Match List-I with List-II
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List-I |
List-II |
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(A) $f(x) =|x| + 1$ |
(I) Not differentiable at $x = 3$ only |
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(B) $f(x)=|x -3|$ |
(II) Not differentiable at $x = -3$ only |
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(C) $f(x) =|x+3|$ |
(III) Not differentiable at $x = 3, -3$ only |
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(D) $f(x) =|x^2-9|$ |
(IV) Not differentiable at $x = 0$ only |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
|
List-I |
List-II |
|
(A) $f(x) =|x| + 1$ |
(IV) Not differentiable at $x = 0$ only |
|
(B) $f(x)=|x -3|$ |
(I) Not differentiable at $x = 3$ only |
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(C) $f(x) =|x+3|$ |
(II) Not differentiable at $x = -3$ only |
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(D) $f(x) =|x^2-9|$ |
(III) Not differentiable at $x = 3, -3$ only |
(A) f(x) = |x| + 1
Here, the function is a vertical shift of |x|. Since |x| is not differentiable at x = 0 (it has a sharp corner there), this function is also not differentiable only at x = 0.
Match: (A) → (IV)
(B) f(x) = |x − 3|
This is a modulus function shifted right by 3 units. It has a corner (non-differentiability) at x = 3 only.
Match: (B) → (I)
(C) f(x) = |x + 3|
This is a modulus function shifted left by 3 units. So, it's not differentiable at x = −3 only.
Match: (C) → (II)
(D) f(x) = |x² − 9|
This expression becomes 0 when x² = 9 ⇒ x = ±3. So, modulus function is not differentiable at both x = 3 and x = −3.
Match: (D) → (III)