Question

The sum of integral values of a, where $aϵ[0,10]$ for which $f\left(x\right)={\text{log}}_{a+1}(2ax-x^2)$ is strictly increasing in $[\frac{1}{2},1]$ is 'b'. The value of 'b/11' is

Answer 1

$f\left(x\right)={\text{log}}_{(a+1)}(2ax-x^2)$

Case (i)

If $a+1>1\Rightarrow a>0$, $f\left(x\right)=2ax-x^2$ must be increasing $\forall xϵ[\frac{1}{2},1]$
$a\ge 1\Rightarrow a\ge 1$
$\Rightarrow a\ge 1$

Case (ii)

If $0<a+1<1\Rightarrow -1<a<0$
$f\left(x\right)=2ax-x^2$ must be decreasing $\forall xϵ[\frac{1}{2},1]$
$a\le \frac{1}{2}$ and $2a>1$
$a\le \frac{1}{2}$ and $a>\frac{1}{2}$
No solution exist.

So by case (i) and (ii)
$a\ge 1$ $\text{Sum}=1+\dots \dots +10=55$