JEE Advanced 2023 · Paper 2, Mathematics question 6

$[4x]=0,1,2,3$ on $(0,\frac14)$, $[\frac14,\frac12)$, $[\frac12,\frac34)$, $[\frac34,1)$. Let $g(x)=(x-\frac14)^2(x-\frac12)$, so $f=[4x]\,g$.

At the jumps of $[4x]$.

• $x=\frac14$: $g$ has a double zero. $f$ is continuous, and both one-sided derivatives are 0, so $f$ is differentiable.
• $x=\frac12$: $g=0$, so $f$ is continuous. But the one-sided derivatives are $1\cdot g'(\frac12)$ and $2\cdot g'(\frac12)$, with $g'(\frac12)=\frac1{16}$. They differ, so $f$ is not differentiable.
• $x=\frac34$: $g(\frac34)=\frac1{16}\ne0$, so $f$ jumps from $\frac2{16}$ to $\frac3{16}$: discontinuous.

(A) True: the only discontinuity is $x=\frac34$.

(B) True: $x=\frac12$ is the only point where $f$ is continuous but not differentiable.

(C) False: $f$ fails to be differentiable only at $\frac12$ and $\frac34$.

(D) False. $f<0$ only on $(\frac14,\frac12)$, where $f=g$. $g'=(x-\frac14)(3x-\frac54)$ vanishes at $x=\frac5{12}$:
$$g\left(\tfrac5{12}\right)=\left(\tfrac16\right)^2\left(-\tfrac1{12}\right)=-\frac1{432}.$$
The minimum is $-\frac1{432}$, not $-\frac1{512}$.