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 200 questions; sign in to attempt the remaining 195 in the exam simulator.

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 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 $\left(i\right)$) ${\left(\dfrac{dS}{dt}\right)}_{h=10}=\dfrac{\sqrt{26}}{5}$
4. Q4JEE Main 2026 (04 Apr, Shift 2)Maxima Minima
   $\max_{0 \leq x \leq \pi}\left(16\sin\left(\dfrac{x}{2}\right)\cos^3\left(\dfrac{x}{2}\right)\right)$ is equal to:
   1. A.$\dfrac{3\sqrt{3}}{2}$
   2. B.$3\sqrt{3}$
   3. C.$4\sqrt{3}$
   4. D.$6\sqrt{3}$
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   Answer: (B)

   Let $y = \dfrac{x}{2}$. Since $0 \leq x \leq \pi$, we have $0 \leq y \leq \dfrac{\pi}{2}$. The given expression becomes $f(y) = 16\sin y \cos^3 y$. Differentiating with respect to $y$: $f'(y) = 16(\cos y \cdot \cos^3 y + \sin y \cdot 3\cos^2 y(-\sin y))$ $f'(y) = 16\cos^2 y(\cos^2 y - 3\sin^2 y)$ For maximum value, $f'(y) = 0$: $\cos^2 y - 3\sin^2 y = 0 \Rightarrow \tan^2 y = \dfrac{1}{3}$ Since $y \in \left[0, \dfrac{\pi}{2}\right]$, $\tan y = \dfrac{1}{\sqrt{3}} \Rightarrow y = \dfrac{\pi}{6}$. Substituting $y = \dfrac{\pi}{6}$ in $f(y)$: $f\left(\dfrac{\pi}{6}\right) = 16\sin\left(\dfrac{\pi}{6}\right)\cos^3\left(\dfrac{\pi}{6}\right)$ $f\left(\dfrac{\pi}{6}\right) = 16 \times \dfrac{1}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^3$ $f\left(\dfrac{\pi}{6}\right) = 8 \times \dfrac{3\sqrt{3}}{8} = 3\sqrt{3}$ The maximum value is $3\sqrt{3}$. Answer: $3\sqrt{3}$
5. Q5JEE Main 2026 (02 Apr, Shift 2)Maxima Minima
   Let $f(x)$ be a polynomial of degree $5$, and have extrema at $x = 1$ and $x = -1$. If $\displaystyle\lim_{x \to 0} \left(\dfrac{f(x)}{x^3}\right) = -5$, then $f(2) - f(-2)$ is equal to:
   1. A.$0$
   2. B.$50$
   3. C.$92$
   4. D.$112$
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   Answer: (D)

   Let the polynomial of degree $5$ be $f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + k$. Given $\displaystyle\lim_{x \to 0} \left(\dfrac{f(x)}{x^3}\right) = -5$, the terms of degree less than $3$ must be zero, and the coefficient of $x^3$ must be $-5$. Thus, $k = 0$, $e = 0$, $d = 0$, and $c = -5$. The polynomial becomes $f(x) = ax^5 + bx^4 - 5x^3$. Differentiating with respect to $x$, we get: $f'(x) = 5ax^4 + 4bx^3 - 15x^2$ Since $f(x)$ has extrema at $x = 1$ and $x = -1$, we have $f'(1) = 0$ and $f'(-1) = 0$. $f'(1) = 5a + 4b - 15 = 0$ $f'(-1) = 5a - 4b - 15 = 0$ Adding both equations, we get $10a - 30 = 0 \Rightarrow a = 3$. Subtracting the equations, we get $8b = 0 \Rightarrow b = 0$. So, the polynomial is $f(x) = 3x^5 - 5x^3$. Now, we find $f(2)$ and $f(-2)$: $f(2) = 3(2)^5 - 5(2)^3 = 3(32) - 5(8) = 96 - 40 = 56$ $f(-2) = 3(-2)^5 - 5(-2)^3 = 3(-32) - 5(-8) = -96 + 40 = -56$ Therefore, $f(2) - f(-2) = 56 - (-56) = 112$. Answer: $112$