If the range of the function f(x) = (5 - x)/(x^2 - 3x + 2), x ≠ 1, 2, is (-∞, α] ∪ [β, ∞), then α^2 + β^2 is equal to :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q2 – JEE-MAIN 2025

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Home Maths Relations and FunctionsQuestion #383

If the range of the function

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

x \neq 1, 2

, is

(- \infty, α] \cup [β, \infty)

, then

α^{2} + β^{2}

is equal to :

Clue 1 of 3: Range of Rational Functions

To find the range of a rational function $y = f (x)$ , express $x$ in terms of $y$ . This often leads to a quadratic equation in $x$ .

🧠 Full Solution Path

Step 1: Express

x

in terms of

y

✦ Active

Let the given function be $y = f (x)$ . We need to rearrange the equation to express $x$ in terms of $y$ . The denominator can be factored as $x^{2} - 3 x + 2 = (x - 1) (x - 2)$ . So, the function is $y = \frac{5 - x}{(x - 1) (x - 2)}$ . Cross-multiplying and rearranging terms to form a quadratic equation in $x$ :

y (x^{2} - 3 x + 2) = 5 - x

y x^{2} - 3 y x + 2 y = 5 - x

y x^{2} + (- 3 y + 1) x + (2 y - 5) = 0

This is a quadratic equation in $x$ of the form $A x^{2} + B x + C = 0$ , where $A = y$ , $B = (- 3 y + 1)$ , and $C = (2 y - 5)$ .

Step 2: Apply the Discriminant Condition to find the range○ Expand

For $x$ to be a real number, the discriminant $D = B^{2} - 4 A C$ must be greater than or equal to zero.

(- 3 y + 1)^{2} - 4 (y) (2 y - 5) \geq 0

(9 y^{2} - 6 y + 1) - (8 y^{2} - 20 y) \geq 0

y^{2} + 14 y + 1 \geq 0

To solve this inequality, find the roots of $y^{2} + 14 y + 1 = 0$ using the quadratic formula $y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

y = \frac{- 14 \pm \sqrt{14^{2} - 4 (1) (1)}}{2 (1)} = \frac{- 14 \pm \sqrt{196 - 4}}{2} = \frac{- 14 \pm \sqrt{192}}{2}

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

The roots are $y_{1} = - 7 - 4 \sqrt{3}$ and $y_{2} = - 7 + 4 \sqrt{3}$ . Since the leading coefficient of $y^{2} + 14 y + 1$ is positive, the inequality $y^{2} + 14 y + 1 \geq 0$ holds when $y \leq y_{1}$ or $y \geq y_{2}$ . Thus, the range is $(- \infty, - 7 - 4 \sqrt{3}] \cup [- 7 + 4 \sqrt{3}, \infty)$ . Comparing this with the given range $(- \infty, α] \cup [β, \infty)$ , we identify $α = - 7 - 4 \sqrt{3}$ and $β = - 7 + 4 \sqrt{3}$ . (Note: The excluded values $x = 1, 2$ do not lead to any specific $y$ values that need to be removed from the range, as substituting $x = 1$ or $x = 2$ into the quadratic in $x$ leads to a contradiction, e.g., $- 4 = 0$ or $- 3 = 0$ ).

💡 Teacher's Secret Hint

Remember to check for any values of $y$ that would make the original denominator zero, but in this case, $x = 1$ and $x = 2$ do not correspond to any real $y$ values from the quadratic equation.

Step 3: Calculate

α^{2} + β^{2}

○ Expand

Now, we calculate $α^{2} + β^{2}$ using the values found for $α$ and $β$ :

α^{2} + β^{2} = (- 7 - 4 \sqrt{3})^{2} + (- 7 + 4 \sqrt{3})^{2}

This expression is of the form $(A - B)^{2} + (A + B)^{2} = 2 (A^{2} + B^{2})$ , where $A = - 7$ and $B = 4 \sqrt{3}$ .

α^{2} + β^{2} = 2 ((- 7)^{2} + (4 \sqrt{3})^{2})

= 2 (49 + 16 \times 3)

= 2 (49 + 48)

= 2 (97)

= 194

💡 Teacher's Secret Hint

Utilize the algebraic identity $(a - b)^{2} + (a + b)^{2} = 2 (a^{2} + b^{2})$ to simplify calculations.

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Maths Question 2 – JEE-MAIN 2025

Clue 1 of 3: Range of Rational Functions

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