JEE Mathematics 2023 Question with Solution

Given: The piecewise function is

$$f(x) = \begin{cases} -2x, & -2 < x < -1, \\ -x, & -1 \leq x < 0, \\ 0, & 0 \leq x < 1, \\ x - 1, & 1 \leq x < 2. \end{cases}$$

Find: The value of $m+n$, where $m$ is the number of points in $(-2,2)$ where $y=|f(x)|$ is not continuous and $n$ is the number of points where it is not differentiable.

Step 1: Check continuity. From the piecewise form, $f(x)$ is discontinuous at $x=-1$. Hence $|f(x)|$ is also discontinuous at $x=-1$. So, $m=1$.

Step 2: Check differentiability. Differentiating on each interval,

$$f'(x) = \begin{cases} -2, & -2 < x < -1, \\ -1, & -1 < x < 0, \\ 0, & 0 < x < 1, \\ 1, & 1 < x < 2. \end{cases}$$

Thus the function is non-differentiable at the junction points $x=-1, 0, 1$. So, $n=3$.

Step 3: Compute the required sum.

$$m+n=1+3=4$$

Therefore, the required value is $4$.

Given:

• For $-2 < x < 0$, $f(x)=x\lfloor x\rfloor$
• For $0 \leq x < 2$, $f(x)=(x-1)\lfloor x\rfloor$

Find: Points where $|f(x)|$ is not continuous and not differentiable.

Use the values of the floor function on subintervals:

• If $-2 < x < -1$, then $\lfloor x\rfloor=-2$, so

$$f(x)=x(-2)=-2x$$

• If $-1 \leq x < 0$, then $\lfloor x\rfloor=-1$, so

$$f(x)=x(-1)=-x$$

• If $0 \leq x < 1$, then $\lfloor x\rfloor=0$, so

$$f(x)=(x-1)\cdot 0=0$$

• If $1 \leq x < 2$, then $\lfloor x\rfloor=1$, so

$$f(x)=(x-1)\cdot 1=x-1$$

Hence,

f(x)=\begin{cases}-2x,& -2