Linear Inequalities Class - 11
Basic concepts based on linear inequalities chapter 5 class XI. Important points and revision notes on linear inequalities.
Definition :
If any two numbers or algebraic
expressions are related with four symbols like < , > , ≤ and ≥ form an inequality.
If
degree of the algebraic expression is 1 then it is called a linear inequality.
For
example 2x - 3 ≤ 5, 2x+ 3 > 5 are the linear inequalities.
Solutions of
linear inequility :-
Real
solutions of inequality are represented in the form of intervals. Intervals are of 4 types
1. Open
interval written as ( )
2. Close
Interval written as [ ]
3. Semi open or semi close interval written as ( ]
4. Semi open or semi close interval written as [
)
* Small
square bracket always denote the open
interval.
* Capital
square bracket represents the close interval.
For example
(2, 8) it means all real numbers between 2 and 8, except 2 and 8
[2, 8] it
means all real numbers from 2 to 8
[2,8) it
means that number 8 is not included in this interval.
(2,8] it
means that number 2 is not included in this interval.
Solution of the Linear
inequality
Solutions
of linear inequality are represented in the form of a set.
Eg. {a, b, c, d, ……..} depending upon the nature
of the solution.
All real
solutions are represented in the form of intervals.
All the
intervals can be expressed on the number line.
Real Number Line
Case 1: Graph of
the line given below is called real number line
Real
numbers start from -∞ and goes upto +∞
Set of real numbers are
represented by (-∞, ∞)
Case II
In the
above graph O represent the open interval
at 1 and this solution is written as
(1, ∞)
Case III
In the
above graph the dark spot represent the close interval at 1 and this solution
is written as [1. ∞)
Case IV
Above
solution can be written as (-2, 3) it
means all terms between -2 and 3
Case V
Above
solution can be written as [-2, 3] it
means all numbers from -2 to 3
If an
inequality is multiplied or divided by negative sign then sign of the
inequality changes.
Eg. 2x < 5 ⇒ -2x > - 5
Graphical solution of
linear equations in two variables
Graph of
the line divide the Cartesian plane into two half planes.
A Vertical
line divides the plane into two half planes called left half plane and right
half plane as shown in the figure.
Non-vertical
line divides the plane into two half planes called upper half plane and lower
half plane as shown in the figure below.
Note :-
- A point in
the Cartesian plane either lie on the line or lie on either of the half plane.
- The region
containing all the solutions of an inequality is called the solution region or
feasible region.
- In order to
identify the half plane represented by an inequality, it is sufficient to take
any point (a, b) (not on the line ) and check whether it satisfies the
inequality or not.
- If it
satisfies the given inequality then the half plane in which the point lie is
called the solution region.
- If the
point does not satisfy the inequality then the region in which the point does
not lie is called the solution region. For convenience the point (0,0) is preferred.
- We should
shade the solution region identified in the above steps.
- For the
inequality with the sign ≤ or ≥ , the points on the
line are also included in the solution region or feasible region. In this case
the graph line is the full line.
- For the
inequality with the sign < or >, the points on the line are not included
in the feasible region or solution region. In this case the graph line is a
dotted line as shown in the figure.
Solution of system of linear inequalities in two variables
- Here we may be given two or three or four equations.
- We find the solution region for all the equations as discussed above.
- The common solution region of all the equations in the given system is
called the solution region or feasible region of the system of equations.
NCERT Miscellaneous Exercise
Q 12 A solution of 8% boric acid is to be diluted
by adding a 2% boric acid solution to it. The resulting mixture is to be more
than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution,
how many litres of the 2% solution will
have to be added ?
Solution
Amount of 8% boric acid solution = 640 litre
Let 2%
boric acid solution added = x litre
Total
solution becomes = (640 + x) litre
According
to the question
4% (640 +
x) ≤ 8% (640) + 2% of x ≤ 6% (640 + x)
\[\frac{4}{100}(640+x)\leq \frac{8}{100}\times 640+\frac{2}{100}\times x\leq \frac{6}{100}\left ( 640+x \right )\]
\[4(640+x)\leq 8\times 640+2\times x\leq 6\left ( 640+x \right )\]
\[2560+4x\leq 5120+2x\leq 3840+6x\]
|
Case I
\[2560+4x\leq 5120+2x\]
\[4x-2x\leq 5120-2560\]
\[2x\leq 2560\]
\[x\leq 1280\]
|
Case II
\[5120+2x\leq 3840+6x\]
\[5120-3840\leq 6x-2x\]
\[1280\leq 4x\]
\[1280\leq 4x\]
\[320\leq x\]
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\[320\leq x\leq 1280\]
Hence amount of 2% acid solution added is ≥ 320 litre but ≤ 1280 litre
|
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Case Study Based Questions
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Ans
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1
|
Marks obtained by
Radhika in quarterly and half yearly examinations of Mathematics are 60 and
70 respectively.
Based on the above
information, answer the following questions
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|
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(i)
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Minimum
marks she should get in the annual exam to have an average of atleast 70
marks is
a)
80 b) 85
c) 75
d)
90
|
a
|
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(ii)
|
Maximum
marks, she should get in the annual exam to have an average of atmost 75
marks is
a)
85 b) 90
c) 95
d) 80
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c
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(iii)
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Range
of marks in annual exam, so that the average marks is atleast 60 and atmost
70 is
a)
[60, 70]
b) [50, 80] c) [50, 70] d) [60, 80]
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b
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(iv)
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If
the average of atleast 60 marks is considered pass, then minimum marks she
need to score in annual exam to pass is
a)
60 b) 65 c) 70 d) 50
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d
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(v)
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If
she scored atleast 20 and atmost 80 marks in annual exam, then the range of
average marks is
a)
[50, 70]
b) [60, 70] c) 50, 60] d [50, 80]
|
a
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