Question

If $x>0$ then greatest value of the expression $\frac{x^{50}}{1+x+x^2+\dots \dots +x^{100}}$ is

• A. $\frac{1}{102}$
• B. $\frac{1}{101}$
• C. $\frac{1}{100}$
• D. none of these

Answer 1

Answer: (B)

$A.M.\ge G.M.$
$\Rightarrow \frac{1+x+x^2+\dots ..+x^{100}}{101}\ge {\left(1.x\cdot x^2\dots \dots ..x^{100}\right)}^{\frac{1}{101}}$
$\Rightarrow \frac{1+x+x^2+\dots \dots ..+x^{100}}{101}\ge x^{\frac{100\times 101}{2}\times \frac{1}{101}}$
$\Rightarrow \frac{1+x+x^2+\dots \dots ...x^{100}}{101}\ge x^{50}$
$\Rightarrow \frac{1}{101}\ge \frac{x^{50}}{1+x+x^2+\dots \dots +x^{100}}$
$\therefore$ Greatest value $=\frac{1}{101}$