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

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 $f^{2}$

Since

\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.