Sets, Relations and FunctionsMCQMTP Nov 19, MTP March 21Question 1976 of 217
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If f(x)=x24x2\displaystyle f(x) = \frac{x^2-4}{x-2}, then f(2)\displaystyle f(2) is

Options

A0
B2
C4
D1
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Correct Answer

Option c4

All Options:

  • A0
  • B2
  • C4
  • D1

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Detailed Solution & Explanation

Let's evaluate the function at x=2\displaystyle x = 2:
f(x)=x24x2f(x) = \frac{x^2 - 4}{x - 2}
Direct substitution of x=2\displaystyle x = 2 gives:
f(2)=22422=4422=00f(2) = \frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{2 - 2} = \frac{0}{0}
This is an indeterminate form. To find the value, we simplify the rational function by factoring the numerator (using the difference of squares a2b2=(ab)(a+b)\displaystyle a^2 - b^2 = (a-b)(a+b)):
x24=(x2)(x+2)x^2 - 4 = (x - 2)(x + 2)
Substitute this back into the function definition:
f(x)=(x2)(x+2)x2f(x) = \frac{(x - 2)(x + 2)}{x - 2}
For x2\displaystyle x \neq 2, we can cancel the common factor (x2)\displaystyle (x-2) in the numerator and denominator:
f(x)=x+2f(x) = x + 2
By extending the function to x=2\displaystyle x = 2 by continuity (which is the standard convention in CA Foundation and algebraic simplification questions):
f(2)=2+2=4f(2) = 2 + 2 = 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.

View Official ICAI Syllabus

Exam Strategy Tip

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

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