CUET · MATHS · PYQ PAPER 2025
Match List-I with List-II
| List - I (Function) | List - II (Points of discontinuity) |
| (A) \(f(x)=\frac{x^2+1}{x}\) | (I) x = 4 |
| (B) \(f(x)=\frac{|x-1|}{x-1}\) | (II) x = 2 |
| (C) \(f(x)=\{x-1\), if \(x<2 ; x+1\), if \(x \geq 2\}\) | (III) x = 0 |
| (D) \(f(x)=\frac{1-x}{x-4}\) | (IV) x = 1 |
Choose the Correct answer from the options given below :
- A (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
- B (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
- C (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
- D (A) - (I), (B) - (III), (C) - (IV), (D) - (II)
Answer & Solution
Correct Answer
(A) (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
Step-by-step Solution
Detailed explanation
(A) For \(f(x)=\frac{x^2+1}{x}\), discontinuity occurs when \(x=0\). So, (A) - (III). (B) For \(f(x)=\frac{|x-1|}{x-1}\), discontinuity occurs when \(x-1=0\), i.e., \(x=1\). So, (B) - (IV). (C) For \(f(x)=\{x-1, \text{ if } x<2 ; x+1, \text{ if } x \geq 2\}\), discontinuity…
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