Question

The sum of real values of $K$ for which the equation $x^3-Kx+K–1=0$ has exactly two distinct real solutions.

• A. $\frac{15}{4}$
• B. $\frac{3}{4}$
• C. $3$
• D. $\frac{15}{2}$

Answer 1

The correct option is A

$\frac{15}{4}$

$(x^3-1)-K(x-1)=0$

$(x-1)(x^2+x+1-K)=0$

$\Rightarrow x=1$ or $x^2+x+(1-K)=0$

For exactly two distinct solutions, $x^2+x+(1-K)=0$ have identical solution.

i.e., Discriminant $=0$

$=1–4(1-K)=0\Rightarrow K=\frac{3}{4}$

OR

For exactly two distinct solutions, $x^2+x+(1-K)=0$ has $x=1$ as one of its solution.

So, $1+1+1-K=0\Rightarrow K=3$

$\therefore$ Sum of the real values of $K=\frac{3}{4}+3=\frac{15}{4}$