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**Hint:**In this question, we have a function \[f\] is defined by \[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\]. We will first find the derivative \[\dfrac{d}{dx}f\left( x \right)\] of the given function. We will then check if \[\dfrac{d}{dx}f\left( x \right)\] greater than zero or it is less than zero. Now using the first derivative test for a function \[f\] which states that if \[\dfrac{d}{dx}f\left( x \right)\] greater than zero, then the function is decreasing and if \[\dfrac{d}{dx}f\left( x \right)\] less than zero, then the function is increasing. We will then get our desired answer.

**Complete step by step answer:**

We are given a function \[f\] is defined by \[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\].

We will now find the derivative \[\dfrac{d}{dx}f\left( x \right)\] of the given function \[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\].

The derivative of the function \[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\] is given by

\[\begin{align}

& \dfrac{d}{dx}f\left( x \right)=\dfrac{d}{dx}\left( {{x}^{3}}-3{{x}^{2}}+5x+7 \right) \\

& =3{{x}^{2}}-6x+5+0 \\

& =3{{x}^{2}}-6x+5

\end{align}\]

Therefore we have \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\].

Now we can see that the derivative \[\dfrac{d}{dx}f\left( x \right)\] of the given function\[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\] is a quadratic polynomial. That is a polynomial of degree 2.

Now for any quadratic polynomial \[a{{x}^{2}}-bx+c\], the discriminant is given by

\[D={{b}^{2}}-4ac\]

On comparing the derivative \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\] with the quadratic polynomial \[a{{x}^{2}}-bx+c\],we will have

\[a=3,b=-6\] and \[c=5\]

Therefore the discriminant of the quadratic equations \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\] is given by

\[\begin{align}

& D={{b}^{2}}-4ac \\

& ={{6}^{2}}-4\left( 3 \right)\left( 5 \right) \\

& =36-60 \\

& =-14

\end{align}\]

Therefore \[D<0\]

Since we know that for a quadratic polynomial \[a{{x}^{2}}-bx+c\], if the discriminant

\[D={{b}^{2}}-4ac\] is greater than zero then the polynomial \[a{{x}^{2}}-bx+c\] is less than zero for all positive real values of \[x\] and if \[D={{b}^{2}}-4ac\] is less than zero then the polynomial \[a{{x}^{2}}-bx+c\] is greater than zero for all positive real values of \[x\].

Now here for the \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\], the discriminant is less than zero.

Therefore we have that \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\] is greater than zero for all \[x\in {{R}^{+}}\].

Also since we know that \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\] is greater than zero for all \[x\in {{R}^{-}}\].

Therefore we have that the derivative \[\dfrac{d}{dx}f\left( x \right)=3{{x}^{2}}-6x+5\] is greater than zero for all \[x\in R\].

Now using the first derivative test for a function \[f\] which states that if \[\dfrac{d}{dx}f\left( x \right)\] greater than zero, then the function is decreasing and if \[\dfrac{d}{dx}f\left( x \right)\] less than zero, then the function is increasing.

We get that the function given by \[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\] is increasing on \[R\].

**So, the correct answer is “Option D”.**

**Note:**In this problem, in order to determine the nature of the given function \[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+5x+7\] we have to actually find the function increasing or decreasing on the real line. And this can be done directly using the first derivative test for a function \[f\] which states that if \[\dfrac{d}{dx}f\left( x \right)\] greater than zero, then the function is decreasing and if \[\dfrac{d}{dx}f\left( x \right)\] less than zero, then the function is increasing.

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