Inverse Trigonometric Functions Question 82
Question: The number of real solutions of $ {{\tan }^{-1}}\sqrt{x(x+1)}+{{\sin }^{-1}}\sqrt{x^{2}+x+1}=\frac{\pi }{2} $ is
[IIT 1999]
Options:
A) Zero
B) One
C) Two
D) Infinite
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\tan }^{-1}}\sqrt{x(x+1)}+{{\sin }^{-1}}\sqrt{x^{2}+x+1}=\frac{\pi }{2} $
$ {{\tan }^{-1}}\sqrt{x(x+1)} $ is defined when $ x(x+1)\ge 0 $ -. .(i)
$ {{\sin }^{-1}}\sqrt{x^{2}+x+1} $ is defined when $ 0\le x(x+1)+1\le 1 $ or $ 0\le x(x+1)\le 0 $ -..(ii)
From (i) and (ii), $ x(x+1)=0 $
Or $ x=0 $ and -1. Hence number of solution is 2.
BETA
