Question

Please solve the question in attachment

Answer 1

Answer:

The  number  of real solutions of  following  equation is  Two

Step-by-step explanation:

Given,

$\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\frac{\pi}{2}\\\;\\\tan^{-1}\sqrt{x^2+x}+\sin^{-1}\sqrt{x^2+x+1}=\frac{\pi}{2}\\\;\\\tan^{-1}\sqrt{x^2+x}=\frac{\pi}{2}-\sin^{-1}\sqrt{x^2+x+1}\\\;\\\tan^{-1}\sqrt{x^2+x}=\cos^{-1}\sqrt{x^2+x+1}\\\;\\\cos^{-1}\frac{1}{\sqrt{x^2+x+1}}=\cos^{-1}\sqrt{x^2+x+1}\\\\\;\\\frac{1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}\\\;\\1=x^2+x+1\\\;\\x^2+x=0\\\;\\x(x+1)=0\\\;\\\textbf{Hence,}\\\;\\\textbf{x=0\;\;or\;\;x= -1}$

Answer 2

Answer:

Option(C)

Step-by-step explanation:

Given Equation is tan⁻¹√x(x + 1) + sin⁻¹√x² + x + 1 = π/2

⇒ tan⁻¹√(x + 1) = π/2 - sin⁻¹√x² + x + 1

⇒ tan⁻¹√(x + 1) = cos⁻¹√x² + x + 1

We know that tan⁻¹(x) = cos⁻¹(1/√x² + 1).

⇒ cos⁻¹(1/√x² + x + 1) = cos⁻¹(√x² + x + 1)

⇒ 1 = (√x² + x + 1)(√x² + x + 1)

⇒ 1 = x² + x + 1

⇒ x² + x + 1 - 1 = 0

⇒ x² + x = 0

⇒ x(x + 1) = 0

⇒ x = 0, -1.

Therefore, the solutions are 0,1.

Hope this helps!