JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

The set of all values of

x

satisfying the inequality

\log _2 \sqrt{x}-2\left(\log _{1 / 4} x\right)^2+1>0

is the interval

View answer and explanation

Correct option: C

\left(\frac{1}{2}, 4\right)

Option C (

\frac{1}{2}, 4

) is correct because it represents the interval of x values that satisfy the given logarithmic inequality after simplification and solving.

More Questions from FUNCTIONS

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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\}