#### Need solution for RD Sharma Maths Class 12 Chapter Function Exercise 2.2 Question 9.

Answer : $\fn_cm h\; o(g\; o\; f)=(h\; o\; g)o\; f$

Hint : $\fn_cm g\; o\; f$ means $\fn_cm f(x)$ function is in $\fn_cm g(x)$ function

$\fn_cm f\; o\; g$ means $\fn_cm g(x)$ function is in $\fn_cm f(x)$ function

For associative property $\fn_cm h\; o(g\; o\; f)=(h\; o\; g)o\; f$

Given :

\fn_cm \begin{aligned} &f: N \rightarrow Z_{0},: Z_{0} \rightarrow Q, R: Q \rightarrow R \\ &f(x)=2 x \\ &g(x)=\frac{1}{x} \\ &h(x)=e^{x} \end{aligned}

Solution :

\fn_cm \begin{aligned} &g \circ f: N \rightarrow Q \text { and } h \circ g: Z_{0} \rightarrow R \\ &h \circ(g \circ f): N \rightarrow R \text { and }(h \circ g) \circ f: N \rightarrow R \end{aligned}

So, both have same domains                                                    $\fn_cm \left [ \text {since}f(x)=2x \right ]$

$\fn_cm \begin{gathered} (g \circ f)(x)=g[f(x)] \\ =g(2 x) \\ =\frac{1}{2 x} \\ (h \circ g)(x)=h[g(x)] \\ =h\left(\frac{1}{x}\right) \Rightarrow e^{\frac{1}{x}} \end{gathered}$                                                                     .....(ii)

Now,

$\fn_cm \begin{gathered} {[h \circ(g \circ f)](x)=h[(g \circ f)(x)]} \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =h\left(\frac{1}{2 x}\right) \end{gathered}$                                       ....[From (i)]

$\fn_cm =e^{\frac{1}{2x}}$                                                           ....[From (ii)]

$\fn_cm [(h \circ g) \circ f(x)]=(h \circ g)[f(x)]$

$\fn_cm =(h \circ g)(2 x)$                                           ...[since $\fn_cm f(x)=2x$]

$\fn_cm =e^{\frac{1}{2x}}$                                                           ....[From (ii)]

$\fn_cm [h \circ(g \circ f)](x)=[(h \circ g) \circ f(x)], \quad \in N$

So,     $\fn_cm h \circ(g \circ f)=(h \circ g) \circ f$

Hence, the associative property has benn verified.