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Explain solution for RD Sharma Class 12 Chapter Functions Exercise 2.3 Question 8 maths textbook solution.

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Answer :

\begin{aligned} &(g o f)(x)=\log _{e} \sqrt{1-x} \\ &(f \circ g)(x)=\log _{e} \sqrt{1-x} \end{aligned}

Hint : Domain of f and g are R

           Range of f=(-\infty ,1)

           Range of f=(0,e)

Given : Here given that f(x)=\sqrt{1-x} \text { and } g(x)=\log _{e} x

Solution :

Clearly Range f\; \; \subset domain of g

\Rightarrow \quad g \circ f exists

And Range g\; \; \subset domain of f

\Rightarrow \quad f \circ g

\begin{aligned} \therefore(g \circ f)(x) &=g(f(x)) \\ &=g(\sqrt{1-x}) \\ &=\log _{e} \sqrt{1-x} \end{aligned}

Again, we will compute f \circ g

\begin{aligned} \therefore(f \circ g)(x) &=f(g(x)) \\ &=f\left(\log _{e} x\right) \\ (f \circ g)(x)=& \log _{e} \sqrt{1-x} \end{aligned}

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