
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
162.3k+ views
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.
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
