Question

If $m$ is the value of $x$ for which ${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$ is greater than zero and $k$ be the digit of the ten's place in the number $\sum _{r=1}^{2000}r!$, then which of the following is true?

• A. $m=2k$
• B. $2m=k$
• C. $m=k$
• D. None of these

Answer 1

Answer: (C)

$m=k$

${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)>0\Rightarrow \frac{1+x^2}{2x}>0\Rightarrow x>0$
For $x>0$, $\frac{1+x^2}{2}\ge x$ (using A.M. $\ge$ G.M.)
$\Rightarrow (x-1{)}^2\ge 0$....(1)
For ${\sin }^{-1}\left(\frac{1+x^2}{2x}\right)$ to be defined ,
$\frac{1+x^2}{2x}\le 1\Rightarrow {(x-1)}^2\le 0$....(2)
Hence, $x=1$ is the only solution.
Hence, $m=1$...(3)
For the series $\sum _{r=1}^{2000}r!$, all the terms starting from $10!$ have at least two zeros in the end. Hence, We just need to find the digit at the ten's place for the series
$\sum _{r=1}^{9}r!=1!+2!+3!+...+9!$
Only taking into account the last two digits (unit's and ten's) of these terms, we get the sum as $1+2+6+24+20+20+40+20+80=2\underset{–}{1}3$
Hence, the digit at the ten's place of the sum is $1$.
The value of $k=1$....(4)
From (3) and (4), $m=k$