AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Consider the equation,
1−2x−5x2
The maximum value of above is given by,
a=−

4+20

4(−5)

=

6

5

The minimum value is given by,
b=−

4−20

4

= 4
Now the quadratic expression 5ax2+bx+7 at a=

6

5

and b=4 must be positive then,
6x2+4x+7>0
The discriminant is given by,
D=16−4(6)(7)
D=−152
The discriminant D<0 and coefficient of x2 is positive. So,
6x2+4x+7>0, ∀x∈R=(−∞,∞)