Question

hhshhdhhdhdhdhdhfhhdhj... solve this 2 question cbse 2015 paper

Answer 1

12. As the equation has equal roots then D=0 [D is discriminant]

D=b2-4ac

Now equation is kx2+1-2(k-1)x+x2=0

=> kx2+x2-2(k+1)+1

=>x2(k+1)-2(k+1)+1 (taking x2 common)

here a = (k+1)

b = -2(k+1)

c = 1

Now, b2-4ac=0

[-2(k-1)]2 - 4 (k+1)(1)=0

(-2k+2)2 - 4k-4=0

4k2-2(2k)(2)+4 - 4k-4=0

4k2-4k+4-4k-4=0

4k2-8k-4k=0

4k2-12k=0

Now here take 4k as common

4k(k-3)=0

Now(case 1)

4k=0

K=0/4

K=0

Or (case 2)

K-3=0

K=3

Hence k=0 or k=3