Question

If $\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}=0$,where $C_0,C_1,C_2$ are all real, the equation $C_2{x}^2+C_{1}x+C_0=0$ has:

• A. atleat one root is $(0,1)$
• B. one root is $(1,2)$ and $(3,4)$
• C. one root is $(-1,1)$ and $(-5,-2)$
• D. both roots imaginary

Answer 1

The correct option is A atleat one root is $(0,1)$

$\begin{array}{c}Letf\left(x\right)=\frac{C_2{x}^3}{3}+\frac{C_1{x}^2}{2}+C_{0}x\\ Now,f\left(0\right)=0,\\ f\left(1\right)=\frac{C_2}{3}+\frac{C_1}{2}+C_0=0\left(given\_eq\right)\\ \therefore f\left(0\right)=f\left(1\right)=0\\ Alsof\left(x\right)iscontinuousin[0,1]\\ and,differentiable,in(0,1)\\ \therefore ByRoll{e}^{\prime }stheorematleastoneCin(0,1)\\ suchthat{f}^{\prime }\left(C\right)=0,0<C<1\\ \Rightarrow C_2{C}^2+C_{1}C+C_0=0[since,f^{\prime }(x)=C_2{x}^2+C_{1}x+C_0]\\ \therefore C_2{x}^2+C_{1}x+C_0=0hasatleastonerealrootin(0,1)\end{array}$