If $\sqrt {9{x^2} + 6x + 1} < \left( {2 - x} \right),$ then

$x \in \left( { - \frac{3}{2},\;\frac{1}{4}} \right)$

$x \in \left[ { - \frac{3}{2},\;\frac{1}{4}} \right]$

JEE MATHEMATICS MCQ: FUNCTIONS

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Application +4 | -1

\sqrt {9{x^2} + 6x + 1} < \left( {2 - x} \right),

then

View answer and explanation

Correct option: D

x < \frac{1}{4}

The correct answer is D (

x < \frac{1}{4}

) because solving the inequality after removing the square root leads to this solution.

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}