
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
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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
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