JEE Mathematics 2025 Question with Solution

Given: $f(x) = \frac{x}{3} + \frac{3}{x} + 3$ with $x \neq 0$.

Find: $\sum_{i=1}^{5} \alpha_i^2$ from the intervals of increase and decrease.

Differentiate the function:

$$f'(x) = \frac{d}{dx}\left(\frac{x}{3} + \frac{3}{x} + 3\right) = \frac{1}{3} - \frac{3}{x^2}$$

Now find the critical and break points. From $f'(x)=0$,

$$\frac{1}{3} - \frac{3}{x^2} = 0$$ $$\frac{1}{3} = \frac{3}{x^2}$$ $$x^2 = 9 \Rightarrow x = \pm 3$$

Also, $f'(x)$ is undefined at $x = 0$, and the function itself is not defined there.

So the real line is divided into the intervals $(-\infty,-3)$, $(-3,0)$, $(0,3)$, and $(3,\infty)$.

Check the sign of $f'(x)$:

• For $x < -3$, $f'(x) > 0$, so the function is increasing.
• For $-3 < x < 0$, $f'(x) < 0$, so the function is decreasing.
• For $0 < x < 3$, $f'(x) < 0$, so the function is decreasing.
• For $x > 3$, $f'(x) > 0$, so the function is increasing.

Hence,

• increasing in $(-\infty,-3) \cup (3,\infty)$,
• decreasing in $(-3,0) \cup (0,3)$.

Matching with the given notation, we get

$$\alpha_1 = -3, \quad \alpha_2 = 3, \quad \alpha_3 = -3, \quad \alpha_4 = 0, \quad \alpha_5 = 3$$

Therefore,

$$\sum_{i=1}^{5} \alpha_i^2 = (-3)^2 + 3^2 + (-3)^2 + 0^2 + 3^2 = 9 + 9 + 9 + 0 + 9 = 36$$

Therefore, the correct option is A.

Given: $f(x) = \frac{x}{3} + \frac{3}{x} + 3$.

Find: the values of $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$ and then compute their squared sum.

Using the derivative,

$$f'(x) = \frac{1}{3} - \frac{3}{x^2}$$

Set it equal to zero:

$$\frac{1}{3} - \frac{3}{x^2} = 0 \Rightarrow x = \pm 3$$

The point $x = 0$ must also be considered because the function is not defined there.

Test values in each interval:

$$f'(-4) = \frac{1}{3} - \frac{3}{16} = \frac{7}{48} > 0$$

So $(-\infty,-3)$ is increasing.

$$f'(-1) = \frac{1}{3} - 3 = -\frac{8}{3} < 0$$

So $(-3,0)$ is decreasing.

$$f'(1) = \frac{1}{3} - 3 = -\frac{8}{3} < 0$$

So $(0,3)$ is decreasing.

$$f'(4) = \frac{1}{3} - \frac{3}{16} = \frac{7}{48} > 0$$

So $(3,\infty)$ is increasing.

Thus the decreasing part must be split at $x = 0$. Therefore,

$$(\alpha_3,\alpha_4) = (-3,0), \quad (\alpha_4,\alpha_5) = (0,3)$$

which gives

$$\alpha_3 = -3, \quad \alpha_4 = 0, \quad \alpha_5 = 3$$

and from the increasing intervals,

$$\alpha_1 = -3, \quad \alpha_2 = 3$$

So,

$$\sum_{i=1}^{5} \alpha_i^2 = 9 + 9 + 9 + 0 + 9 = 36$$

The correct option is A.

The first solution text on the page briefly shows a discrepancy before re-evaluation, but the interval split at $x=0$ resolves it and gives the correct result $36$.