
The solution set of \[{(x)^2} + {(x + 1)^2} = 25\]; where \[(x)\] denotes the nearest integer greater than or equal to x.
(A) \[\left[ { - 4, - 3} \right)\bigcup {x \in \left[ {3,4} \right)} \]
(B) \[\left( { - 5, - 4} \right]\bigcup {x \in \left( {2,3} \right]} \]
(C) \[(2,4)\]
(D) None of these
Answer
557.7k+ views
Hint: Nearest Integer functions are the functions that come after rounding it off to the nearest integer.
Solving a quadratic equation: \[a{x^2} + bx + c = 0\]by using middle term splitting or using discriminant method.
Complete step-by-step answer:
Let, \[x = y\], where y is an integer.
The given equation can be written as:
\[ \Rightarrow {y^2} + {(y + 1)^2} = 25\]
On simplifying above equation we get,
\[ \Rightarrow {y^2} + {y^2} + 1 + 2y = 25\]
\[ \Rightarrow 2{y^2} + 2y = 25 - 1\]
\[ \Rightarrow 2{y^2} + 2y - 24 = 0\]
\[ \Rightarrow {y^2} + y - 12 = 0\]
\[ \Rightarrow {y^2} + 4y - 3y - 12 = 0\]
\[ \Rightarrow y(y + 4) - 3(y + 4) = 0\]
Taking \[(y + 4)\] common we get,
\[ \Rightarrow (y + 4)(y - 3) = 0\]
\[ \Rightarrow y = - 4;y = 3\]
\[ \Rightarrow x = - 4;x = 3\]
If \[x = y + s\]; where y is an integer and \[0 < s < 1\].
The equation can be written as:
\[ \Rightarrow {(y + 1)^2} + {(y + 2)^2} = 25\]
On simplifying above equation, we get:
\[ \Rightarrow {y^2} + 1 + 2y + {y^2} + 4 + 4y = 25\]
\[ \Rightarrow 2{y^2} + 6y = 25 - 5\]
\[ \Rightarrow 2{y^2} + 6y = 20\]
\[ \Rightarrow 2{y^2} + 6y - 2 = 0\]
\[ \Rightarrow {y^2} + 3k - 10 = 0\]
\[ \Rightarrow {y^2} + 5k - 2k - 10 = 0\]
\[ \Rightarrow y(y + 5) - 2(y + 5) = 0\]
Taking common;
\[ \Rightarrow (y + 5)(y - 2) = 0\]
\[ \Rightarrow y = 2, - 5\]
\[x = 2 + s\]and \[x = - 5 + s\].
\[ \Rightarrow x = - 5 + s\] and
\[ \Rightarrow x \in \left( { - 5, - 4} \right]\]
\[ \Rightarrow x = 2 + s\] and \[x = 3\]
\[ \Rightarrow x \in \left( {2,3} \right]\]
Required solution set= \[\left( { - 5, - 4} \right]\bigcup {x \in \left( {2,3} \right]} \].
Option (B) is correct.
Note: Nearest Integer functions include rounding of seven different types of functions.
They all deal with the separation of integer or fractional parts from real and complex number: the floor functions , the nearest integer function (round), the ceiling function (least integer), integer part of the quotient etc
Solving a quadratic equation: \[a{x^2} + bx + c = 0\]by using middle term splitting or using discriminant method.
Complete step-by-step answer:
Let, \[x = y\], where y is an integer.
The given equation can be written as:
\[ \Rightarrow {y^2} + {(y + 1)^2} = 25\]
On simplifying above equation we get,
\[ \Rightarrow {y^2} + {y^2} + 1 + 2y = 25\]
\[ \Rightarrow 2{y^2} + 2y = 25 - 1\]
\[ \Rightarrow 2{y^2} + 2y - 24 = 0\]
\[ \Rightarrow {y^2} + y - 12 = 0\]
\[ \Rightarrow {y^2} + 4y - 3y - 12 = 0\]
\[ \Rightarrow y(y + 4) - 3(y + 4) = 0\]
Taking \[(y + 4)\] common we get,
\[ \Rightarrow (y + 4)(y - 3) = 0\]
\[ \Rightarrow y = - 4;y = 3\]
\[ \Rightarrow x = - 4;x = 3\]
If \[x = y + s\]; where y is an integer and \[0 < s < 1\].
The equation can be written as:
\[ \Rightarrow {(y + 1)^2} + {(y + 2)^2} = 25\]
On simplifying above equation, we get:
\[ \Rightarrow {y^2} + 1 + 2y + {y^2} + 4 + 4y = 25\]
\[ \Rightarrow 2{y^2} + 6y = 25 - 5\]
\[ \Rightarrow 2{y^2} + 6y = 20\]
\[ \Rightarrow 2{y^2} + 6y - 2 = 0\]
\[ \Rightarrow {y^2} + 3k - 10 = 0\]
\[ \Rightarrow {y^2} + 5k - 2k - 10 = 0\]
\[ \Rightarrow y(y + 5) - 2(y + 5) = 0\]
Taking common;
\[ \Rightarrow (y + 5)(y - 2) = 0\]
\[ \Rightarrow y = 2, - 5\]
\[x = 2 + s\]and \[x = - 5 + s\].
\[ \Rightarrow x = - 5 + s\] and
\[ \Rightarrow x \in \left( { - 5, - 4} \right]\]
\[ \Rightarrow x = 2 + s\] and \[x = 3\]
\[ \Rightarrow x \in \left( {2,3} \right]\]
Required solution set= \[\left( { - 5, - 4} \right]\bigcup {x \in \left( {2,3} \right]} \].
Option (B) is correct.
Note: Nearest Integer functions include rounding of seven different types of functions.
They all deal with the separation of integer or fractional parts from real and complex number: the floor functions , the nearest integer function (round), the ceiling function (least integer), integer part of the quotient etc
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
Who composed the song Vande Mataram A RabindraNath class 10 social science CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

The revolutionary who died after 63 days of the hunger class 10 social science CBSE

The slogan of Bande Mataram was first adopted during class 10 social science CBSE

Why is Sardar Vallabhbhai Patel called the Iron man class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

