A) 1
B) 0
C) -1/2
D) 1/2
Correct Answer: B
Solution :
Given equation is \[2{{\cos }^{-1}}\sqrt{\left( \frac{1+x}{2} \right)}=\frac{\pi }{2}\] Þ \[{{\cos }^{-1}}\sqrt{\left( \frac{1+x}{2} \right)}=\frac{\pi }{4}\Rightarrow \cos \frac{\pi }{4}=\frac{\sqrt{1+x}}{\sqrt{2}}\] Þ \[\frac{1}{\sqrt{2}}=\frac{\sqrt{1+x}}{\sqrt{2}}\Rightarrow 1=\sqrt{1+x}\Rightarrow x=0\].
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