If ${3^x} + {2^{2x}} \ge {5^x},$ then the solution set for $x$ is

JEE MATHEMATICS MCQ: FUNCTIONS

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{3^x} + {2^{2x}} \ge {5^x},

then the solution set for

x

View answer and explanation

Correct option: A

\left( { - \infty ,2} \right]

Option A (

(- \infty, 2]

) is correct because the inequality holds true for x less than or equal to 2.

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\}