Question

If $x-y$ and $y-2x$ are two factor of the expression $x^3-3x^{2}y+\lambda x{y}^2+\mu y^3$, then

• A. $\lambda =11,\mu =-3$
• B. $\lambda =3,\mu =-11$
• C. $\lambda =\frac{11}{4},\mu =-\frac{3}{4}$
• D. none of these

Answer 1

The correct option is C $\lambda =\frac{11}{4},\mu =-\frac{3}{4}$

Let $f\left(x\right)=x^3-3x^{2}y+\lambda x{y}^2+\mu y^3$
As $x-y$ is factor
$\Rightarrow f\left(y\right)=y^3-3y^{2}y+\lambda y{y}^2+\mu y^3=0$
$\Rightarrow -2y^3+(\lambda +\mu )y^3=0\Rightarrow \lambda +\mu =2$ ...(1)
And as $y-2x$ is factor
$\Rightarrow f\left(\frac{y}{2}\right)={\left(\frac{y}{2}\right)}^3-3{\left(\frac{y}{2}\right)}^{2}y+\lambda \left(\frac{y}{2}\right)y^2+\mu y^3=0\Rightarrow \frac{y^3}{8}-\frac{3y^3}{4}+\frac{\lambda y^3}{2}+\mu y^3=0\Rightarrow y^3-6y^3+4\lambda y^3+8\mu y^3=0$
$\Rightarrow 4(\lambda +2\mu )=5\Rightarrow \lambda +2\mu =\frac{5}{4}$ ...(2)
Solving (1) and (2), we get
$\lambda =\frac{11}{4},\mu =-\frac{3}{4}$
Hence, option 'C' is correct.