The solution set of ${x^2} + 2 \le 3x \le 2{x^2} - 5,$ is

$\left( { - \infty , - 1} \right] \cup \left[ {5/2,\;\infty } \right)$

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

The solution set of

{x^2} + 2 \le 3x \le 2{x^2} - 5,

View answer and explanation

Correct option: A

\mathrm{\Phi }

Option A (

\Phi

) is correct because there is no value of x that satisfies both inequalities

{x^2} + 2 \le 3x

and

3x \le 2{x^2} - 5

simultaneously.

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)