Applications Of Derivatives Question 27
Question: If an equation of a tangent to the curve, $ y=\cos (x+y), $ $ -1 \le x\le 1 + \pi $ , is $ x+2y=k $ then k is equal to:
Options:
A) 1
B) 2
C) $ \frac{\pi }{4} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: D
Solution:
[d] Let $ y=\cos (x+y) $
$ \Rightarrow \frac{dy}{dx}=-\sin (x+y)( 1+\frac{dy}{dx} ) $ - (1) Now, given equation of tangent is $ x+2y=k $
$ \Rightarrow ,Slope=\frac{-1}{2} $ So, $ =\frac{dy}{dx}=\frac{-1}{2} $ put this value in (1), we get $ \frac{-1}{2}=-\sin (x+y)( 1-\frac{1}{2} )\Rightarrow \sin (x+y)=1 $
$ \Rightarrow x+y=\frac{\pi }{2}\Rightarrow y=\frac{\pi }{2}-x $ Now, $ \frac{\pi }{2}-x=\cos (x+y)\Rightarrow x=\frac{\pi }{2} $ and $ y=0 $ Thus $ x+2y=k\Rightarrow \frac{\pi }{2}=k $
BETA
