ExamBro
ExamBro
enEnglishguગુજરાતી
GUJCET · Maths · Inverse Trigonometric Functions

સમીકરણ \(\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}\) ના વાસ્તવિક ઉકેલોની સંખ્યા ___________ છે.

  1. A 1
  2. B 3
  3. C 2
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 2

Step-by-step Solution

Detailed explanation

\( \tan^{-1} \sqrt{x(x+1)} \): \( x(x+1) \ge 0 \implies x \in (-\infty, -1] \cup [0, \infty) \) \( \sin^{-1} \sqrt{x^2+x+1} \): \( 0 \le \sqrt{x^2+x+1} \le 1 \)
From GUJCET
Explore more questions on app