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
- -14
- 26
- -16
- 36
The Correct Option is B
Solution and Explanation
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 \( \alpha + \beta = 26 \).
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