Match List-I with List-II
|
List-I Function |
List-II Property |
|
(A) $f(x) =\left\{\begin{matrix}\frac{x}{|x|}&:x≠0\\0&:x=0\end{matrix}\right.$ |
(I) continuous but not differentiable at $x= 0$ |
|
(B) $f(x) = |x|$ |
(II) continuous but not differentiable at $x=1$ |
|
(C) $f(x) = |x^2 – 1|$ |
(III) discontinuous at $x = 0$ |
|
(D) $f(x) = |x – 1|$ |
(IV) continuous but not differentiable at $x =1, -1$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
|
List-I Function |
List-II Property |
|
(A) $f(x) =\left\{\begin{matrix}\frac{x}{|x|}&:x≠0\\0&:x=0\end{matrix}\right.$ |
(III) discontinuous at $x = 0$ |
|
(B) $f(x) = |x|$ |
(I) continuous but not differentiable at $x= 0$ |
|
(C) $f(x) = |x^2 – 1|$ |
(IV) continuous but not differentiable at $x =1, -1$ |
|
(D) $f(x) = |x – 1|$ |
(II) continuous but not differentiable at $x=1$ |
(A) $f(x)=\begin{cases}\frac{x}{|x|},&x\ne 0\\0,&x=0\end{cases}$
Left and right limits at $0$ are $-1$ and $1$, so $f$ is discontinuous at $0$.
So (A) → (III).
(B) $f(x)=|x|$ is continuous everywhere but not differentiable at $x=0$.
So (B) → (I).
(C) $f(x)=|x^{2}-1|$ is continuous everywhere but not differentiable where $x^{2}-1=0$, i.e., $x=1,-1$.
So (C) → (IV).
(D) $f(x)=|x-1|$ is continuous everywhere but not differentiable at $x=1$.
So (D) → (II).
Final answer: (A)–(III), (B)–(I), (C)–(IV), (D)–(II)