જો g(x) = int_{sin x}^{sin(2x)} sin^{-1}(t) , | vedclass

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(C) આપેલ છે $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$. લેબનીઝના નિયમનો ઉપયોગ કરતા: $g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot \frac{d}{dx}(

Vedclass · Accepted Answer

$g^{\prime}\left(\frac{\pi}{2}\right) = 0$

Vedclass · Answer

(C) આપેલ છે $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$. લેબનીઝના નિયમનો ઉપયોગ કરતા: $g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot \frac{d}{dx}(\sin(2x)) - \sin^{-1}(\sin x) \cdot \frac{d}{dx}(\sin x)$. $g^{\prime}(x) = \sin^{-1}(\sin(2x)) \cdot (2\cos(2x)) - \sin^{-1}(\sin x) \cdot (\cos x)$. $x = \frac{\pi}{2}$ માટે: $g^{\prime}\left(\frac{\pi}{2}\right) = \sin^{-1}(\sin(\pi)) \cdot (2\cos(\pi)) - \sin^{-1}(\sin(\frac{\pi}{2})) \cdot (\cos(\frac{\pi}{2}))$. $\sin(\pi) = 0$ હોવાથી,$\sin^{-1}(0) = 0$. $\cos(\frac{\pi}{2}) = 0$ હોવાથી,બીજું પદ પણ $0$ થશે. આમ,$g^{\prime}\left(\frac{\pi}{2}\right) = 0 \cdot (-2) - (\frac{\pi}{2}) \cdot 0 = 0$. તેથી,વિકલ્પ $C$ સાચો છે.

જો $g(x) = \int_{\sin x}^{\sin(2x)} \sin^{-1}(t) \, dt$ હોય,તો

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