JEE Main 2022 — Definite Integration Question with Solution

$$f(x)={\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt$$

$$f^{\prime }(x)=2x\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)=0$$

$$x=0$$, or $$(x^2-4)(x^2-1)=0$$

$$x=0,$$ $$x=\pm 2,\pm 1$$

Now, $$f^{\prime }(x)=\frac{2x(x+1)(x-1)(x+2)(x-2)}{\left(e^{x^2}+2\right)}$$

f'(x) changes sign from positive to negative at x = $$-$$1, 1 So, number of local maximum points = 2

f'(x) changes sign from negative to positive at x = $$-$$2, 0, 2 So, number of local minimum points = 3

$$\therefore$$ m = 2, n = 3