Application of Derivatives JEE Main previous year questions with solutions

5 solved JEE Main questions on Application of Derivatives, free to read — no sign-in needed. The full chapter has 211 questions; sign in to attempt the remaining 206 in the exam simulator.

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1. Q1JEE Main 2026 (05 Apr, Shift 1)Maxima Minima
   Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
   1. A.$3\left(\dfrac{e+1}{e}\right)$
   2. B.$3\left(\dfrac{e-1}{e}\right)$
   3. C.$\dfrac{3-e}{e}$
   4. D.$3e$
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   Answer: (B)

   Given $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ Substituting $x = 0$ and $y = 0$, we get: $f(0) = \dfrac{2f(0)}{3} \Rightarrow f(0) = 0$ Differentiating the given equation partially with respect to $x$, treating $y$ as a constant: $f'\left(\dfrac{x+y}{3}\right) \cdot \dfrac{1}{3} = \dfrac{f'(x)}{3}$ $\Rightarrow f'\left(\dfrac{x+y}{3}\right) = f'(x)$ Substituting $x = 0$, we get: $f'\left(\dfrac{y}{3}\right) = f'(0) = 3$ Since this is true for all $y \in \mathbb{R}$, $f'(x) = 3$ for all $x \in \mathbb{R}$. Integrating both sides with respect to $x$: $f(x) = 3x + C$ Using $f(0) = 0$, we get $C = 0$. Thus, $f(x) = 3x$. Now, the function $g(x)$ is given by: $g(x) = 3 + e^x f(x) = 3 + 3x e^x$ To find the minimum value, we differentiate $g(x)$ with respect to $x$: $g'(x) = 3(e^x + x e^x) = 3e^x(1 + x)$ Setting $g'(x) = 0$ gives $x = -1$. For $x \lt -1$, $g'(x) \lt 0$ and for $x \gt -1$, $g'(x) \gt 0$. Therefore, $x = -1$ is a point of global minimum. The minimum value of $g(x)$ is: $g(-1) = 3 + 3(-1)e^{-1} = 3 - \dfrac{3}{e} = 3\left(\dfrac{e-1}{e}\right)$ Answer: $3\left(\dfrac{e-1}{e}\right)$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Monotonicity
   Let the function $f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0$ be strictly increasing in $\left(-\infty, \alpha_1\right) \mathrm{U}\left(\alpha_2, \infty\right)$ and strictly decreasing in $\left(\alpha_3, \alpha_4\right) \mathrm{U}\left(\alpha_4, \alpha_5\right)$. Then $\sum_{\mathrm{i}=1}^5 \alpha_{\mathrm{i}}^2$ is equal to :-
   1. A.48
   2. B.28
   3. C.40
   4. D.36
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   Answer: (D)

   $\begin{aligned} & f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0 \\ & f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}=0 \quad \Rightarrow x= \pm 3 \\ & f^{\prime}(x)=\frac{x^2-3}{3 x^2}\end{aligned}$ $\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x}) \gt 0 \forall(-\infty,-3) \cup(3, \infty) \rightarrow \text { increasing } \\ & \mathrm{f}^{\prime}(\mathrm{x}) \lt 0 \forall(-3,0) \cup(0,3) \rightarrow \text { decreasing }\end{aligned}$ $\sum_{i=1}^5 \alpha_i^2=(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2$ $=36$
3. Q3JEE Main 2023 (31 Jan, Shift 2)Arithmetic Progression
   Let ${a}_{1},{a}_{2},{a}_{3},\ldots \ldots$. be an A.P. If ${a}_{7}=3$, the product $\left({a}_{1}{a}_{4}\right)$ is minimum and the sum of its first $n$ terms is zero then $n!-4{a}_{n\left(n+2\right)}$ is equal to
   1. A.$\dfrac{381}{4}$
   2. B.$9$
   3. C.$\dfrac{33}{4}$
   4. D.$24$
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   Answer: (D)

   We know the ${n}^{th}$ term of an A.P. is given by, ${a}_{n}=a+\left(n-1\right)d$ Given, ${a}_{7}=3$ $\Rightarrow a+6d=3$ $\Rightarrow a=3-6d$ And, ${a}_{1}{a}_{4}=a\left(a+3d\right)$ $=\left(3-6d\right)\left(3-3d\right)$ $=18{d}^{2}-27d+9$ Given product $\left({a}_{1}{a}_{4}\right)$ is minimum then, Let $f(d)=18{d}^{2}-27d+9$ ${f}^{'}(d)=36d-27$ Product to be minimum, ${f}^{'}\left(d\right)=0$ $\Rightarrow 36d-27=0$ $\Rightarrow d=\dfrac{27}{36}=\dfrac{3}{4}$ So, $a=3-\dfrac{9}{2}=\dfrac{-3}{2}$ Given, ${S}_{n}=0$ ${S}_{n}=\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]=0$ $-3+\left(n-1\right)\dfrac{3}{4}=0$ $\Rightarrow n=5$ Now $n!-4{a}_{n(n+2)}=5!-4{a}_{35}$ $=120-4\left(a+34d\right)$ $=120-4\left(\dfrac{-3}{2}+34\times \dfrac{3}{4}\right)$ $=120+6-102=24$
4. Q4JEE Main 2022 (25 Jun, Shift 2)Rate Measure Error and Approximation
   Water is being filled at the rate of $1{cm}^{3}{\sec }^{-1}$ in a right circular conical vessel (vertex downwards) of height $35cm$ and diameter $14cm$. When the height of the water level is $10cm$, the rate (in ${cm}^{2}{\sec }^{-1}$) at which the wet conical surface area of the vessel increases is
   1. A.$5$
   2. B.$\dfrac{\sqrt{21}}{5}$
   3. C.$\dfrac{\sqrt{26}}{5}$
   4. D.$\dfrac{\sqrt{26}}{10}$
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   Answer: (C)

   Let the volume of the cone be $Vc.c.$ Given $\dfrac{dv}{dt}=1cc/\sec ,h=35cm,r=7cm$ i.e. $\dfrac{h}{r}=5$ We know for a cone ${l}^{2}={r}^{2}+{h}^{2}$ Lateral surface area, $S=\pi r\sqrt{{r}^{2}+{h}^{2}}$ $S=\pi \dfrac{h}{5}\sqrt{\dfrac{{h}^{2}}{25}+{h}^{2}}=\pi \dfrac{\sqrt{26}}{25}{h}^{2}$ Also $V=\dfrac{1}{3}\pi {r}^{2}h=\dfrac{1}{3}\pi {\left(\dfrac{h}{5}\right)}^{2}h=\dfrac{\pi }{75}{h}^{3}$ $\Rightarrow \dfrac{dV}{dt}=\dfrac{\pi }{25}{h}^{2}\dfrac{dh}{dt}\Rightarrow \dfrac{\pi }{25}{h}^{2}\dfrac{dh}{dt}=1$ $\Rightarrow \dfrac{dh}{dt}=\dfrac{25}{\pi {h}^{2}}\ldots \left(i\right)$ Now $\dfrac{dS}{dt}=\dfrac{\pi \sqrt{26}}{25}\times 2h\dfrac{dh}{dt}=\dfrac{2\sqrt{26}}{h}$ (from $\left(i\right)$) ${\left(\dfrac{dS}{dt}\right)}_{h=10}=\dfrac{\sqrt{26}}{5}$
5. Q5JEE Main 2020 (09 Jan, Shift 1)Differentiation of composite functions
   Let $f$ be any function continuous on $\left[a,b\right]$ and twice differentiable on $\left(a,b\right)$ . If all $x\in \left(a,b\right),{f}^{'}\left(x\right)\gt 0$ and ${f}^{''}\left(x\right)\lt 0$ , then for any $c\in \left(a,b\right),\dfrac{f\left(c\right)-f\left(a\right)}{f\left(b\right)-f\left(c\right)}$
   1. A.$\dfrac{b+a}{b-a}$
   2. B.$1$
   3. C.$\dfrac{b-c}{c-a}$
   4. D.$\dfrac{c-a}{b-c}$
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   Answer: (D)

   Let’s use LMVT for $x\in \left[a,c\right]$ $\dfrac{f\left(c\right)-f(a)}{c-a}={f}^{'}\left(\alpha \right),\alpha \in \left(a,c\right)$ Also use LMVT for $x\in \left[c,b\right]$ $\dfrac{f\left(b\right)-f(c)}{b-c}={f}^{'}\left(\beta \right),\beta \in \left(c,b\right)$ $∵{f}^{''}\left(x\right)\lt 0\Rightarrow {f}^{'}\left(x\right)$ is decreasing ${f}^{'}\left(\alpha \right)\gt {f}^{'}\left(\beta \right)$ $\dfrac{f\left(c\right)-f(a)}{c-a}\gt \dfrac{f\left(b\right)-f(c)}{b-c}$ $\dfrac{f\left(c\right)-f(a)}{f\left(b\right)-f(c)}\gt \dfrac{c-a}{b-c}$ ( $∵f(x)$ is increasing)