Let $A$ be the set of values of $k$ for which $2$ lies between the roots of the quadratic equation $x^2+\left(k+2\right)x-\left(k+3\right)=0,$ then $A$ is given by

• Correct answer
  A

  $\left(-\mathrm{\infty },-5\right)$
• B

  $\left(5,\mathrm{\infty }\right)$
• C

  $\left(-\mathrm{\infty },-5\right]$
• D

  $\left[5,\mathrm{\infty }\right)$

Solution

Opening upward parabola as the coefficient of $x^2$ is $+ve$
$2$ lies between the roots $\Rightarrow f\left(2\right)<0$
$f\left(2\right)=4+2\left(k+2\right)-\left(k+3\right)<0$
$4+2k+4-k-3<0$
$k+5<0\Rightarrow k<-5$

Hence, the correct answer is option: 'A'.