Question

Let 4 $x^2$ - 4(a - 2) x + a - 2 = 0, a R be a quadratic equation with real roots. Atleast one root of this equation lies in (0, 0.5) if

• A. (, 2)(3, )
• B. (, 4)(5, )
• C. (4,5)
• D. (2,3)

Answer 1

The correct option is A

(, 2)(3, )

For the given equation to have real roots,

$b^2$ - 4ac > 0

${[-4(a-2\left)\right]}^2$ - 4(4) (a - 2) > 0

16 [ ${(a-2)}^2$ - (a - 2)] > 0

(a - 2) (a - 3) > 0

a < 2 (or) a > 3 -------- (1)

As a > 0, graph is upward parabola

Atleast one root in (0, 0.5) gives rise to the 2 cases mentioned above. Values of a can be the either of the values obtained in these 2 cases.

Case (i)

Here f(0) > 0 and $f\left(\frac{1}{2}\right)$ < 0

f(0) $f\left(\frac{1}{2}\right)$ < 0

In question, it is given that f(x) =4 $x^2$ - 4(a - 2) x + a - 2

f(0) = a - 2

$f\left(\frac{1}{2}\right)$ = 1 - 2(a - 2) + (a - 2) = 3 - a

$\Rightarrow$ (a - 2) (3 - a) < 0

(a - 2) (a - 3) > 0

a < 2 (or) a > 3 ------------- (2)

From (1) & (2)

a $ϵ$ (-$\infty$, 2)$\cup$(3, $\infty$)

Case (ii):

Here f(0) > 0 and $f\left(\frac{1}{2}\right)$ > 0

$\Rightarrow$ a > 2 and a < 3

a $ϵ$ (2, 3) ------------- (3)

From (1) & (3), a $ϵ$ null set

So, answer is ($-\infty$, 2) $\cup$ (3, $\infty$)