JEE Mathematics 2025 Question with Solution

Given:

$$f(x) = \begin{cases} 3x, & \text{if } x < 0\\ \min(1+x+\lfloor x \rfloor, 2+x\lfloor x \rfloor), & \text{if } 0 \leq x \leq 2\\ 5, & \text{if } x > 2 \end{cases}$$

Find: The value of $\alpha + \beta$, where $\alpha$ is the number of points of discontinuity and $\beta$ is the number of points of non-differentiability.

For $0 \leq x < 1$, $\lfloor x \rfloor = 0$, so

$$f(x) = \min(1+x, 2)$$

Hence, for $0 \leq x < 1$,

$$f(x) = 1+x$$

For $1 \leq x < 2$, $\lfloor x \rfloor = 1$, so

$$f(x) = \min(1+x+1, 2+x) = \min(x+2, x+2) = x+2$$

Thus the possible critical points are $x = 0, 1, 2$.

Check continuity:

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3x = 0$$ $$\lim_{x \to 0^+} f(x) = 1$$

So $f$ is not continuous at $x = 0$.

At $x = 1$,

$$\lim_{x \to 1^-} f(x) = 1+1 = 2$$ $$\lim_{x \to 1^+} f(x) = 1+2 = 3$$

So $f$ is not continuous at $x = 1$.

At $x = 2$,

$$f(2) = \min(1+2+2, 2+2\cdot 2) = \min(5, 6) = 5$$

For $x > 2$, $f(x) = 5$, so the right-hand limit is also $5$. Hence $f$ is continuous at $x = 2$.

Therefore,

$$\alpha = 2$$

with discontinuities at $x = 0, 1$.

Now check differentiability. Every discontinuity point is automatically a point of non-differentiability, so $x = 0$ and $x = 1$ are included. Also, at $x = 2$, the slope changes from the left piece $x+2$ to the right constant piece $5$, so $f$ is not differentiable at $x = 2$.

Thus,

$$\beta = 3$$

with non-differentiability at $x = 0, 1, 2$.

Therefore,

$$\alpha + \beta = 2 + 3 = 5$$

The required answer is $5$.

Given: The middle part of the function contains the greatest integer term $\lfloor x \rfloor$.

Find: Count discontinuity points and non-differentiability points.

1. For $x < 0$,

$$f(x) = 3x$$

which is continuous and differentiable.

2. For $0 \leq x < 1$,

$$\lfloor x \rfloor = 0$$

so

$$f(x) = \min(1+x+0, 2+x\cdot 0) = \min(1+x, 2)$$

Since $1+x < 2$ for $x < 1$,

$$f(x) = 1+x$$

3. For $1 \leq x < 2$,

$$\lfloor x \rfloor = 1$$

so

$$f(x) = \min(1+x+1, 2+x\cdot 1) = \min(x+2, x+2) = x+2$$

4. At $x = 0$,

• left limit is $0$,
• right limit is $1$. So there is a discontinuity.

5. At $x = 1$,

• left limit from $1+x$ is $2$,
• right limit from $x+2$ is $3$. So there is a discontinuity.

6. At $x = 2$,

$$f(2) = 5$$

and for $x > 2$ also $f(x) = 5$, so the function is continuous there.

Hence the discontinuity points are $x = 0, 1$, giving

$$\alpha = 2$$

For differentiability:

• discontinuity at $x = 0$ and $x = 1$ already implies non-differentiability,
• at $x = 2$, left derivative of $x+2$ is $1$ and right derivative of $5$ is $0$. So $x = 2$ is also non-differentiable.

Therefore,

$$\beta = 3$$

and

$$\alpha + \beta = 5$$

So the final answer is $5$.