Question

Match List-I with List-II: 

List-I		List-II
(A)	Domain of \(f(x)=\dfrac{1}{\sqrt{x^2-1}}\)	(II)	\((-\infty,-1)\cup(1,\infty)\)
(B)	Range of \(f(x)=\dfrac{1}{\sqrt{x^2-1}}\)	(III)	\((0,\infty)\)
(C)	Domain of \(f(x)=\sqrt{x-2}\)	(I)	\([2,\infty)\)
(D)	Range of \(f(x)=\sqrt{x-2}\)	(IV)	\([0,\infty)\)

• A. A-II, B-III, C-I, D-IV
• B. A-III, B-II, C-I, D-IV
• C. A-III, B-II, C-IV, D-I
• D. A-II, B-III, C-IV, D-I

Hint:

For \(\sqrt{g(x)}\), use \(g(x)\geq0\). For \(\frac{1}{\sqrt{g(x)}}\), use \(g(x)>0\).

Answer 1

The Correct Option is A

Concept:
For square root functions, the expression inside square root must be non-negative. If the square root is in denominator, it must be strictly positive.
Step 1: Domain of \(f(x)=\dfrac{1{\sqrt{x^2-1}}\).} \[ x^2-1>0 \] \[ x^2>1 \] \[ x1 \] So, \[ A\rightarrow (-\infty,-1)\cup(1,\infty) \] \[ A\rightarrow II \]
Step 2: Range of \(f(x)=\dfrac{1{\sqrt{x^2-1}}\).}
Since denominator is positive, function is always positive. Also, it can become very large and can approach \(0\). So, \[ \text{Range}=(0,\infty) \] \[ B\rightarrow III \]
Step 3: Domain of \(f(x)=\sqrt{x-2\).} \[ x-2\geq 0 \] \[ x\geq 2 \] So, \[ \text{Domain}=[2,\infty) \] \[ C\rightarrow I \]
Step 4: Range of \(f(x)=\sqrt{x-2\).}
Square root always gives a non-negative value. So, \[ \text{Range}=[0,\infty) \] \[ D\rightarrow IV \] Therefore, \[ A-II,\ B-III,\ C-I,\ D-IV \] \[ \therefore \text{Correct Answer is (A)} \]