CBSE Sample Papers for Class 11 Maths Set 10 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 10 with Solutions

Time Allowed : 3 hours
Maximum Marks : 80

General Instructions:

• All questions are compulsory.
• The question paper consists of 36 questions divided into 4 Sections A, B, C and D.
• Section A comprises of 20 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 6 questions of 4 marks each and Section D comprises of 4 questions of 6 marks each.
• There is no overall choice. However internal choice has been provided in 6 questions of 1 mark, 2 questions of 2 marks, 2 questions of 4 marks and 2 questions of 6 marks. You have to attempt only one of the alternatives in all such questions.
• Write the serial number of questions before attempting.
• Use of a calculator is not permitted.

Section – A
Question numbers 1 to 10 carries 1 mark each. For each of these questions, four alternative choices have been provided of which only one is correct. Select the correct choice :

Question 1.
If x₁, x₂, x₃ …… x_n be n observations and x be their arithmetic mean. Then, formula for the standard the deviation is given by
(A) Σ(x_i – x̄)²
(B) \(\frac{\sum\left(x_i-\bar{x}\right)^2}{n}\)
(C) \(\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}\)
(D) \(\sqrt{\frac{\sum x_i^2}{n}-\bar{x}^2}\)
Answer:
(C) \(\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}\)

Explanation:
We know that, For n observations,
Standard deviation = \(\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}\)

Question 2.

is equal to :
(A) 2
(B) \(\frac{2}{3}\)
(C) \(\frac{-3}{2}\)
(D) 1
Answer:
(A) 2

Explanation:
We have,
On evaluating the limit,

Question 3.
The locus of a point on which x = 0 is
(A) XY-plane
(B) YZ-plane
(C) ZX-plane
(D) None of these
Answer:
(B) YZ-plane

Explanation:
We know that,
The locus of point for which x = 0 is YZ plane in 3D- geometry.

Question 4.
The standard deviation of data 6, 5,9,13,12,8 and 10 is
(A) \(\sqrt{\frac{52}{7}}\)
(B) \(\frac{52}{7}\)
(C) √6
(D) 6
Answer:
(A) \(\sqrt{\frac{52}{7}}\)

Explanation:
We have,
Standard Deviation of the given data:
SD = \(\sqrt{\frac{52}{7}}\)

Question 5.
If the focus of a parabola is (0, – 3) and its directrix is y = 3, then its equation is
(A) x² = – 12y
(B) x² = 12y
(C) y² = – 12x
(D) y² = 12x
Answer:
(A) x² = – 12y

Explanation:
Given that,
Equation of directrix, y = 3
Focus of parabola (0, -3)
Equation of the parabola is:
⇒ x² = – 12y

Question 6.
If f(x) = – cos x then f'(0) = ?
(A) 0
(B) 1
(C) 2
(D) 4
Answer:
(A) 0

Explanation:
Given that,
f(x) = – cos x
f ‘(x) = sin x
f ‘(0) = 0

Question 7.
The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1: 4 are
(A) 3rd and 4th
(B) 4th and 5th
(C) 5th and 6th
(D) 6th and 7th
OR
The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is
(A) 150
(B) 15
(C) 175
(D) 135
Answer:
(C) 5th and 6th

Explanation:
Given that,
The terms whose ratio is 1 : 4 in the expansion of (1+x)²⁴ are 5^th and 6^th

OR

(D) 135

Explanation:
The number of triangles formed
= ¹²C₃ – ⁷C₃ = 135

Question 8.
The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at the time is
(A) 432
(B) 108
(C) 36
(D) 18
OR
Every body in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
(A) 11
(B) 12
(C) 13
(D) 14
Answer:
(A) 432

Explanation:
The sum of the digits in the unit place of all the numbers formed with 3, 4, 5, 6 taken all at a time = 4! × (3 + 4 + 5 + 6) = 432

OR

(B) 12

Explanation:
We have,
ⁿC₂ = 66
On simplifying we get
⇒ n = 12
So there are 12 persons in room.

Question 9.
If sets A and B are defined as
A = {(x, y) | y = \(\frac{1}{x}\), x ≠ 0, x ∈ R}, then
(A) A ∩ B = A
(B) A ∩ B = B
(C) A ∩ B = Φ
(D) A ∪ B = A
Let F₁ be the set of parallelograms, F₂ the set of rectangles, F₃ the set of rhombuses, F₄ the set of squares and F₅ the set of trapeziums in a plane. Then F₁ may be equal to
(A) Φ₂ ∩ Φ₃
(B) F₃ ∩ F₄
(C) Φ₂ ∩ Φ₃
(D) F₂ ∪ F₃ ∪ F₄ ∩ F₁
Answer:
(C) A ∩ B = Φ

Explanation:
We have,
From the given sets, there is no element common in them
So A ∩ B = Φ

OR

(D) F₂ ∪ F₃ ∪ F₄ ∩ F₁

Explanation:
We have,
From the given information, we get
F₂ ∪ F₃ ∪ F₄ ∩ F₁

Question 10.
The distance of the point P(0, 1, 0) from the origin is
(A) 0 units
(B) 1 units
(C) 2 units
(D) 5 units
Answer:
(B) 1 units

Explanation:
We have,
Point P(0, 1, 0)
Distance from origin = [0 + 1 + 0]
= 1 units

Question numbers 11 to 15 carry 1 mark each.
Write whether the statement is true or false.

Question 11.
If tan.A = \(\frac{1-\cos B}{\sin B}\), than tan 2A = tan B.
Answer:
True

Explanation:
Given that,
⇒ tan A = \(\frac{1-\cos B}{\sin B}\)
then we have
⇒ tan 2A = tan B

Question 12.
The ordered pair (5, 2) belongs to the relation
R = {(x, y): y = x – 5, x, y ∈ Z}
OR
(x : x is an even natural number less than 6} and {x : x is a natural number which is a factor of 36}. These two sets have some common elements.
Answer:
False

Explanation:
We have,
From the given Relation
when x = 5 we get
y = 0
So the given statement is False

OR

True

Explanation:
{x : x is an even natural number less than 6} = {2, 4} and {x : x a natural number which divides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}.
So there are some common elements Hence, the given statement is true.

Question 13.
If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im(z²) = 0.
Answer:
True

Explanation:
For the given complex number z we have, Re(z) = 0 and Im(z²) = 0
Hence the given statement is true.

Question 14.
If some or all of n objects are taken at a time, the number of combinations is 2ⁿ – 1.
Answer:
True

Explanation:
If some or all of n objects are taken at a time, the number of combinations is 2ⁿ – 1
Hence, the given statement is true.

Question 15.
If x > – 5, then 4x < – 20. Answer: False Explanation: If x > – 5 then 4x > -20
So the given statement is False.

Question numbers 16 to 20 carry 1 mark each.

Question 16.
Find the distance between the directrices of the ellipse \(\frac{x^2}{36}+\frac{y^2}{20}\) = 1.
Answer:
Given equation of the ellipse is

Question 17.
Evaluate

OR
Find the value of \(\lim _{x \rightarrow 0} \frac{e^x-1}{x}\)
Answer:

OR

Question 18.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is ……………
Answer:
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive.
So we have
P(A ∪ B) = P(A) + P(B)
Required probability = P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – (0.5 + 0.3) = 0.2

Question 19.
What is the value of y, so that the line through (3, y) and (2,7) is parallel to the line through (-1, 4) and (0, 6)?
OR
Find the equation of lines passing through (1,2) and making angle 30° with Y-axis.
Answer:
Slope of the line passing through (3, y) and (2, 7),
m₁ = \(\frac{7-y}{2-3}\) = y – 7
and slope of the line passing through (-1, 4) and (0, 6), m₂ = \(\frac{6-4}{0+1}\) = 2
Since, the lines are parallel,
so m₁ = m₂
⇒ \(\frac{y-7}{3-2}\) = 2
⇒ So we get, y = 9

OR

Given that, angle with Y – axis = 30°
and angle with X – axis = 60°
∴ Slope of the line, m = tan 60° = √3

So, the equation of a line passing through (1, 2) and having slope x√3 , is
y – 2 = √3 (x – 1)
⇒ y – 2 = √3x – x√3
⇒ y – x√x – 2 + √3 = 0 is the required equation of line.

Question 20.
Solve, -3 ≤ – 3x + 2 < 4, x ∈ R.
Answer:
We have, – 3 ≤ – 3x + 2 < 4
⇒ – 3 – 2 ≤ – 3x + 2 – 2 < 4 – 2
(Adding – 2 in inequality)
⇒ – 5 ≤ – 3x < 2 ⇒ \(\frac{-5}{-3} \geq \frac{-3 x}{-3}>\frac{2}{-3}\) (Dividing by – 3)
⇒ \(\frac{5}{3}\) ≥ x > – \(\frac{2}{3}\)
or – \(\frac{2}{3}\) < x ≤ \(\frac{5}{3}\)
or x ∈ \(\left(-\frac{2}{3}, \frac{5}{3}\right]\)

Commonly Made Errors:
Students make simplification errors by putting wrong sign in inequalities.

Answering Tips:
Students must remember the signs of inequalities while simplifying them.

Section – B
Question numbers 21 to 26 carry 2 marks each.

Question 21.
Qut of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a) You both enter the same sections ?
(b) You both enter the different sections ?
Answer:
My friend and I are among the 100 students.
Total number of ways of selecting 2 students out of 100 students = ¹⁰⁰C₂

(a)The two of us will enter the same section if both of us are among 40 students or among 60 students.
∴ Number of ways in which both of us enter the same section = ⁴⁰C₂ + ⁶⁰C₂
Probability that both us enter the same section [40 [60

(b) P (we enter different sections)
= 1 – P (we enter the same section)
= 1 – \(\frac{17}{33}\) = \(\frac{16}{33}\)

Question 22.
Differentiate the function with respect to x.
\(\frac{x^4+x^3+x^2+1}{x}\)
Answer:

Question 23.
In a group of 70 people, 37 like coffee, 50 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea ?
Answer:
Let C and T denote the people who like coffee and tea, respectively.
Then, n(C ∪ T) = 70, n(C) = 37, n(T) = 52
Using the identity,
n(C ∪ T) = n(C) + n(T) – n(C ∩ T),
we have 70 = 37 + 52 – n(C ∩ T)
70 = 89 – n(C ∩ T)
n(C ∩ T) = 89 – 70 = 19
So 19 persons like both Tea and Coffee.

Question 24.
Evaluate

OR
Find the value of n, if \(\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}\) = 80, n ∈ N.
Answer:

OR

⇒ n(2)^{n – 1} = 5 × 16
⇒ n × 2^{n – 1} = 5 × (2)⁴
⇒ n × 2^{n – 1} = 5 × (2)^{5 – 1}

Question 25.
If m sin θ = n sin (θ + 2α), then prove that tan(θ + α)cot α = \(\frac{m+n}{m-n}\)
OR
If tan θ = \(\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}\), then show that sin α + cos α = √2 cos θ
Answer:
Given, m sin θ = n sin (θ + 2α)

OR

Hence proved

Question 26.
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Answer:
Let ON be perpendicular to AB.
The point N is (-2, 9)
∴ Slope of ON = \(\frac{9-0}{-2-0}=-\frac{9}{2}\)
⇒ Slope of AB which is perpendicular to ON = \(\frac{2}{9}\)
Now, AB passes through (-2, 9) and has slope \(\frac{2}{9}\)
∴ Equation of AB is
y – 9 = \(\frac{2}{9}\) (x + 2)
⇒ 2x – 9y + 85 = 0

Section – C
Question numbers 27 to 32 carry 4 marks each.

Question 27.
Find the minimum and maximum value of sin⁴x + cos²x, x ∈ R.
OR
Find the value of \(\left|(1+i) \frac{(2+i)}{(3+i)}\right|\)
Answer:
We have,
A = cos²θ + sin⁴θ = cos²θ + sin²θ sin² θ ≤ cos² θ + sin² θ
Therefore
A ≤ 1
Also, A = cos²θ + sin⁴θ = (1 – sin²θ) + sin⁴θ

∴ \(\frac{3}{4}\) ≤ A ≤ 1.
Hence, minimum value = \(\frac{3}{4}\)
and maximum value = 1

OR

Question 28.
Find the value of r, if the coefficient of (2r + 4)^th and (r – 2)^th terms in the expansion of (1 + x)¹⁸ are equal.
Answer:
Given expansion is (1 + x)¹⁸.
Now, (2r + 4)^th term i.e., T_2r+3+1
∴ T_2r+3+1 = ¹⁸C_2r+3 (1)^{18 – 2r – 3} (x)^{2r + 3}
= ¹⁸C_2r+3x^{2r + 3}
Now, (r – 2)^thterm i.e., T_{r – 3 + 1}
∴ T_2r+3+1 = ¹⁸C_r-3x^{r – 3}
As, ¹⁸C_2r+3 = ¹⁸C_r-3 [∵ ⁿC_x = ⁿC_y ⇒ x + y = n]
⇒ 2r + 3 + r – 3 = 18
⇒ 3r = 18
⇒ r = 6

Question 29.
Find the value of 0 such that, \(\frac{1+i \cos \theta}{1-2 i \cos \theta}\) is a real number.
Answer:

Question 30.
Find the equation of a line drawn perpendicular to the line \(\frac{x}{4}+\frac{y}{6}\) = 1 through the point, where it meets the Y-axis
OR
Find the equation of a circle of radius 5 which is touching another circle x² + y² – 2x – 4y – 20 = 0 at (5,5).
Answer:
The equation of the given line is \(\frac{x}{4}+\frac{y}{6}\) = 1
The equation can also be written as 3x + 2y – 12 = 0
y = \(\frac{-3}{2}\)x + 6 , which is of the form y = mx + C
∴ Slope of the given line = \(\frac{-3}{2}\)
∴ Slope of the perpendicular to the given line
= – \(\frac{1}{\left(-\frac{3}{2}\right)}=\frac{2}{3}\)
let the given line intersect the y-axis at (0, y).
On substituting X with O in the equation of the
given line, we obtain \(\frac{y}{6}\) = 1 ⇒ y = 6
∴ The given line intersects the Y-axis at (0, 6).
The equation of the line that has a slope of and
passes through point (0,6) is
(y – 6) = \(\frac{2}{3}\)(x – 0)
3y – 18 = 2x
2x – 3y + 18 = 0
Thus, the required equation of the line is 2x – 3y + 18 = 0

OR

Let the coordinates of centre of the required circle are (h, k), then the centre of another circle is (1, 2).

Radius = \(\sqrt{1+4+20}\) = 5
P is the mid-point of C₁ C₂.
∴ 5 = \(\frac{1+h}{2}\) ⇒ h = 9
and 5 = \(\frac{2+k}{2}\) ⇒ k = 8
the equation of required circle is,
(x – 9)² + (y – 8)² = 25
⇒ x² – 18x + 81 + y² – 16 y + 64 = 25
⇒ x² + y² – 18x -16y +120 = 0

Question 31.
A man running a racecourse notes that the sum of the distance from the two flag posts to him is always 10 m and the distance between the flag posts are 8 m. Find the equation of the posts traced by the man.
Answer:
Let A and B be the position of the two flag posts and P(x, y) be the position of the man.
Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points are constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis, is 10 m, while points A and B are the foci.
Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, where a is the semi-major axis
Accordingly, 2a = 10 ⇒ a = 5
Distance between the foci ⇒ 16 = 25 – c²
⇒ c = 4
On using the relation c = \(\sqrt{a^2-b^2}\) , we obtain
4 = \(\sqrt{25-b^2}\)
⇒ 16 = 25 – b²
⇒ b² = 25 – 16 = 9
⇒ =
Thus,the equation of the path traced by the man is
\(\frac{x^2}{25}+\frac{y^2}{9}\) = 1

Commonly Made Errors:
Some students do not make a figure for the problem which leads to deduction of marks.

Answering Tips:
Students must make a well labelled figure for the required problem.

Question 32.
The mean life of a sample of 60 bulbs was 650 h and the standard deviation was 8 h. If a second sample of 80 bulbs has a mean life of 660 h and standard deviation 7 h, then find the overall standard deviation.
Answer:
Here, n₁ = 60, x̄₁ = 650, σ₁ = 8 and n₂ = 80, x̄₂ = 660, σ₂ = 7

Section – D
Question numbers 33 to 36 carry 6 marks each.

Question 33.
In a group of 50 students, the number of students studying French, English Sanskrit were found to be as follows :
French = 17, English = 13, Sanskrit = 15 French and English = 09, English and Sanskrit = 4 French and Sanskrit = 5, English, French and Sanskrit = 3. Find the number of students who study
(i) French only
(ii) English only
(iii) Sanskrit only
Answer:
Let, total no. of students = 50 ⇒ n(U) = 50
No. of students who study French = 17 ⇒ n(F) = 17
No. of students who study English = 13 ⇒ n(E) = 13
No. of students who study Sanskrit = 15 ⇒ n(S) = 15
No. of students who study French and English = 9
⇒ n(F ∩ E) = 9
Number of students who study English and Sanskrit = 4
⇒ n(E ∩ S) = 4
Number of students who study French and Sanskrit = 5
⇒ n(F ∩ S) = 5
Number of students who study French, English and Sanskrit = 3
⇒ n(F ∩ E ∩ S) = 3

n(F) = 17
a + b + d + e = 17
n(E) = 13
b + c + e + f = 13
n(S) = 15
d + e + f + g = 15
n(F ∩ E) = 9
∴ b + e = 9
n(E ∩ S) = 4
∴ e + f = 4
n(F ∩ S) = 5
∴ d + e = 5
n(F ∩ E ∩ S) = 3 ∴ e = 3 (vii)
From (iv), b + 3 = 9 ⇒ b = 9 – 3 = 6
From (v), 3 + f = 4 ⇒ f = 4 – 3 = 1
From (vi), d + 3 = 5 ⇒ d = 5 – 3 = 2
Now, from eq. (i), a + 6 + 2 + 3 = 17 ⇒ a = 17 – 11 ⇒ a = 6
Now, from eq. (ii), 6 + c + 3 + 1 = 13 ⇒ c = 13 – 10 ⇒ c = 3
From eq. (iii), 2 + 3 + 1 + g = 15 ⇒ g = 15 – 6 ⇒ g = 9.
(i) No. of students who study French only, a = 6
(ii) No. of students who study English only, c = 3
(iii) No. of students who study Sanskrit only, g = 9

Question 34.
Suppose x and y are two real numbers such that the r^th mean between x and 2y is equal to the r^th mean between 2x and y when n arithmetic means are inserted between them in both the cases.
Show that \(\frac{n+1}{r}-\frac{y}{x}\) = 1
OR
If |z₁| = |z₂| =… = |z_n| = 1, then show that |z₁ + z₂ + z₃ +…+z_n| = \(\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\ldots+\frac{1}{z_n}\right|\)
Answer:
Let A₁. A₂,…. A_n be n arithmetic paeans between x and 2y. Then,
x,A₁,A₂, …A_n,2y are in A.P with common difference d1 given by
d₁ = \(\frac{2 y-x}{n+1}\)
∴ rth mean = A_r = x + rd₁ = x + r\(\left(\frac{2 y-x}{n+1}\right)\)

Let A₁¹. A₂¹,…. A_n¹ be n arithmetic means between 2x and y. Then 2x, A₁¹. A₂¹,…. A_n¹ n,y are in A.P with common difference d₂ given by
∴ rth mean = A_r¹ = x + rd₂ = 2x + r\(\left(\frac{y-2 x}{n+1}\right)\)

It is given that A_r = A_r¹
⇒ x + r\(\frac{2 y-x}{n+1}\) = 2x + r\(\left(\frac{y-2 x}{n+1}\right)\)
⇒ (n + 1)x + r(2y – x) = (n + 1)2x + r(y – 2x)
⇒ (n + 1)x – ry = rx
⇒ \(\frac{n+1}{r}-\frac{y}{x}\) = 1
Hence Proved.

OR

Question 35.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Particular	Firm A	Firm B
No. of wage earners	586	648
Mean of monthly wages	₹ 5253	₹ 5253
Variance of the distribution of wages	100	121

(1) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm A or B shows greater variability in individual wages?
Answer:
For Firm A :
No. of wages earners = 586
Mean of monthly wages, x = ₹ 5253
Amount paid by firm A = ₹ (586 × 5253)
= ₹ 3078258
Variance of distribution of wages, σ² = 100
Standard deviation, σ = \(\sqrt{\sigma^2}\)
= \(\sqrt{100}\)
= 10
coefficient of Variation = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{10}{5,253}\) × 100
= 0.19

For Firm B :
No. of wage earners = 648
Mean of monthly wages, x =₹ 5253
Amount paid by firm B = 648 × 5253 = ₹ 3403944
Standard deviation, σ = \(\sqrt{\sigma^2}\) = \(\sqrt{121}\) = 11
coefficient of Variation = \(\frac{\sigma}{\bar{x}}\) × 100
= \(\frac{11}{5,253}\) × 100
= 0.21
Monthly wages paid by firm A = ₹ 3078258
Monthly wages paid by firm B = ₹ 3403944
Firm B pays larger amount as monthly wages.
coefficient of variation of wages, of firm A = 0.19
coefficient of variation of wages, of firm B = 0.21
Firm B shows greater variability in individual wages.

Question 36.
Solve the following system of inequalities graphically
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3,y ≥ 0.
OR
Solve the following system of inequalities graphically
2x + y ≥ 6,3x + 4y ≤ 12.
Answer:

The inequalities are
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3,y ≥ 0.
(a) The line 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB.
Putting x = 0, y = 0, we have
0 + 0 = 0 < 60,
which is true, therefore, origin lies in this region. 2
Thus, the region below AB and the points lying on AB represents the inequality 4x + 3y ≤ 60.

(b) The line y ≥ 2x passes through (0, 0), (5, 10). It is represented by CD.
Putting x = 0, y = 5 in y – 2x ≥ 0
5 ≥ 0, which is true (0, 5) lies in this region.
The region lying above the line CD and including the points on CD represents y ≥ 2x.

(c) x ≥ 3 is the region lying on the right of the line x = 3, which is represented EF and points lying on x = 3 represent the inequality x ≥ 3.
The shaded area is the APQR in which x ≥ 0 and y ≥ 0 is true for each point, is the solution of given inequalities.

OR
Inequalities are :
2x + y ≥ 6, 3x + 4y ≤ 12.
(a) The line 2x + y = 6 passed through the points (3, 0), (0, 6).
Putting x = 0, y = 0, in 2x + y ≥ 6, we have
0 ≥ 6, which is not true.
So, origin does not lies in the region.
∴ The region lying above AB and all the points on AB represent the inequality 2x + y ≥ 6.

(b)The line 3x + 4y = 12 passes through the points (4, 0) and (0, 3) and represented by the line CD.
Putting x = 0, y =0 in 3x + 4y ≤ 12, we have 0 ≤ 12, which is true.
3x + 4y ≤ 12 represents the region below CD (towards origin) and all the points lying on it.
The common region is the solution of given inequalities, represented by the shaded region in the graph.

Commonly Made Errors
Students don’t highlight the feasible region of the graph which leads to deduction of marks.

Answering Tips
Students must highlight and mark the required portion of the graph.