Question

If \[\left| {z + 4} \right| \le 3\]. Then find the greatest and least value of \[\left| {z + 1} \right|\].
A. \[6 , - 6\]
B. \[6 , 0\]
C. \[7 , 2\]
D. \[0 , - 1\]

Answer 1

Hint In the given problem, absolute value inequality is given. First, use the formula \[\left| p \right| \le a \Rightarrow - a \le p \le a\] in the given absolute inequality and simplify it. Then subtract 3 from each term of the inequality to calculate the maxima and minima value of \[z + 1\]. For the negative value of modulus, multiply the inequality by \[ - 1\]. Check both equations of the absolute inequality of \[z + 1\] and get the required answer.


Formula used
Absolute inequality:
If \[\left| p \right| < a\] and \[a > 0\], then \[ - a < p < a\].
If \[\left| p \right| \le a\] and \[a > 0\], then \[ - a \le p \le a\].

Complete step by step solution:
The given absolute value inequality is \[\left| {z + 4} \right| \le 3\].
Let’s simplify the above inequality.
Now apply the rule If \[\left| p \right| \le a\] and \[a > 0\], then \[ - a \le p \le a\].
\[ - 3 \le z + 4 \le 3\]
Subtract \[3\] from each term of the above inequality.
\[ - 3 - 3 \le z + 4 - 3 \le 3 - 3\]
Simplify the above inequality.
\[ - 6 \le z + 1 \le 0\] \[.....\left( 1 \right)\]
Now multiply each term of equation \[\left( 1 \right)\] by \[ - 1\].
\[6 \ge - \left( {z + 1} \right) \ge 0\] \[.....\left( 2 \right)\]
From the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[0 \le \left| {z + 1} \right| \le 6\]
Thus, the greatest value of \[\left| {z + 1} \right|\] is 6 and the least value is 0.
Hence the correct option is B.

Note: Some students take the inequality \[0 \le z + 4 \le 3\]. Since they thought absolute value was always greater than or equal to zero. But it is a wrong concept.