Question

Please help me in this question. Explaination also.....

Answer 1

$\textbf{Given:}$

$\text{ \alpha,\,\beta,\,\gamma are zeros of 2x^3-3x^2-5x+6k }$

$\textbf{To find:}$

$\text{The value of \dfrac{\alpha\beta\gamma+\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha+\beta+\gamma} }$

$\textbf{Solution:}$

$\text{Since 2x^3-9x^2+kx+12 is divided by x+2 , the remainder is -42}$

$\text{By remainder theorem, we get}$

$2(-2)^3-9(-2)^2+k(-2)+12=-42$

$-16-36-2k+12=-42$

$-52-2k+12=-42$

$-40-2k=-42$

$-2k=-2$

$\implies\bf\,k=1$

$\text{Since \alpha,\,\beta,\,\gamma are zeros of 2x^3-3x^2-5x+6 , we have}$

$\alpha+\beta+\gamma=\frac{3}{2}$

$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{-5}{2}$

$\alpha\beta\gamma=\frac{-6}{2}=-3$

$\text{Now,}$

$\dfrac{\alpha\beta\gamma+\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha+\beta+\gamma}$

$=\dfrac{-3+(\frac{-5}{2})}{\frac{3}{2}}$

$=\dfrac{\frac{-11}{2}}{\frac{3}{2}}$

$=\dfrac{-11}{3}$

$\therefore\textbf{The value of \dfrac{\alpha\beta\gamma+\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha+\beta+\gamma} is \bf\,\dfrac{-11}{3} }$

$\implies\textbf{Option (3) is correct}$