Question

If the equation $f(x)=x^3+3x^2-9x+a=0\forall a\le R$ has one distinct & two identical roots, then the total number of values of a equals

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

Answer: (D)

Given $f(x)=x^3+3x^2-9x+a=0\dots (i)$
$\therefore f^{\prime }(x)=3x^2+6xc-9=3[(x^2+2x-3)]$,
here $a>0,c<0$
$\therefore D>0$
$\Rightarrow f^{\prime }(x)=\alpha$ has two distinct roots $\alpha ,\beta$ where $(\alpha =-3,\beta =1)$, therefore $f(\alpha )f(\beta )=0$ gives the values of a.
Now $f(-3)f(1)=0$
$\Rightarrow (a-5)(a+27)=0$
$\therefore a=5$ or $27$
$\therefore$ Number of values $=2$