# Algebra

The maximum value M of 3x+5x-9x+15x-25x,as x varies over reals,satisfies
A.3<M<5
B.9<M<25
C.0<M<2
D.5<M<9

\hspace{-16}$Given$\bf{\mathbb{M} = 3^x+5^x-9^x+15^x-25^x}$\\\\\\ Let$\bf{3^x = a>0\forall x\in \mathbb{R}}$and$\bf{5^x = b>0\forall x\in \mathbb{R}}$\\\\\\ So$\bf{\mathbb{M} = a+b-a^2+ab-b^2=\frac{1}{2}\left(2a+2b-2a^2+2ab-2b^2\right)}$\\\\\\$\bf{\mathbb{M} = \frac{1}{2}\left\{2-(a-1)^2-(b-1)^2-(a-b)^2\right\}\leq \frac{1}{2}\cdot 2}$\\\\\\ and equality hold when$\bf{(a-1)=(b-1)=(a-b) = 0\Leftrightarrow a=b=1}$\\\\\\ means at$\bf{x=0}$\\\\\\ So from four options, we get$\bf{0<\mathbb{M}<2}$• Anik Chatterjee can u explain the 4th step?? 1708 man111 singh · \hspace{-16}\bf{\mathbb{M}=\frac{1}{2}\left\{2a+2b-2a^2+2ab-2b^2\right\}}$\\\\\\ $\bf{\mathbb{M} = \frac{1}{2}\left\{2-\left\{2a^2+2b^2-2ab+2a+2b-2\right\}\right\}}$\\\\\\ $\bf{\mathbb{M} = \frac{1}{2}\left\{2-\left\{(a-b)^2+(a-1)^2+(b-1)^2\right\}\right\}}$\\\\\\ Now we know that $\bf{square\;quantity\geq 0}$\\\\\\ So $\bf{(a-b)^2+(a-1)^2+(b-1)^2\geq 0...............(1)}$\\\\\\ and equality hold when $\bf{(a-b)^2=0\Rightarrow a=b}$\\\\\\ and $\bf{(a-1)^2=0\Rightarrow a=1}$ and $\bf{(b-1)^2=0\Rightarrow b=1}$\\\\\\ So $\bf{-\left\{(a-b)^2+(a-1)^2+(b-1)^2\right\}\leq 0}$ from $\bf{(1)}$\\\\\\ So $\bf{2-\left\{(a-b)^2+(a-1)^2+(b-1)^2\right\}\leq 2-0}$\\\\\\ So $\bf{\frac{1}{2}\cdot \left\{2-\left\{(a-b)^2+(a-1)^2+(b-1)^2\right\}\leq \frac{1}{2}\left(2-0\right)=1}$ \\\\\\ So $\bf{\mathbb{M}\leq 1}$\\\\\\ So from $\bf{4}$ given options , we get $\bf{0<\mathbb{M}<2}$\\\\\\ bcz $\bf{Max.(M) = 1}$