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Match List-I with List-II

List-IList-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)=∣x2−4∣(III) Not differentiable at x=2 only(D) f(x)=∣x−2∣(IV) Not differentiable at x=2,−2 only\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = |x| & \text{(I) Not differentiable at } x=-2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x+2| & \text{(II) Not differentiable at } x=0 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = |x^2-4| & \text{(III) Not differentiable at } x=2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x-2| & \text{(IV) Not differentiable at } x=2,-2 \text{ only} \\[1.2ex] \hline \end{array}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

A function with absolute value ∣g(x)∣|g(x)| is not differentiable where g(x)=0g(x) = 0, as this creates a sharp corner.


For f(x)=∣x∣f(x) = |x|:

The expression inside is xx.

Setting x=0x = 0 gives the point where the function is not differentiable.

Not differentiable at x=0x = 0 only.

(A) matches with (II)


For f(x)=∣x+2∣f(x) = |x + 2|:

The expression inside is x+2x + 2.

x+2=0x + 2 = 0

x=−2x = -2

Not differentiable at x=−2x = -2 only.

(B) matches with (I)


For f(x)=∣x2−4∣f(x) = |x^2 - 4|:

The expression inside is x2−4x^2 - 4.

x2−4=0x^2 - 4 = 0

x2=4x^2 = 4

x=±2x = \pm 2

Not differentiable at x=2,−2x = 2, -2 only.

(C) matches with (IV)


For f(x)=∣x−2∣f(x) = |x - 2|:

The expression inside is x−2x - 2.

x−2=0x - 2 = 0

x=2x = 2

Not differentiable at x=2x = 2 only.

(D) matches with (III)


Final matching:

(A) → (II)

(B) → (I)

(C) → (IV)

(D) → (III)

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