Question

The values of a for which the equation $2x^2$ - 2(2a +1 ) x + a ( a-1 ) = 0 has

roots $\alpha \&\beta$ satisfying the condition $\alpha$ < a < $\beta$ , are

• A. (-3 , 0)
• B. ( 0 , )
• C. ( - , - 3 ) ( 0 , )
• D. ( 3 , )

Answer 1

The correct option is C

( - , - 3 ) ( 0 , )

Since 'a' lies between the roots of the given equation,

$\therefore$ 2f(a) < 0 $\Rightarrow$ f(a) < 0

Here f(x) = 2$x^2$ - 2 ( 2a +1) x + a (a-1)

$\therefore$ f(a) = $2a^2$ - 2(2a + 1)a + a(a-1) < 0

$\Rightarrow$ - $a^2-3a$ < 0 $\Rightarrow$ a (a+3) > 0

$\Rightarrow$ a $\in$ ( - $\infty$ , -3 ) $\cup$ ( 0 , $\infty$)