Question

If one root of the quadratic equation $2x^2-2kx+k-4=0$ is smaller than $1$ & other is greater than $2$, then the complete set of values of $k$ is

• A. $(-2,\infty )$
• B. $(\frac{4}{3},\infty )$
• C. $(-2,\frac{4}{3})$
• D. $(-\infty ,-2)$

Answer 1

Answer: (B)

$(\frac{4}{3},\infty )$

Let $f\left(x\right)=2x^2-2kx+k-4$,
For real roots to exist for $f\left(x\right)=0$, the discriminant must be positive.
$\Delta =4k^2-8(k-4)$
$\Delta =4(k^2-2k+8)$
$\Delta$ is always positive, since the discriminant of the expression $k^2-2k+8$ is negative and the coefficient of $k^2$ is also positive.
Hence for all values of $k$, $f\left(x\right)=0$ has real and distinct roots.
Now $f\left(x\right)$ is an upward opening parabola and one of the roots is greater than $2$ and the other root is less than $1$.
So, $f\left(1\right)<0$ and $f\left(2\right)<0$.
$f\left(1\right)=2-2k+k-4<0$
$\Rightarrow k>-2$
$f\left(2\right)<0$,
$\Rightarrow 2(2{)}^2-4k+k-4<0$
$\Rightarrow -3k+4<0$
$\Rightarrow k>\frac{4}{3}$
Hence, the common set for $k$ is $\left(\frac{4}{3},\infty \right)$.