JEE Main 2025 Question solution.

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JEE Main 2025 (22 Jan Shift-2) – Question 8

Question:

\[ f(x)=\int_{0}^{x^{2}} \frac{t^{2}-8t+15}{e^{t}}\,dt,\quad x\in\mathbb{R}. \]

Then the numbers of local maximum and local minimum points of \(f\), respectively, are:

Solution:

Using Fundamental Theorem of Calculus and chain rule,

\[ f'(x)=\frac{t^{2}-8t+15}{e^{t}}\Big|_{t=x^{2}}\cdot 2x\] \[ =2x\,\frac{x^{4}-8x^{2}+15}{e^{x^{2}}}. \]

Since \(e^{x^{2}}>0\) for all \(x\), critical points are obtained from

\[ 2x(x^{4}-8x^{2}+15)=0. \]

\[ x=0 \] and \[ x^{4}-8x^{2}+15=0. \]

Let \(y=x^{2}\). Then \[ y^{2}-8y+15=0 \] \[\Rightarrow (y-3)(y-5)=0. \]

\[ x=\pm\sqrt{3},\;\pm\sqrt{5}. \]

Thus the critical points are \[ x=0,\;\pm\sqrt{3},\;\pm\sqrt{5}. \]

Now, \[ f'(x)=\frac{2x(x^{2}-3)(x^{2}-5)}{e^{x^{2}}}. \]

From sign analysis of \(f'(x)\):

• Local minima at \(x=-\sqrt{5},\,-\sqrt{3},\,\sqrt{3}\)
• Local maxima at \(x=0,\,\sqrt{5}\)

Number of local maxima = 2
Number of local minima = 3

Correct Answer: (1) 2 and 3