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MHT CET 2020 Math Solved Paper

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Question : 4 of 50

Marks: +1, -0

If

f: R \rightarrow R

be a function defined by

f(x)=4 x^{3}-7

. Then

$f$ is one-one -into
fis many-one - into
fis many-one onto
$f$ is bijective

Solution:

We have

f(x)=4 x^{3}-7, x \in R

.

fis one-one. Let

x_{1}, x_{2} \in R

and

f(x_{1})=f(x_{2})

.

\Rightarrow 4 x_{1}^{3}-7=4 x_{2}^{3}-7 \Rightarrow 4 x_{1}^{3}=4 x_{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})\left[\left(x_{1}+\frac{x_{2}}{2}\right)^{2}+\frac{3 x_{2}^{2}}{4}\right]=0

\Rightarrow x_{1}-x_{2}=0

, because the other factor is nonzero.

\Rightarrow x_{1}=x_{2}

\therefore f

is one-one.

f is onto. Let

k \in R

any real number.

f(x)=k \Rightarrow 4 x^{3}-7=k \Rightarrow x=\left(\frac{k+7}{4}\right)^{1/3}

\text{Now} \left(\frac{k+7}{4}\right)^{1/3} \in R

, because

k \in 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.

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