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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

સમીકરણ \(\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}\) નાં વાસ્તવિક બીજની સંખ્યા મેળવો.

  1. A \(0\)
  2. B \(4\)
  3. C \(1\)
  4. D \(2\)
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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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