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JEE Advanced 2023 Paper 1 · Mathematics · Differentiation and Application of Derivatives · medium

Match each function in List-I with the number of real points at which it is not differentiable (List-II). For the function in (S), consider only the…

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Match each function in List-I with the number of real points at which it is not differentiable (List-II). For the function in (S), consider only the open interval (0,2π)(0,2\pi); the others are defined on R\mathbb R.

List-I
(P) f(x)=∣x2−4∣f(x)=|x^{2}-4|
(Q) f(x)=(x2−1) ∣x−1∣f(x)=(x^{2}-1)\,|x-1|
(R) f(x)=∣x−1∣+∣x2−4∣f(x)=|x-1|+|x^{2}-4|
(S) f(x)=∣cos⁡2x∣f(x)=|\cos 2x|

List-II
(1) 00
(2) 11
(3) 22
(4) 33
(5) 44
  1. A. P→3, Q→2, R→4, S→5
  2. B. P→3, Q→1, R→4, S→5
  3. C. P→2, Q→1, R→4, S→5
  4. D. P→3, Q→1, R→3, S→4
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Answer: (B) P→3, Q→1, R→4, S→5

A function ∣g(x)∣|g(x)| fails to be differentiable exactly at simple zeros of gg (where g′≠0g'\ne0).
The criterion is that the corner appears only where the inner function changes sign linearly. (P) x2−4x^2-4 has simple zeros at ±2\pm2: 22 points, (3).
(Q) f(x)=(x−1)(x+1)∣x−1∣f(x)=(x-1)(x+1)|x-1|. At x=1x=1 the factor (x−1)∣x−1∣(x-1)|x-1| is differentiable (derivative 00) and multiplying by the smooth factor x+1x+1 keeps it so; elsewhere ff is smooth. 00 points, (1).
(R) The kinks of ∣x−1∣|x-1| at 11 and of ∣x2−4∣|x^2-4| at ±2\pm2 cannot cancel: 33 points, (4).
(S) cos⁡2x=0\cos2x=0 in (0,2π)(0,2\pi) at x=π4,3π4,5π4,7π4x=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4, all simple zeros: 44 points, (5).
The correct code is P→3, Q→1, R→4, S→5.
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