A) \[y-\sqrt{2}=2\sqrt{2}\left( x-\frac{\pi }{4} \right)\]
B) \[y+\sqrt{2}=\sqrt{2}\left( x+\frac{\pi }{4} \right)\]
C) \[y-\sqrt{2}=-\sqrt{2}\left( x-\frac{\pi }{4} \right)\]
D) \[y-\sqrt{2}=\sqrt{2}\left( x-\frac{\pi }{4} \right)\]
Correct Answer: C
Solution :
\[y=2\cos x\] At \[x=\frac{\pi }{4},\ \ y=\frac{2}{\sqrt{2}}=\sqrt{2}\]and \[\frac{dy}{dx}=-2.\sin x\] \[\therefore {{\left( \frac{dy}{dx} \right)}_{x=\pi /4}}=-\sqrt{2}\] \[\therefore \] Equation of tangent at \[\left( \frac{\pi }{4},\sqrt{2} \right)\] is \[y-\sqrt{2}=-\sqrt{2}\left( x-\frac{\pi }{4} \right)\].
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