Mathematics - Application of Derivative Question with Solution | TestHub

Let $f(x)=a{x}^3+b{x}^2+cx+d$ Put $x=0$ and $x=1$ Then, we get $f(0)=-1$ and $f(1)=0$ $\Rightarrow \mathrm{d}=-1$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=0$ $\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1$ It is given that $x=0$ is a stationary point of $\mathrm{f}(\mathrm{x})$, but it is not a point of extremum. Therefore, $f(0)=0=f^{\prime \prime }(0)$ and $f^{\prime \prime \prime }(0)=0$ Now, $\mathrm{f}(\mathrm{x})={\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$ $\Rightarrow f(x)=3a{x}^2+2bx+c$ $$\begin{array}{l}{f}^{\prime \prime }(x)=6ax+2b\text{ and }{f}^{\prime \prime }(x)=6a\\ f^{\prime }=0,f^{\prime \prime }(0)=0\text{ and }{f}^{\prime \prime \prime }(0)=0\ne 0\end{array}$$ $\Rightarrow c=0,b=0$ and $a\ne 0$ From Eqs. (i) and (ii), we get $\mathrm{a}=1,\mathrm{b}=\mathrm{c}=0$ and $\mathrm{d}=-1$ Put these values in $\mathrm{f}(\mathrm{x})$ we get $f(x)=x^3-1$ Hence, $\int \frac{\mathrm{f}(\mathrm{x})}{{\mathrm{x}}^3-1}\mathrm{dx}=\int \frac{{\mathrm{x}}^3-1}{{\mathrm{x}}^3-1}\mathrm{d}\mathrm{x}=\int 1\mathrm{d}\mathrm{x}=\mathrm{x}+\mathrm{C}$