The value of $x$ for which ${\cos ^{ - 1}}(\cos 4) > 3{x^2} - 4x\;$ is

$\left( {0,\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)$

$\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},0} \right)$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

The value of

x

for which

{\cos ^{ - 1}}(\cos 4) > 3{x^2} - 4x\;

View answer and explanation

Correct option: D

\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)

Option D (

\left( {\frac{{2 - \sqrt {6\pi - 8} }}{3},\frac{{2 + \sqrt {6\pi - 8} }}{3}} \right)

) is correct because it represents the interval of x values that satisfy the inequality after correctly evaluating

{\cos ^{ - 1}}(\cos 4)