If ϕ(x)=1√x∫(π)/4x4√2sint-3ϕ^(')(t)dt,x>0 then ϕ^(')(π)/4 is equal to

Correct answer: (B) 8/(6+√(π)). Given, ϕ(x)=(1√x∫π/4x4√2sint-3ϕ^(')(t)dtDifferentiate both sides w.r.t. x we get,ϕ^(')(x)=-12x^(3/2)∫π/4x4√2sint-3ϕ^(')(t)dt+1√x4√2sinx-3ϕ^(')(x)Put x=π4 ϕ^(')(π4)=-12π4^(3/2)×0+√(4π)4√2×1√2-3ϕ^(')(π4)⇒√(π4)ϕ^(')(π4)+3ϕ^(')(π4)=4⇒ϕ^(')π43+√(π4)=4⇒ϕ^(')π4=8)/(6+√(π))

Q44 / 90 · JEE Main 2023 (31 Jan Shift 2) · Mathematics

If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t, x > 0$ then $ϕ^{'} (\frac{π}{4})$ is equal to

A
$\frac{4}{6 + \sqrt{π}}$
Correct answer
B
$\frac{8}{6 + \sqrt{π}}$
C
$\frac{8}{\sqrt{π}}$
D
$\frac{4}{6 - \sqrt{π}}$

Solution

Given,

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

Differentiate both sides w.r.t. $x$ we get,

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

Put $x = \frac{π}{4}$

$ϕ^{'} (\frac{π}{4}) = \frac{- 1}{2 {(\frac{π}{4})}^{3 / 2}} \times 0 + \sqrt{\frac{4}{π}} (4 \sqrt{2} \times \frac{1}{\sqrt{2}} - 3 ϕ^{'} (\frac{π}{4}))$

$\Rightarrow \sqrt{\frac{π}{4}} ϕ^{'} (\frac{π}{4}) + 3 ϕ^{'} (\frac{π}{4}) = 4$

$\Rightarrow ϕ^{'} (\frac{π}{4}) (3 + \sqrt{\frac{π}{4}}) = 4$

$\Rightarrow ϕ^{'} (\frac{π}{4}) = \frac{8}{6 + \sqrt{π}}$

Hence, the correct answer is option: 'B'.

If ϕ(x)=1x∫π4x42sint-3ϕ'(t)dt,x>0 then ϕ'π4 is equal to

Solution

Relevant Questions

If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{\frac{π}{4}}^{x} (4 \sqrt{2} \sin t - 3 ϕ^{'} (t)) d t, x > 0$ then $ϕ^{'} (\frac{π}{4})$ is equal to