Functions JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Functions, free to read — no sign-in needed. The full chapter has 23 questions; sign in to attempt the remaining 18 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $\mathbb{N}$ denote the set of all positive integers. 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 $f: A \rightarrow B$ such that $f(2) \neq 2$ and $f(4) \neq 4$. Consider the set $T = \{f \in S : \text{there exists a function } g : B \rightarrow \mathbb{N} \text{ such that } g(f(x)) = 2^x \text{ for all } x \in A\}$. Then the number of elements in the set $T$ is ___________.
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   Answer: 1860

   The condition $g(f(x)) = 2^x$ for all $x \in A$ implies that if $f(x_1) = f(x_2)$, then $g(f(x_1)) = g(f(x_2))$, which gives $2^{x_1} = 2^{x_2} \Rightarrow x_1 = x_2$. Thus, $f$ must be an injective (one-to-one) function. Conversely, if $f$ is injective, such a function $g$ can always be constructed. Therefore, $T$ is the set of all injective functions $f: A \rightarrow B$ such that $f(2) \neq 2$ and $f(4) \neq 4$. The total number of injective functions from $A$ to $B$ is $^{7}P_{5} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$. Let $E_2$ be the set of injective functions where $f(2) = 2$. The number of such functions is $^{6}P_{4} = 6 \times 5 \times 4 \times 3 = 360$. Let $E_4$ be the set of injective functions where $f(4) = 4$. The number of such functions is $^{6}P_{4} = 360$. Let $E_2 \cap E_4$ be the set of injective functions where both $f(2) = 2$ and $f(4) = 4$. The number of such functions is $^{5}P_{3} = 5 \times 4 \times 3 = 60$. Using the Principle of Inclusion-Exclusion, the number of injective functions where $f(2) = 2$ or $f(4) = 4$ is: $|E_2 \cup E_4| = |E_2| + |E_4| - |E_2 \cap E_4| = 360 + 360 - 60 = 660$ The number of elements in the set $T$ is the number of injective functions where $f(2) \neq 2$ and $f(4) \neq 4$, which is: $|T| = 2520 - 660 = 1860$ Answer: $1860$
2. Q2JEE Advanced Adv 2025 (Paper 1)
   Let $\mathbb{N}$ denote the set of all natural numbers, and $\mathbb{Z}$ denote the set of all integers. Consider the functions $f: \mathbb{N} \rightarrow \mathbb{Z}$ and $g: \mathbb{Z} \rightarrow \mathbb{N}$ defined by $f(n)= \begin{cases}(n+1) / 2 & \text { if } n \text { is odd } \\ (4-n) / 2 & \text { if } n \text { is even }\end{cases}$ And $g(n)=\left\{\begin{array}{cc}3+2 n & \text { if } n \geq 0 \\ -2 n & \text { if } n \lt 0\end{array}\right.$ Define $(g \circ f)(n)=g(f(n))$ for all $n \in \mathbb{N}$, and $(f \circ g)(n)=f(g(n))$ for all $n \in \mathbb{Z}$. Then which of the following statements is (are) TRUE?
   1. A.$g \circ f$ is NOT one-one and $g \circ f$ is NOT onto
   2. B.$f \circ \mathrm{~g}$ is NOT one-one but $f \circ g$ is onto
   3. C.$g$ is one-one and $g$ is onto
   4. D.$f$ is NOT one-one but $f$ is onto
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   Answer: A,D

   $\begin{aligned} & f(n)= \begin{cases}(n+1) / 2 & \text { if } n \text { is odd } \\ (4-n) / 2 & \text { if } n \text { is even }\end{cases} \\ & f(n)=\{(1,1),(2,1),(3,2),(4,0),(5,3),(6,-1), \ldots\} \end{aligned}$ $\therefore \mathrm{f}(\mathrm{n})$ is many one and onto function $\begin{aligned} & \mathrm{g}(\mathrm{n})=\left\{\begin{array}{cc} 3+2 \mathrm{n} & \text { if } \mathrm{n} \geq 0 \\ -2 \mathrm{n} & \text { if } \mathrm{n} \lt 0 \end{array}\right. \\ & \mathrm{g}(\mathrm{n})=\{(-3,6),(-2,4),(-1,2),(0,3),(1,5),(2,7),(3,9),(4,15), \ldots\} \end{aligned}$ $\therefore \mathrm{g}(\mathrm{n})$ is one-one and into function $\mathrm{f}(\mathrm{~g}(\mathrm{n}))=2+\mathrm{n}, \mathrm{n} \in \mathrm{~N}$ fog is one-one and into $g(f(n))= \begin{cases}4+n & \text { if } n \text { is odd natural number } \\ 7-n & \text { if } n=2,4 \\ n-4 & \text { if } n \text { is even natural number and } n \geq 6\end{cases}$ $g(f(2))=g(f(1))=5$ $\therefore$ gof is many one and into
3. Q3JEE Advanced Adv 2025 (Paper 1)
   Let $\mathbb{R}$ denote the set of all real numbers. Let $a_{\mathrm{i}}, b_{\mathrm{i}} \in \mathbb{R}$ for $\mathrm{i} \in\{1,2,3\}$. Define the functions $f: \mathbb{R} \rightarrow \mathbb{R}, g: \mathbb{R} \rightarrow \mathbb{R}$, and $h: \mathbb{R} \rightarrow \mathbb{R}$ by $\begin{aligned} & f(x)=a_1+10 x+a_2 x^2+a_3 x^3+x^4, \\ & g(x)=b_1+3 x+b_2 x^2+b_3 x^3+x^4, \\ & h(x)=f(x+1)-g(x+2) .\end{aligned}$ If $f(x) \neq g(x)$ for every $x \in \mathbb{R}$, then the coefficient of $x^3$ in $h(x)$ is
   1. A.8
   2. B.2
   3. C.-4
   4. D.-6
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   Answer: (C)

   $\begin{aligned} & h(x)=f(x+1)-g(x+2) \\ & =a_1+10(x+1)+a_2(x+1)^2+a_3(x+1)^3+(x+1)^4-b_1-3(x+2)-b_2(x+2)^2-b_3(x+2)^3-(x+2)^4 \end{aligned}$ Coeff. of $x^3$ in $h(x)$ $\begin{aligned} & =a_3-b_3-4 \\ & f(x)-g(x) \neq 0 \quad \forall x \in R \\ & \Rightarrow a_1+10 x+a_2 x^2+a_3 x^3+x^4-b_1-3 x-b_2 x^2-b_3 x^3-x^4 \neq 0 \\ & \Rightarrow x^3\left(a_3-b_3\right)+x^2\left(a_2-b_2\right)+7 x+\left(a_1-b_1\right) \neq 0 \end{aligned}$ Cubic Eq. will become zero at atleast are value of $x$ So it will be quadratic $\Rightarrow \mathrm{a}_3-\mathrm{b}_3=0$
4. Q4JEE Advanced Adv 2023 (Paper 1)
   Let $S=\left(0,1\right)\cup \left(1,2\right)\cup \left(3,4\right)$ and $T=\left\{0,1,2,3\right\}$. Then which of the following statements is(are) true?
   1. A.There are infinitely many functions from $S$ to $T$
   2. B.There are infinitely many strictly increasing functions from $S$ to $T$
   3. C.The number of continuous functions from $S$ to $T$ is at most $120$
   4. D.Every continuous function from $S$ to $T$ is differentiable
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   Answer: A,C,D

   Given, $S=\left(0,1\right)\cup \left(1,2\right)\cup \left(3,4\right)$ and $T=\left\{0,1,2,3\right\}$ Now, let domain and co-domain of a function $y=f\left(x\right)$ are $S$ and $T$ respectively. Now solving option, $\left(A\right)$ There are infinitely many elements in domain and four elements in co-domain. $\Rightarrow$ There are infinitely many functions from $S$ to $T$. $\Rightarrow$ Option $\left(A\right)$ is correct $\left(B\right)$ If number of elements in domain is greater than number of elements in co-domain, then number of strictly increasing function is zero. { Assume $\sin x$ function its elements in domain is greater than its range hence, it is not an strictly increasing function.} $\Rightarrow$ Option $\left(B\right)$ is incorrect $\left(C\right)$ Maximum number of continuous functions $=4\times 4\times 4=64$ (Every subset $\left(0,1\right),\left(1,2\right),\left(3,4\right)$ has four choices) $∵64\lt 120$ $\Rightarrow$ Option $\left(C\right)$ is correct. $\left(D\right)$ For every point at which $f\left(x\right)$ is continuous, ${f}^{'}\left(x\right)=0$ {as derivative of constant is always zero} $\Rightarrow$ Every continuous function from $S$ to $T$ is differentiable. $\Rightarrow$ Option $\left(D\right)$ is correct
5. Q5JEE Advanced Adv 2020 (Paper 2)
   Let the function $f:\left[0,1\right]\rightarrow ℝ$ be defined by $f\left(x\right)=\dfrac{{4}^{x}}{{4}^{x}+2}$. Then the value of $f\left(\dfrac{1}{40}\right)+f\left(\dfrac{2}{40}\right)+f\left(\dfrac{3}{40}\right)+\cdots +f\left(\dfrac{39}{40}\right)-f\left(\dfrac{1}{2}\right)$ is_________
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   Answer: 19

   $f\left(x\right)=\dfrac{{4}^{x}}{{4}^{x}+2},f\left(1-x\right)=\dfrac{{4}^{1-x}}{{4}^{1-x}+2}=\dfrac{2}{{4}^{x}+2}$ $\Rightarrow f\left(x\right)+f\left(1-x\right)=1$ $f\left(\dfrac{1}{40}\right)+f\left(\dfrac{2}{40}\right)+\ldots ..+f\left(\dfrac{39}{40}\right)-f\left(\dfrac{1}{2}\right)$ $=\left(f\left(\dfrac{1}{40}\right)+f\left(\dfrac{39}{40}\right)\right)+\left(f\left(\dfrac{2}{40}\right)+f\left(\dfrac{38}{40}\right)\right)+.....\left(f\left(\dfrac{19}{40}\right)+f\left(\dfrac{21}{40}\right)\right)+f\left(\dfrac{20}{40}\right)-f\left(\dfrac{1}{2}\right)$ $=19$.