Question

For the equation $x^2-(a-3)x+a=0,a\in R$, find the values of ${}^{\prime }{a}^{\prime }$ such that exactly one root lies between $1$ and $2$.

• A. $a\in (-\infty ,-1)$
• B. $a\in (10,\infty )$
• C. $a\in (-\infty ,-10)$
• D. None of these

Answer 1

The correct option is B

$a\in (10,\infty )$

Consider the graph of $f\left(x\right)=x^2-(a-3)x+a$

For roots to be real

$D\ge 0$

$\Rightarrow b^2-4ac\ge 0$

$\Rightarrow$ $a\le 1$ or $a\ge 9$ . . . $\left(1\right)$

For exactly $1$ root to lie in between $1$ and $2$, $f\left(1\right)$ and $f\left(2\right)$ should be of opposite signs.

$f\left(1\right).f\left(2\right)<0$

$f\left(x\right)=x^2-(a-3)x+a$ (Given)

$f\left(1\right)=1-(a-3)+a=4$

$f\left(2\right)=4-2(a-3)+a=4-2a+6+a$

$=10-a$

$f\left(1\right).\left(2\right)<0$

$\Rightarrow 4(10-a)<0$

$a\in (10,\infty )$ . . . $\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$, $a\in (10,\infty )$.