If \( f(x)=\sqrt{x^{2}-25} \) is a function then which of the following are true ?
(A) \( \mathrm{Range}\,(f(x))=[0,\infty) \)
(B) \( \mathrm{Range}\,(f(x))=\mathrm{R} \), R is set of real number
(C) \( \mathrm{Domain}\,(f(x))=\mathrm{R} \)
(D) \( \mathrm{Domain}\,(f(x))=(-\infty,-5]\cup[5,\infty) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
A square root is defined only for non-negative values and is itself never negative.
Domain. We need
\( x^{2}-25\ge 0 \)
\( x\le -5 \) or \( x\ge 5 \)
Range. The expression \( x^{2}-25 \) takes every value from 0 upwards, so \( f(x) \) takes every value in \( [0,\infty) \).
So the domain is \( (-\infty,-5]\cup[5,\infty) \) and the range is \( [0,\infty) \).
Checking the statements
(A) \( \mathrm{Range}\,(f(x))=[0,\infty) \) — Correct. The range is \( [0,\infty) \).
(B) \( \mathrm{Range}\,(f(x))=\mathrm{R} \), R is set of real number — Incorrect. A square root is never negative.
(C) \( \mathrm{Domain}\,(f(x))=\mathrm{R} \) — Incorrect. For example \( f(0)=\sqrt{-25} \) is not real.
(D) \( \mathrm{Domain}\,(f(x))=(-\infty,-5]\cup[5,\infty) \) — Correct. These are exactly the \( x \) with \( x^{2}\ge 25 \).