Question

The sum of all integral values of $a$ for which one root of equation $(a-5)x^2-2ax+a-4=0$ is smaller than $1$ and the other greater than $2$ must be equal to

• A. $260$
• B. $261$
• C. $263$
• D. $265$

Answer 1

The correct option is B $261$

Given equation is ${(a-5)x}^2-2ax+a-4=0$. It has one root smaller than $1$ and other greater than $2$.

Then the product of roots will be less than $2$. So putting values we get,

$\frac{(a-4)}{(a-5)}<2\Rightarrow a>6$.....(1)

Now since root is greater than $2$, then the function should have a negative value at a value of $2$.

This means, ${(a-5)2}^2-2a\times 2+a-4<0\Rightarrow a<24$....(2)

Now we will check the roots for $a=6$ and $a=24$

1. For $a=6$, we have $x^2-12x+2=0\Rightarrow x=6\pm \sqrt{34}$

So, we can see that one of the roots here is less than $1$ and other is greater than $2$. So $a=6$ is accepted value.

2. For $a=24$, we get ${19x}^2-48x+20=0\Rightarrow x=2,0$

In the obtained roots, one root is equal to $2$, so $a=24$ is not a acceptable value.

So, then acceptable integral values of a will be $6,7,8,9,...$ up to $23$.

Sum of these values $=261$