Question

The roots of the equation $x^2-ax+a+5=0$ are real and each is one more than the respective roots of the equation, $x^2+px+q=0$. 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
$\therefore p^2-4q\ge 0\Rightarrow p^2\ge 24$
$\therefore k=5$
Now, $P(x)<k$
$\therefore x^2-ax+a+5<5\Rightarrow x^2-ax+a<0\text{ for some }x\Rightarrow D>0$
$\Rightarrow a^2-4a>0\Rightarrow a(a-4)>0\Rightarrow a\in (-\infty ,0)\cup (4,\infty )$ and roots of $P(x)$ must also be real
$\therefore a^2-4a\ge 20$, hence from the options ' $a$ ' can be 10 only