If $n_{1}$ and $n_{2}$ are the numbers of real solutions of $x = s in^{- 1} x$ and $x = s in x$ respectively, then the value of $n_{2} - n_{1}$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — 0

Detailed Solution

$x = s in x$ : for $x \neq = 0$ , $∣ s in x ∣ < ∣ x ∣$ , so the only solution is $x = 0$ . Hence $n_{2} = 1$ .

$x = s in^{- 1} x$ : for $x < 0$ the left side is negative and the right side is not, so there is no solution. For $0 < x \leq 1$ , $s in^{- 1} x > x$ . Only $x = 0$ works, so $n_{1} = 1$ .

$n_{2} - n_{1} = 1 - 1 = 0$ .

✓ Check

x = 0.5

s in^{- 1} 0.5 = \frac{π}{6} \approx 0.524 > 0.5

and

s in 0.5 \approx 0.479 < 0.5

, so neither equation holds there.

⚠ Common trap

(A) 1 is the source paper's key, but its own working ends with

n_{2} - n_{1} = 0

; the key is corrected to (B).

Quick tip

$∣ sin^{- 1} x ∣ > ∣ x ∣$ and $∣ sin x ∣ < ∣ x ∣$ for $x \neq = 0$ , so $n_{1} = n_{2} = 1$

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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.

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If $n_{1}$ and $n_{2}$ are the numbers of real solutions of $x = s in^{- 1} x$ and $x = s in x$ respectively, then the value of $n_{2} - n_{1}$ is

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If n1​ and n2​ are the numbers of real solutions of x=​sin−1x​ and x=sinx respectively, then the value of n2​−n1​ is

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If $n_{1}$ and $n_{2}$ are the numbers of real solutions of $x = s in^{- 1} x$ and $x = s in x$ respectively, then the value of $n_{2} - n_{1}$ is