Question

The largest natural number ${}^{\prime }{a}^{\prime }$ for which the maximum value of $f\left(x\right)=a-1+2x-x^2$ is always smaller than the minimum value of $g\left(x\right)=x^2-2ax+10-2a$ is

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is A $1$

$f\left(x\right)=a-1+2x-x^2$
$f(x{)}_{max}=\frac{4(-1)(a-1)-4}{-4}=a$
$g\left(x\right)=x^2-2ax+10-2a$
$g(x{)}_{min}=\frac{4\left(1\right)(10-2a)-4a^2}{4}=10-2a-a^2$

Given that,
$f(x{)}_{max}<g(x{)}_{min}\Rightarrow a<10-2a-a^2\Rightarrow a^2+3a-10<0\Rightarrow (a+5)(a-2)<0\Rightarrow a\in (-5,2)$

Hence, the largest natural number value of $a=1$