Question

If x is an integer greater than -10 but less than 10 and $\left| x-2 \right| < 3$ then the values of x are
$\begin{align}
  & a)-10,-9,...3,4 \\
 & b)0,1,2,....10 \\
 & c)0,1,2,3,4 \\
 & d)1,0,1 \\
\end{align}$

Answer 1

Hint: Now to solve the given problem we will first solve the inequality $\left| x-2 \right| < 3$. Now we know that if $\left| x \right| < a$ then we can say that $-a < x < a$. Hence using this and the given condition we will find the possible values of x by taking intersection of both the intervals.

Complete step by step answer:
Now we are given that x is an integer greater than -10 and less than 10.
Hence we have $-10 < x < 10$
Hence we get $x\in \left( -10,10 \right)$
Now consider the inequality $\left| x-2 \right| < 3$
Now we know that if $\left| x \right| < a$ then $-a < x < a$
Hence we have $-3 < x-2 < 3$
Now adding 2 on each side we get,
$\begin{align}
  & \Rightarrow -3+2 < x-2+2 < 3+2 \\
 & \Rightarrow -1 < x < 5 \\
\end{align}$
Hence we get that $x\in \left( -1,5 \right)$
Now we have $x\in \left( -10,10 \right)$ and $x\in \left( -10,10 \right)$
Hence to satisfy both the conditions we can say that
$\begin{align}
  & \Rightarrow x\in \left( -10,10 \right)\cap \left( -1,5 \right) \\
 & \Rightarrow x\in \left( -1,5 \right) \\
\end{align}$
Hence we get the values of x can be $0,1,2,...,4$

So, the correct answer is “Option c”.

Note: Now note that when we have strictly less than sign then we use open brackets which are parenthesis. When we have less than or equal to sign we use square brackets. Hence the interval $\left( -1,5 \right)$ does not include points -1 and 5 but the interval $\left[ -1,5 \right]$ includes -1 and 5.