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Ever felt the need to count calories or manage an allocation of time and schedule? Linear inequalities help us cover this aspect. Linear inequalities is an important chapter of NCERT which helps students to solve real-life problems using mathematical constraints. This chapter introduces students to the concept of inequalities, helps them understand their graphical representation, and provides methods to solve and interpret inequalities in one or two variables.
NCERT Solutions for Class 11 Maths is an excellent resource for students to dive deep into this chapter. It not only helps in practising the various types of linear inequalities but also guides students through step-by-step solutions and methods to understand the concepts better.
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JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy
Suggested: JEE Main: high scoring chapters | Past 10 year's papers
Class 11 Maths Chapter 6 Exemplar Solutions Exercise: 6.3 Page number: 107-113 Total questions: 32 |
Question:1
Solve for x the inequalities in $\frac{4}{x+1}\leq 3\leq \frac{6}{x+1},\left ( x> 0 \right )$
Answer:
Given: $\frac{4}{x+1}\leq 3\leq \frac{6}{x+1},\left ( x> 0 \right )$
Now, let us multiply all the terms by $\left ( x+1 \right )$, we get,
$4\leq 3\left ( x+1 \right )\leq 6$
i.e., $4\leq 3x+3 \leq 6$
Now, by subtracting each term by 3, we get,
$1\leq 3x\leq 3$
At last, we will divide each term by 3,
We get,
$\left ( 1/3 \right )\leq x\leq 1$
Question:2
Solve for x, the inequalities in $\frac{\left | x-2 \right | -1}{\left | x-2 \right | -2}\leq 0$
Answer:
Given: $\frac{\left |x-2 \right |-1}{\left |x-2 \right |-2}\leq 0$
Now, let us assume that,
y = |x-2|
thus, $\frac{ y-1}{y-2}\leq 0$
if $y< 1$,
y-1<0 & y-2<0 , & thus ,
$y-1 /y-2> 0,$ viz. Not needed
now,
if $1\leq y< 2,$
$y-1\geq 0$ & $y-2< 0,$
& hence,
$y-1/y-2< 0$… thus, we get the answer we need.
Now, $1\leq y< 2,$
Thus, $1\leq \left | x-2 \right |< 2$
Now, from this, we will get 2 cases, which are-
$1\leq x-2< 2= 3\leq x< 4$
& $1\leq -\left (x-2 \right )< 2= 1\leq -x+2< 2$
On multiplying each term by -1, we get,
$-2\leq x-2< -1$
Now, we’ll add 2 to each term, we get,
$0\leq x< 1$
Therefore, $\left [ 0,1 \right ]\upsilon \left [ 3,4 \right ]$
Question:3
Solve for x the inequalities in $\frac{1}{\left | x \right |-3}\leq \frac{1}{2}$
Answer:
Given: $\frac{1}{\left | x \right |-3}\leq \frac{1}{2}$
From this, we get,
$\frac{1}{\left | x \right |-3}- \frac{1}{2}\leq 0$
$\frac{1}{\left | x \right |-3}- \frac{1}{2}\leq 0$
$\\\frac{2-|x|+3}{2(|x|-3)} \leq 0 \\\\ \frac{5-|x|}{|x|-3} \leq 0$
Thus, we get,
$5-\left | x \right |\leq 0$ & $\left | x \right |-3\geq 0$ or $5-\left | x \right |\geq 0$ & $\left | x \right |-3< 0$
Thus $\left | x \right |\geq 5$ & $\left | x \right | > 3$ or $\left | x \right | \leq 5$ & $\left | x \right | < 3$
Thus,
$x \epsilon \left ( -\infty , -5 \right ]$ or $\left [ 5, -\infty \right )$ or $x \epsilon \left ( -3, 3 \right )$
Therefore,
$x \epsilon \left ( -\infty , -5 \right ] \upsilon \left ( -3,3 \right ) \upsilon \left [5,\infty \right )$
Question:4
Solve for x, the inequalities in $\left | x -1 \right |\leq 5,\left | x \right |\geq 2$
Answer:
$\left | x-1 \right |\leq 5........(given)$
(i) Now, there will be two cases –
$x-1 \leq 5,$
Adding 1 on both the sides, we get,
$x \leq 6$
(ii) $-\left (x-1 \right ) \leq 5$
i.e., $-x+1 \leq 5$
Subtract 1 from both sides, we will get,
$-x \leq 4$ i.e $x \geq -4$
Now, from (i) & (ii), we get,
$-4 \leq x\leq 6.........(a)$
& $\left | x \right |\geq 2$'
Thus, $x \geq 2$ & $-x \geq 2$
Thus $x \leq -2$
i.e $x \epsilon(-4,-2] \cup[2,6]$
Question:5
Solve for x, the inequalities in $-5\leq \frac{2-3x}{4}\leq 9$
Answer:
Given $-5\leq \frac{2-3x}{4}\leq 9$
On multiplying all the terms by 4, we get,
$-20\leq 2-3x\leq 36$
Now, add -2 to each term,
$-22\leq -3x\leq 34$
And now, divide each term by 3,
$-22/3\leq -x\leq 34/3$
Now, we will multiply each term by -1 to invert the inequality, we get,
$-34/3\leq x\leq 22/3$
Question:6
Solve for x, the inequalities in $4x+3\geq 2x+17, 3x-5< -2$
Answer:
$4x+3\geq 2x+17 .......$ given
Thus, $4x-2x\geq 17-3$
Thus $2x\geq 14$
Thus $x\geq 7$ .........(i)
Now,
$3x-5< -2$............. given
Thus $x< 1$.........(ii)
Therefore, x has no solution as eq. (i) & eq. (ii) cannot be possible simultaneously.
Question:7
Answer:
Given: Revenue, R(x) = 43x
Cost, C(x) = 26,000 + 30x,
Where ‘x’ is the no. of cassettes.
Requirement: profit > 0
Solution: We know that,
Profit = revenue – cost
= 43x – 26000 – 30x > 0
= 13x – 26000 > 0
= 13x > 26000
= x > 2000
Therefore, 2000 more cassettes should be sold by the company to realize the profit.
Question:8
Answer:
Given: First reading = 8.48
Second reading = 8.35
To find: Third reading
Solution:
Let ‘x’ be the third reading,
Now, the average pH should be between 8.2 & 8.5
Average pH = (8.48 + 8.35 + x)/3
Thus,
8.2 < (8.48 + 8.35 + x)/3 < 8.5
On multiplying each term by 3, we get,
24.6 < 16.83 + x < 25.5
Now, subtracting 16.83 from each term, we get,
7.77 < x < 8.67
Therefore, the third reading should be between 7.77 & 8.67.
Question:9
Answer:
Let us assume that ‘x’ litres of 3% solution is added to 460 L of 9% solution.
Thus, total solution = (460 + x)L
& total acid content in resulting solution = (460 × 9/100 + x × 3/100)
= (41.4 + 0.03x) %
Now, according to the question,
The resulting mixture we get should be less than 7% acidic & more than 5% acidic
Thus, we get,
5% of (460 + x) < 41.4 + 0.03x < 7% 0f (460 + x)
= 23 + 0.05x < 41.4 + 0.03x < 32.2 + 0.07x
Now,
23 + 0.05x < 41.4 + 0.03x & 41.4 + 0.03x < 32.2 + 0.07x
= 0.02x < 18.4 & 0.04x > 9.2
Thus, 2x < 1840 & 4x > 920
= 230 < x < 920
Therefore, the solution between 230 l & 920 l should be added.
Question:10
Answer:
Given: The solution should be kept between 400C & 450C
Now, let C be the temp in Celsius & F be the temp in Fahrenheit.
Thus, 40 < C < 45
On multiplying each term by 9/5,
72 < 9/5 C < 81
Now, add 32 to each term,
104 < 9/5 C + 32 < 113
Thus, 104 < F < 113
Therefore, F, i.e., temp in Fahrenheit, should be between 1040F & 1130F.
Question:11
Answer:
Let us assume that the length of the shortest side of the triangle is ‘x’ cm
Thus, length of the largest side = 2x …. (given)
& length of the third side = (x + 2) cm …. (given)
Now, we know that,
The perimeter of a triangle = sum of all three sides
= x + 2x + x + 2
= 4x + 2 cm
Now, it is given that the perimeter of the triangle is more than 166 cm,
Thus, $4x+2\geq 166$
$4x\geq 164$
Thus,$x\geq 41$
Therefore, the minimum length of the shortest side should be = 41 cm.
Question:12
Answer:
Given:
T = 30 + 25 (x – 3), $3 \leq x\leq 15$
Where, T = temperature
x = depth inside earth
It is also given that T should be between 155? C & 205? C
Thus,
155 < T < 205
155 < 30 + 25 (x – 3) < 205
155 < 30 + 25x – 75 < 205
155 < 25x – 45 < 205
Now, we will add 45 to each term,
We get,
200 < 25x < 250
Now, on dividing each term by 25, we get,
8 < x < 10
Therefore, at a depth of 8-10 km, temperature varies from 1550C to 2050C.
Question:13
Solve the following system of inequalities $\frac{2x+1}{7x-1}> 5,\frac{x+7}{x-8}>2$
Answer:
Given: 2x+1/7x-1 > 5
(2x+1)/(7x - 1) – 5 > 0 …….. (on subtracting 5 from both the sides)
(2x + 1 – 35x + 5)/(7x – 1) > 0
(6 – 33x)/(7x – 1) >0
Now, either the numerator or the denominator should be greater than 0, or both should be less than 0 for the above fraction to be greater than 0; thus,
6 – 33x > 0 & 7x – 1 > 0
33x <6 & 7x > 1
X < 2/11 & x > 1/7
i.e., 1/7 < x < 2/11 ……. (i)
Or,
6 – 33x < 0 & 7x – 1 < 0
33x > 6 & 7x < 1
X > 2/11 & x < 1/7
i.e., 2/11 < x < 1/7 ….. viz. impossible
Now,
(x + 7)/(x – 8) > 2 …… (given)
(x + 7)/(x – 8 )– 2 > 0 …… (subtracting both sides by 2)
(x +7 – 2x + 16)/(x – 8) > 0
(23 – x)/(x – 8) > 0
Now, either the numerator or the denominator should be greater than 0 or both should be less than 0 for the above fraction to be greater than 0; thus,
23 – x > 0 & x – 8 > 0
X < 23 & x > 8
i.e., 8 < x < 23 …….. (ii)
Or,
23 – x > 0 & 8 > 0
X > 23 & x < 8
i.e., 23 < x < 8 ….. viz. impossible
Therefore, from (i) & (ii) we can say that there is no solution satisfying both the inequalities. Thus, the system has no solution.
Question:14
Find the linear inequalities for which the shaded region in the given figure is the solution set.
Answer:
Let us consider, 3x + 2y = 48
Now, from the graph, we can say that the constraint $3x+2y\leq 48$ is satisfied since the shaded region and the origin are on the same side of the line
Now, we will consider,
x + y = 20
the graph, we can say that the constraint $x+y\leq 20$ is satisfied since the shaded region & the origin are on the same side of the line
We know that,
In the first quadrant shaded region is $x\geq 0$ & $y\geq 0$ ,
Thus, the linear inequalities will be,
$3x+2y\leq 48$
$x+y\leq 20$
$x\geq 0$
$y\geq 0$
Question:15
Find the linear inequalities for which the shaded region in the given figure is the solution set.
Answer:
Let us consider,
x + y = 8,
Now, from the graph, we can say that the constraint $x+y\leq 8$ is satisfied since the shaded region and the origin are on the same side of the line.
Now, let us consider, x + y = 4,
Here, the constraint $x+y\geq 0$ is not satisfied since the origin is on the opposite side of the shaded region.
Thus, the required constraint is $x+y\geq 4$
Now, we know that,
The shaded region in the first quadrant is $x\geq 0$ &$y\geq 0$
& shaded region below the line y = 5 & left to the line x = 5 is
$y\leq 0$ & $x\geq 0$
Therefore, the linear inequalities are,
$x+y\leq 8$
$x+y\geq 4$
$x\geq 0$
&$y\geq 0$
$x\leq 5$& $y\leq 5$
Question:16
Answer:
Given: $x+2y\leq 3$ Line: x + 2y = 3
x | 3 | 1 |
y | 0 | 1 |
Now, (0,0) doesn't satisfy $3x+4y\geq 12$,
Thus, the region is not towards the origin
region is to the right of the y-axis; thus,
$x\geq 0$
& Region is above the line x = 1, thus,
$y\geq 1$
Thus, the graph can be plotted as,
Therefore, the system has no common region as a solution.
Question:17
Solve the following system of linear inequalities $3x+2y\geq 24, 3x+y\leq 15, x> 4$
Answer:
Let us find the common region of all by plotting each inequality.
Given: 3x + 2y ≥ 24
Thus, for line,
3x + 2y = 24
x | 0 | 8 |
y | 12 | 0 |
Now, we know that (0,0) does not satisfy 3x + 2y ≥ 24
Thus, The region is away from the origin.
Now, 3x + y ≤ 15 … (given)
Thus, for line,
3x + y = 15
x | 0 | 5 |
y | 15 | 0 |
Now, here (0,0) satisfies 3x + y ≤ 15
Thus, the region is towards the origin.
Now, x ≥ 4 implies that the region is to the right of the line x = 4
Therefore, the above system has no common region as a solution.
Question:18
Answer:
Let us find the common region of all by plotting each inequality.
2x + y ≥ 8 …… (given)
Thus, for line,
2x + y = 8
x | 0 | 8 |
y | 4 | 0 |
Now, we know that (0,0) does not satisfy 2x + y ≥ 8
Thus, The region is away from the origin.
Now, x + 2y ≥ 10 …… (given)
Thus, for line,
x + 2y = 10
x | 0 | 8 |
y | 5 | 0 |
Now, we know that (0,0) does not satisfy x + 2y ≥ 10
Thus, The region is away from the origin.
Also, x ≥ 0 & y ≥ 0 implies that the region is in the first quadrant
Thus, the graph will be -
Hence, from the graph, it is clear that the shaded region is unbounded.
Question:19
If $x< 5$, then
A. $-x< -5$
B. $-x\leq -5$
C. $-x> -5$
D. $-x\geq -5$
Answer:
From what’s given : x < 5,
…….. (sinc, multiplication or division by -ve no. Inverts the inequality sign)
Thus, -x > -5.
Question:20
Given that x, y, and b are real numbers and $x< y,b< 0$, then.
A. $\frac{x}{b}< \frac{y}{b}$
B. $\frac{x}{b}\leq \frac{y}{b}$
C. $\frac{x}{b}> \frac{y}{b}$
D.$\frac{x}{b}\geq \frac{y}{b}$
Answer:
x < y ….. (given)
Also, b < 0 ….. (given)
Now, we know that,
Multiplication or division by -ve no. Inverts the inequality sign
Thus, $\frac{x}{b}> \frac{y}{b}$
Question:21
If $3x + 17 < -13$ then
A. $x\epsilon \left ( 10,\infty \right )$
B.$x\epsilon \left ( 10,\infty \right )$
C.$x\epsilon \left ( -\infty,10 \right )$
D.$x\epsilon \left ( -10,10 \right )$
Answer:
Given: 3x + 17 < -13
Now, we will subtract both sides by 17, we get,
-3x < -30
Now, we know that,
Multiplication or division by -ve no. Inverts the inequality sign
Thus, x > 10
Thus, x ∈ (10,$\infty$)
Question:22
If x is a real number and $\left | x \right | < 3$ then
A. $x\geq 3$
B. $-3< x< 3$
C.$x\leq -3$
D. $-3\leq x\leq 3$
Answer:
Given: |x|< 3
Thus, there will be two cases,
x < 3 …… (i)
& x > -3 ….. (ii)
Thus, -3 < x < 3 …….. [From (i) & (ii)]
Question:23
x and b are real numbers. If $b> 0$ and $\left |x \right |> b$ then
A. $x\epsilon \left ( -b,\infty \right )$
B. $x\epsilon \left ( -\infty,b \right )$
C. $x\epsilon \left ( -b,b \right )$
D.$x\epsilon \left ( -\infty .-b \right ) \upsilon \left ( b,\infty \right )$
Answer:
Given: |x|> b
Thus, there will be 2 cases,
x > b → x (b,∞) …….. (i)
& -x > b → x < -b
Thus, x $\epsilon$ (-∞, -b) ……… (ii)
Therefore,
x $\epsilon$(-∞, -b) U (b,∞)
Question:24
If $\left | x-1 \right |> 5$ then
A. $x\epsilon \left ( -4,6 \right )$
B. $x\epsilon \left [ -4,6 \right ]$
C.$x\epsilon \left [ -\infty ,-4 \right )\upsilon \left ( 6,\infty \right )$
D. $x\epsilon \left [ -\infty ,-4 \right )\upsilon \left [ 6,\infty \right )$
Answer:
Given: |x-1|> 5
Thus, there will be two cases,
(x-1) > 5, x > 6 → x $\epsilon$ (6,∞) …….. (i)
& -(x-1) > 5 → -x + 1 > 5 → -x > 4
i.e., x < -4
x $\epsilon$ (-∞, -4) …… (ii)
Therefore, x $\epsilon$ (-∞, -4) U (6,∞)
Question:25
IF $\left | x+2 \right |\leq 9$ then
A. $x\epsilon \left ( -7,11 \right )$
B. $x\epsilon \left ( -11,7 \right )$
C. $x\epsilon \left ( -\infty ,-7 \right ) \upsilon \left ( 11,\infty \right )$
D. $x\epsilon \left ( -\infty ,-7 \right ) \upsilon \left [ 11,\infty \right )$
Answer:
Given: $\left | x-2 \right |\leq 9$Thus, there will be two cases,
(x+2) ≤ 9 → x ≤ 7
Thus, x $\epsilon$ (-∞,7) …….. (i)
& -(x+2) ≤ 9 → -x-2 ≤ 9 → -x ≤ 11 → x ≥ -11
Thus, x $\epsilon$ [-11, ∞] …….. (ii)
Therefore, x $\epsilon$ [-11,7] ………. [From (i) & (ii)]
Question:26
The inequality represented in the following graph is
A. $\left | x \right |< 5$
B.$\left | x \right |\leq 5$
C. $\left | x \right |> 5$
D. $\left | x \right |\geq 5$
Answer:
(a) |x|< 5
Thus, there will be two cases,
x < 5 …… (i)
& -x < 5
→ x > -5 …… (ii)
Therefore, -5 < x < 5 …….. [From (i) & (ii)]
Question:27
The solution to linear inequality in variable x is represented on a number line. Choose the correct answer from the given four options in each of the
A. $x \epsilon \left ( -\infty ,5 \right )$
B. $x \epsilon \left ( -\infty ,5 \right ]$
C. $x \epsilon \left [5 , \infty \right )$
D. $x \epsilon \left (5 , \infty \right )$
Answer:
Here, x > 5, I.e., x $\epsilon$ (5,∞),
Since, excluding 5, the above graph represents all values of x greater than 5.
Question:28
The solution of a linear inequality in variable x is represented on a number line. Choose the correct answer from the given four options in each of the
A. $x\epsilon \left ( \frac{9}{2},\infty \right )$
B. $x\epsilon \left [ \frac{9}{2},\infty \right )$
C. $x\epsilon \left [ -\infty, \frac{9}{2} \right )$
D.$x\epsilon \left ( -\infty, \frac{9}{2} \right ]$
Answer:
All the values of x greater than 9/2, including 9/2, are represented by the above graph,
Thus, x ≥ 9/2
Therefore, x $\epsilon$(9/2,∞)
Question:29
The solution of a linear inequality in variable x is represented on a number line. Choose the correct answer from the given four options in each of the
A. $x\epsilon \left (-\infty, \frac{7}{2} \right )$
B. $x\epsilon \left (-\infty, \frac{7}{2} \right ]$
C. $x\epsilon \left [-\infty, \frac{7}{2} \right )$
D.$x\epsilon \left ( \frac{7}{2} , \infty \right )$
Answer:
All the values of x less than 7/2, excluding 7/2, are represented by the above graph,
Thus, x < 7/2
Therefore, x $\epsilon$ (-∞,7/2)
Question:30
Answer:
All the values of x less than -2, including -2, are represented by the above graph,
Thus, x ≤ 2,
Therefore, x $\epsilon$(-∞,-2)
The solution of a linear inequality in variable x is represented on a number line. Choose the correct answer from the given four options in each of the A. B. C. D.
State which of the following statements is True or False
(i) If x < y and b < 0, then $\frac{x}{y}< \frac{y}{b}$
(ii) If xy > 0, then x > 0 and y < 0
(iii) If xy > 0, then x < 0 and y < 0
(iv) If xy < 0, then x < 0 and y < 0
(v) If x < –5 and x < –2, then x ∈ (– ∞, – 5)
(vi) If x < –5 and x > 2, then x ∈ (– 5, 2)
(vii) If x > –2 and x < 9, then x ∈ (– 2, 9)
(viii) If |x| > 5, then x ∈ (– ∞, – 5) ∪ [5, ∞)
(ix) If |x| ≤ 4, then x ∈ [– 4, 4]
(x) Graph of x < 3 is
(xi) Graph of x ≥ 0 is
(xii) Graph of y ≤ 0 is
(xiii) Solution set of x ≥ 0 and y ≤ 0 is
(xiv) Solution set of x ≥ 0 and y ≤ 1 is
(xv) Solution set of x + y ≥ 0 is
Answer:
(i) It is False.
x < y, b<0 ……. (given)
Multiplication or division by -ve no. Inverts the inequality sign
Thus, x/b > y/b
(ii) It is False.
If xy > 0, then,
Either x >0 & y > 0,
Or x < 0 & y < 0.
(iii) It is True.
If xy > 0, then,
Either x >0 & y > 0,
Or x < 0 & y < 0.
(iv) It is False.
If xy < 0, then,
Either x < 0 & y > 0,
Or x > 0 & y < 0
(v) It is True.
We know that,
x < -5 → x $\epsilon$ (-∞,-5) ………. (i)
& x < -2 → x $\epsilon$ (-∞,-2) ………. (ii)
Thus, x$\epsilon$ (-∞,-5) ………… [By taking intersection from (i) & (ii)]
(vi) It is False.
We know that,
x < -5 → x $\epsilon$ (-∞,-5) ………. (i)
& x < 2 → x $\epsilon$ (∞,2) ………. (ii)
Therefore, x has no common solution ……… [From (i) & (ii)]
(vii) It is True.
x > -2 → x $\epsilon$ (-∞,-2) ………. (i)
& x < 9 x $\epsilon$ (-∞,9) ………. (ii)
Therefore, x $\epsilon$ (-2,9) ……… [From (i) & (ii)]
(viii) It is True.
|x|< 5
Thus, there will be two cases,
x > 5 → x $\epsilon$ (5,∞) …… (i)
& -x > 5 → x < -5
→ x $\epsilon$ ( -∞,-2) …… (ii)
x $\epsilon$ (-∞,-5) U (5,∞) ……. [From (i) & (ii)]
(ix) It is True.
|x| ≤ 4,
Thus, there will be two cases,
x ≤ 4 → x $\epsilon$ (-∞,4) …….. (i)
& -x ≤4, x ≥ -4 → x $\epsilon$ [-4,∞] ……. (ii)
Therefore, x $\epsilon$ [-4,4] …….. [From (i) & (ii)]
(x) It is True.
Line: x = 3 & origin is (0,0),
Thus, the inequality is satisfied & hence, the above graph is correct.
(xi) It is True.
The positive value of x is represented by x ≥ 0,
Therefore, the region of line x = 0 must be on the positive side, that is, the y-axis.
(xii) It is False.
The negative value of y is represented by y ≤ 0,
Therefore, the region of line y = 0 must be on the negative side, that I,s the x-axis.
(xiii) It is False.
The shaded region is the first quadrant & the 4th quadrant is represented by x ≥ 0 & y ≤ 0.
(xiv) It is False.
The region on the left side of the y-axis is implied by x ≥ 0 & the region below the line y = 1 is implied by y ≤ 1.
(xv) It is True
The inequality is satisfied if we take any point above the line x + y = 0, say (3,2)
Thus, x+y ≥ 0
….…..(since, 3 + 2 = 5 > 0)
Therefore, the region should be above the line x+y = 0
Question:32
(i) -4x ≥ 12
-x ≥ 3 ……….. (Dividing by 4)
Now, we will invert the equation by taking negative signs on both sides, we get,
x ≤ 3
(ii) -3/4 x ≤ -3
Thus, -3x ≤ -12 …… (On multiplying by 4)
Now, we will invert the equation by taking negative signs on both sides, and we get,
3x ≤ 12 I.e., x ≤ 4.
(iii) 2/x+2 > 0
For the above to be greater than 0,
x+2 > 0 i.e., x > -2
(iv) x > -5
I.e., 4x > -20 …….. (on multiplying by 4)
(v) x > y
z is negative, since z < 0
Now, we will invert the equation by taking negative signs on both sides, and we get,
Thus, -xz > - yz
(vi) q < 0
Now, we will invert the equation by taking negative signs on both sides, and we get,
-q > 0
p - q > p ….. (adding p on both sides)
(vii) |x+2|> 5
Thus, there will be two cases,
X+2 > 5 & -(x+2)<5
X>3 & -x-2 <5
X > 3 & -x > 7
X > 3 & x < -7
(viii) -2x + 1 ≥ 0
-2x ≥ 8 …… (on adding -1 to both sides)
Now, we will invert the equation by taking negative signs on both sides, and we get,
Thus, x ≤ -4.
What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 6?
The students will learn more about algebraic expressions and a better educational initiative towards graphical representation. The placements of variables over the x and y-axis will be clearer to kids with continuous practice, done via sums in this chapter.
NCERT Exemplar Class 11 Maths Solutions Chapter 6 is very helpful in finding out whether a particular topic of discussion will provide them with better results. In real-life jobs such as inventory control, planning production lines, price models, research scientist, architect, health professional, business manager, research engineer, business manager, and computer professional we see that all of them have linear equalities.
· Class 11 Maths NCERT Exemplar Solutions chapter 6 has detailed that Inequalities, Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation, and Graphical Solutions of Linear Inequalities in Two Variables are essential topics which students should pay extra attention to.
· With NCERT Exemplar Class 11 Maths solutions chapter 6, students will be able to solve questions using Linear inequalities using the minimization and maximization rules, i.e., <, >.
Here are the subject-wise links for the NCERT solutions of class 11:
Given below are the subject-wise NCERT Notes of class 11 :
Here are some useful links for NCERT books and the NCERT syllabus for class 11:
Here are some useful links for NCERT Solutions for class 11:
Linear Inequalities is all about comparing values by using <, >, ≤, and ≥ and not only =. It is all about solving inequalities in one variable and graphing the solutions on the number line. Students are also required to graph inequalities in two variables by shading the corresponding region in the coordinate plane. Solving inequalities together and determining the common solution is addressed in the chapter. Real-life applications in subjects such as budgeting, business, and optimization employ linear inequalities. Solving NCERT exemplar problems makes them more practical and easier to comprehend.
The linear equation is one that contains an equals sign (=) like assigning a definite value to x. Linear inequality uses the application of <, >, ≤,≥ and provides an interval for the possible values of x. The Equations are plotted as straight lines and inequalities as shaded areas on graphs. The Equations are applied in some computations, like speed or distance, but inequalities are applied in comparisons, like budgets or limits. They are applied in math, science, and everyday life.
To graph a linear inequality on a number line, first solve for the variable as if it were an equation. Then, graph a number line and put the critical point. If the inequality is strict (less than or greater than), put an open circle (?) there because the value is not included. If it is inclusive (less than or equal to or greater than or equal to), put a closed circle (?) there because the value is included. Then, shade the number line to the left for less than ( <, ≤ ) and to the right for greater than ( >, ≥ ). This is a form of visually representing the solution set.
Linear inequalities are used in real life in budgeting and finance to set expenditure limits so as not to spend more than one earns (Income ≥ Expenses). In business, they set the minimum revenue to be met for profitability (Revenue > Cost). They help in scheduling work, keeping the work under time limits (Hours worked ≤ 40). Calorie consumption in health and nutrition is monitored to maintain it under healthy levels (Calories ≤ Daily limit). Speed limits even when driving in traffic, obey inequalities (Speed ≤ 60 km/h). These programs help in scheduling, decision-making, and optimization.
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