Question

For all $x$, $x^2-2ax+(8-2a)>0$, then the interval in which '$a$' lies is

• A. $-4<a<2$
• B. $4<a<-2$
• C. $-4\le a\le 2$
• D. None of these

Answer 1

Answer: (A)

$-4<a<2$

$y=f\left(x\right)=x^2-2ax+(8-2a)=0$

$y>0$ it means no real root for $f\left(x\right)$

$△<0$

$4a^2-4(8-2a)<0$

$a^2+2a-8<0$

$a^2+4a-2a-8<0$

$(a+4)(a-2)<0$

$-4<a<2$