QJEEM25Mad6ab: JEE Main 2025 Mathematics - Class 12 JEEM25Mad6ab Solution: Let . Then the numbers of local maximum and local minimum…

We are given

$f(x)=\int_0^{x^2} \frac{t^2-8t+15}{e^t}\,dt, \quad x\in\mathbb{R}.$

To find the local extrema, we first compute the derivative using the Fundamental Theorem of Calculus and the chain rule.

Step 1. Rewrite the derivative:

Let

$F(u)=\int_0^u \frac{t^2-8t+15}{e^t}\,dt,$

so that

$f(x)=F(x^2).$

Then by the chain rule,

$f'(x)=F'(x^2)\cdot 2x.$

Since

$F'(u)=\frac{u^2-8u+15}{e^u},$

we substitute $u=x^2$ to get

$f'(x)=\frac{(x^2)^2-8x^2+15}{e^{x^2}}\cdot 2x = \frac{2x\,(x^4-8x^2+15)}{e^{x^2}}.$

Step 2. Factor and Identify Critical Points:

Factor the polynomial $x^4-8x^2+15$ by writing it in terms of $x^2$ . Let $y=x^2$ , then

$y^2-8y+15=(y-3)(y-5)$

so that

$x^4-8x^2+15=(x^2-3)(x^2-5).$

Thus,

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

Since e^{x^2}>0 for all $x$ , the zeros of $f'(x)$ are determined by

$2x\,(x^2-3)(x^2-5)=0.$

That gives the critical points:

$2x=0 \Rightarrow x=0$ ,

$x^2-3=0 \Rightarrow x=\pm\sqrt{3}$ ,

$x^2-5=0 \Rightarrow x=\pm\sqrt{5}$ .

Step 3. Analyzing the Sign of $f'(x)$ :

We need to determine the nature (maximum or minimum) by looking at the sign changes of $f'(x)$ on the intervals determined by the critical points $x=-\sqrt{5}$ , $x=-\sqrt{3}$ , $x=0$ , $x=\sqrt{3}$ , and $x=\sqrt{5}$ . Notice that the factor $2x\,(x^2-3)(x^2-5)$ will dictate the sign.

Let’s define:

$h(x)=x\,(x^2-3)(x^2-5).$

Examine the sign of $h(x)$ in each interval:

For x < -\sqrt{5} :

$x$ is negative.

x^2>5 so x^2-3>0 and x^2-5>0 .

Product: negative $\times$ positive $\times$ positive = negative.

Thus, f'(x)<0 .

For -\sqrt{5} < x < -\sqrt{3} :

$x$ is negative.

$x^2$ is between 3 and 5 so x^2-3>0 while x^2-5<0 .

Product: negative $\times$ positive $\times$ negative = positive.

Thus, f'(x)>0 .

For -\sqrt{3} < x < 0 :

$x$ is negative.

x^2<3 so both x^2-3<0 and x^2-5<0 .

Product: negative $\times$ negative $\times$ negative = negative.

Thus, f'(x)<0 .

For 0 < x < \sqrt{3} :

$x$ is positive.

x^2<3 so both x^2-3<0 and x^2-5<0 .

Product: positive $\times$ negative $\times$ negative = positive.

Thus, f'(x)>0 .

For \sqrt{3} < x < \sqrt{5} :

$x$ is positive.

$x^2$ is between 3 and 5 so x^2-3>0 while x^2-5<0 .

Product: positive $\times$ positive $\times$ negative = negative.

Thus, f'(x)<0 .

For x > \sqrt{5} :

$x$ is positive.

x^2>5 so both x^2-3>0 and x^2-5>0 .

Product: positive $\times$ positive $\times$ positive = positive.

Thus, f'(x)>0 .

Step 4. Classify the Critical Points:

At $x=-\sqrt{5}$ : $f'(x)$ changes from negative to positive → local minimum.

At $x=-\sqrt{3}$ : $f'(x)$ changes from positive to negative → local maximum.

At $x=0$ : $f'(x)$ changes from negative to positive → local minimum.

At $x=\sqrt{3}$ : $f'(x)$ changes from positive to negative → local maximum.

At $x=\sqrt{5}$ : $f'(x)$ changes from negative to positive → local minimum.

Step 5. Conclusion:

There are local maximum points at $x=-\sqrt{3}$ and $x=\sqrt{3}$ (2 points in total), and local minimum points at $x=-\sqrt{5}$ , $x=0$ , and $x=\sqrt{5}$ (3 points in total).

Thus, the numbers of local maximum and local minimum points of $f$ are 2 and 3, respectively.

The correct option is Option C.