Question

If roots of the equation $f\left(x\right)=x^6-12x^5+{\mathrm{bx}}^4+{\mathrm{cx}}^3+{\mathrm{dx}}^2+\mathrm{ex}+64=0$ are positive, then the remainder when f(x) is divided by x-1 is

• A. 2
• B. 1
• C. 3
• D. 10

Answer 1

The correct option is B 1

Let the roots of the equation $x^6-12x^5+{\mathrm{bx}}^4+{\mathrm{cx}}^3+{\mathrm{dx}}^2+\mathrm{ex}+64=0\mathrm{be}{x}_i,i=1,2,3,4,5,6.\mathrm{Now},\mathrm{Sum}\mathrm{of}\mathrm{roots}=-\frac{\mathrm{Coefficient}\mathrm{of}{x}^5}{\mathrm{Coefficient}\mathrm{of}{x}^6}\Rightarrow x_1+x_2+x_3+x_4+x_5+x_6=12\mathrm{and}\mathrm{Product}\mathrm{of}\mathrm{roots}=\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coefficient}\mathrm{of}{x}^6}\Rightarrow x_1{x}_2{x}_3{x}_4{x}_5{x}_6=64\mathrm{Thus},A.M.(x_1,x_2,x_3,x_4,x_5,x_6)=\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}=2G.M.(x_1,x_2,x_3,x_4,x_5,x_6)={\left(x_1{x}_2{x}_3{x}_4{x}_5{x}_6\right)}^{\frac{1}{6}}={\left(64\right)}^{\frac{1}{6}}=2\because A.M.=G.M.\Rightarrow x_1=x_2=x_3=x_4=x_5=x_6=2\mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{is}\mathrm{equivalent}\mathrm{to}{\left(x-2\right)}^6=0\mathrm{or},x^6-12x^5+60x^4-160x^3+240x^2-192x+64=0\therefore f\left(1\right)=1-12+60-160+240-192+64=1$