JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
The number of real roots of the equation \(\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}\) is:
- A \(0\)
- B \(4\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1} \sqrt{x^{2}+x}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}\) For equation to be defined, \(x^{3}+x \geq 0\) \(\Rightarrow x^{2}+x+1 \geq 1\) \(\therefore\) Only possibility that the equation is defined \(x^{2}+x=0 \Rightarrow x=0 ; x=-1\) None of these values satisfy…
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