The number of real solutions of ^ -1 x(x + 4) + ^ -1 x^ 2 + 4x + 1 = (π)/(2) is equal to

Correct answer: C. 2 Let, f (x) = ^ -1 x (x + 4) + ^ -1 x^ 2 + 4x + 1 For f(x) to be defined, we have x^ 2 + 4x ≥ 0 and 0 ≤ x^ 2 + 4x + 1 ≤ 1 Which is only possible if x^ 2 + 4x = 0 ⇒ x = 0, - 4 Also these values satisfies the equation ∴ Number of solutions = 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The number of real solutions of ^ -1 x(x + 4) + ^ -1 x^ 2 + 4x + 1...

The number of real solutions of

\cot^{-1} \sqrt{x\left(x + 4\right)} + \cos^{-1} \sqrt{x^{2} + 4x + 1} = \dfrac{\pi}{2}

is equal to

Options

A $0$
B $1$
C $2$
DInfinite

Correct answer

C. $2$

Step-by-step solution

Let,

f

\left(x\right) = \cot^{-1} \sqrt{x \left(x + 4\right)} + \cos^{-1}

\sqrt{x^{2} + 4x + 1}<br />For

f\left(x\right)

to be defined, we have<br />

x^{2} + 4x ≥ 0

and

0 ≤ x^{2} + 4x + 1 ≤ 1

<br />Which is only possible if<br />

x^{2} + 4x = 0 ⇒ x = 0, - 4

<br />Also these values satisfies the equation<br />

∴

Number of solutions

= 2

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