Question

If all the roots of the equation $2y^3-9y^2+20-a=0$ are real, then the number of integral values of $a$ for which the equation $x+\frac{1}{x}=y$ gives two real and distinct values of $x$ is

• A. $0$
• B. $4$
• C. $26$
• D. $28$

Answer 1

The correct option is A $0$

Let $f\left(y\right)=2y^3-9y^2+20-a=0$
$f^{\prime }\left(y\right)=6y^2-18y$
$f^{\prime }\left(y\right)=0$ gives $y=0,3$
$f^{\prime \prime }\left(y\right)=12y-18$
$f^{\prime \prime }\left(0\right)=-18,f^{\prime \prime }\left(3\right)=18$
$\Rightarrow y=0$ is a point of local maxima and $y=3$ is a point of local minima.
$\Rightarrow f\left(0\right)>0\&f\left(3\right)<0\Rightarrow a\in (-7,20)\cdots \left(1\right)$

Now, using $AM\ge GM$ inequality,
$x+\frac{1}{x}\ge 2,x>0$ and
$x+\frac{1}{x}\le -2,x<0$
$\Rightarrow |y|\ge 2$
Therefore, no root lies between $[-2,2]$
$\Rightarrow f(-2)>0$ and $f\left(2\right)>0$
$\Rightarrow -32-a>0$ and $-a>0$
$\Rightarrow a<-32$ and $a<0$
$\Rightarrow a<-32\cdots \left(2\right)$
From eqn(1) and (2)
There is no real value of $a$.