Question

If f: $R\to R$ be a function defined by f(x) = $4x^3-7$. Then

• A. f is one-one -into
• B. f is many-one -into
• C. f is many-one onto
• D. f is bijective

Answer 1

The correct option is D f is bijective

We have f(x) $=4x_7^3,xϵR$. f is one- one . Let $x_1,x_2ϵR$ and $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 4x_1^3-7=4x_2^3-7\Rightarrow 4x_1^3=4x_2^3$
$\Rightarrow x_1^3=x_2^3\Rightarrow x_1^3-x_2^3=0$
$\Rightarrow (x_1-x_2)(x_1^2+x_1{x}_2+x_2^2)=0$
$\Rightarrow (x_1-x_2)[{(x_1+\frac{x_2}{2})}^2+\frac{3x_2^2}{4}]=0$
$\Rightarrow x_1-x_2=0$, because the other factor is non-zero. $\Rightarrow x_1=x_2\therefore$ f is one-one f is onto. Let k $\epsilon$ R any real number
f(x) = $k\Rightarrow 4x^3-7=k\Rightarrow x={\left[\frac{k+7}{4}\right]}^{1/3}$
Now ${\left[\frac{k+7}{4}\right]}^{1/3}\epsilon R$, because $k\epsilon R$ and $f\left[{\left(\frac{k+7}{4}\right)}^{1/3}\right]=4{\left[{\left(\frac{k+7}{4}\right)}^{1/3}\right]}^3-7$
$=4\left[\frac{k+7}{4}\right]-7=k$
$\therefore$ k is the image of ${\left[\frac{k+7}{4}\right]}^{1/3}$
$\therefore$ f is onto. $\therefore$ f is a bijective function.