Let \(y=f(x)\) be a thrice differentiable function in ( \(-5,5\) ). Let the tangents to the curve \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) at \((1, f(1))\) and ( \(3, f(3)\) ) make angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\), respectively with positive x -axis. If \(27 \int_{1}^{3}\left(\left(f^{\prime}(t)\right)^{2}+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}\) where \(\alpha, \beta\) are integers, then the value of \(\alpha+\beta\) equals

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Definite Integrals

Let

y = f (x)

be a thrice differentiable function in (

- 5, 5

). Let the tangents to the curve

y = f (x)

(1, f (1))

and (

3, f (3)

) make angles

\frac{π}{6}

and

\frac{π}{4}

, respectively with positive x -axis. If

27 \int_{1}^{3} ((f^{'} (t))^{2} + 1) f^{''} (t) d t = α + β 3

where

α, β

are integers, then the value of

α + β

equals

Answer: (B)

The tangent slopes give

f^{'} (1) = tan \frac{π}{6} = \frac{1}{3}

and

f^{'} (3) = tan \frac{π}{4} = 1

.

Put

z = f^{'} (t)

, so

d z = f^{''} (t) d t

. Then

27 \int_{1}^{3} ((f^{'} (t))^{2} + 1) f^{''} (t) d t = 27 \int_{1/ 3}^{1} (z^{2} + 1) d z = 27 [\frac{z ^{3}}{3} + z]_{1/ 3}^{1}

.

At

z = 1

\frac{1}{3} + 1 = \frac{4}{3}

.
At

z = \frac{1}{3}

\frac{1}{9 3} + \frac{1}{3} = \frac{1 + 9}{9 3} = \frac{10}{9 3} = \frac{10 3}{27}

.

So the integral equals

\frac{4}{3} - \frac{10 3}{27}

, and multiplying by 27:

27 \times \frac{4}{3} - 103 = 36 - 103

.

Hence

α = 36

β = - 10

, and

α + β = 26

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option