If $\phi(\mathrm{x})=\frac{1}{\sqrt{\mathrm{x}}} \int_{\frac{\pi}{4}}^{\mathrm{x}}\left(4 \sqrt{2} \sin \mathrm{t}-3 \phi^{\prime}(\mathrm{t})\right) \mathrm{dt}, \mathrm{x}>0$, then $\phi^{\prime}\left(\frac{\pi}{4}\right)$ is equal to :

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Definite Integrals

ϕ (x) = \frac{1}{x} \int_{\frac{π}{4}}^{x} (42 sin t - 3 ϕ^{'} (t)) dt, x > 0

, then

ϕ^{'} (\frac{π}{4})

is equal to :

Answer: (C)

Writing

ϕ (x) = x^{- 1/2} \int_{π /4}^{x} (42 sin t - 3 ϕ^{'} (t)) d t

and differentiating by the product and Leibniz rules,

ϕ^{'} (x) = x^{- 1/2} (42 sin x - 3 ϕ^{'} (x)) - \frac{1}{2} x^{- 3/2} \int_{π /4}^{x} (42 sin t - 3 ϕ^{'} (t)) d t .

x = \frac{π}{4}

the integral is zero and

(\frac{π}{4})^{- 1/2} = \frac{2}{π}

, while

42 sin \frac{π}{4} = 4

, so

ϕ^{'} (\frac{π}{4}) = \frac{2}{π} (4 - 3 ϕ^{'} (\frac{π}{4})) .

Collecting the term,

(1 + \frac{6}{π}) ϕ^{'} (\frac{π}{4}) = \frac{8}{π}

, hence

ϕ^{'} (\frac{π}{4}) = \frac{8}{6 + π} .

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option