Question

If $1+2x+3x^2+4x^3+\cdots +\infty \ge 4,$ $(|x|<1)$, then

• A. least value of $x$ is $\frac{1}{2}$
• B. greatest value of $x$ is $\frac{4}{3}$
• C. least value of $x$ is $\frac{2}{3}$
• D. greatest value of $x$ deos not exist

Answer 1

Answer: (A), (D)

The correct options are
A least value of $x$ is $\frac{1}{2}$
D greatest value of $x$ deos not exist

Let,
$S=1+2x+3x^2+4x^3+\cdots +\infty xS=1x+2x^2+3x^3+\cdots +\infty \Rightarrow (1-x)S=1+x+x^2+x^3+\cdots +\infty \Rightarrow (1-x)S=\frac{1}{1-x}\Rightarrow S=\frac{1}{(1-x{)}^2}\ge 4\Rightarrow (x-1{)}^2\le \frac{1}{4}\Rightarrow -\frac{1}{2}\le (x-1)\le \frac{1}{2}\Rightarrow \frac{1}{2}\le x\le \frac{3}{2}$
Also,
$|x|<1$
So,
$\frac{1}{2}\le x<1$