Applications-Of-Derivatives Question 449
Question: The equation of tangent to the curve $ y=2\cos x $ at $ x=\frac{\pi }{4} $ is
[RPET 1997]
Options:
A) $ y-\sqrt{2}=2\sqrt{2}( x-\frac{\pi }{4} ) $
B) $ y+\sqrt{2}=\sqrt{2}( x+\frac{\pi }{4} ) $
C) $ y-\sqrt{2}=-\sqrt{2}( x-\frac{\pi }{4} ) $
D) $ y-\sqrt{2}=\sqrt{2}( x-\frac{\pi }{4} ) $
Show Answer
Answer:
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 {{( \frac{dy}{dx} )}_{x=\pi /4}}=-\sqrt{2} $
$ \therefore $ Equation of tangent at $ ( \frac{\pi }{4},\sqrt{2} ) $ is $ y-\sqrt{2}=-\sqrt{2}( x-\frac{\pi }{4} ) $ .
BETA
