If set of values of 'a' for which f(x)=a x^2-(3+2 a) x+6, a 0 is positive for exactly three distinct negative integral values of x is (c, d] then the value of (c^2+16 d^2 ) is equal to

Step-by-step solution

aligned f(x) & =a x^2-(3+2 a) x+6 & =(a x-3)(x-2) aligned Here, roots of the equation f(x)=0 are 2 and 3 a , and f(0)=6f(x) should be positive for exactly three negative integral values of x . Therefore, graph of f(x) must be a downward parabola passing through x=2 and x= 3 a aligned & and -4 3 a -3 & a (-1,- 3 4 ] & c=-1, d=- 3 4 & c^2+16 d^2=1+16 9 16 =1+9=10 aligned