Question

The set of value(s) of k in R for which Column I (a) kx^(2)-(k+1)x+2k-1=0 has no real roots Column II (b) x^(2)-2(4k-1)x+15k^(2)-2k-7>0 for each x (p){1 -2} (c) Sum of the roots of x^(2)+(2-k-k^(2))x-k^(2)=0 is zero (q)(-oo -1/7)uu(1 oo) (d) The roots of x^(2)+(2k-1)x+k^(2)+2=0 (s)(-4} are in the ratio 1:2 (A) (a-s) (b-q) (c-p) (d-r) (C)(a-p) (b-r) (c-s) (d-q) (B) (a-q) (b-s) (c-p) (d-r) (D)(a-r) (b-q) (c-p) (d-s)​

Answer 1

Answer:

Explanation:

x  

2

+kx+k=0 has only one solution.

∴ The discriminant b  

2

−4ac=0

∴k  

2

−4k=0

⇒k(k−4)=0

∴k=0 or k=4

So for k=0 or k=4 the equation x  

2

+4x+4=0 has only one solution.