Question

Let f (x) = ax2 + bx + cabc Ïµ R. if f (x) takes real values for real values of x and non – real values for non – real values of x, then.

Solution

Correct option is

a = 0

Suppose a ≠ 0. We rewrite f (x) as follows:

                         

            

                                  

Which is a real number This is against the hypothesis. Therefore,

 a = 0.

SIMILAR QUESTIONS

Q1

The real values of a for which the quadratic equation 3x2 + 2 (a2 + 1) x+(a2 – 3a + 2) = 0 possesses roots opposite signs lie in

Q2

If abc Ïµ R and a + b + c = 0, then the quadratic equation 3 ax2 + 3ax2 + 2bx + c = 0 has

Q3

Let f (x) = ax2 + bx + cac, Ïµ R and a ≠ 0. Suppose (x) > 0 for all x Ïµ R.

Let g (x) = f (x) + f ’ (x) + f ”(x). Then

Q4

If b < 0, then the roots x1 and x2 of the equation 2x2 + 6x + b = 0, satisfy the condition (x1/x2) < k where k is equal to.

Q5

If ax2 + bx + cabc Ïµ R has no real zeros, and if c < 0, then,

Q6

If x is real, then the maximum value of    

                                                                           

Q7

If both the roots of the equation x2 – 6ax + 2 – 2a + 9a2 = 0 exceed 3, then

Q9

Let α, β be the roots of the equation x2 – ax + b = 0 and An = αn + βn

Q10

If α, β, γ are such that α + β + γ = 2, α2 + β2 + γ2 = 6, α3 + β3 + γ3 = 8, then α4 + β4 + γ4 is