Application of Derivatives JEE Main PYQs (2020–2026)

Solved Application of Derivatives PYQs

3 recent questions with answers. The full set of 157 is free to practise.

1. JEE Main 2026MediumMaxima and Minima (Local & Absolute)
   Q1

   Let $f(x)$ be a polynomial of degree 5, and have extrema at $x=1$ and $x=-1$ . If ${\lim }_{x\to 0}\left(\frac{f(x)}{x^3}\right)=-5$ , then $f(2)-f(-2)$ is equal to:

   • A0
   • B50
   • C92
   • D112

   Show answer & solution

   Correct answer: (D)

   Given $f(x)$ is a polynomial of degree $5$
   
   Also, $f^{\prime }(1)=0$; $f^{\prime }(-1)=0$.
   
   Also $\because {\lim }_{x\to 0}\frac{f(x)}{x^3}=-5$ (fixed and finite)
   
   $\therefore f(0)=0$; $f^{\prime }(0)=0$; $f^{\prime \prime }(0)=0$
   
   $$\frac{f^{\prime \prime \prime }(0)}{6}=-5\to f^{\prime \prime \prime }(0)=-30$$
   
   Hence, let
   
   $$f^{\prime }(x)=(ax+b)(x-0)(x-1)(x+1)$$
   
   $$\to f^{\prime }(x)=a{x}^4+b{x}^3-a{x}^2-bx$$
   
   $$f^{\prime \prime }(x)=4a{x}^3+3b{x}^2-2ax-b$$
   
   $$\therefore f^{\prime \prime }(0)=0\to b=0$$
   
   $$f^{\prime \prime \prime }(x)=12a{x}^2+6bx-2a$$
   
   $$\therefore f^{\prime \prime \prime }(0)=-2a=-30$$
   
   $$\to a=15$$
   
   $$\therefore f^{\prime }(x)=15x.x.(x-1)(x+1)$$
   
   $$\to f^{\prime }(x)=15x^4-15x^2$$
   
   $$\to f(x)=3x^5-5x^3+C$$
   
   $$\therefore f(0)=0\to C=0$$
   
   $$\therefore f(x)=3x^5-5x^3$$
   
   $$\therefore f(2)-f(-2)=112$$
2. JEE Main 2026MediumIncreasing and Decreasing Functions (Monotonicity)
   Q2

   Let $f:\mathbb{R}\to \mathbb{R}$ be a twice differentiable function such that $f^{\prime \prime }(x)>0$ for all $x\in \mathbb{R}$ and $f^{\prime }(a-1)=0$, where $a$ is a real number.
   
   Let $g(x)=f({\tan }^{2}x-2\tan x+a),0<x<\frac{\pi }{2}$.
   
   Consider the following two statements:
   
   g is increasing in $\left(0,\frac{\pi }{4}\right)$
   
   g is decreasing in $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$
   
   Then,

   • ABoth (I) and (II) are True
   • BNeither (I) nor (II) is True
   • COnly (I) is True
   • DOnly (II) is True

   Show answer & solution

   Correct answer: (B)

   $$\begin{array}{rl}g(x)& =f\left((tanx-1{)}^2+a-1\right)\\ g^{\prime }(x)=f^{\prime }\left((tanx-1{)}^2+a-1\right)\cdot & \\ 2(tanx-1)\cdot {\sec }^{2}x....(i)& \end{array}$$
   
   Given $f^{{}^{\prime \prime }}(x)>0\forall x\in R$
   
   $\therefore f^{\prime }(x)$ is increasing function
   
   Case (i) $0<x<\frac{\pi }{4}$
   
   $$\begin{array}{rl}& 0<tanx>1\\ & -1<tanx-1<0\\ & 0<(tanx-1{)}^2<1\\ & a-1<(tanx-1{)}^2+(a-1)<a\\ & \therefore f^{\prime }\left((tanx-1{)}^2+a-1\right)>f^{\prime }(a-1)=0\end{array}$$
   
   From (i) $g^{\prime }(x)=(+)(-)(+)=-ve$
   
   $g(x)$ is decreasing in $\left(0,\frac{\pi }{4}\right)$
   
   Similarly we can prove that $g(x)$ is increasing in $\left(\frac{\pi }{4},\frac{\pi }{2}\right)$
3. JEE Main 2025HardGeometric and Word Problems on Optimization
   Q3

   The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is:

   • A$2\sqrt{3}-1$
   • B$2\sqrt{2}-1$
   • C$3\sqrt{2}-1$
   • D$\sqrt{2}$

   Show answer & solution

   Correct answer: (B)

   Equation of the Normal to the Parabola:
   
   The equation for the normal to the parabola $y^2=8x$ can be expressed as:
   
   $$y=mx-2am-a{m}^3\text{where }a=2$$
   
   Simplifying, we have:
   
   $$y=mx-4m-2m^3$$
   
   Center and Radius of the Circle:
   
   The given second equation $x^2+y^2+12y+35=0$ can be rewritten to find the center and radius of the circle:
   
   Rewrite it as $x^2+(y+6{)}^2=1$ .
   
   Thus, the center of the circle is $C(0,-6)$ and the radius $r=1$ .
   
   Finding the Slope of the Normal:
   
   To find the point $P$ on the parabola where the normal meets, equate:
   
   $$-6=-4m-2m^3$$
   
   Solving for $m$ gives:
   
   $$m=1$$
   
   Determine Point $P$ on the Parabola:
   
   Substituting $m=1$ back to find $P\left(a{m}^2,-2am\right)$ :
   
   $$P(2,-4)$$
   
   Calculate the Shortest Distance:
   
   The shortest distance from point $P(2,-4)$ to the center $C(0,-6)$ minus the radius is:
   
   $$CP-r=\sqrt{(2-0{)}^2+(-4+6{)}^2}-1=\sqrt{4+4}-1$$
   
   $$=\sqrt{8}-1=2\sqrt{2}-1$$
   
   Thus, the shortest distance between the given curves is $2\sqrt{2}-1$ .

Topic-wise question count

Where JEE Main has asked from Application of Derivatives, most frequent first.

• Maxima and Minima (Local & Absolute)56
• Increasing and Decreasing Functions (Monotonicity)38
• Tangents and Normals to Curves27
• Geometric and Word Problems on Optimization18
• Rate of Change of Quantities9
• Rolle's and Lagrange's Mean Value Theorems6
• Point of Inflection, Concavity & Curve Tracing2

Year-wise Application of Derivatives questions in JEE Main

Number of questions from this chapter in each year's paper(s).