JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the value of real number $a > 0$ for which $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2\beta}}$ , then $\beta$ is equal to ______.

Choose the correct answer:

Explanation

Step 1: $\alpha$ ki value nikalne ke liye subtract karein

Equation (1) mein se Equation (2) ko ghatane par:

(\alpha^2 - 5a\alpha + 1) - (\alpha^2 - a\alpha - 5) = 0

-5a\alpha + a\alpha + 1 + 5 = 0

-4a\alpha + 6 = 0

4a\alpha = 6 \implies \alpha = \frac{6}{4a} = \frac{3}{2a}

Step 2: $\alpha$ ki value Equation (2) mein rakhein

Ab $\alpha = \frac{3}{2a}$ ko $x^2 - ax - 5 = 0$ mein rakhte hain:

\left(\frac{3}{2a}\right)^2 - a\left(\frac{3}{2a}\right) - 5 = 0

\frac{9}{4a^2} - \frac{3}{2} - 5 = 0

\frac{9}{4a^2} - \frac{13}{2} = 0

Step 3: $a$ ki value find karein

\frac{9}{4a^2} = \frac{13}{2}

18 = 52a^2

a^2 = \frac{18}{52} = \frac{9}{26}

a = \sqrt{\frac{9}{26}} = \frac{3}{\sqrt{26}}

Step 4: $\beta$ ki value nikalne ke liye compare karein

Sawal mein diya gaya hai ki $a = \frac{3}{\sqrt{2\beta}}$ .

Hamari nikal li gayi value hai $a = \frac{3}{\sqrt{26}}$ .

Dono ko compare karne par:

\sqrt{2\beta} = \sqrt{26}

2\beta = 26

\beta = 13

Answer: $\beta$ ki value 13 hai.

Choose the correct answer:

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