x$ and

$$g(x)=\frac{x^4-2x^3+3x^2-2x+2}{2x^2-2x+1}$$

Then the domain of $f\circ g$ is

▸ SHOW SOLUTION

Step 1. Condition

$f\circ g$ needs $g(x)$ defined and $g(x)>0$.

Step 2. Denominator

$2x^2-2x+1$ has discriminant $-4$, so it is always positive.

Step 3. Numerator

$$x^4-2x^3+3x^2-2x+2$$$$=(x^2-x+1)^2+1>0$$

So $g(x)>0$ for every real $x$.

Answer: $\mathbb{R}$, option (1)

Example 36 (JEE Main 2025, 4 April Shift 1). Let $f,g:(1,\infty)\to\mathbb{R}$ be defined as $f(x)=\dfrac{2x+3}{5x+2}$ and $g(x)=\dfrac{2-3x}{1-x}$. If the range of $f\circ g:[2,4]\to\mathbb{R}$ is $[\alpha,\beta]$, then $\dfrac{1}{\beta-\alpha}$ is equal to

▸ SHOW SOLUTION

Step 1. Composite

$g(x)=3+\dfrac{1}{x-1}>3$ for $x>1$, so $f(g(x))$ is defined.

$$f(g(x))=\frac{2(2-3x)+3(1-x)}{5(2-3x)+2(1-x)}$$$$=\frac{9x-7}{17x-12}$$

Step 2. Monotone

$$\frac{9x-7}{17x-12}=\frac9{17}-\frac{11}{17(17x-12)}$$

For $x\in[2,4]$, $17x-12>0$ and grows, so $f\circ g$ increases.

Step 3. Range

$$(f\circ g)(2)=\frac{11}{22}=\frac12,\quad (f\circ g)(4)=\frac{29}{56}$$$$\beta-\alpha=\frac{29}{56}-\frac{28}{56}=\frac1{56}$$

Answer: $56$, option (4)

Inverse of a function

If $f:A\to B$ is a bijection, the rule that sends each $b\in B$ back to the unique $a$ with $f(a)=b$ is the inverse $f^{-1}:B\to A$.

$$f^{-1}(f(x))=x\quad\text{for } x\in A$$ $$f(f^{-1}(y))=y\quad\text{for } y\in B$$

A function that is not one-one has no inverse; one that is not onto has an inverse only on its range.

Finding $f^{-1}$. Put $y=f(x)$, solve for $x$ in terms of $y$, then rename $y$ as $x$.

Graph. $(a,b)$ lies on $y=f(x)$ exactly when $(b,a)$ lies on $y=f^{-1}(x)$, so the two graphs are mirror images in the line $y=x$.

Two graphs side by side: on the left the half parabola y equals x squared for x at least 0 and its mirror image y equals root x across the dashed line y equals x; on the right the curve y equals x minus 1 to the power 4 plus 1 for x at least 1 and its inverse curve, mirror images in the dashed line y equals x, meeting it at the two marked points 1 comma 1 and 2 comma 2

The last line holds because if $f(a)=b$ and $f^{-1}(a)=b$, then $f(b)=a$; with $f$ increasing, $a<b$ would give $b=f(a)<f(b)=a$, and $a>b$ fails the same way. So $a=b$.

Example 37 (IITianForum). Let $f:[2,\infty)\to[1,\infty)$, $f(x)=x^2-4x+5$. Show that $f$ is invertible and find $f^{-1}$.

▸ SHOW SOLUTION

Step 1. Bijective

$f(x)=(x-2)^2+1$. On $[2,\infty)$, $x-2\ge0$ grows, so $f$ increases from $1$ without bound: one-one, and onto $[1,\infty)$.

Step 2. Solve for x

$$y=(x-2)^2+1\Rightarrow x-2=\sqrt{y-1}$$

The positive root is taken since $x\ge2$.

Answer: $f^{-1}(x)=2+\sqrt{x-1}$, $x\ge1$

Example 38 (IITianForum). Find $k$ so that $f(x)=\dfrac{2x+3}{x+k}$, $x\ne-k$, is its own inverse.

▸ SHOW SOLUTION

Step 1. Condition

$f=f^{-1}$ means $f(f(x))=x$ for every $x$ in the domain.

$$f(f(x))=\frac{2\cdot\frac{2x+3}{x+k}+3}{\frac{2x+3}{x+k}+k}$$$$=\frac{7x+6+3k}{(2+k)x+3+k^2}$$

Step 2. Compare

$f(f(x))=x$ for all $x$ gives

$$(2+k)x^2+(k^2-4)x-(6+3k)=0$$

for all $x$, so every coefficient is zero: $k=-2$ (and then $k^2-4=0$ also holds).

Answer: $k=-2$. In general $\dfrac{ax+b}{cx+d}$ is self-inverse when $a+d=0$; its graph is then symmetric about $y=x$.

Example 39 (JEE Main 2026, 4 April Shift 2). For the function $f:[1,\infty)\to[1,\infty)$ defined by $f(x)=(x-1)^4+1$, consider the two statements:

(I) The set $S=\{x\in[1,\infty):f(x)=f^{-1}(x)\}$ contains exactly two elements.

(II) The set $S=\{x\in[1,\infty):f(x)=f^{-1}(x+1)\}$ is empty.

▸ SHOW SOLUTION

Step 1. Statement I

$f$ is increasing on $[1,\infty)$, so $f(x)=f^{-1}(x)\iff f(x)=x$:

$$(x-1)^4=x-1$$$$(x-1)\bigl[(x-1)^3-1\bigr]=0\Rightarrow x=1,2$$

Exactly two elements. True.

Step 2. The inverse

$y-1=(x-1)^4$ gives $f^{-1}(y)=1+(y-1)^{1/4}$, so

$$f^{-1}(x+1)=1+x^{1/4}$$

Step 3. Statement II

$f(x)=f^{-1}(x+1)$ becomes $(x-1)^4=x^{1/4}$, that is

$$x=(x-1)^{16}$$

Let $h(x)=(x-1)^{16}-x$. Then $h(1)=-1<0$ and $h(3)=2^{16}-3>0$. $h$ is continuous, so it is zero somewhere in $(1,3)$ (near $x\approx2.05$). The set is not empty. False.

Answer: option (1)

Example 40 (JEE Main 2026, 6 April Shift 1). Let $f:\{1,2,3,4\}\to\{1,e,e^2,e^3\}$ and $g:\{1,e,e^2,e^3\}\to\left\{1,\tfrac12,\tfrac13,\tfrac14\right\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If

$$\phi(x)=\left[f^{-1}\left\{g^{-1}\left(\tfrac12\right)\right\}\right]^x,$$

then the area of the region $\{(x,y):x^2\le y\le\phi(x),\ 0\le x\le1\}$ is

▸ SHOW SOLUTION

Step 1. g increasing

The smallest input goes to the smallest output:

$$g(1)=\tfrac14,\ g(e)=\tfrac13$$$$g(e^2)=\tfrac12,\ g(e^3)=1$$

So $g^{-1}\left(\tfrac12\right)=e^2$.

Step 2. f decreasing

$$f(1)=e^3,\ f(2)=e^2$$$$f(3)=e,\ f(4)=1$$

So $f^{-1}(e^2)=2$ and $\phi(x)=2^x$.

Step 3. Area

On $[0,1]$, $2^x\ge1\ge x^2$.

$$\int_0^1(2^x-x^2)\,dx=\frac{1}{\ln2}-\frac13$$$$=\frac{3-\ln2}{3\ln2}$$

Answer: option (1)

Even and odd functions

Let the domain be symmetric about $0$ (with $x$, it contains $-x$).

$$\text{even: } f(-x)=f(x)$$ $$\text{odd: } f(-x)=-f(x)$$

The graph of an even function is symmetric about the $y$ axis; the graph of an odd function is symmetric about the origin.

Two graphs side by side: on the left an even function, a W shaped curve symmetric about the y axis, with two marked points at the same height at x equals minus a and x equals a; on the right an odd function, an S shaped curve through the origin, with two marked points at x equals minus a and x equals a whose heights are equal and opposite

Standard members: $x^2$, $\cos x$, $\lvert x\rvert$, $e^x+e^{-x}$ are even; $x^3$, $\sin x$, $\tan x$, $\operatorname{sgn}x$, $e^x-e^{-x}$ are odd. Most functions are neither, but every function on a symmetric domain splits into an even part and an odd part:

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

An even function on a symmetric domain is never one-one ($f(-a)=f(a)$), so it has no inverse there.

Example 41 (IITianForum). Write $e^x$ as the sum of an even function and an odd function.

▸ SHOW SOLUTION

Step 1. Use the split

$$e^x=\frac{e^x+e^{-x}}{2}+\frac{e^x-e^{-x}}{2}$$

Step 2. Check

The first part is unchanged when $x\to-x$; the second changes sign.

Answer: even part $\dfrac{e^x+e^{-x}}{2}$, odd part $\dfrac{e^x-e^{-x}}{2}$

Example 42 (IITianForum). Which of the following are odd functions?

(A) $\dfrac{x^3}{3^{\lvert x\rvert}+\cos x}$

(B) $x^2e^{\sqrt{x^2+1}}$

(C) $\dfrac{2^x-1}{2^x+1}\bigl(h(x)+h(-x)\bigr)$, for any function $h$ defined on $\mathbb{R}$

(D) $p(q(r(x)))$, where $p$, $q$, $r$ are all odd

▸ SHOW SOLUTION

Step 1. (A)

$x^3$ is odd, $3^{\lvert x\rvert}+\cos x$ is even (and positive). Odd over even: odd.

Step 2. (B)

Every factor is even. Even.

Step 3. (C)

$$\frac{2^{-x}-1}{2^{-x}+1}=\frac{1-2^x}{1+2^x}$$

So the first factor is odd. $h(x)+h(-x)$ is even. Odd times even: odd.

Step 4. (D)

$p(q(r(-x)))=p(q(-r(x)))=p(-q(r(x)))=-p(q(r(x)))$. Odd.

Answer: (A), (C), (D)

Periodic functions

$f$ is periodic if there is a $T>0$ with $f(x+T)=f(x)$ for every $x$ in the domain. The smallest such $T$ is the fundamental period (usually just "the period").

Rules.

1. If $f$ has period $T$, then $f(ax+b)$ has period $\dfrac{T}{\lvert a\rvert}$.
2. If $f$ and $g$ have periods $T_1$ and $T_2$, then $f+g$ repeats after the LCM of $T_1$ and $T_2$. For fractions, LCM $=\dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}$.
3. The LCM is a period but not always the smallest one. Check smaller values when the two parts interact, as in $\lvert\sin x\rvert+\lvert\cos x\rvert$ or $\sin^4x+\cos^4x$ (period $\tfrac\pi2$ for both).

Example 43 (IITianForum). Find the period of $\sin\dfrac{2x}{3}+\cos\dfrac x2$.

▸ SHOW SOLUTION

Step 1. Each part

$$\sin\frac{2x}{3}:\ \frac{2\pi}{2/3}=3\pi$$$$\cos\frac x2:\ \frac{2\pi}{1/2}=4\pi$$

Step 2. LCM

LCM of $3\pi$ and $4\pi$ is $12\pi$. The smaller candidates fail: $f(0)=1$, but $f(6\pi)=\sin4\pi+\cos3\pi=-1$ and $f(4\pi)=\sin\tfrac{8\pi}{3}+1\ne1$.

Answer: $12\pi$

Example 44 (IITianForum). Find the period and the range of $f(x)=\lvert\sin x\rvert+\lvert\cos x\rvert$.

▸ SHOW SOLUTION

Step 1. π/2 is a period

$$f\left(x+\tfrac\pi2\right)=\lvert\cos x\rvert+\lvert\sin x\rvert=f(x)$$

Step 2. Nothing smaller

On $\left[0,\tfrac\pi2\right]$,

$$f(x)=\sin x+\cos x=\sqrt2\sin\left(x+\tfrac\pi4\right)$$

For $0<T<\tfrac\pi2$ this gives $f(T)>1=f(0)$, so no such $T$ is a period.

Step 3. Range

On one period $\sqrt2\sin\left(x+\tfrac\pi4\right)$ runs from $1$ up to $\sqrt2$ and back.

Answer: period $\dfrac\pi2$, range $\left[1,\sqrt2\right]$

The graph of y equals modulus of sin x plus modulus of cos x over several periods, a row of identical arches between heights 1 and root 2, with one arch from 0 to pi over 2 shaded to mark a single period and dashed horizontal guides at heights 1 and root 2

Example 45 (IITianForum). A function $f:\mathbb{R}\to\mathbb{R}$ satisfies

$$f(x+1)+f(x-1)=\sqrt3\,f(x)$$

for all real $x$, and $f(4)=2$. Find $\displaystyle\sum_{r=1}^{10}f(4+12r)$.

▸ SHOW SOLUTION

Step 1. Write the values one step apart

Fix $x$ and let $a_n=f(x+n)$. The relation says $a_{n+1}=\sqrt3\,a_n-a_{n-1}$. Start with $a_0=p$, $a_1=q$:

$$a_2=\sqrt3q-p,\quad a_3=2q-\sqrt3p$$$$a_4=\sqrt3q-2p,\quad a_5=q-\sqrt3p$$$$a_6=\sqrt3(q-\sqrt3p)-(\sqrt3q-2p)=-p$$

Step 2. Period

So $f(x+6)=-f(x)$ for every $x$, and applying it twice,

$$f(x+12)=f(x)$$

Step 3. Sum

Each term $f(4+12r)=f(4)=2$.

Answer: $10\times2=20$

Example 46 (IITianForum). A real function $f$ satisfies, for every real $x$,

$$f(x+1)=2+\sqrt[3]{u(x)},\ \text{where}$$ $$u(x)=9-12f(x)+6f(x)^2-f(x)^3$$

and $f(1)=3$. Find $f(2026)+f(2027)$.

▸ SHOW SOLUTION

Step 1. Shift by 2

Let $g(x)=f(x)-2$. Expanding shows $u(x)=1-(f(x)-2)^3$, so

$$g(x+1)^3=1-g(x)^3$$

Step 2. Period 2

Replace $x$ by $x+1$: $g(x+2)^3=1-g(x+1)^3=g(x)^3$, so $g(x+2)=g(x)$ and $f$ has period $2$.

Step 3. Values

$g(1)=1$, so $g(2)^3=1-1=0$ and $f(2)=2$.

$$f(2026)=f(2)=2$$$$f(2027)=f(1)=3$$

Answer: $5$

Functional equations

A functional equation gives a relation that $f$ satisfies for all $x$ (and $y$). The usual plan:

1. Put special values: $x=y=0$, $x=y=1$, $y=-x$, $y=\dfrac1x$, to find $f(0)$, $f(1)$ and symmetry.
2. When $f(x)$ appears with $f\left(\dfrac1x\right)$ or $f(a-x)$, replace $x$ by $\dfrac1x$ or $a-x$ and solve the two linear equations.
3. When a form is given (polynomial of degree $2$, say), substitute it and compare coefficients.
4. Recognise the standard forms below. In JEE the function is given to be continuous, differentiable or a polynomial, which is what makes these forms the only answers.

Example 47 (JEE Main 2026, 22 January Shift 2). Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$ and $g(x+y)=g(xy)$, with $f(1)=7$ and $g(1)=1$, for all $x,y\in\mathbb{N}$. If

$$\sum_{x=1}^{n}\frac{f(x)}{g(x)}=19607,$$

then $n$ is equal to

▸ SHOW SOLUTION

Step 1. f

$y=1$: $f(x+1)=7f(x)$, so $f(x)=7^x$.

Step 2. g

$y=1$: $g(x+1)=g(x)$, so $g$ is constant, $g(x)=1$.

Step 3. Sum

$$7+7^2+\cdots+7^n=\frac{7(7^n-1)}{6}=19607$$$$7^n=1+\frac{6\cdot19607}{7}=16807=7^5$$

Answer: $n=5$, option (2)

Example 48 (JEE Main 2026, 4 April Shift 2). Let for some $\alpha\in\mathbb{R}$, $f:\mathbb{R}\to\mathbb{R}$ satisfy

$$f(x+y)=f(x)+2y^2+y+\alpha xy$$

for all $x,y\in\mathbb{R}$. If $f(0)=-1$ and $f(1)=2$, then the value of $\displaystyle\sum_{n=1}^{5}\bigl(\alpha+f(n)\bigr)$ is

▸ SHOW SOLUTION

Step 1. Put x = 0

$$f(y)=f(0)+2y^2+y=2y^2+y-1$$

Step 2. Find α

Substitute $f(x)=2x^2+x-1$ back. The left side has $4xy$, the right side has $\alpha xy$; all other terms match. So $\alpha=4$. (Check: $f(1)=2$.)

Step 3. Sum

$$\alpha+f(n)=2n^2+n+3$$$$\sum_{n=1}^{5}(2n^2+n+3)=2(55)+15+15$$

Answer: $140$, option (2)

Example 49 (JEE Main 2026, 24 January Shift 2). Let $f$ be a function such that

$$3f(x)+2f\left(\frac{m}{19x}\right)=5x,\quad x\ne0,$$

where $m=\displaystyle\sum_{i=1}^{9}i^2$. Then $f(5)-f(2)$ is equal to

▸ SHOW SOLUTION

Step 1. The constant

$m=\dfrac{9\cdot10\cdot19}{6}=285$, so $\dfrac{m}{19}=15$:

$$3f(x)+2f\left(\frac{15}{x}\right)=5x\qquad(1)$$

Step 2. Replace x by 15/x

$$3f\left(\frac{15}{x}\right)+2f(x)=\frac{75}{x}\qquad(2)$$

Step 3. Eliminate

$3\times(1)-2\times(2)$:

$$5f(x)=15x-\frac{150}{x}\Rightarrow f(x)=3x-\frac{30}{x}$$

Step 4. Value

$f(5)=9$, $f(2)=-9$.

Answer: $18$, option (3)

Example 50 (JEE Main 2026, 5 April Shift 2). Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x)+3f\left(\dfrac\pi2-x\right)=\sin x$. Let the maximum value of $f$ be $\alpha$. If the area of the region bounded by the curves $g(x)=x^2$ and $h(x)=\beta x^3$, $\beta>0$, is $\alpha^2$, then $30\beta^3$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Pair of equations

Replace $x$ by $\dfrac\pi2-x$:

$$f\left(\tfrac\pi2-x\right)+3f(x)=\cos x$$

Eliminate $f\left(\tfrac\pi2-x\right)$:

$$8f(x)=3\cos x-\sin x$$

Step 2. Maximum

$3\cos x-\sin x$ has maximum $\sqrt{10}$, so $\alpha=\dfrac{\sqrt{10}}{8}$ and $\alpha^2=\dfrac{5}{32}$.

Step 3. Area

The curves meet at $x=0$ and $x=\dfrac1\beta$.

$$\int_0^{1/\beta}(x^2-\beta x^3)\,dx=\frac{1}{12\beta^3}$$$$\frac{1}{12\beta^3}=\frac{5}{32}\Rightarrow\beta^3=\frac{8}{15}$$

Answer: $30\beta^3=16$

Example 51 (JEE Main 2025, 28 January Shift 2). Let $f:\mathbb{R}-\{0\}\to(-\infty,1)$ be a polynomial of degree $2$ satisfying

$$f(x)f\left(\tfrac1x\right)=f(x)+f\left(\tfrac1x\right)$$

If $f(K)=-2K$, then the sum of squares of all possible values of $K$ is

▸ SHOW SOLUTION

Step 1. Factor the equation

Bring every term to one side and add $1$:

$$\bigl(f(x)-1\bigr)\left(f\left(\tfrac1x\right)-1\right)=1$$

Step 2. The form of f

$g(x)=f(x)-1$ is a polynomial with $g(x)g\left(\tfrac1x\right)=1$; the only such polynomials are $\pm x^n$. Degree $2$ gives $f(x)=1\pm x^2$.

Step 3. Use the co-domain

$1+x^2\ge1$ is outside $(-\infty,1)$, so $f(x)=1-x^2$.

Step 4. Solve f(K) = −2K

$$K^2-2K-1=0$$

Sum of squares $=(K_1+K_2)^2-2K_1K_2=4+2$.

Answer: $6$, option (2)

Example 52 (IITianForum). A function $f:\mathbb{R}\to\mathbb{R}$ with $f(0)\ne0$ satisfies

$$f(x+y)+f(x-y)=2f(x)f(y)$$

for all real $x,y$, and $f(1)=2$. Show that $f$ is even and find $f(2)$ and $f(3)$.

▸ SHOW SOLUTION

Step 1. f(0)

$x=y=0$: $2f(0)=2f(0)^2$. Since $f(0)\ne0$, $f(0)=1$.

Step 2. Even

$x=0$: $f(y)+f(-y)=2f(y)$, so $f(-y)=f(y)$.

Step 3. f(2)

$x=y=1$: $f(2)+f(0)=2f(1)^2$.

$$f(2)=2\cdot4-1=7$$

Step 4. f(3)

$x=2$, $y=1$: $f(3)+f(1)=2f(2)f(1)$.

$$f(3)=2\cdot7\cdot2-2=26$$

Answer: $f$ is even, $f(2)=7$, $f(3)=26$. In general $f(n+1)=2kf(n)-f(n-1)$ with $k=f(1)$, which gives $f(2)=2k^2-1$.

Example 53 (IITianForum). Let $f:(0,\infty)\to\mathbb{R}$ with $f(1)=1$ satisfy, for all $x,y>0$,

$$f(x)f(y)+f\left(\tfrac5x\right)f\left(\tfrac5y\right)=2f(xy)$$

Find $f(2026)$.

▸ SHOW SOLUTION

Step 1. x = y = 1

$1+f(5)^2=2$, so $f(5)^2=1$.

Step 2. y = 1

$$f(x)+f\left(\tfrac5x\right)f(5)=2f(x)$$$$f(x)=f(5)\,f\left(\tfrac5x\right)\qquad(1)$$

Step 3. y = 5/x

$$2f(x)f\left(\tfrac5x\right)=2f(5)$$$$f(x)f\left(\tfrac5x\right)=f(5)\qquad(2)$$

Step 4. Square of f

Multiply (1) by $f(x)$ and use (2): $f(x)^2=f(5)\cdot f(5)=1$.

Step 5. Sign

Put $x=y=\sqrt t$: $f(\sqrt t)^2+f\left(\tfrac5{\sqrt t}\right)^2=2f(t)$, and both squares are $1$, so $f(t)=1$ for every $t>0$.

Answer: $f(2026)=1$

Which method to use

A decision chart: from the question asked, arrows lead to eight situations, each paired with its method: domain, one condition per part then intersect; range, solve for x or use the discriminant or a substitution; one-one, test monotonicity or find two equal outputs; onto, compare range with co-domain; counting functions, use n to the power m, the falling product or inclusion-exclusion; composite, split where the inner function crosses the outer break points; inverse, solve y equals f of x for x and use y equals x for increasing f; functional equation, put special values or replace x by 1 over x then solve

What JEE asks

Across the 2025 and 2026 papers the chapter appears in five forms, roughly in this order of frequency.

1. Domain of a formula with logs, roots, inverse trigonometric parts or $[x]$, answered as a sum of interval end points (Examples 3 to 11). This is the single most common functions question in JEE Main 2026.
2. Counting functions: all, one-one, onto, not onto, with conditions (Examples 28 to 32).
3. Functional equations of the substitution type, often followed by a sum or an area (Examples 47 to 51).
4. One-one and onto decided from monotonicity and range (Examples 21 to 24).
5. Range, composite and inverse, usually inside a larger question (Examples 13, 15, 34 to 40).

JEE Advanced 2026 tested the link between a composite and one-one functions (Example 32).

Quick recall

Practise these methods on real exam questions in the doMath Functions chapter. Functions are the language of calculus: the Limits note and the Continuity and Differentiability note build directly on everything here.