Set, Relations and FunctionsPYQ July 21Question 1966 of 217
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The range of the function f\displaystyle f defined by f(x)=16−x2\displaystyle f(x) = \sqrt{16-x^2} is

Options

A(−4,0)\displaystyle (-4, 0)
B(−4,4)\displaystyle (-4, 4)
C[0,4]\displaystyle [0, 4]
D(+4,4)\displaystyle (+4, 4)
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Correct Answer

✅ Option c — [0,4]\displaystyle [0, 4]

All Options:

  • A(−4,0)\displaystyle (-4, 0)
  • B(−4,4)\displaystyle (-4, 4)
  • C[0,4]\displaystyle [0, 4]
  • D(+4,4)\displaystyle (+4, 4)

Detailed Solution & Explanation

We are given the function:
f(x)=16−x2f(x) = \sqrt{16-x^2}
Let's find the domain of the function first. For the square root to be defined, the expression inside must be non-negative:
16−x2≥0  ⟹  x2≤16  ⟹  −4≤x≤416 - x^2 \ge 0 \implies x^2 \le 16 \implies -4 \le x \le 4
Thus, the domain of f(x)\displaystyle f(x) is [−4,4]\displaystyle [-4, 4].
Now, let's determine the range. For x∈[−4,4]\displaystyle x \in [-4, 4]:
- The minimum value of 16−x2\displaystyle 16 - x^2 occurs when x2\displaystyle x^2 is maximized, i.e., x2=16\displaystyle x^2 = 16 (when x=±4\displaystyle x = \pm 4):
f(±4)=16−16=0f(\pm 4) = \sqrt{16 - 16} = 0
- The maximum value of 16−x2\displaystyle 16 - x^2 occurs when x2\displaystyle x^2 is minimized, i.e., x2=0\displaystyle x^2 = 0 (when x=0\displaystyle x = 0):
f(0)=16−0=4f(0) = \sqrt{16 - 0} = 4
Since f(x)\displaystyle f(x) is continuous on the interval [−4,4]\displaystyle [-4, 4], it takes all values in the interval [0,4]\displaystyle [0, 4].
Therefore, the range of the function is [0,4]\displaystyle [0, 4].
Hence, **Option C** is the correct answer.

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