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 / 6 and / 4 , respectively with the positive x -axis. If 27 1 3 ( ( f ' ( t )

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Solution

Q.
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 $π / 6$ and $π / 4,$ respectively with the positive $x$ -axis. If $27 \int_{1}^{3} ((f' {(t))}^{2} + 1) f'' (t) d t = α + β \sqrt{3}$ where $α, β$ are integers, then the value of $α + β$ equals [2024]

1 - 14

2 36

3 - 16

4 26

Ans.
(4)

We have, $f' (1) = \tan \frac{π}{6} = \frac{1}{\sqrt{3}}$

$f' (3) = \tan \frac{π}{4} = 1$

Now, let $I = \int_{1}^{3} ((f' {(t))}^{2} + 1) f'' (t) d t$

where $f' (t) = u$

$\Rightarrow f'' (t) d t = d u$

$∴ I = \int_{1 / \sqrt{3}}^{1} (u^{2} + 1) d u$

$= {(\frac{u^{3}}{3} + u)}_{1 / \sqrt{3}}^{1} = \frac{4}{3} - \frac{10}{9 \sqrt{3}} = \frac{4}{3} - \frac{10 \sqrt{3}}{27}$

$\Rightarrow 27 I = 36 - 10 \sqrt{3}$

$∴ α = 36 and β = - 10$

$∴ α + β = 36 - 10 = 26$

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