If -1sin xt2ftdt-1-sin x - where x -0-2- then the value of f1-3 is

∫^1_sin xt^2f(t)dt=1 sin x Differentiating both sides using Leibnitz theorem, we get: 1^2× f(1)0 sin^2 xf(sin x)cos x= cos x ⇒ f(sin x)=1sin^2x nbsp; ∴ f(1√(3 ...

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Standard XII

Mathematics

Property 5

If ∫1sin xt2f...

Question

If $1 \int sin x t^{2} f (t) d t = 1 - sin x,$ where $x \in (0, \frac{π}{2}),$ then the value of $f (\frac{1}{\sqrt{3}})$ is

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Solution

The correct option is D $3$
$1 \int sin x t^{2} f (t) d t = 1 - sin x$
Differentiating both sides using Leibnitz theorem, we get:
$1^{2} \times f (1) 0 - {sin}^{2} x f (sin x) cos x = - cos x$
$\Rightarrow f (sin x) = \frac{1}{{sin}^{2} x} ∴ f (\frac{1}{\sqrt{3}}) = 3$

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Property 5

Standard XII Mathematics

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If 1∫sinxt2f(t)dt=1−sinx, where x∈(0,π2), then the value of f(1√3) is

The correct option is D 31∫sinxt2f(t)dt=1−sinx Differentiating both sides using Leibnitz theorem, we get: 12×f(1)0−sin2xf(sinx)cosx=−cosx ⇒f(sinx)=1sin2x∴f(1√3)=3

If $1 \int sin x t^{2} f (t) d t = 1 - sin x,$ where $x \in (0, \frac{π}{2}),$ then the value of $f (\frac{1}{\sqrt{3}})$ is

The correct option is D $3$
$1 \int sin x t^{2} f (t) d t = 1 - sin x$
Differentiating both sides using Leibnitz theorem, we get:
$1^{2} \times f (1) 0 - {sin}^{2} x f (sin x) cos x = - cos x$
$\Rightarrow f (sin x) = \frac{1}{{sin}^{2} x} ∴ f (\frac{1}{\sqrt{3}}) = 3$