Question

If a real valued function \(f:A\rightarrow B\) defined by \[ f(x)=\sin^{-1}\!\left(\sqrt{x^{2}-4x+5}\right) \] is a bijection, then \(A\cup B=\)

• A. \(\mathbb R\)
• B. \([0,1]\cup\left[0,\frac{\pi}{2}\right]\)
• C. \([-1,1]\cup\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)
• D. \(\left\{2,\frac{\pi}{2}\right\}\) \bigskip

Hint:

Whenever an inverse trigonometric function appears, first check whether its argument lies in the permitted interval. Most questions become simple after applying this restriction.

Answer 1

The Correct Option is D

Concept: For the inverse sine function, \[ \sin^{-1}(u) \] is defined only when \[ -1\le u\le 1. \] Since a square root is always non-negative, \[ 0\le \sqrt{x^2-4x+5}\le1. \] We first determine the admissible values of \(x\).

Step 1: Simplify the expression inside the square root.
\[ x^2-4x+5 \] can be written as \[ (x-2)^2+1. \] Thus, \[ f(x)=\sin^{-1}\left(\sqrt{(x-2)^2+1}\right). \]

Step 2: Apply the condition for inverse sine.
For \(\sin^{-1}\) to exist, \[ 0\le \sqrt{(x-2)^2+1}\le1. \] Squaring throughout, \[ (x-2)^2+1\le1. \] Hence \[ (x-2)^2=0. \] Therefore, \[ x=2. \] Thus \[ A=\{2\}. \]

Step 3: Find the range.
Substituting \(x=2\), \[ f(2)=\sin^{-1}(1). \] Therefore, \[ f(2)=\frac{\pi}{2}. \] Hence \[ B=\left\{\frac{\pi}{2}\right\}. \]

Step 4: Find \(A\cup B\).
Therefore, \[ A\cup B= \left\{2,\frac{\pi}{2}\right\}. \] Hence, \[ \boxed{A\cup B= \left\{2,\frac{\pi}{2}\right\}} \] and the correct option is \[ \boxed{(D)}. \]