Question

What will be the value of $k$ for which the number $3$ lies between the roots of the below mentioned equation?
$x^2+(1-2k)x+(k^2-k-2)=0$.

• A. $k\in (-\infty ,2)\cup (5,\infty )$
• B. $k=\frac{1}{2}$
• C. $k\in (2,5)$
• D. $k\in (-1,2)$

Answer 1

The correct option is C $k\in (2,5)$

Let $f\left(x\right)=x^2+(1-2k)x+k^2-K-2$

Since $a>0$ and $x$ is real i.e $D>0$, the number $3$ lies between the roots of the given equation, if $f\left(3\right)<0$.

Now, $f\left(3\right)=9+(1-2k)3+K^2-K-2$

$=10-7K+K^2=K^2-7K+10$

Hence, $f\left(3\right)<0\Rightarrow k^2-7K+10<0$

$\Rightarrow (k-2)(k-5)<0$

$\Rightarrow 2<k<5$