Functions JEE Main previous year questions with solutions

5 solved JEE Main questions on Functions, free to read — no sign-in needed. The full chapter has 227 questions; sign in to attempt the remaining 222 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Composite Function
   Let $f:(1,\infty)\to\mathbb{R}$ be a function defined as $f(x) = \dfrac{x-1}{x+1}$. Let $f^{i+1}(x) = f(f^i(x))$, $i=1, 2, \ldots, 25$, where $f^1(x)=f(x)$. If $g(x) + f^{26}(x) = 0$, $x \in (1, \infty)$, then the area of the region bounded by the curves $y=g(x)$, $2y=2x-3$, $y=0$ and $x=4$ is:
   1. A.$\dfrac{1}{8} + \log_e 2$
   2. B.$\dfrac{1}{4} + \log_e 2$
   3. C.$\dfrac{5}{6} + 3\log_e 2$
   4. D.$\dfrac{5}{6} + \log_e 2$
   Show answer & solution

   Answer: (A)

   Given $f(x) = \dfrac{x-1}{x+1}$. Let us find the first few compositions of $f(x)$: $f^2(x) = f(f(x)) = \dfrac{\dfrac{x-1}{x+1}-1}{\dfrac{x-1}{x+1}+1} = \dfrac{x-1-x-1}{x-1+x+1} = \dfrac{-2}{2x} = -\dfrac{1}{x}$ $f^3(x) = f(f^2(x)) = f\left(-\dfrac{1}{x}\right) = \dfrac{-\dfrac{1}{x}-1}{-\dfrac{1}{x}+1} = \dfrac{-1-x}{-1+x} = \dfrac{x+1}{1-x}$ $f^4(x) = f(f^3(x)) = f\left(\dfrac{x+1}{1-x}\right) = \dfrac{\dfrac{x+1}{1-x}-1}{\dfrac{x+1}{1-x}+1} = \dfrac{x+1-1+x}{x+1+1-x} = \dfrac{2x}{2} = x$ Since $f^4(x) = x$, the sequence of functions is periodic with a period of $4$. Therefore, $f^{26}(x) = f^{4 \times 6 + 2}(x) = f^2(x) = -\dfrac{1}{x}$. We are given $g(x) + f^{26}(x) = 0$, which implies: $g(x) - \dfrac{1}{x} = 0 \Rightarrow g(x) = \dfrac{1}{x}$ We need to find the area of the region bounded by the curves $y = \dfrac{1}{x}$, $y = x - \dfrac{3}{2}$ (from $2y = 2x - 3$), $y = 0$, and $x = 4$. First, find the point of intersection of $y = \dfrac{1}{x}$ and $y = x - \dfrac{3}{2}$: $\dfrac{1}{x} = x - \dfrac{3}{2} \Rightarrow 2x^2 - 3x - 2 = 0$ $(2x+1)(x-2) = 0$ Since $x \in (1, \infty)$, we get $x = 2$. The line $y = x - \dfrac{3}{2}$ intersects the x-axis ($y = 0$) at $x = \dfrac{3}{2}$. The required area $A$ is bounded by $y = x - \dfrac{3}{2}$ from $x = \dfrac{3}{2}$ to $x = 2$, and by $y = \dfrac{1}{x}$ from $x = 2$ to $x = 4$. $A = \int_{3/2}^{2} \left(x - \dfrac{3}{2}\right) dx + \int_{2}^{4} \dfrac{1}{x} dx$ Evaluating the first integral: $\int_{3/2}^{2} \left(x - \dfrac{3}{2}\right) dx = \left[ \dfrac{1}{2}\left(x - \dfrac{3}{2}\right)^2 \right]_{3/2}^{2} = \dfrac{1}{2} \left(2 - \dfrac{3}{2}\right)^2 - 0 = \dfrac{1}{2} \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{8}$ Evaluating the second integral: $\int_{2}^{4} \dfrac{1}{x} dx = [\log_e x]_{2}^{4} = \log_e 4 - \log_e 2 = \log_e 2$ Total Area = $\dfrac{1}{8} + \log_e 2$
2. Q2JEE Main 2025 (28 Jan, Shift 1)Defintion of Function
   If $f(x)=\frac{2^x}{2^x+\sqrt{2}}, \mathrm{x} \in \mathbb{R}$, then $\sum_{\mathrm{k}=1}^{81} f\left(\frac{\mathrm{k}}{82}\right)$ is equal to
   1. A.$_{1 .} 81 \sqrt{2}$
   2. B.41
   3. C.82
   4. D.$\frac{81}{2}$
   Show answer & solution

   Answer: (D)

   $\begin{aligned} & f(x)=\frac{2^x}{2^x+\sqrt{2}} \\ & f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} \\ & =\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1\end{aligned}$ $\begin{aligned} & \text { Now, } \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\ & =\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{1}{82}\right)+\ldots \ldots+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\end{aligned}$ $\begin{aligned} & {\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots .40 \text { cases }+\mathrm{f}\left(\frac{41}{82}\right)} \\ & (1+1+\ldots .+1) 40 \text { times }+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\ & 40+\frac{1}{2}=\frac{81}{2}\end{aligned}$
3. Q3JEE Main 2024 (04 Apr, Shift 1)Domain
   If the domain of the function $\sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right)$ is $(\alpha, \beta]$, then $3 \alpha+10 \beta$ is equal to:
   1. A.100
   2. B.95
   3. C.97
   4. D.98
   Show answer & solution

   Answer: (C)

   $\begin{array}{ll} -1 \leq \frac{3 x-22}{2 x-19} \leq 1 & \frac{3 x^2-8 x+5}{x^2-3 x-10}\gt 0 \\ x \in\left(5, \frac{41}{5}\right] & \\ 3 \alpha+10 \beta=97 & \end{array}$
4. Q4JEE Main 2023 (01 Feb, Shift 2)Functional Equation
   Let $f:R-\left\{0,1\right\}\rightarrow R$ be a function such that $f\left(x\right)+f\left(\dfrac{1}{1-x}\right)=1+x$. Then $f\left(2\right)$ is equal to :
   1. A.$\dfrac{9}{2}$
   2. B.$\dfrac{9}{4}$
   3. C.$\dfrac{7}{4}$
   4. D.$\dfrac{7}{3}$
   Show answer & solution

   Answer: (B)

   Given, $f\left(x\right)+f\left(\dfrac{1}{1-x}\right)=1+x$ $x=2\Rightarrow f\left(2\right)+f\left(-1\right)=3\ldots \left(1\right)$ $x=-1\Rightarrow f\left(-1\right)+f\left(\dfrac{1}{2}\right)=0\ldots \left(2\right)$ $x=\dfrac{1}{2}\Rightarrow f\left(\dfrac{1}{2}\right)+f\left(2\right)=\dfrac{3}{2}\ldots \left(3\right)$ On subtracting equation $\left(3\right)$ and equation $\left(2\right)$ we get, $f(2)-f(-1)=\dfrac{3}{4}........(4)$ On adding equation $\left(1\right)$ and equation $\left(4\right)$ we get, $2f\left(2\right)=\dfrac{9}{2}$ $\Rightarrow f\left(2\right)=\dfrac{9}{4}$
5. Q5JEE Main 2022 (25 Jul, Shift 2)Range
   The sum of the maximum and minimum values of the function $f\left(x\right)=\left|5x-7\right|+\left[{x}^{2}+2x\right]$ in the interval $\left[\dfrac{5}{4},2\right]$, where $\left[t\right]$ is the greatest integer $\leq t$, is ______.
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   Answer: 15

   Given, $f\left(x\right)=\left|5x-7\right|+\left[{x}^{2}+2x\right]$ $=\left|5x-7\right|+\left[{\left(x+1\right)}^{2}-1\right]$ $=\left|5x-7\right|+\left[{\left(x+1\right)}^{2}\right]-1$ $\left(\text{as}\left[x-1\right]\text{=}\left[x\right]\text{-1 where}\left[.\right]\text{is greatest integer function}\right)$ Now critical points of $f\left(x\right)=\dfrac{7}{5},\sqrt{5}-1,\sqrt{6}-1,\sqrt{7}-1,\sqrt{8}-1,2$ $∴$ Maximum or minimum value of $f\left(x\right)$ occur at critical points or boundary points. So, $f\left(\dfrac{5}{4}\right)=\dfrac{3}{4}+4=\dfrac{19}{4}$ and $f\left(\dfrac{7}{5}\right)=0+4=4$ As both $\left|5x-7\right|$ and ${x}^{2}+2x$ are increasing in nature after $x=\dfrac{7}{5}$ So, $f\left(2\right)=\left|10-7\right|+\left[4+4\right]=3+8=11$ $∴f{\left(\dfrac{7}{5}\right)}_{\min }=4$ and $f{\left(2\right)}_{\max }=11$ So sum of minimum and maximum value is $4+11=15$