ધારો કે f(x) = frac{sin x + cos x - sqrt{2}}{ | vedclass

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(B) આપેલ છે $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$.અંશ અને છેદને $\sqrt{2}$ વડે ભાગતા,$f(x) = \frac{\frac{1}{\sqrt{2}}\sin x + \f

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(B) આપેલ છે $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$.અંશ અને છેદને $\sqrt{2}$ વડે ભાગતા,$f(x) = \frac{\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x - 1}{\frac{1}{\sqrt{2}}\sin x - \frac{1}{\sqrt{2}}\cos x} = \frac{\sin(x + \frac{\pi}{4}) - 1}{\sin(x - \frac{\pi}{4})}$.$\sin 	heta - 1 = -2 \sin^2(\frac{\pi}{4} - \frac{	heta}{2})$ અને $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$ નો ઉપયોગ કરીને,$f(x) = 	an(\frac{\pi}{8} - \frac{x}{2})$ મળે.તેથી,$f(x) = -	an(\frac{x}{2} - \frac{\pi}{8})$.$f'(x) = -\frac{1}{2} \sec^2(\frac{x}{2} - \frac{\pi}{8})$.$f''(x) = -\frac{1}{2} \cdot 2 \sec(\frac{x}{2} - \frac{\pi}{8}) \cdot \sec(\frac{x}{2} - \frac{\pi}{8}) 	an(\frac{x}{2} - \frac{\pi}{8}) \cdot \frac{1}{2} = -\frac{1}{2} \sec^2(\frac{x}{2} - \frac{\pi}{8}) 	an(\frac{x}{2} - \frac{\pi}{8})$.$x = \frac{7\pi}{12}$ માટે,$\frac{x}{2} - \frac{\pi}{8} = \frac{7\pi}{24} - \frac{3\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$.$f(\frac{7\pi}{12}) = -	an(\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$.$f''(\frac{7\pi}{12}) = -\frac{1}{2} \sec^2(\frac{\pi}{6}) 	an(\frac{\pi}{6}) = -\frac{1}{2} \cdot (\frac{2}{\sqrt{3}})^2 \cdot \frac{1}{\sqrt{3}} = -\frac{1}{2} \cdot \frac{4}{3} \cdot \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}}$.તેથી,$f(\frac{7\pi}{12}) f''(\frac{7\pi}{12}) = (-\frac{1}{\sqrt{3}}) \cdot (-\frac{2}{3\sqrt{3}}) = \frac{2}{3 \cdot 3} = \frac{2}{9}$.

ધારો કે $f(x) = \frac{\sin x + \cos x - \sqrt{2}}{\sin x - \cos x}$,$x \in [0, \pi] - \{\frac{\pi}{4}\}$. તો $f(\frac{7\pi}{12}) f''(\frac{7\pi}{12})$ ની કિંમત શોધો.

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