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

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

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

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

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

Given :

         \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 :

          \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                                                    \left [ \text {since}f(x)=2x \right ]

        \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)


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

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

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

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

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

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

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

Hence, the associative property has benn verified.      

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