JEE Advanced 2023 — Paper 2 — MATH-II solution

JEE Advanced2023Paper 2MATH-II

Let $f : (0, 1) \to R$ be the function defined as $f (x) = [4 x] (x - \frac{1}{4})^{2} (x - \frac{1}{2})$ , where $[x]$ denotes the greatest integer less than or equal to $x$ . Then which of the following statements is(are) true?

A
The function $f$ is discontinuous exactly at one point in $(0, 1)$
B
There is exactly one point in $(0, 1)$ at which the function $f$ is continuous but NOT differentiable
C
The function $f$ is NOT differentiable at more than three points in $(0, 1)$
D
The minimum value of the function $f$ is $- \frac{1}{512}$

Solution

Split $(0, 1)$ using the value of $[4 x]$ :

$f (x) = ⎩ ⎨ ⎧ 0, 0 < x < 1/4 (x - 1/4)^{2} (x - 1/2), 1/4 \leq x < 1/2 2 (x - 1/4)^{2} (x - 1/2), 1/2 \leq x < 3/4 3 (x - 1/4)^{2} (x - 1/2), 3/4 \leq x < 1.$

Continuity: at $x = 1/4$ and $x = 1/2$ both pieces vanish, so $f$ is continuous there. At $x = 3/4$ , the left limit is $2 (1/2)^{2} (1/4) = 1/8$ while the value is $3 (1/2)^{2} (1/4) = 3/16$ , so $f$ is discontinuous only at $x = 3/4$ . Option (A) is true.

Differentiability: at $x = 1/4$ the two-sided derivatives are both $0$ , so $f$ is differentiable there. At $x = 1/2$ the left derivative is $1/16$ while the right derivative is $1/8$ , so $f$ is continuous but not differentiable. Together with the discontinuity at $x = 3/4$ , the non-differentiable points are ${1/2, 3/4}$ , only one of which is also a point of continuity. Option (B) is true; option (C) is false.

Minimum: on $[1/4, 1/2)$ , set the derivative $(x - 1/4) [3 x - 5/4] = 0$ , giving the interior critical point $x = 5/12$ . There $f (5/12) = (1/6)^{2} (- 1/12) = - 1/432$ . On the other pieces $f \geq 0$ . So $min f = - 1/432 \neq = - 1/512$ . Option (D) is false.

$∴$ Correct options are (A) and (B).

← Back to full paper

Practice more MATH-II

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →