Question

The maximum value M of $3^x+5^x-9^x+{15}^x-{25}^x,$ as x varies over reals , occurs at $x=$

• A. $3$
• B. $0$
• C. $9$
• D. $5$

Answer 1

Answer: (B)

$0$

Let $a=3^{x}b=5^x$,then expression convert into $f(a,b)=a+b-a^2+ab-b^2$

So

$f(a,b)=-\frac{1}{2}[2a^2+2b^2-2ab-2a-2b]$

So

$f(a,b)=-\frac{1}{2}\left[\right(a^2-2a+1)+(b^2-2b+1)+(a^2+b^2-2ab-2\left)\right]$

So

$f(a,b)=-\frac{1}{2}[2-(a-1{)}^2-(b-1{)}^2-(a-b{)}^2-2]$

So

$f(a,b)=\frac{1}{2}[2-(a-1{)}^2-(b-1{)}^2-(a-b{)}^2]\le 1$

And enquality hold when

$(a-1{)}^2=0\Rightarrow a=1\Rightarrow 3^x=1=3^0$

And

$(b-1{)}^2=0\Rightarrow b=1\Rightarrow 5^x=1=5^0$

And

$(a-b{)}^2=0\Rightarrow a=b=1\Rightarrow 3^x=5^x$

So from above we get $x=0$for which $f\left(x\right)=3^x+5^x-4^x+{15}^x-9^x$ is maximum And max

$f\left(x\right)=1$ at $x=0$ Answer.