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The composition of functions MCQs for JEE

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Master The composition of functions through exam-style multiple-choice questions. This page features 5+ JEE MCQs covering The composition of functions, each with verified answers and clear explanations.

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Part of Sets, Relations and Functions in JEE Main Mathematics.

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Q1MathsUnit 1: Sets, Relations and Functions
Let f(x)=2100x+1\boldsymbol{f}(\boldsymbol{x})=\mathbf{2}^{100} \boldsymbol{x}+\mathbf{1} g(x)=3100x+1\boldsymbol{g}(\boldsymbol{x})=\boldsymbol{3}^{100} \boldsymbol{x}+\mathbf{1} Then the set of real numbers x such that f(g(x))=x\boldsymbol{f}(\boldsymbol{g}(\boldsymbol{x}))=\boldsymbol{x} is
Q2MathsUnit 1: Sets, Relations and Functions
Assertion Let ff and gg be increasing and decreasing functions respectively from [0,∞][0, \infty] to [0,∞].Let⁡h(x)=f(g(x)).[0, \infty] . \operatorname{Let} h(x)=f(g(x)) . If h(0)=0,h(0)=0, then h(x)h(x) is always zero Reason h(x)h(x) is an increasing function of xx
Q3MathsUnit 1: Sets, Relations and Functions
If f:R→R\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R} and g:R→R\boldsymbol{g}: \boldsymbol{R} \rightarrow \boldsymbol{R} are defined by f(x)=x−[x]\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}-[\boldsymbol{x}] and g(x)=[x]\boldsymbol{g}(\boldsymbol{x})=[\boldsymbol{x}] for x∈R,\boldsymbol{x} \in \boldsymbol{R}, where [x][\boldsymbol{x}] is the greatest integer not exceeding x,x, then for every x∈x \in R,f(g(x))=\boldsymbol{R}, \boldsymbol{f}(\boldsymbol{g}(\boldsymbol{x}))=
Q4MathsUnit 1: Sets, Relations and Functions
Read the following information and answer the three items that follow: Let f(x)=x2+2x−5\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{x}^{2}+\boldsymbol{2} \boldsymbol{x}-\boldsymbol{5} and g(x)=\boldsymbol{g}(\boldsymbol{x})= 5x+305 x+30 What are the roots of the equation g[f(x)]=0?\boldsymbol{g}[\boldsymbol{f}(\boldsymbol{x})]=\mathbf{0} ?
Q5MathsUnit 1: Sets, Relations and Functions
If f:R→R\boldsymbol{f}: \boldsymbol{R} \rightarrow \boldsymbol{R} is defined by f(x)=2x−\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{2} \boldsymbol{x}- 2, then (f∘f)(x)+2=(f \circ f)(x)+2=

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