Applications-Of-Derivatives Question 448
Question: The equation of the normal to the curve $ y=\sin \frac{\pi x}{2} $ at (1, 1) is
[AMU 1999]
Options:
A) $ y=1 $
B) $ x=1 $
C) $ y=x $
D) $ y-1=\frac{-2}{\pi }(x-1) $
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Answer:
Correct Answer: B
Solution:
$ y=\sin \frac{\pi x}{2} $
Þ $ \frac{dy}{dx}=\frac{\pi }{2}\cos \frac{\pi }{2}x $
Þ $ {{( \frac{dy}{dx} )}_{(1,1)}}=0 $ \ Equation of normal is $ y-1=\frac{1}{0}(x-1) $
Þ $ x=1 $ .
BETA
