Question

Let $y = f(x)$ be a thrice differentiable function in $(-5, 5)$. Let the tangents to the curve $y = f(x)$ at $(1, f(1))$ and $(3, f(3))$ make angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$, respectively, with the positive x-axis. If $2 \int_{\frac{1}{\sqrt{3}}}^{1} \left( \left( f'(t) \right)^2 + 1 \right) f''(t) \, dt = \alpha + \beta \sqrt{3}$ where $\alpha$, $\beta$ are integers, then the value of $\alpha + \beta$ equals

• A. -14
• B. 26
• C. -16
• D. 36

Answer 1

The Correct Option is B

From the tangents, we find:

$$f'(1) = \frac{1}{\sqrt{3}}, \quad f'(3) = 1.$$

Assume $f'(t) = t$ (consistent with the slopes), then $f''(t) = 1$.

Substitute into the integral:

$$2 \int_{\frac{1}{\sqrt{3}}}^{1} \left( t^2 + 1 \right) dt = \alpha + \beta \sqrt{3}.$$

$$\alpha + \beta \sqrt{3} = 27 \left( \frac{4}{3} - \frac{10}{27} \sqrt{3} \right) = 36 - 10\sqrt{3}.$$

Here $\alpha = 36, \beta = -10$.

$$\alpha + \beta = 36 - 10 = 26$$

Compute the integral to find $\alpha + \beta = 26$.