JEE Mathematics 2025 Question with Solution

Given: $f(x) = \frac{5 - x}{x^2 - 3x + 2}$ with $x \neq 1, 2$.

Find: $\alpha^2 + \beta^2$, where the range is $(-\infty, \alpha] \cup [\beta, \infty)$.

Let

$$y = \frac{5 - x}{(x - 1)(x - 2)}$$

Then

$$5 - x = y(x^2 - 3x + 2)$$

which gives the quadratic in $x$:

$$yx^2 - (3y + 1)x + (2y + 5) = 0$$

For real values of $x$, its discriminant must satisfy

$$(3y + 1)^2 - 4y(2y + 5) \ge 0$$

Now,

$$(3y + 1)^2 = 9y^2 + 6y + 1$$

and

$$4y(2y + 5) = 8y^2 + 20y$$

So,

$$9y^2 + 6y + 1 - (8y^2 + 20y) \ge 0$$ $$y^2 - 14y + 1 \ge 0$$

The corresponding equation

$$y^2 - 14y + 1 = 0$$

has roots

$$y = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}$$

Hence,

$$\alpha = 7 - 4\sqrt{3}, \qquad \beta = 7 + 4\sqrt{3}$$

Now,

$$\alpha + \beta = 14, \qquad \alpha\beta = (7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 1$$

Using

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$

we get

$$\alpha^2 + \beta^2 = 14^2 - 2 \cdot 1 = 196 - 2 = 194$$

Therefore, the correct option is D.](streamdown:incomplete-link)

Given: $f(x) = \frac{5 - x}{x^2 - 3x + 2}$.

Find: $\alpha^2 + \beta^2$.

First factorize the denominator:

$$x^2 - 3x + 2 = (x - 1)(x - 2)$$

So the function is undefined at $x = 1$ and $x = 2$.

Its derivative is

$$f'(x) = \frac{(-1)(x^2 - 3x + 2) - (5 - x)(2x - 3)}{(x^2 - 3x + 2)^2}$$

Simplifying the numerator,

$$-x^2 + 3x - 2 - (10x - 15 - 2x^2 + 3x) = x^2 - 10x + 13$$

Thus,

$$f'(x) = \frac{x^2 - 10x + 13}{(x^2 - 3x + 2)^2}$$

Setting $f'(x) = 0$ gives

$$x^2 - 10x + 13 = 0$$ $$x = \frac{10 \pm \sqrt{100 - 52}}{2} = \frac{10 \pm \sqrt{48}}{2} = 5 \pm 2\sqrt{3}$$

These are the critical points where boundary values of the range occur.

Now let

$$y = \frac{5 - x}{(x - 1)(x - 2)}$$

Then

$$y(x^2 - 3x + 2) = 5 - x$$

so

$$yx^2 - (3y + 1)x + (2y + 5) = 0$$

For real $x$, the discriminant must be non-negative:

$$(3y + 1)^2 - 4y(2y + 5) \ge 0$$

which simplifies to

$$y^2 - 14y + 1 \ge 0$$

Thus the limiting values are

$$y = 7 \pm 4\sqrt{3}$$

So,

$$\alpha = 7 - 4\sqrt{3}, \qquad \beta = 7 + 4\sqrt{3}$$

and hence

$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 14^2 - 2(1) = 194$$

Therefore, the value of $\alpha^2 + \beta^2$ is $194$, so the correct option is D.

Note: The second provided approach contains an algebraic sign error while forming the quadratic in $x$, but it still concludes with the same numerical value $194$.