JEE Main 2019 (Online) 10th April Evening Slot | Quadratic Equation and Inequalities Question 110 | Mathematics

All the pairs (x, y) that satisfy the inequality
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation

sin x = |sin y|

sin x = 2sin y

2 sin x = sin y

2 |sin x | = 3 sin y

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

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 :-

$$4\sqrt 3 $$

$$8\sqrt 3 $$

$$8\sqrt 5 $$

$$10\sqrt 5 $$

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