JEE Main 2019 (Online) 10th April Morning Slot | Quadratic Equation and Inequalities Question 112 | Mathematics

If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
x² + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to :

$${{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}$$

$${{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}$$

$${{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$$

$${{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}$$

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

If m is chosen in the quadratic equation

(m² + 1) x² – 3x + (m² + 1)² = 0

such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-

$$4\sqrt 3 $$

$$8\sqrt 3 $$

$$8\sqrt 5 $$

$$10\sqrt 5 $$

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

Let p, q $$ \in $$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic equation, x² + px + q = 0, then :

p² – 4q – 12 = 0

q² – 4p – 16 = 0

q² + 4p + 14 = 0

p² – 4q + 12 = 0