If ( x ) = 1 x / 4 x ( 4 2 sin t - 3 ( t ) ) d t , x 0 , then ( 4 ) is equal to [2023]

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Q.
If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t, x > 0,$ then $ϕ' (\frac{π}{4})$ is equal to [2023]

1 $\frac{4}{6 + \sqrt{π}}$

2 $\frac{8}{6 + \sqrt{π}}$

3 $\frac{8}{\sqrt{π}}$

4 $\frac{4}{6 - \sqrt{π}}$

Ans.
(2)

By Leibnitz rule,

$ϕ' (x) = \frac{1}{\sqrt{x}} [(4 \sqrt{2} \sin x - 3 ϕ' (x)) \cdot 1 - 0] - \frac{1}{2} x^{- 3 / 2} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t$

Put $x = \frac{π}{4}$ , we get $ϕ' (\frac{π}{4}) = \frac{2}{\sqrt{π}} [4 - 3 ϕ' (\frac{π}{4})] + 0$

$\Rightarrow (1 + \frac{6}{\sqrt{π}}) ϕ' (\frac{π}{4}) = \frac{8}{\sqrt{π}}$ $= \frac{8}{\sqrt{π}} \times (\frac{\sqrt{π}}{6 + \sqrt{π}}) = \frac{8}{6 + \sqrt{π}}$

Solution

Q.
If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t, x > 0,$ then $ϕ' (\frac{π}{4})$ is equal to [2023]

1 $\frac{4}{6 + \sqrt{π}}$

2 $\frac{8}{6 + \sqrt{π}}$

3 $\frac{8}{\sqrt{π}}$

4 $\frac{4}{6 - \sqrt{π}}$

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Solution

Q. If ϕ(x)=1x∫π/4x(42sint-3ϕ'(t))dt,x>0, then ϕ'(π4) is equal to [2023]

1 46+π

2 86+π

3 8π

4 46-π

Ans. (2) By Leibnitz rule, ϕ'(x)=1x[(42sinx-3ϕ'(x))·1-0]-12x-3/2∫π/4x(42sint-3ϕ'(t))dt Put x=π4, we get ϕ'(π4)=2π[4-3ϕ'(π4)]+0 ⇒(1+6π)ϕ'(π4)=8π=8π×(π6+π)=86+π

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Q.
If $ϕ (x) = \frac{1}{\sqrt{x}} \int_{π / 4}^{x} (4 \sqrt{2} \sin t - 3 ϕ' (t)) d t, x > 0,$ then $ϕ' (\frac{π}{4})$ is equal to [2023]

1 $\frac{4}{6 + \sqrt{π}}$

2 $\frac{8}{6 + \sqrt{π}}$

3 $\frac{8}{\sqrt{π}}$

4 $\frac{4}{6 - \sqrt{π}}$