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 \( \alpha + \beta = 26 \).