Let f : (-1, 1) to R be a continuous function | vedclass

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(D) Given that $f : (-1, 1) 	o R$ is a continuous function and $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$.Applying the Leibniz rule for

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$\sqrt{3}$

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(D) Given that $f : (-1, 1) 	o R$ is a continuous function and $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$.Applying the Leibniz rule for differentiation under the integral sign on both sides with respect to $x$:$\frac{d}{dx} \left( \int\limits_0^{\sin x} {f(t)dt} ight) = \frac{d}{dx} \left( \frac{\sqrt{3}}{2}x ight)$$f(\sin x) \cdot \frac{d}{dx}(\sin x) = \frac{\sqrt{3}}{2}$$f(\sin x) \cdot \cos x = \frac{\sqrt{3}}{2}$To find $f\left(\frac{\sqrt{3}}{2}ight)$,we set $\sin x = \frac{\sqrt{3}}{2}$.This implies $x = \frac{\pi}{3}$ (since $x \in (-1, 1)$ is satisfied by the domain of $\sin x$ and the integral).Substituting $x = \frac{\pi}{3}$ into the equation:$f\left(\sin \frac{\pi}{3}ight) \cdot \cos \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$f\left(\frac{\sqrt{3}}{2}ight) \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$Multiplying both sides by $2$,we get:$f\left(\frac{\sqrt{3}}{2}ight) = \sqrt{3}$

Let $f : (-1, 1) \to R$ be a continuous function. If $\int\limits_0^{\sin x} {f(t)dt} = \frac{\sqrt{3}}{2}x$,then $f\left(\frac{\sqrt{3}}{2}\right)$ is equal to

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