JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the system of linear equations: $7x + 11y + \alpha z = 13$ $5x + 4y + 7z = \beta$ $175x + 194y + 57z = 361$ has infinitely many solutions, then $\alpha + \beta + 2$ is equal to:

Choose the correct answer:

Explanation

Solution

Infinitely many solutions ke liye, determinant $\Delta = 0$ hona chahiye aur equations linear combination mein honi chahiye.

Maan lijiye Equation 3, pehli do equations ka combination hai:

E_3 = k_1 E_1 + k_2 E_2

$x$ aur $y$ ke coefficients ko compare karne par:

$7k_1 + 5k_2 = 175 \quad \dots(i)$

$11k_1 + 4k_2 = 194 \quad \dots(ii)$

Inhe solve karne par humein $k_1 = 10$ aur $k_2 = 21$ milta hai. Ab yahi values $z$ aur constant term ke liye use karenge:

For $\alpha$ : $10(\alpha) + 21(7) = 57 \implies 10\alpha + 147 = 57 \implies 10\alpha = -90 \implies \alpha = -9$
For $\beta$ : $10(13) + 21(\beta) = 361 \implies 130 + 21\beta = 361 \implies 21\beta = 231 \implies \beta = 11$

Final Value:

\alpha + \beta + 2 = -9 + 11 + 2 = 4

Choose the correct answer:

Explore More