JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The number of real roots of the equation $\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$ is:

Choose the correct answer:

Explanation

Solution:

Sabse pehle terms ko factorize karte hain:

\sqrt{(x-1)(x-3)} + \sqrt{(x-3)(x+3)} = \sqrt{2(2x^2 - 7x + 3)}

\sqrt{(x-1)(x-3)} + \sqrt{(x-3)(x+3)} = \sqrt{2(2x-1)(x-3)}

Yahan se clear hai ki $x = 3$ ek solution hai kyunki ye sabhi square roots ko zero kar deta hai. Ab baaki roots ke liye $\sqrt{x-3}$ se divide karte hain (assuming $x \neq 3$ ):

\sqrt{x-1} + \sqrt{x+3} = \sqrt{2(2x-1)}

Dono taraf square karne par:

(x-1) + (x+3) + 2\sqrt{(x-1)(x+3)} = 2(2x-1)

2x + 2 + 2\sqrt{x^2 + 2x - 3} = 4x - 2

2\sqrt{x^2 + 2x - 3} = 2x - 4

\sqrt{x^2 + 2x - 3} = x - 2

Phir se square karne par:

x^2 + 2x - 3 = x^2 - 4x + 4

6x = 7 \implies x = \frac{7}{6}

Lekin, humein domain check karni hogi. $x = \frac{7}{6}$ par pehla term $\sqrt{x^2-4x+3}$ negative value dega square root ke andar ( $(\frac{7}{6}-1)(\frac{7}{6}-3) < 0$ ), jo real nahi hai.

Isliye sirf $x = 3$ hi ekmatra real root hai.

Correct Option: (2) 1

Choose the correct answer:

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