 # 11 years ago

sol-1

here we have to use  three conditions,

1)   f(k) > 0                      (K=3 in this question)

2)   D > = 0

from 1 , f(3) > 0

a2 - 5a + 6 > 0

(a-3)(a-2) > 0

from this we can say , value of a should be

a > 3  , a < 2

from 2 , D> = 0

4a2 > = 4(a2+a-3)

a-3 < = 0

a < =  3                  ..................2

these all conditions are satisfied if a < 2

so option  a is correct

11 years ago

ax2 + bx +c = 0

sum of roots = -b/a

product = c/a

we have eq , 2x2 - 3x - 6 = 0

let x,y are the roots of this eq then

x+y = -b/a = 3/2

xy = c/a = -3

x2+y2 = (x+y)2 - 2xy

x2+y2 = (3/2)2 - 2(-3) = 9/4+6 = 33/4          ........................1

now , eq whose roots are (x2+2 , y2+2) will be

[X - (x2+2)][X-(y2+2)] = 0

X2 - X(x2+y2+4) + x2y2 + 2(x2+y2) + 4 = 0

x2+y2 = 33/4 & xy = -3  so

X2 - X(49/4) + 59/2 = 0

4X2 - 49X + 59 = 0

this is the required quadratic equation ...

approve if u like my ans

11 years ago

Dear student,

for real roots b^2 - 4ac > 0

=> 4a^2 - 4a^2 - 4a + 12 > 0

=> a<3

for a = 3 roots of equation are 3,3

so option a is correct.