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Need solution for RD Sharma Class 12 Chapter Functions Exercise 2.3 Question 4 Sub question (iv) maths textbook solution.

Answers (1)

Answer : f^{2}(x)=4 x^{2}+20 x+25

Given : f(x)=2 x+5 \text { and } g(x)=x^{2}+1

Here we have to find out f^{2}

Hint : If we want to f^{2} then first we compute f (x)\times f(x).

Solution :

Since f (x)=2x-5 is a polynomial.

          f:R\rightarrow R

          \begin{aligned} f^{2}(x) &=f(x) \times f(x) \\ &=(2 x+5)(2 x+5) \\ &=(2 x+5)^{2} \\ f^{2}(x) &=4 x^{2}+20 x+25 \end{aligned}

In this question we have to prove that

         f\; o\; f=f^{2}

Then first we have to find out f\; o\; f

For f\; o\; f

               \begin{aligned} &\operatorname{fof}(x)=f(f(x)) \\ &=f(2 x+5) \\ &=2(2 x+5)+5 \\ &=4 x+10+5 \\ &\operatorname{fof}(x)=4 x+15 \end{aligned}                                                                  ...(1)

Again we will compute f^{2}


        \begin{aligned} &f^{2}(x)=f(x) \times f(x) \\ &=(2 x+5)(2 x+5) \\ &=4 x^{2}+20 x+25 \end{aligned}                                                                 ...(2)

From (1) and (2) we have

           (f\; o\; f)(x)=f^{2}(x)

Hence proved.

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