Differentiation Question 157
Question: The differential coefficient of $ {{\cos }^{-1}}{ \sqrt{\frac{1+x}{2}} } $ with respect to x is
[MP PET 1993]
Options:
A) $ -\frac{1}{2\sqrt{1-x^{2}}} $
B) $ \frac{1}{2\sqrt{1-x^{2}}} $
C) $ \frac{1}{\sqrt{1-x}} $
D) $ {{\sin }^{-1}}{ \sqrt{\frac{1+x}{2}} } $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{\cos }^{-1}}{ \sqrt{\frac{1+x}{2}} } $
Let $ \sqrt{\frac{1+x}{2}}=\cos \theta $ or $ x=2{{\cos }^{2}}\theta -1=\cos 2\theta $ ;
$ \therefore \theta =\frac{1}{2}{{\cos }^{-1}}x $ . So, $ y=\frac{1}{2}{{\cos }^{-1}}x $
Therefore $ -\frac{1}{2\sqrt{1-x^{2}}} $ .
BETA
