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NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

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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.

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 \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, on 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 the 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

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?

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

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.

Answer:

Given: First reading = 8.48
Second reading = 8.35
To find: Third reading
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?

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

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 F=\frac{9}{5}C+32

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

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.

Answer:

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, 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

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?

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 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

Find the linear inequalities for which the shaded region in the given figure is the solution set.
fireshot-capture-003-ncertnicin

Answer:

Let us consider, 3x + 2y = 48
Now, from the graph we can say that the constraint 3x+2y\leq 48 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 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
q14

Answer:

Let us consider,
x + y = 8,
Now, from the graph we can say that the constraint x+y\leq 8 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 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,
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

Show that the following system of linear inequalities has no solution x+2y\leq 3, 3x+4y\geq 12, x\geq 0, y\geq 1

Answer:

Given: x+2y\leq 3 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,
x\geq 0
& Region is above the line x = 1, thus,
y\geq 1
Thus, the graph can be plotted as,
ms-paint-_-microsoft-paint-online
Therefore, the system has no common region as 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
fireshot-capture-006-desmos-graphing-calculator-wwwdesmoscom
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 2x+y \geq 8, x+2y\geq 10, x\geq 0, y\geq 0

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 -
a18
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,
…….. (since, 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 9Thus, 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 representing the following graph is
q26

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

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
q27
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

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
q28

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

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
q29
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

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
q30
A. x\epsilon \left ( -\infty ,-2 \right )
B. x\epsilon \left ( -\infty ,-2 \right ]
C. x\epsilon \left ( -2, \infty \right ]
D.x\epsilon \left [ -2, \infty \right )

Answer:

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 \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
q31x
(xi) Graph of x ≥ 0 is

fireshot-capture-012-ncertnicin
(xii) Graph of y ≤ 0 is
fireshot-capture-015-ncertnicin
(xiii) Solution set of x ≥ 0 and y ≤ 0 is
q31xiii
(xiv) Solution set of x ≥ 0 and y ≤ 1 is

fireshot-capture-018-ncertnicin
(xv) Solution set of x + y ≥ 0 is

fireshot-capture-023-ncertnicin

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 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.
a31
(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

Fill in the blanks of the following:
(i) If – 4x ≥ 12, then x ... – 3.
(ii) If -3x/4 ≤ –3, then x…..4.
(iii) If \frac{2}{x+2}> 0, then x …. –2.
(iv) If x > –5, then 4x ... –20.
(v) If x > y and z < 0, then – xz ... – yz.
(vi) If p > 0 and q < 0, then p – q ... p.
(vii) If |x+ 2|> 5, then x ... – 7 or x ... 3.
(viii) If – 2x + 1 ≥ 9, then x ... – 4.

Answer:

(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_________________

Answer:

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.

Also read - NCERT Solutions for Class 11 Chemistry Chapter 6

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., <, >.

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Frequently Asked Question (FAQs)

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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

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

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Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
ITSM Manager
3 Jobs Available
.NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

2 Jobs Available
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