Question

The points of extremum of \[ \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \, dt \] are:

• A. ±1
• B. ±2
• C. ±3
• D. \(±\sqrt2\)

Hint:

For integrals with variable limits, apply the Leibniz rule to find derivatives and set the resulting equation to zero for extremum points.

Answer 1

The Correct Option is A, B

1. The given function is defined as:

\[F(x) = \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} dt.\]

2. Differentiate \(F(x)\) with respect to \(x\) using the Leibniz rule:

\[F'(x) = \frac{d}{dx}\left(\int_{0}^{x^2} f(t) dt\right) = f(x^2) \cdot \frac{d}{dx}(x^2) = f(x^2) \cdot 2x.\]

Where,

\[f(t) = \frac{t^2 - 5t + 4}{2 + e^t}.\]

3. To find the extremum, set \(F'(x) = 0\):

\[f(x^2) \cdot 2x = 0.\]

This leads to two cases: \(x = 0\) (which is not a valid extremum as it's on the boundary) and \(f(x^2) = 0.\)

4. Solve \(f(x^2) = 0\):

\[\frac{t^2 - 5t + 4}{2 + e^t} = 0 \Rightarrow t^2 - 5t + 4 = 0.\]

5. Factorize \(t^2 - 5t + 4 = 0\):

\[(t - 1)(t - 4) = 0 \Rightarrow t = 1, t = 4.\]

6. Since \(t = x^2\), we obtain:

\[x^2 = 1 \Rightarrow x = \pm 1\]

\[x^2 = 4 \Rightarrow x = \pm 2.\]

Therefore, the points of extremum are \(\pm 1\) and \(\pm 2\).