#### Explain solution RD Sharma class 12 chapter Functions exercise 2.1 question 11 maths

f  is bijection.

Given:

$f:R \rightarrow R$  is a function defined by $f(x)=4x^3+7$.

To prove:

f  is bijection.

Hint:

Any function to be bijection. The given function should be one-one and onto.

Solution:

First we will check whether the function is one-one or not.

Let x and y be any two element in the domain R such that

\begin{aligned} & f(x)=f(y) \\ \Rightarrow & 4 x^{3}+7=4 y^{3}+7 \\ \Rightarrow & \quad 4 x^{3}=4 y^{3} \\ \Rightarrow & x^{3}=y^{3} \\ \Rightarrow & x=y \end{aligned}

So, f is one-one.

Now, we will check if the given function is onto or not.

Let y be any element in the co-domain R such that $f(x)=y$ for some element x in R (domain).

$f(x)=y$

\begin{aligned} &\Rightarrow \quad 4 x^{3}+7=y \\ &\Rightarrow \quad 4 x^{3}=y-7 \\ &\Rightarrow \quad x=\sqrt[3]{y-7} \end{aligned}

So, for every element in the co-domain there exists some pre-image in the domain.

So, f is onto.

Hence  f is bijection.