Mathematics - Application of Derivative Question with Solution | TestHub

$\because f(x)$ has extremum values at $x=1$ and $x=2$ $\because f^{\prime }(1)=0$ and $f^{\prime }(2)=0$ As, $f(x)$ is a polynomial of degree 4 . Suppose $f(x)=A{x}^4+B{x}^3+C{x}^2+Dx+E$ $$\begin{array}{rl}& \because \underset{x\to 0}{\lim }\left(\frac{f(x)}{x^2}+1\right)=3\\ & \Rightarrow \underset{x\to 0}{\lim }\left(\frac{A{x}^4+B{x}^3+C{x}^2+Dx+E}{x^2}+1\right)=3\\ & \Rightarrow \underset{x\to 0}{\lim }\left(A{x}^2+Bx+C+\frac{D}{x}+\frac{E}{x^2}+1\right)=3\end{array}$$ As limit has finite value, so $D=0$ and $E=0$ Now $A(0{)}^2+B(0)+C+0+0+1=3$ $$\begin{array}{rl}& \Rightarrow c+1=3\Rightarrow c=2\\ & f^{\prime }(x)=4A{x}^3+3B{x}^2+2Cx+D\\ & f^{\prime }(1)=0\Rightarrow 4A(1)+3B(1)+2C(1)+D=0\end{array}$$ $$\begin{array}{rl}& \Rightarrow 4A+3B=-4\\ f^{\prime }(2)=0& \Rightarrow 4A(8)+3B(4)+2C(2)+D=0\\ & \Rightarrow 8A+3B=-2\end{array}$$ From equations (1) and (2), we get $A=\frac{1}{2}$ and $B=-2$ So, $f(x)=\frac{x^4}{2}-2x^3+2x^2$ Therefore, $f(-1)=\frac{(-1{)}^4}{2}-2(-1{)}^3+2(-1{)}^2$ $$=\frac{1}{2}+2+2=\frac{9}{2}$$ Hence $f(-1)=\frac{9}{2}$