JEE Main 2024 (Online) 6th April Morning Shift | Quadratic Equation and Inequalities Question 6 | Mathematics

Let $$\alpha, \beta$$ be the distinct roots of the equation $$x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R}$$ and $$a_n=\alpha^n+\beta^n$$. Then the minimum value of $$\frac{a_{2023}+a_{2025}}{a_{2024}}$$ is

$$-1 / 2$$

$$-1 / 4$$

$$1 / 4$$

$$1 / 2$$

JEE Main 2024 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

If 2 and 6 are the roots of the equation $$a x^2+b x+1=0$$, then the quadratic equation, whose roots are $$\frac{1}{2 a+b}$$ and $$\frac{1}{6 a+b}$$, is :

$$x^2+8 x+12=0$$

$$2 x^2+11 x+12=0$$

$$4 x^2+14 x+12=0$$

$$x^2+10 x+16=0$$

JEE Main 2024 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

-1

Let $\alpha$ and $\beta$ be the roots of the equation $p x^2+q x-r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^2$ is :

$\frac{20}{3}$

$\frac{80}{9}$

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

Let $\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is :