Sets, Relations and FunctionsPYQ Jan 26Question 4266 of 145
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The function f(x)=x2−25x−5\displaystyle f(x) = \frac{x^2 - 25}{x - 5} is undefined at x=5\displaystyle x=5. What value must be assigned to f(5) if f(x) is to be continuous at x=5\displaystyle x=5?

Options

A0
B1
C10
D100
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Correct Answer

✅ Option c — 10

All Options:

  • A0
  • B1
  • C10
  • D100

Detailed Solution & Explanation

For a function f(x)\displaystyle f(x) to be continuous at x=a\displaystyle x = a, the function value f(a)\displaystyle f(a) must equal the limit of the function as x\displaystyle x approaches a\displaystyle a: f(a)=lim⁡x→af(x)f(a) = \lim_{x \to a} f(x)
Here, we want f(x)\displaystyle f(x) to be continuous at x=5\displaystyle x = 5. Thus: f(5)=lim⁡x→5x2−25x−5f(5) = \lim_{x \to 5} \frac{x^2 - 25}{x - 5}
We can simplify the expression inside the limit by factoring the numerator: x2−25=(x−5)(x+5)x^2 - 25 = (x - 5)(x + 5) For x≠5\displaystyle x \neq 5: x2−25x−5=(x−5)(x+5)x−5=x+5\frac{x^2 - 25}{x - 5} = \frac{(x - 5)(x + 5)}{x - 5} = x + 5
Now we take the limit as x\displaystyle x approaches 5\displaystyle 5: lim⁡x→5f(x)=lim⁡x→5(x+5)=5+5=10\lim_{x \to 5} f(x) = \lim_{x \to 5} (x + 5) = 5 + 5 = 10
Therefore, the value that must be assigned to f(5)\displaystyle f(5) is 10\displaystyle 10. 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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