Explain solution for RD Sharma Class 12 Chapter Function Exercise 2.3 Question 5 maths textbook solution.

Answer : $f \circ g(x)=\sin 2 x ; g \circ f(x)=2 \sin x$ and they are not equal functions.

Given : Here given that $f(x)=\sin x \; \text {and} \; g(x)=2 x$

Here we have to compute $g\; o\; f$ and $f\; o\; g$

Hint : First we will compute $g\; o\; f$

we know that $f: R \rightarrow[-1,1] \text { and } g: R \rightarrow R$

Clearly the range of f is a subset of the domain of g.

Second, we will compute $f\; o\; g$

Clearly the range of g is a subset of the domain of f.

Solution :

\begin{aligned} &g \circ f: R \rightarrow R \\ &(g \circ f)(x)=g(f(x)) \\ &\qquad \begin{aligned} &=g(\sin x) \\ &=2 \sin x \end{aligned} \end{aligned}                                                               ....(1)

\begin{aligned} &f \circ g: R \rightarrow R \\ &\text { So, } f \circ g(x)=f(g(x)) \\ &\qquad \begin{aligned} &=f(2 x) \\ &=\sin 2 x \end{aligned} \end{aligned}                                                         ....(2)

From (1) and (2) clearly $f\; o\; g\neq g\; o\; f$

Hence, they are not equal functions.