Question

If m and n respectively are the number of local maximum and local minimum points of the function
$f(x)=∫_0 ^{x^2} \frac{t^2–5t+4}{2+et}dt$, then the ordered pair (m, n) is equal to

• A. (3, 2)
• B. (2, 3)
• C. (2, 2)
• D. (3, 4)

Answer 1

Answer: (B)

(2, 3)

Answer 2

$f(x)=∫_0 ^{x^2} \frac{t^2–5t+4}{2+et}dt$

$f'(x)=2x\bigg(\frac{x^4–5x^2+4}{2+ex^2}\bigg)=0$

$x=0$, or $(x^2–4)(x^2–1)=0$

$x = 0, x = ±2, ±1$

Now,

$f'(x)=\frac{2x(x+1)(x-1)(x+2)(x-2)}{(ex^2+2)}$

changes sign from positive to negative at

$x = –1$, $1$ So, number of local maximum points = $2$

changes sign from negative to positive at

$x = –2, 0, 2$

Hence, number of local minimum points = $3$

$∴ m = 2, n = 3$