Inverse Trigonometric Functions Question 21
Question: If $ 2{{\cos }^{-1}}\sqrt{\frac{1+x}{2}}=\frac{\pi }{2}, $ then $ x= $
Options:
A) 1
B) 0
C) -1/2
D) 1/2
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation is $ 2{{\cos }^{-1}}\sqrt{( \frac{1+x}{2} )}=\frac{\pi }{2} $
Therefore $ {{\cos }^{-1}}\sqrt{( \frac{1+x}{2} )}=\frac{\pi }{4}\Rightarrow \cos \frac{\pi }{4}=\frac{\sqrt{1+x}}{\sqrt{2}} $
Therefore $ \frac{1}{\sqrt{2}}=\frac{\sqrt{1+x}}{\sqrt{2}}\Rightarrow 1=\sqrt{1+x}\Rightarrow x=0 $ .
BETA
