CUET · MATHS · PYQ PAPER 2025
Match List-I with List-II
| List-I (Function) | List-II (Property) |
| (A) \(f(x)=\left\{\begin{array}{ll}\frac{x}{|x|} & : x \neq 0 \\ 0 & : x=0\end{array}\right.\) | (I) continuous but not differentiable at \(x=0\) |
| (B) \(f ( x )=| x |\) | (II) continuous but not differentiable at \(x=1\) |
| (C) \(f(x)=\left|x^2-1\right|\) | (III) discontinuous at \(x=0\) |
| (D) \(f(x)=|x-1|\) | (IV) continuous but not differentiable at \(x=1,-1\) |
- A (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
- B (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
- C (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
- D (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Answer & Solution
Correct Answer
(A) (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Step-by-step Solution
Detailed explanation
(A) \(f(x)=\left\{\begin{array}{ll}\frac{x}{|x|} & : x \neq 0 \\ 0 & : x=0\end{array}\right.\) \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{-x} = -1\) \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{x} = 1\) \(-1 \neq 1\), so \(f(x)\) is discontinuous at \(x=0\).…
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