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
						

							
ax2 + bx +c = 0
in this quadratic expression
 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
 
 
						

							
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.