jee-main 2024 Q71

jee-main · India · session1_29jan_shift2 Solving quadratics and applications Finding roots or coefficients of a quadratic using Vieta's relations

Let $x = \frac { m } { n } \left( m , n \right.$ are co-prime natural numbers) be a solution of the equation $\cos \left( 2 \sin ^ { - 1 } x \right) = \frac { 1 } { 9 }$ and let $\alpha , \beta ( \alpha > \beta )$ be the roots of the equation $m x ^ { 2 } - n x - m + n = 0$. Then the point $( \alpha , \beta )$ lies on the line
(1) $3 x + 2 y = 2$
(2) $5 x - 8 y = - 9$
(3) $3 x - 2 y = - 2$
(4) $5 x + 8 y = 9$

Let $x = \frac { m } { n } \left( m , n \right.$ are co-prime natural numbers) be a solution of the equation $\cos \left( 2 \sin ^ { - 1 } x \right) = \frac { 1 } { 9 }$ and let $\alpha , \beta ( \alpha > \beta )$ be the roots of the equation $m x ^ { 2 } - n x - m + n = 0$. Then the point $( \alpha , \beta )$ lies on the line\\
(1) $3 x + 2 y = 2$\\
(2) $5 x - 8 y = - 9$\\
(3) $3 x - 2 y = - 2$\\
(4) $5 x + 8 y = 9$

Paper Questions

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