JAMIA 2023 — Mathematics PYQ

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Question

JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

The values of the parameter a such that the roots α,β of the equation 2x²+6x+a=0 satisfy The inequality $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}<2$ are

Choose the correct answer:

Explanation

Equation: $2x^2+6x+a=0$

1. Roots ka Relation:

\alpha + \beta = \frac{-6}{2} = -3

\alpha\beta = \frac{a}{2}

2. Inequality solve karein:

\frac{\alpha}{\beta} + \frac{\beta}{\alpha} < 2

\frac{\alpha^2 + \beta^2}{\alpha\beta} < 2

3. $(\alpha+\beta)^2$ ka use karein:

\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} < 2

\frac{(-3)^2 - 2(\frac{a}{2})}{\frac{a}{2}} < 2

4. Simplify karein:

\frac{9 - a}{\frac{a}{2}} < 2

\frac{2(9 - a)}{a} < 2

\frac{9 - a}{a} < 1

5. Rational Inequality solve karein:

\frac{9 - a}{a} - 1 < 0

\frac{9 - a - a}{a} < 0

\frac{9 - 2a}{a} < 0

6. Critical Points ( $a=0$ aur $a=\frac{9}{2}$ ) se intervals check karein:

Case 1 ( $a < 0$ ): $\frac{(+)}{(-)} < 0$ (Satisfied)
Case 2 ( $0 < a < \frac{9}{2}$ ): $\frac{(+)}{(+)} > 0$ (Not Satisfied)
Case 3 ( $a > \frac{9}{2}$ ): $\frac{(-)}{(+)} < 0$ (Satisfied)

Final Answer:

a < 0 \quad \text{or} \quad a > \frac{9}{2}