NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

# NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

Edited By Ravindra Pindel | Updated on Sep 12, 2022 06:02 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 6 covers linear inequalities and its application. Linear inequalities mean comparing two objects, or in the form of greater or lower, i.e., ‘<’ or ‘>.’ The comparison is made for two algebraic expressions, and above symbols are the denominations of answers. NCERT Exemplar Class 11 Maths chapter 6 solutions help the students in understanding the application of Linear Equalities in real life. There are specific terms that we will find in the NCERT Exemplar solutions for Class 11 Maths chapter 6 such as regional inequality, which means a region or a place that contains all the solutions for the inequalities.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

Also read - NCERT Solutions for Class 11 Maths

## NCERT Exemplar Class 11 Maths Solutions Chapter 6: Exercise-1.3

Question:1

Solve for x the inequalities in

Given:
Now, let us multiply all the terms by , we get,

i.e.,
Now, on subtracting each term by 3, we get,

At last, we will divide each term by 3,
We get,

Question:2

Solve for x, the inequalities in $\frac{\left | x-2 \right | -1}{\left | x-2 \right | -2}\leq 0$

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<0 & y-2<0 , & thus ,
viz. Not needed
now,
if
&
& hence,
… thus, we get the answer we need.
Now,
Thus,
Now, from this, we will get 2 cases, which are-

&
On multiplying each term by -1, we get,

Now, we’ll add 2 to each term, we get,

Therefore,

Question:3

Solve for x the inequalities in

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,
& or &
Thus & or &
Thus,
$x \epsilon \left ( -\infty , -5 \right ]$ or $\left [ 5, -\infty \right )$ or
Therefore,
$x \epsilon \left ( -\infty , -5 \right ] \upsilon \left ( -3,3 \right ) \upsilon \left [5,\infty \right )$

Question:4

(i) Now, there will be two cases –

Adding 1 on both the sides, we get,

(ii)
i.e.,
Subtract 1 from both the sides, we will get,
i.e
Now, from (i) & (ii), we get,

& '
Thus, &
Thus
i.e

Question:5

Solve for x, the inequalities in

Given
On multiplying all the terms by 4, we get,

Now, add -2 to each term,

And now, divide each term by 3,

Now, we will multiply each term by -1 to invert the inequality, we get,

Question:6

Solve for x, the inequalities in

given
Thus, $4x-2x\geq 17-3$
Thus
Thus .........(i)
Now,
............. given
Thus .........(ii)
Therefore, x has no solution as eq. (i) & eq. (ii) cannot be possible simultaneously.

Question:7

A company manufactures cassettes. Its cost and revenue functions are C(x) = 26,000 + 30x and R(x) = 43x, respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit?

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

The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of pH value for the third reading that will result in the acidity level being normal.

Solution:
Let ‘x’ be the third reading,
Now, 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

A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 litres of the 9% solution, how many litres of 3% solution will have to be added?

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

A solution is to be kept between 40°C and 45°C. What is the range of temperature in degree Fahrenheit, if the conversion formula is

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

The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.

Let us assume that 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,
Perimeter of a triangle = sum of all three sides
= x + 2x + x + 2
= 4x + 2 cm
Now, it is given that perimeter of the triangle is more than 166 cm,
Thus,

Thus,$x\geq 41$
Therefore, the minimum length of the shortest side should be = 41 cm

Question:12

In drilling world’s deepest hole it was found that the temperature T in degree Celsius, x km below the earth’s surface was given by T = 30 + 25 (x – 3), 3 ≤ x ≤ 15. At what depth will the temperature be between 155°C and 205°C?

Given:
T = 30 + 25 (x – 3),
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

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 grater 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 …… (on subracting 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 grater 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

Let us consider, 3x + 2y = 48
Now, from the graph we can say that the constraint is satisfiedsince the shaded region and the origin are on the same side of the line
Now, we will consider,
x + y = 20
Again, the graph we can say that the constraint is satisfied since the shaded region & the origin are on the same side of the line
We know that,
Thus, the linear inequalities will be,

&

Question:15

Let us consider,
x + y = 8,
Now, from the graph we can say that the constraint is satisfiedsince the shaded region and the origin are on the same side of the line.
Now, let us consider, x + y = 4,
Here, the constraint is not satisfied since the origin is on the opposite side of the shaded region.
Thus, the required constraint is
Now, we know that,
& shaded region below the line y = 5 & left to the line x = 5 is
&
Therefore, the linear inequalities are,

&
&

Question:16

Show that the following system of linear inequalities has no solution

Given: Line: x + 2y = 3

 x 3 1 y 0 1

Now, (0,0) doesn't satisfies $3x+4y\geq 12$,
Thus, the region is not towards the origin
region is to the right of the y axis, thus,

& Region is above the line x = 1, thus,

Thus, the graph can be plotted as,

Therefore, the system has no common region as solution.

Question:17

Solve the following system of linear inequalities

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 solution.

Question:18

Show that the solution set of the following system of linear inequalities is an unbounded region

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 , then
A.
B.
C.
D.

From what’s given : x < 5,
…….. (since, multiplication or division by -ve no. Inverts the inequality sign)
Thus, -x > -5.

Question:20

x < y ….. (given)
Also, b < 0 ….. (given)
Now, we know that,
Multiplication or division by -ve no. Inverts the inequality sign
Thus,

Question:21

If then
A.
B.
C.
D.

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,)

Question:22

If x is a real number and then
A.
B.
C.
D.

Given: |x|< 3
Thus, there will be two cases,
x < 3 …… (i)
& x > -3 ….. (ii)
Thus, -3 < x < 3 …….. [From (i) & (ii)]

Question:23

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 then
A.
B.
C.
D.

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 then
A.
B.
C.
D.

Given: Thus, there will be two cases,
(x+2) ≤ 9 → x ≤ 7
Thus, x (-∞,7) …….. (i)
& -(x+2) ≤ 9 → -x-2 ≤ 9 → -x ≤ 11 → x ≥ -11
Thus, x [-11, ∞] …….. (ii)
Therefore, x [-11,7] ………. [From (i) & (ii)]

Question:26

(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

Here, x > 5, I.e., x $\epsilon$ (5,∞),
Since, excluding 5, the above graph represents all values of x greater than 5.

Question:28

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

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

Solution of a linear inequality in variable x is represented on number line. Choose the correct answer from the given four options in each of the

A.
B.
C.
D.

All the values of x less than -2, including -2 are represented by the above graph,
Thus, x ≤ 2,
Therefore, x $\epsilon$(-∞,-2)

Solution of a linear inequality in variable x is represented on number line. Choose the correct answer from the given four options in each of the A. B. C. D.

Question:31

State which of the following statements is True or False

(i) If x < y and b < 0, then
(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

(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 is 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 the 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 the sides, we get,
3x ≤ 12 I.e., x ≤ 4.
(iii) 2/x+2 > 0
In order 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 the sides, we get,
Thus, -xz > - yz
(vi) q < 0
Now, we will invert the equation by taking negative signs on both the sides, 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 the sides, we get,
Thus, x ≤ -4.

Question:26

The number of terms in the expansion of (x + y + z)n_________________

We know that,
(x+y+z)n = [x+(y+z)]n
= nC0xn + nC1xn-1 (y+z) + nC2xn-2 (y+z)2 + …… + nCn (y + z)n
Therefore, the no. of terms in the expansion will be –
1+2+3+ …. +n+(n+1)
= (n+1)(n+2)/2

## Highlights

Invented by mathematician Thomas Harriot in 1631, now linear inequalities are an essential part of academics, thus being taught from a young age Linear inequalities deals with a variable such as x,y,z, etc.

In the past, we have learnt about singular or double variable formulas. But in linear inequalities, we will put an entire problem into a single variable which could be solved through the variable formulas. Class 11 Maths NCERT Exemplar solutions chapter 6 will help students in deriving a variable formula, to solve the sum and procure answers.

With the help of NCERT Exemplar Class 10 Maths solutions chapter 6, students can easily understand these topics. The confusion will be cleared for students, and they can score better in exams.

All the concepts are detailed in the NCERT Exemplar Solutions for Class 11 Maths chapter 6. By using the NCERT Exemplar Class 11 Maths chapter 6 solutions PDF Download, students get access to quality study material effectively constructed by experts for the best learning experience.

## Main Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 6

• Introduction
• Inequalities
• Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation
• Graphical Solution of Linear Inequalities in Two Variables

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 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, plan 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.

## Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 6

· Class 11 Maths NCERT Exemplar solutions chapter 6 has detailed that Inequalities, Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation, Graphical Solution 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 Equalities using the minimization and maximization rules, i.e., <, >.

Check Chapter-Wise NCERT Solutions of Book

 Chapter-1 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

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1. What all topics are added in this chapter?

Class 11 Maths NCERT Exemplar solutions chapter 6 covers the topics related to linear inequalities, graphical representation of inequalities using one and two variables.

2. Who has prepared these solutions?

These NCERT Exemplar Class 11 Maths solutions chapter 6 prepared by our highly experienced Maths teachers with years of CBSE teaching background.

3. Are these solutions helpful in entrance exams?

Practising NCERT questions for the entrance exam will help in understanding the basics in a much better way.

4. Are all questions solved in these NCERT exemplar solutions for Class 12 Maths chapter 6?

Yes, we have covered questions that are given in various exercises in the NCERT book along with the miscellaneous questions as well.

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With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9