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

The Correct Option is B

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

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

\(x=0\), 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 = \(3\)

\(∴ m = 2, n = 3\)