Question

Consider $f(x)=(a-1) x^2-8 x+1, a \in R, a \neq 1, x \in R$ and $g(x)=x^2-8 x+13, x \in R$
The least integral value of ' $a$ ' for which minimum finite value of $f(x)>$ minimum value of $g(x)$ is

• A. 0
• B. 1
• C. 5
• D. 6

Answer 1

Answer: (D)

$f(x)=(a-1) x^2-8 x+1, g(x)=x^2-8 x+13 $
$-\frac{64-4(a-1)}{4(a-1)} >-3 \Rightarrow \frac{a-1-16}{a-1} >-3 \Rightarrow \frac{a-17+3 a-3}{a-1} >0$
$\Rightarrow \frac{a-5}{a-1}>0 \Rightarrow a \in(-\infty, 1) \cup(5, \infty)$
For the existance of minimum finite value of $f ( x )$
$a-1 >0 \Rightarrow a >1$
$\therefore $ least possible integral value of ' $a$ ' is 6 .