Match List-I with List-II
|
List-I |
List-II |
|
(A) $f(x) = |x|$ |
(I) Not differentiable at $x=-2$ only |
|
(B) $f(x) = |x+2|$ |
(II) Not differentiable at $x = 0$ only |
|
(C) $f(x) = |x^2-4|$ |
(III) Not differentiable at $x = 2$ only |
|
(D) $f(x)=|x-2|$ |
(IV) Not differentiable at $x = 2,-2$ only |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
|
List-I |
List-II |
|
(A) $f(x) = |x|$ |
(II) Not differentiable at $x = 0$ only |
|
(B) $f(x) = |x+2|$ |
(I) Not differentiable at $x=-2$ only |
|
(C) $f(x) = |x^2-4|$ |
(IV) Not differentiable at $x = 2,-2$ only |
|
(D) $f(x)=|x-2|$ |
(III) Not differentiable at $x = 2$ only |
$(A)\ f(x)=|x|$
$|x|=\begin{cases}x,&x\ge 0\\-x,&x<0\end{cases}$
Left derivative at $x=0=-1$, right derivative at $x=0=1$
$f'(0^-)\ne f'(0^+)$
Not differentiable at $x=0$ only
$(A)\rightarrow(II)$
$(B)\ f(x)=|x+2|$
Corner point when $x+2=0$
$x=-2$
Not differentiable at $x=-2$ only
$(B)\rightarrow(I)$
$(C)\ f(x)=|x^2-4|$
$x^2-4=0 \Rightarrow x=\pm2$
Sign of $x^2-4$ changes at $x=2,-2$
Derivative is discontinuous at both points
Not differentiable at $x=2,-2$
$(C)\rightarrow(IV)$
$(D)\ f(x)=|x-2|$
Corner point when $x-2=0$
$x=2$
Not differentiable at $x=2$ only
$(D)\rightarrow(III)$
Final Matching: (A)-(II), (B)-(I), (C)-(IV), (D)-(III).