Let $\mathrm{f}:(0,1) \rightarrow \mathbb{R}$ be the function defined as $\mathrm{f}(\mathrm{x})=[4 \mathrm{x}]\left(\mathrm{x}-\frac{1}{4}\right)^{2}\left(\mathrm{x}-\frac{1}{2}\right)$, 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 $-\frac{1}{512}$

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Continuity and Differentiability

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

Answer: (A), (B)

Solution

The function is:

f (x) = ⎩ ⎨ ⎧ 0 (x - \frac{1}{4})^{2} (x - \frac{1}{2}) 2 (x - \frac{1}{4})^{2} (x - \frac{1}{2}) 3 (x - \frac{1}{4})^{2} (x - \frac{1}{2}); 0 < x < \frac{1}{4}; \frac{1}{4} \leq x < \frac{1}{2}; \frac{1}{2} \leq x < \frac{3}{4}; \frac{3}{4} \leq x < 1

x = \frac{1}{4}

f (\frac{1}{4}^{-}) = 0 = f (\frac{1}{4}^{+}) = 0

, continuous.

At

x = \frac{1}{2}

f (\frac{1}{2}^{-}) = 0 = f (\frac{1}{2}^{+}) = 0

, continuous.

At

x = \frac{3}{4}

f (\frac{3}{4}^{-}) = 2 (\frac{1}{2})^{2} (\frac{1}{4}) = \frac{1}{16}

f (\frac{3}{4}^{+}) = 3 (\frac{1}{2})^{2} (\frac{1}{4}) = \frac{3}{16}

. Discontinuous at

x = \frac{3}{4}

only. (A) is true.

For differentiability: f is not differentiable at

x = \frac{1}{2}

(derivative changes from one formula to another) and at

x = \frac{3}{4}

(discontinuous). So exactly 2 points, not more than 3. (B) is true, (C) is false.

For minimum: on

[\frac{1}{4}, \frac{1}{2})

f^{'} (x) = 0

gives

x = \frac{5}{12}

f (\frac{5}{12}) = - \frac{1}{432} \neq = - \frac{1}{512}

. (D) is false.

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Continuity and Differentiability

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