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

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

cdquestions Admin · Accepted Answer

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 $.

Answer

Answer

Answer

Answer

The Correct Option is B

Solution and Explanation

Top Questions on integral

Questions Asked in JEE Main exam