The number of solutions of tan^-1√(x(x+1))+sin^-1x^2+x+1=π /2 is:

C. two. Step 1: sin^-1x^2+x+1 needs x^2+x+1≤ 1, i.e. x(x+1)≤ 0; tan^-1√(x(x+1)) needs x(x+1)≥ 0. Step 2: Both hold only when x(x+1)=0: x=0 or x=-1. Step 3: For both values the equation reads tan^-10+sin^-11=0+π /2=π /2, which is true. So there are two solutions. Check: At x=1, x^2+x+1=√(3)>1 and sin^-1 is undefined; every x other than 0 and -1 fails Step 1 in the same way. Trap: (A) zero (the source key) throws away the boundary values; at x=0 and x=-1 both terms are defined and the equation holds.

Question 15 of 21 · Inverse Trigonometric Functions

Questionmedium

This question was previously asked in

NIMCET 2009

The number of solutions of $t an^{- 1} x (x + 1) + s in^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is:

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — two

Detailed Solution

$s in^{- 1} x^{2} + x + 1$ needs $x^{2} + x + 1 \leq 1$ , i.e. $x (x + 1) \leq 0$ ; $t an^{- 1} x (x + 1)$ needs $x (x + 1) \geq 0$ .

Both hold only when $x (x + 1) = 0$ : $x = 0$ or $x = - 1$ .

For both values the equation reads $t an^{- 1} 0 + s in^{- 1} 1 = 0 + \frac{π}{2} = \frac{π}{2}$ , which is true. So there are two solutions.

✓ Check

x = 1

x^{2} + x + 1 = 3 > 1

and

s in^{- 1}

is undefined; every

x

other than

0

and

- 1

fails Step 1 in the same way.

⚠ Common trap

(A) zero (the source key) throws away the boundary values; at

x = 0

and

x = - 1

both terms are defined and the equation holds.

Quick tip

$x (x + 1) \geq 0 and x^{2} + x + 1 \leq 1 \Rightarrow x (x + 1) = 0 \Rightarrow x = 0, - 1$

💬 Discussions on this question

Ask a doubt ›

Still stuck on this question? Ask our faculty and fellow students — answers appear here.

Ask a doubt

Inverse Trigonometric Functions — Preparation

This question tests your understanding of Inverse Trigonometric Functions, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Inverse trigonometric functions, such as sin^-1 x, cos^-1 x and tan^-1 x, return the angle whose trigonometric ratio is a given value. Because trigonometric functions repeat, each inverse is defined only on a principal value branch. For example, sin^-1 x takes values between -pi/2 and pi/2. This topic covers domains and principal value ranges, finding principal values, and key properties such as sin^-1 x + cos^-1 x = pi/2 and the formulas for tan^-1 x + tan^-1 y. Exam questions test simplifying expressions, solving equations, and evaluating compositions such as sin(cos^-1 x). The most common mistake is ignoring the principal range, for example assuming sin^-1(sin x) always equals x. Practise the inverse trigonometric functions questions below, each with a step-by-step solution.

‹ All Inverse Trigonometric Functions questions Browse all topics ›

The number of solutions of $t an^{- 1} x (x + 1) + s in^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is:

💬 Discussions on this question

More Inverse Trigonometric Functions Questions

More Quantitative Aptitude Questions

Inverse Trigonometric Functions — Preparation

The number of solutions of tan−1x(x+1)​+sin−1x2+x+1​=2π​ is:

💬 Discussions on this question

More Inverse Trigonometric Functions Questions

More Quantitative Aptitude Questions

Inverse Trigonometric Functions — Preparation

The number of solutions of $t an^{- 1} x (x + 1) + s in^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is: