Let f:(0,1) arrow \mathbbR be the function defined as f(x)=[4 x](x-\frac14)^2(x-\frac12) , where [x] denotes the greatest integer less than or equal to x . Then which of the following statements is(are) true?

The function f is discontinuous exactly at one point in (0,1), There is exactly one point in (0,1) at which the function f is continuous but NOT differentiable

The function f is discontinuous exactly at one point in (0,1)

There is exactly one point in (0,1) at which the function f is continuous but NOT differentiable

The function f is NOT differentiable at more than three points in (0,1)

The minimum value of the function f is -\frac1512

Limits Continuity and Differentiability PYQ — JEE Advanced 2023 · Shift 2 · Q23 with Solution

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Correct answer: A, B

Understand the floor term

Given

f(x)=[4x]\left(x-\frac14\right)^2\left(x-\frac12\right), \qquad x\in(0,1).

Since $x\in(0,1)$ , we have $4x\in(0,4)$ . Therefore,

[4x]=\begin{cases} 0, & 0<x<\frac14,\\[4pt] 1, & \frac14\le x<\frac12,\\[4pt] 2, & \frac12\le x<\frac34,\\[4pt] 3, & \frac34\le x<1. \end{cases}

So $f$ is piecewise:

f(x)=\begin{cases} 0, & 0<x<\frac14,\\[4pt] \left(x-\frac14\right)^2\left(x-\frac12\right), & \frac14\le x<\frac12,\\[4pt] 2\left(x-\frac14\right)^2\left(x-\frac12\right), & \frac12\le x<\frac34,\\[4pt] 3\left(x-\frac14\right)^2\left(x-\frac12\right), & \frac34\le x<1. \end{cases}

Potential trouble points are where $[4x]$ changes value: $x=\frac14,\ \frac12,\ \frac34.$

Check continuity at the jump points

Let $g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$ Then $f(x)=[4x]g(x)$ .

At $x=\frac14$

For $x<\frac14$ , $f(x)=0$ , so $\lim_{x\to(1/4)^-}f(x)=0.$
For $x\ge \frac14$ , $f(x)=g(x)$ near $1/4$ , and $g\left(\frac14\right)=0.$ Hence $\lim_{x\to(1/4)^+}f(x)=0=f\left(\frac14\right).$

So $f$ is continuous at $x=\frac14$ .

At $x=\frac12$

For $x<\frac12$ , $f(x)=g(x)$ , so $\lim_{x\to(1/2)^-}f(x)=g\left(\frac12\right)=0.$
For $x\ge \frac12$ , $f(x)=2g(x)$ , so $\lim_{x\to(1/2)^+}f(x)=2g\left(\frac12\right)=0.$ Also, $f\left(\frac12\right)=2\left(\frac14\right)^2(0)=0.$

So $f$ is continuous at $x=\frac12$ .

At $x=\frac34$

For $x<\frac34$ , $f(x)=2g(x)$ , so $\lim_{x\to(3/4)^-}f(x)=2g\left(\frac34\right).$
For $x>\frac34$ , $f(x)=3g(x)$ , so $\lim_{x\to(3/4)^+}f(x)=3g\left(\frac34\right).$ Now, $g\left(\frac34\right)=\left(\frac34-\frac14\right)^2\left(\frac34-\frac12\right)=\left(\frac12\right)^2\left(\frac14\right)=\frac1{16}.$ Thus, $\lim_{x\to(3/4)^-}f(x)=\frac18, \qquad \lim_{x\to(3/4)^+}f(x)=\frac3{16}.$ These are unequal, so $f$ is discontinuous at $x=\frac34$ .

Hence there is exactly one discontinuity in $(0,1)$ .

So A is true.

Check differentiability

Inside each open interval $\left(0,\frac14\right),\ \left(\frac14,\frac12\right),\ \left(\frac12,\frac34\right),\ \left(\frac34,1\right),$ $f$ is a polynomial (or zero), hence differentiable.

So only $x=\frac14,\frac12,\frac34$ need checking.

At $x=\frac14$

Since $f(x)=0$ for $x<\frac14$ , left derivative is $0$ .

For $x>\frac14$ , near $1/4$ , $f(x)=g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$ Then

=\lim_{x\to1/4^+}\left(x-\frac14\right)\left(x-\frac12\right)=0.$$ So both one-sided derivatives are $0$. Hence $f$ is **differentiable at $x=\frac14$**. ### At $x=\frac12$ Left side: $f(x)=g(x)$, right side: $f(x)=2g(x)$. Since $g(1/2)=0$, $$f'_-(1/2)=g'(1/2), \qquad f'_+(1/2)=2g'(1/2).$$ Compute $g'(1/2)$: $$g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$$ Differentiate: $$g'(x)=2\left(x-\frac14\right)\left(x-\frac12\right)+\left(x-\frac14\right)^2.$$ At $x=\frac12$, $$g'\left(\frac12\right)=0+\left(\frac14\right)^2=\frac1{16}.$$ Thus $$f'_-(1/2)=\frac1{16}, \qquad f'_+(1/2)=\frac18.$$ They are unequal, so $f$ is **not differentiable at $x=\frac12$**. But it is continuous there. Therefore, there is exactly one point where $f$ is continuous but not differentiable. So **B is true**. ### At $x=\frac34$ Since $f$ is discontinuous at $x=\frac34$, it is automatically **not differentiable** there. Therefore points of non-differentiability are exactly $$x=\frac12,\ \frac34,$$ which are only $2$ points. So **C is false** because it says non-differentiable at more than three points. --- 4. **Find the minimum value of $f$** We inspect each interval. ### On $\left(0,\frac14\right)$ $$f(x)=0.$$ Minimum here is $0$. ### On $\left[\frac14,\frac12\right)$ $$f(x)=g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$$ Here $(x-1/2)<0$ and $(x-1/4)^2\ge 0$, so $f\le 0$. We find its minimum by critical points. Using $$g'(x)=2\left(x-\frac14\right)\left(x-\frac12\right)+\left(x-\frac14\right)^2 =\left(x-\frac14\right)\left[2\left(x-\frac12\right)+\left(x-\frac14\right)\right],$$ $$g'(x)=\left(x-\frac14\right)\left(3x-\frac54\right).$$ Critical points are $$x=\frac14,\quad x=\frac5{12}.$$ In the interval $(1/4,1/2)$, relevant interior point is $x=5/12$. Evaluate: $$g\left(\frac5{12}\right)=\left(\frac5{12}-\frac14\right)^2\left(\frac5{12}-\frac12\right) =\left(\frac16\right)^2\left(-\frac1{12}\right) =-\frac1{432}.$$ So minimum on this piece is $-\frac1{432}$. ### On $\left[\frac12,\frac34\right)$ $$f(x)=2g(x),$$ and now $g(x)\ge 0$ since $(x-1/2)\ge 0$. So minimum here is $0$ at $x=1/2$. ### On $\left[\frac34,1\right)$ $$f(x)=3g(x)>0,$$ so minimum here is positive. Hence the global minimum on $(0,1)$ is $$-\frac1{432},$$ not $-\frac1{512}$. So **D is false**. --- 5. **Final conclusion** The true statements are: $$\boxed{A,\ B}$$

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: A, B

Understand the floor term

Given

f(x)=[4x]\left(x-\frac14\right)^2\left(x-\frac12\right), \qquad x\in(0,1).

Since $x\in(0,1)$ , we have $4x\in(0,4)$ . Therefore,

[4x]=\begin{cases} 0, & 0<x<\frac14,\\[4pt] 1, & \frac14\le x<\frac12,\\[4pt] 2, & \frac12\le x<\frac34,\\[4pt] 3, & \frac34\le x<1. \end{cases}

So $f$ is piecewise:

f(x)=\begin{cases} 0, & 0<x<\frac14,\\[4pt] \left(x-\frac14\right)^2\left(x-\frac12\right), & \frac14\le x<\frac12,\\[4pt] 2\left(x-\frac14\right)^2\left(x-\frac12\right), & \frac12\le x<\frac34,\\[4pt] 3\left(x-\frac14\right)^2\left(x-\frac12\right), & \frac34\le x<1. \end{cases}

Potential trouble points are where $[4x]$ changes value: $x=\frac14,\ \frac12,\ \frac34.$

Check continuity at the jump points

Let $g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$ Then $f(x)=[4x]g(x)$ .

At $x=\frac14$

For $x<\frac14$ , $f(x)=0$ , so $\lim_{x\to(1/4)^-}f(x)=0.$
For $x\ge \frac14$ , $f(x)=g(x)$ near $1/4$ , and $g\left(\frac14\right)=0.$ Hence $\lim_{x\to(1/4)^+}f(x)=0=f\left(\frac14\right).$

So $f$ is continuous at $x=\frac14$ .

At $x=\frac12$

For $x<\frac12$ , $f(x)=g(x)$ , so $\lim_{x\to(1/2)^-}f(x)=g\left(\frac12\right)=0.$
For $x\ge \frac12$ , $f(x)=2g(x)$ , so $\lim_{x\to(1/2)^+}f(x)=2g\left(\frac12\right)=0.$ Also, $f\left(\frac12\right)=2\left(\frac14\right)^2(0)=0.$

So $f$ is continuous at $x=\frac12$ .

At $x=\frac34$

For $x<\frac34$ , $f(x)=2g(x)$ , so $\lim_{x\to(3/4)^-}f(x)=2g\left(\frac34\right).$
For $x>\frac34$ , $f(x)=3g(x)$ , so $\lim_{x\to(3/4)^+}f(x)=3g\left(\frac34\right).$ Now, $g\left(\frac34\right)=\left(\frac34-\frac14\right)^2\left(\frac34-\frac12\right)=\left(\frac12\right)^2\left(\frac14\right)=\frac1{16}.$ Thus, $\lim_{x\to(3/4)^-}f(x)=\frac18, \qquad \lim_{x\to(3/4)^+}f(x)=\frac3{16}.$ These are unequal, so $f$ is discontinuous at $x=\frac34$ .

Hence there is exactly one discontinuity in $(0,1)$ .

So A is true.

Check differentiability

Inside each open interval $\left(0,\frac14\right),\ \left(\frac14,\frac12\right),\ \left(\frac12,\frac34\right),\ \left(\frac34,1\right),$ $f$ is a polynomial (or zero), hence differentiable.

So only $x=\frac14,\frac12,\frac34$ need checking.

At $x=\frac14$

Since $f(x)=0$ for $x<\frac14$ , left derivative is $0$ .

For $x>\frac14$ , near $1/4$ , $f(x)=g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$ Then

=\lim_{x\to1/4^+}\left(x-\frac14\right)\left(x-\frac12\right)=0.$$ So both one-sided derivatives are $0$. Hence $f$ is **differentiable at $x=\frac14$**. ### At $x=\frac12$ Left side: $f(x)=g(x)$, right side: $f(x)=2g(x)$. Since $g(1/2)=0$, $$f'_-(1/2)=g'(1/2), \qquad f'_+(1/2)=2g'(1/2).$$ Compute $g'(1/2)$: $$g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$$ Differentiate: $$g'(x)=2\left(x-\frac14\right)\left(x-\frac12\right)+\left(x-\frac14\right)^2.$$ At $x=\frac12$, $$g'\left(\frac12\right)=0+\left(\frac14\right)^2=\frac1{16}.$$ Thus $$f'_-(1/2)=\frac1{16}, \qquad f'_+(1/2)=\frac18.$$ They are unequal, so $f$ is **not differentiable at $x=\frac12$**. But it is continuous there. Therefore, there is exactly one point where $f$ is continuous but not differentiable. So **B is true**. ### At $x=\frac34$ Since $f$ is discontinuous at $x=\frac34$, it is automatically **not differentiable** there. Therefore points of non-differentiability are exactly $$x=\frac12,\ \frac34,$$ which are only $2$ points. So **C is false** because it says non-differentiable at more than three points. --- 4. **Find the minimum value of $f$** We inspect each interval. ### On $\left(0,\frac14\right)$ $$f(x)=0.$$ Minimum here is $0$. ### On $\left[\frac14,\frac12\right)$ $$f(x)=g(x)=\left(x-\frac14\right)^2\left(x-\frac12\right).$$ Here $(x-1/2)<0$ and $(x-1/4)^2\ge 0$, so $f\le 0$. We find its minimum by critical points. Using $$g'(x)=2\left(x-\frac14\right)\left(x-\frac12\right)+\left(x-\frac14\right)^2 =\left(x-\frac14\right)\left[2\left(x-\frac12\right)+\left(x-\frac14\right)\right],$$ $$g'(x)=\left(x-\frac14\right)\left(3x-\frac54\right).$$ Critical points are $$x=\frac14,\quad x=\frac5{12}.$$ In the interval $(1/4,1/2)$, relevant interior point is $x=5/12$. Evaluate: $$g\left(\frac5{12}\right)=\left(\frac5{12}-\frac14\right)^2\left(\frac5{12}-\frac12\right) =\left(\frac16\right)^2\left(-\frac1{12}\right) =-\frac1{432}.$$ So minimum on this piece is $-\frac1{432}$. ### On $\left[\frac12,\frac34\right)$ $$f(x)=2g(x),$$ and now $g(x)\ge 0$ since $(x-1/2)\ge 0$. So minimum here is $0$ at $x=1/2$. ### On $\left[\frac34,1\right)$ $$f(x)=3g(x)>0,$$ so minimum here is positive. Hence the global minimum on $(0,1)$ is $$-\frac1{432},$$ not $-\frac1{512}$. So **D is false**. --- 5. **Final conclusion** The true statements are: $$\boxed{A,\ B}$$

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