Question

If $f\left(x\right)=\{\begin{array}{cc}2x+3,& \text{if}x<0\\ x^2-1,& \text{if}x\ge 0\end{array}$

and $g\left(x\right)=\{\begin{array}{cc}{x}^3,& \text{if}x<1\\ 3x-5,& \text{if}x\ge 1\end{array}$,

then the function $f\left(g\right(x\left)\right)$ is

• A. one-one and onto
• B. many one and onto
• C. one-one and into
• D. many one and into

Answer 1

The correct option is B many one and onto

$f\left(g\right(x\left)\right)=$$\{\begin{array}{cc}2\left(g\right(x\left)\right)+3& \text{if}g\left(x\right)<0\\ \left(g\right(x){)}^2-1& \text{if}g\left(x\right)\ge 0\end{array}$

$=\{\begin{array}{cc}2x^3+3& ,x^3<0,x<1\\ 2(3x-5)+3& ,3x-5<0,x\ge 1\\ x^6-1& ,x^3\ge 0,x<1\\ (3x-5{)}^2-1& ,3x-5\ge 0,x\ge 1\end{array}$

$=\{\begin{array}{cc}2x^3+3& ,x\in (-\infty ,0)\\ 6x-7& ,x\in [1,\frac{5}{3})\\ x^6-1& ,x\in [0,1)\\ (3x-5{)}^2-1& ,x\in [\frac{5}{3},\infty )\end{array}$

$=\{\begin{array}{c}2x^3+3\in (-\infty ,3)\\ 6x-7\in [-1,3)\\ x^6-1\in [-1,0)\\ (3x-5{)}^2-1\in [-1,\infty )\end{array}$

So, the range of $f\left(x\right)$ is $R$ and function is many-one.