Mathematics - Application of Derivative Question with Solution | TestHub

Let $p(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$ $$\Rightarrow p^{\prime }(x)=4a{x}^3+3b{x}^2+2cx+d$$ $$\therefore p^{\prime }(1)=4a+3b+2c+d=0$$ and $p^{\prime }(2)=32a+12b+4c+d=0$ Since, ${\lim }_{x\to 0}\left(1+\frac{p(x)}{x^2}\right)=2$ [given] $$\underset{x\to 0}{\lim }\frac{a{x}^4+b{x}^3+(c+1)x^2+dx+e}{x^2}=2$$ $$\begin{array}{rlrl}\Rightarrow & & x+1=2,d& =0,e=0\\ \Rightarrow & c& =1& \end{array}$$ From Eqs. (i) and (ii) $$\begin{array}{rlrl}& & 4a+3b=-2& \text{ and }32a+12b=-4\\ \Rightarrow & & a=\frac{1}{4}\text{ and }b=-1& \\ \therefore & & p(x)=\frac{x^4}{4}-x^3+x^2& \\ \Rightarrow & & p(2)=\frac{16}{4}-8+4& \end{array}$$