Question

The roots of the equation $x^2-a x+a+5=0$ are real and each is one more than the respective roots of the equation, $x^2+p x+q=0$. Also minimum positive integral value of $p$ is $k$ where a, p, $q \in R$.
If $x ^2- ax + a +5< k$ for atleast one real $x$, then ' $a$ ' can be

• A. -2
• B. 2
• C. 5
• D. 10

Answer 1

Answer: (D)

Also root are real
$\therefore p ^2-4 q \geq 0 \Rightarrow p ^2 \geq 24 $
$\therefore k =5$
Now, $P ( x )< k$
$\therefore x ^2- ax + a +5<5 \Rightarrow x ^2- ax + a <0 \text { for some } x \Rightarrow D >0$
$\Rightarrow a ^2-4 a >0 \Rightarrow a ( a -4)>0 \Rightarrow a \in(-\infty, 0) \cup(4, \infty)$ and roots of $P ( x )$ must also be real
$\therefore a^2-4 a \geq 20$, hence from the options ' $a$ ' can be 10 only