Question

# The sum of real values of K for which the equation x3−Kx+K–1=0 has exactly two distinct real solutions. 34  3 154  152

Solution

## The correct option is C 154  (x3−1)−K(x−1)=0 (x−1)(x2+x+1−K)=0 ⇒x=1 or x2+x+(1−K)=0 For exactly two distinct solutions, x2+x+(1−K)=0 have identical solution. i.e., Discriminant =0 =1–4(1−K)=0⇒K=34 OR For exactly two distinct solutions, x2+x+(1−K)=0 has x=1 as one of its solution. So, 1+1+1−K=0⇒K=3 ∴ Sum of the real values of K=34+3=154

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