Question

If \[min.(2{x^2} - ax + 2) > \max .(b - 1 + 2x - {x^2})\] then roots of the equation\[2{x^2} + ax + (2 - b) = 0\], are
(A) Positive and distinct
(B) negative and distinct
(C) Opposite in sign
(D) imaginary

Answer 1

Hint: We will find the root of the given equation and then compare it with the following properties we get the required answer.

Formula used:
The roots of a quadratic equation \[a{x^2} + bx + c = 0\] is given by,
     \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
The type of the roots of the quadratic equation is calculated by its discriminant\[{b^2} - 4ac\].
1) \[{b^2} - 4ac\]>0, then two roots are different real numbers;
2) \[{b^2} - 4ac\]=0, then two roots are equal real numbers;
3) \[{b^2} - 4ac\]<0, then two roots are imaginary numbers.

Complete step-by-step answer:
It is given that \[min.(2{x^2} - ax + 2) > \max .(b - 1 + 2x - {x^2})\].
Let us consider, \[u = (2{x^2} - ax + 2)\]
Differentiating both sides of the above equation with respect to x we get,
\[\dfrac{{du}}{{dx}} = 2 \times 2x - a = 4x - a\]
Let differentiate both sides again with respect to x,
     \[\dfrac{{{d^2}u}}{{d{x^2}}} = 4 > 0\]
So we have found that u is minimum.
 Now “u” is minimum at the point where \[\dfrac{{du}}{{dx}}\] is zero,
That is \[4x - a = 0\] ,
Which in turn imply that \[x = \dfrac{a}{4}\]
At \[x = \dfrac{a}{4}\],
\[u = \{ 2{(\dfrac{a}{4})^2} - a \times \dfrac{a}{4} + 2\} \]
On solving the above equation we get,
\[u = \dfrac{{{a^2}}}{8} - \dfrac{{{a^2}}}{4} + 2\]
\[u = \dfrac{{{a^2} - 2{a^2} + 16}}{8} = \dfrac{{16 - {a^2}}}{8}\]
So, \[min.(2{x^2} - ax + 2)\]=\[\dfrac{{16 - {a^2}}}{8}\]
Let us consider, \[u = b - 1 + 2x - {x^2}\]
Differentiating both sides of the above equation with respect to x we get,
\[\dfrac{{du}}{{dx}} = 2 - 2x\]
Let us again differentiate both sides again with respect to x,
     \[\dfrac{{{d^2}u}}{{d{x^2}}} = - 2 < 0\]
So u is maximum.
 Now u is maximum at the point where \[\dfrac{{du}}{{dx}}\]is zero,
That is \[2 - 2x{\rm{ }} = 0\]
Which in turn imply that \[x = 1\]
At x=1,
\[u = b - 1 + 2 \times 1 - {1^2}\]
\[u = b - 1 + 2 - 1\]
\[u = b\]
So, \[\max .(b - 1 + 2x - {x^2})\]=b
Then by given condition we have,
\[\dfrac{{16 - {a^2}}}{8} > b\]….(1)
The discriminant of the quadratic equation \[2{x^2} + ax + (2 - b) = 0\]is found using the formula given in the hint, we get,
Here \[b = a,a = 2,c = (2 - b)\]
\[D = {a^2} - 4 \times 2 \times (2 - b)\]
\[D = {a^2} + 8b - 16\]
From (1) we get,
 \[16 - {a^2} > 8b\]
That is \[{a^2} + 8b - 16 < 0\]
So D<0 so by the property the roots of the quadratic equations are imaginary.

Hence (D) is the correct option.

Note:To determine the maxima or minima of a function we can apply the second derivative test,
When a function's slope that is first derivative is zero at x, and the second derivative at x is:
1) Less than 0, it is a local maximum
2) Greater than 0, it is a local minimum
3) Equal to 0, then the test fails (there may be other ways of finding out though)