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

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

Answers (1)

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)

Now,

             \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.      

Posted by

infoexpert23

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