Sets, Relations and FunctionsMCQPYQ July 21Question 1966 of 217
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The range of the function f\displaystyle f defined by f(x)=16x2\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)

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Detailed Solution & Explanation

We are given the function:
f(x)=16x2f(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:
16x20    x216    4x416 - 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 16x2\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)=1616=0f(\pm 4) = \sqrt{16 - 16} = 0
- The maximum value of 16x2\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)=160=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.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

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Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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