Application of Derivatives JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Application of Derivatives, free to read — no sign-in needed. The full chapter has 68 questions; sign in to attempt the remaining 63 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   Let $P$ be the point on the parabola $y = x^2$ such that the slope of the tangent to the parabola at the point $P$ is $4$. Let $Q$ be the point in the first quadrant lying on the circle $x^2 + y^2 = 2$ such that the slope of the tangent to the circle at the point $Q$ is $-1$. Let $R$ be the point in the first quadrant lying on the ellipse $x^2 + 4y^2 = 8$ such that the slope of the tangent to the ellipse at the point $R$ is $-\dfrac{1}{2}$. Then the radius of the circle passing through the points $P, Q$ and $R$ is
   1. A.$\sqrt{10}$
   2. B.$\sqrt{5}$
   3. C.$\sqrt{\dfrac{5}{2}}$
   4. D.$2\sqrt{5}$
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   Answer: (C)

   For the point $P$ on the parabola $y = x^2$, the slope of the tangent is $\dfrac{dy}{dx} = 2x$. Given $2x = 4 \Rightarrow x = 2$. Substituting $x = 2$ in $y = x^2$, we get $y = 4$. Thus, the coordinates of $P$ are $(2, 4)$. For the point $Q$ on the circle $x^2 + y^2 = 2$, differentiating with respect to $x$ gives $2x + 2y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{y}$. Given $-\dfrac{x}{y} = -1 \Rightarrow x = y$. Since $Q$ lies in the first quadrant, substituting $x = y$ in $x^2 + y^2 = 2$ gives $2x^2 = 2 \Rightarrow x = 1, y = 1$. Thus, the coordinates of $Q$ are $(1, 1)$. For the point $R$ on the ellipse $x^2 + 4y^2 = 8$, differentiating with respect to $x$ gives $2x + 8y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{4y}$. Given $-\dfrac{x}{4y} = -\dfrac{1}{2} \Rightarrow x = 2y$. Since $R$ lies in the first quadrant, substituting $x = 2y$ in $x^2 + 4y^2 = 8$ gives $4y^2 + 4y^2 = 8 \Rightarrow 8y^2 = 8 \Rightarrow y = 1, x = 2$. Thus, the coordinates of $R$ are $(2, 1)$. The points are $P(2, 4)$, $Q(1, 1)$, and $R(2, 1)$. The line segment $PR$ lies on the vertical line $x = 2$, and the line segment $QR$ lies on the horizontal line $y = 1$. Since $PR$ is perpendicular to $QR$, the angle $\angle PRQ = 90^{\circ}$. Therefore, the triangle $PQR$ is a right-angled triangle with the hypotenuse $PQ$. The circumcircle of $\triangle PQR$ has $PQ$ as its diameter. The length of the diameter is $PQ = \sqrt{(2 - 1)^2 + (4 - 1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$. The radius of the circle is $\dfrac{PQ}{2} = \dfrac{\sqrt{10}}{2} = \sqrt{\dfrac{5}{2}}$. Answer: $\sqrt{\dfrac{5}{2}}$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   Consider the function $f: (0, \infty) \rightarrow (-\infty, \infty)$ given by $f(x) = \sqrt{x}\, \log_e(x) - x + 1$. Then which one of the following statements is TRUE?
   1. A.The derivative of the function $f$ is decreasing in the interval $(0, 1)$
   2. B.The function $f$ has a local maximum at some point $a \in (0, \infty)$
   3. C.The function $f$ has a local minimum at some point $b \in (0, \infty)$
   4. D.The function $f$ has NEITHER a point of local maximum NOR a point of local minimum in the interval $(0, \infty)$
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   Answer: (D)

   Given $f(x) = \sqrt{x} \log_e(x) - x + 1$ Differentiating with respect to $x$: $f'(x) = \dfrac{1}{2\sqrt{x}} \log_e(x) + \sqrt{x} \left(\dfrac{1}{x}\right) - 1$ $f'(x) = \dfrac{\log_e(x)}{2\sqrt{x}} + \dfrac{1}{\sqrt{x}} - 1$ Differentiating again with respect to $x$: $f''(x) = \dfrac{d}{dx} \left( \dfrac{1}{2} x^{-1/2} \log_e(x) + x^{-1/2} - 1 \right)$ $f''(x) = \dfrac{1}{2} \left( -\dfrac{1}{2} x^{-3/2} \log_e(x) + x^{-1/2} \cdot \dfrac{1}{x} \right) - \dfrac{1}{2} x^{-3/2}$ $f''(x) = -\dfrac{\log_e(x)}{4x^{3/2}} + \dfrac{1}{2x^{3/2}} - \dfrac{1}{2x^{3/2}}$ $f''(x) = -\dfrac{\log_e(x)}{4x^{3/2}}$ For $x \in (0, 1)$, $\log_e(x) \lt 0$, which implies $f''(x) \gt 0$. Thus, $f'(x)$ is strictly increasing in $(0, 1)$. For $x \in (1, \infty)$, $\log_e(x) \gt 0$, which implies $f''(x) \lt 0$. Thus, $f'(x)$ is strictly decreasing in $(1, \infty)$. Therefore, $f'(x)$ attains its maximum value at $x = 1$. Maximum value of $f'(x) = f'(1) = \dfrac{\log_e(1)}{2} + 1 - 1 = 0$. Since the maximum value of $f'(x)$ is $0$, we have $f'(x) \le 0$ for all $x \in (0, \infty)$, with equality holding only at $x = 1$. This means $f(x)$ is a strictly decreasing function on $(0, \infty)$. Hence, $f(x)$ has neither a point of local maximum nor a point of local minimum in the interval $(0, \infty)$. Answer: The function $f$ has NEITHER a point of local maximum NOR a point of local minimum in the interval $(0, \infty)$
3. Q3JEE Advanced Adv 2025 (Paper 1)
   Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f: \mathbb{R \rightarrow \mathbb { R }}$ by $f(\mathrm{x})=\left\{\begin{array}{cc}2-2 x^2-x^2 \sin \frac{1}{x} & \text { if } x \neq 0 \\ 2 & \text { if } x=0\end{array}\right.$ Then which one of the following statements is TRUE ?
   1. A.$\text { The function } f \text { is NOT differentiable at } x=0$
   2. B.There is a positive real number $\delta$, such that $f$ is a decreasing function on the interval $(0, \delta)$
   3. C.For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$
   4. D.$x=0$ is a point of local minima of $f$
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   Answer: (C)

   (A) RHD at $x=0: \lim _{h \rightarrow 0} \frac{\left(2-2 h^2-h^2 \sin \frac{1}{h}\right)-2}{h}=0$ Similarly LHD at $\mathrm{x}=0$ is also equal to 0 . $\therefore$ Differentiable at $\mathrm{x}=0$ (B) $f^{\prime}(x)=-4 x-2 x \sin \frac{1}{x}-x^2\left(\cos \frac{1}{x}\right)\left(\frac{-1}{x^2}\right)$ $\begin{aligned} & f^{\prime}(x)=-\left(4 x+2 x \sin \frac{1}{x}\right)+\cos \frac{1}{x} \\ & f^{\prime}(x)=-\left(2 x\left(4-\sin \frac{1}{x}\right)\right)+\cos \frac{1}{x} \end{aligned}$ for $\mathrm{x} \in(0, \delta)$ $\Rightarrow$ We can't say $f(x)$ is decreasing on $(0, \delta)$ as $\cos \frac{1}{x}$ oscillates. (C) for $x \in(-\delta, 0)$, for any $\delta\gt 0$ $\Rightarrow f(x)$ is not increasing on $(-\delta, 0)$ as $\cos \frac{1}{x}$ oscillates from -1 to 1. (D) $f(0)=2$ $\begin{aligned} & f(0+h) \lt 2 \\ & f(0-h) \lt 2 \end{aligned}$ $\therefore \mathrm{x}=0$ is local maxima
4. Q4JEE Advanced Adv 2024 (Paper 2)
   Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{\sin x}{e^{\pi x}} \frac{\left(x^{2023}+2024 x+2025\right)}{\left(x^2-x+3\right)}+\frac{2}{e^{\pi x}} \frac{\left(x^{2023}+2024 x+2025\right)}{\left(x^2-x+3\right)}$. Then the number of solutions of $f(x)=0$ in $\mathbb{R}$ is
   Show answer & solution

   Answer: 1

   $f(x)=\frac{\left(x^{2023}+2024 x+2025\right)}{e^{\pi x}\left(x^2-x+3\right)}(\sin x+2)$ $\because(\sin x+2)$ is never zero $\therefore$ for $x^{2023}+2024 x+2025=0$ $\begin{aligned} & \text { let } \phi(\mathrm{x})=\mathrm{x}^{2023}+2024 \mathrm{x}+2025 \\ & \phi^{\prime}(\mathrm{x})=2023 \mathrm{x}^{2022}+2024\gt 0 \forall \mathrm{x} \in \mathrm{R}\end{aligned}$ $\therefore \phi(\mathrm{x})$ is an Strictly Increasing function $\therefore \phi(\mathrm{x})=0$ for exactly one value of $\mathrm{x}$ $\therefore \mathrm{f}(\mathrm{x})=0$ has one solution
5. Q5JEE Advanced Adv 2021 (Paper 2)
   Paragraph: Let $\psi_{1}:[0, \infty) \rightarrow \mathbb{R}, \psi_{2}:[0, \infty) \rightarrow \mathbb{R}, f:[0, \infty) \rightarrow \mathbb{R}$ and $g:[0, \infty) \rightarrow \mathbb{R}$ be functions such that $f(0)=g(0)=0$, $\psi_{1}(x)=e^{-x}+x, \quad x \geq 0$, $\psi_{2}(x)=x^{2}-2 x-2 e^{-x}+2, \quad x \geq 0$, $f(x)=\int_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} d t, \quad x\gt 0$ and $g(x)=\int_{0}^{x^{2}} \sqrt{t} e^{-t} d t, \quad x\gt 0$ Question: Which of the following statements is TRUE ?
   1. A.${\psi }_{1}\left(x\right)\leq 1$ for all $x\gt 0$
   2. B.${\psi }_{2}\left(x\right)\leq 0,$ for all $x\gt 0$
   3. C.$f\left(x\right)\geq 1-{e}^{-{x}^{2}}-\dfrac{2}{3}{x}^{3}+\dfrac{2}{5}{x}^{5}$, for all $x\in \left(0,\dfrac{1}{2}\right)$
   4. D.$g\left(x\right)\leq \dfrac{2}{3}{x}^{3}-\dfrac{2}{5}{x}^{5}+\dfrac{1}{7}{x}^{7}$ for all $x\in \left(0,\dfrac{1}{2}\right)$
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   Answer: (D)

   (A) Given ${\psi }_{1}\left(x\right)={e}^{-x}+x,x\geq 0$ ${{\psi }^{'}}_{1}\left(x\right)=-{e}^{-x}+1,x\geq 0$ ${{\psi }^{'}}_{1}\left(x\right)$ is increasing function. ${\psi }_{1}\left(x\right)\gt {\psi }_{1}\left(0\right)\forall x\geq 0$ ${\psi }_{1}\left(x\right)\geq 1$ So, ${e}^{-x}+x\lt 1$ for $x\in (0,\infty )$ is incorrect. LHS is increasing and unbounded function. (B) Given${\psi }_{2}\left(x\right)={x}^{2}-2x-2{e}^{-x}+2,x\geq 0$ ${{\psi }^{'}}_{2}\left(x\right)=2x-2+2{e}^{-x},x\geq 0$ ${{\psi }^{'}}_{2}\left(x\right)=2{\psi }_{1}\left(x\right)-2\geq 0,x\geq 0$ So, ${\psi }_{2}\left(x\right)$ is increasing function. ${x}^{2}-2x-2{e}^{-x}+2\lt 1$ for $x\in (0,\infty )$ is incorrect because LHS $\rightarrow \infty$ when $x\rightarrow \infty$ (C) $f\left(x\right)=2{\int }_{0}^{x}\left(t-{t}^{2}\right){e}^{-{t}^{2}}dt$ $f\left(x\right)={\left(-{e}^{-{t}^{2}}\right)}_{0}^{x}-2{\int }_{0}^{x}{t}^{2}{e}^{-{t}^{2}}dt$ $=1-{e}^{-{x}^{2}}-2{\int }_{0}^{x}{t}^{2}\left(1-{t}^{2}+\dfrac{{t}^{4}}{2!}\cdots ..\right)$ $=f\left(x\right)-1+{e}^{-{x}^{2}}+\dfrac{2{x}^{3}}{3}-\dfrac{2{x}^{5}}{5}\lt 0$ in $\left(0,\dfrac{1}{2}\right)$ $g\left(x\right)={\int }_{0}^{{x}^{2}}\sqrt{t}{e}^{-1}dt$ Put $t={z}^{2}$ $g\left(x\right)={\int }_{0}^{x}2{z}^{2}{e}^{-{z}^{2}}dz$ $={\int }_{0}^{x}2{z}^{2}\left(1-{z}^{2}+\dfrac{{z}^{4}}{2!}\ldots ..\right)dz$ $g\left(x\right)=\dfrac{2{x}^{3}}{3}-\dfrac{2{x}^{5}}{5}+\dfrac{2{x}^{7}}{2.7}-\dfrac{2\cdot {x}^{9}}{9.6}-\ldots \ldots$ $g\left(x\right)-\dfrac{2{x}^{3}}{3}+\dfrac{2{x}^{5}}{5}-\dfrac{2{x}^{7}}{2.7}\leq 0$ $=1-{e}^{-{x}^{2}}-\dfrac{2{x}^{3}}{3}+\dfrac{2{x}^{5}}{5}-\dfrac{2{x}^{7}}{2}+\dfrac{2{x}^{9}}{6}\ldots \ldots \ldots$ Now $f\left(x\right)+g\left(x\right)=1-{e}^{-{x}^{2}}$ $\Rightarrow f\left(x\right)=1-{e}^{-{x}^{2}}-g\left(x\right)$ $\Rightarrow f\left(x\right)\leq 1-{e}^{-{x}^{2}}-\left(\dfrac{2}{3}{x}^{3}-\dfrac{2}{3}{x}^{5}\right)$ (D) $g\left(x\right)={\int }_{0}^{{x}^{2}}\sqrt{t}{e}^{-t}dt$ $\Rightarrow g\left(x\right)\leq {\int }_{0}^{{x}^{2}}\sqrt{t}\left(1-t+\dfrac{{t}^{2}}{2}\right)dt\Rightarrow g\left(x\right)\leq \dfrac{2}{3}{x}^{3}-\dfrac{2}{5}{x}^{5}+\dfrac{1}{7}{x}^{7}$ $g\left(x\right)\geq {\int }_{0}^{{x}^{2}}\sqrt{t}\left(1-t\right)dt\Rightarrow g\left(x\right)\geq \dfrac{2}{3}{x}^{3}-\dfrac{2}{5}{x}^{5}$