CBSE Sample Papers for Class 11 Maths Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 2 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 the probabilities For A to fail in an examination is 0.2 and that for B is 0.3. then the probability that either A or B fails is
(A) >0.5
(B) 0.5
(C) <0.5
(D) 0
Answer:
Option (C) is correct

Explanation:
We know that
P(A) = 0.2
P(B) = 0.3
P(A ∩ B) = P(A).P(B)
= 0.2 × 0.3 7.
= 0.06
Required probability = P(A ∪ B)
P(A ∪ B)= 0.2 + 0.3 – P(Both A & B fail)
P(A ∪ B) =0.5-0.06
P(A ∪ B)= 0.44

Question 2.
The standard deviations for first ten natural number is
(A) 5.5
(B) 3.87
(C) 2.97
(D) 2.87
Answer:
Option (D) is correct

Explanation:
We know that the standard deviation for first ten natural numbers is 2.87.

Question 3.
If M and N are any two events, the probability that atleast one of them occurs is
(A) P(M) + P(N) – 2P(M∩N)
(B) P{M) + P(N) – P(M ∩ N)
(C) P(M) + P(N) + P(M ∩ N)
(D) P(M) + P(N) + 2P(M ∩ N)
Answer:
Option (B) is correct.

Explanation:
The probability of occurrence of at least one of them is
P(M) + P(N) – P(M ∩ N)

Question 4.
If x₁, x₂, x₃ x₄ and x₅ be the observation with mean m and standard deviation s then, the standard deviation of the observations kx₁, kx₂ kx₃, kx₄ and kx₅ is
(A) k + 5
(B) \(\frac{s}{k}\)
(C) ks
(D) s
Answer:
Option (C) is correct.

Explanation:
We have,
Standard deviation of the observations kx₁ kx₂, kx₃ kx₄ and kx₅ will be ks i.e. product of original mean and standard deviation.

Question 5.
A plane is parallel to YZ-plane, so it is perpendicular to
(A) X-axis
(B) Y-axis
(C) Z-axis
(D) None of these
Answer:
Option (A) is correct

Explanation:
We know that,
The X-axis is perpendicular to plane parallel to YZ plane.

Question 6.
A line passes through (2,2) and is perpendicular to the line 3x + y = 3. its y-intercept is
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{3}\)
(C) 1
(D) \(\frac{4}{3}\)
OR
If the line \(\frac{x}{a}+\frac{y}{b}\) = 1 passes through the points (2, – 3) and (4, – 5) then (a, b) equals a b
(A) (1,1)
(B) (-1,1)
(C) (1,-1)
(D) (-1,-1)
Answer:
Option (D) is correct

Explanation:
We have,
Equation of required line as
⇒ (y – 2) = \(\frac{1}{3}\)(x -2)
⇒ 3y – 6 = x – 2
⇒ x – 3y + 4 = 0
⇒ \(\frac{x}{4}+\frac{3 y}{4}\) = 1
So y intercept is \(\frac{4}{3}\)

OR

Option (D) is correct

Explanation:
We know that
The points (2, – 3) and (4, – 5) must satisfy the given equation of line
On simplifying the equations we get
(a, b) = (-1,-1).

Question 7.
If the coefficient of variation of two distributions are 50,60 and their arithmetic means are 30 and 25 respectively, then the difference of their standard deviation is
(A) 0
(B) 1
(C) 1.5
(D) 2.5
Answer:
Option (A) is correct.

Explanation:
We have,
Coefficient of variation=(SD/Mean) × 100
So we have on putting values
Difference of SD = 15 -15 = 0

Question 8.
If y = \(\frac{\sin x+\cos x}{\sin x-\cos x}\), then \(\frac{d y}{d x}\) at x = 0 is equal to:
(A) -2
(B) 0
(C) 1
(D) Does not exist
OR
\(\lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3}\) is equal to:
(A) \(\frac{1}{10}\)
(B) –\(\frac{1}{10}\)
(C) 1
(D) None of these
Answer:
Option (A) is correct

Explanation:
We know that
y = \(\frac{\sin x+\cos x}{\sin x-\cos x}\)
So \(\frac{d y}{d x}\) at x = 0
\(\frac{d y}{d x}\) = -2

OR
Option (C) is correct

Explanation:
We have,
On factorizing the denominator and applying the limits we get
\(\lim _{x \rightarrow 1}\) f(x) = 1

Question 9.
The value of cos 12° + cos 84° + cos 156° + cos 132° is
(A) \(\frac{1}{2}\)
(B) 1
(C) –\(\frac{1}{2}\)
(D) \(\frac{1}{8}\)
OR
The value of sin 50° – sin 70° + sin 10° is equal to
(A) 1
(B) 0
(C) \(\frac{1}{2}\)
(D) 2
Answer:
Option (C) is correct

Explanation:
We have cos12° + cos84° + cos156° + cos132°
= (cos12° + cos132°) + (cos 156° + cos 84°)
= 2 cos72°cos60° + 2cos120° cos36°
= cos72° – cos36°
= sin 18° – cos36°
= – \(\frac{1}{2}\)

OR

Option (B) is correct

Explanation:
We know that, sin50° – sin70° + sin10°
⇒ -2cos60°sin10° + sin100 = sin10° – sin10° = 0

Question 10.
The distance between the foci of a hyperbola is 16 and its eccentricity is √2 . Its equation is
(A) x² – y² = 32
(B) \(\frac{x^2}{4}-\frac{y^2}{9}\) = 1
(C) 2x – 3y² = 7
(D) None of these
Answer:
Option (A) is correct

Explanation:
We have
Foci of hyperbola = 16
eccentricity = (2)^1/2

Equation of hyperbola will be
⇒ x² – y² = 32

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

Question 11.
The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84 and the probability that he will see both is 0.52.
Answer:
False

Explanation:
We have,
P(giraffe) = 0.72
P(bears) = 0.84
P(both giraffe and bear) must not be equal to 0.52
Hence, given statement is False.

Question 12.
Ar candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.
OR
In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is ^n-mP_r-m × ^rP_m.
Answer:
False

Explanation:
Required number of ways
= 2C(6, 4) × C(6, 3) + 2C(6, 5) × C(6,2)
= 600 + 1440
= 2040 ways
Hence given statement is False.

OR

False

Explanation:
In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is not equal to given expression.
Hence it is False.

Question 13.
Given that M = {1, 2, 3,4,5,6, 7, 8, 9 } and if B = {1, 2, 3,4,5, 6, 7, 8,9}, then B ⊄ M.
Answer:
False

Explanation:
Set B is equal to Set C so it will also be 0 a subset.
So given statement is FALSE.

Question 14.
Line joining the points (3, – 4) and (- 2,6) is perpendicular to the line joining the points (- 3, 6) and (9, -18).
Answer:
False

Explanation:
We have the product of slopes not equal to -1
So they are not perpendicular.

Question 15.
If P = {1, 2}, then P × P × P= {(1,1,1),(2, 2, 2), (1,2,2), (2,1,1)}.
Answer:
False

Explanation:
The given cross product is wrong as it will contain 8 elements.
So it is a False statement.

Question numbers 16 to 20 carry 1 mark each.

Question 16.
Show the graph of the solution of 2x – 3 > x – 5
Answer:
We have; 2x – 3 > x – 5
Transposing x to LHS and -3 to RHS,
we get 2x – x > -5 + 3
⇒ x > – 2 =£ (- 2, ∞)

Commonly Made Error
Errors are made while making circle on the number line.

Answering Tip
Take precaution while making circle (o)/dark circle (•) to avoid mistakes.

Question 17.
Find the sum of the coefficient in (x + y)⁸.
Answer:
To get the sum of coefficients of (x + y)⁸, put x = 1 and y = 1, we get
(1 + 1)⁸ = 28 = 256

Commonly Made Error
Some students don’t know how to calculate the sum of coefficients using formula.

Answering Tip
Revise the properties and formulae related to Binomial Theorem.

Question 18.
Find the value of ‘p’ so that the equation x² + y² – 2px + 4y -12 = 0 may represent a circle of radius 5 units.
OR
Find the equation of the parabola with vertex at (0, 0) and focus at (0,2).
Answer:
Given equation of the circle is
x² + y² -2px + 4y – 12 = 0
⇒ (x² – 2px + p²) + (y² + 4y + 4) – p² – 4 – 12 = 0
⇒ (x – p)² + (y + 2)² = p² +16
⇒ (x-p)² + {y-(-2)}² = \(\left(\sqrt{p^2+16}\right)^2\)
⇒ The centre of the circle is (p,-2) and radius \(\sqrt{p^2+16}\) units.
According to the question
\(\sqrt{p^2+16}\) = 5
⇒ p² + 16 = 25
⇒ p² = 9
⇒ p = ± 3
∴ Values of p are 3 and -3

Commonly Made Error
When converting the given equation into standard equation, some students commit algebraic errors which gives incorrect result.

Answering Tip
Follow all the steps while solving the problem to avoid errors.

OR

Equation of the parabola is
x² = 4ay
⇒ x² = 4 . 2 . y [∵ focus (0, a) = (0,2)]
⇒ x² = 8y

Question 19.
Find the value of \(\lim _{x \rightarrow 0} \frac{e^x-1}{x}\)
Answer:

Question 20.
Write the argument of (1 + √3i)(cos θ + i sin θ).
OR
Find the value of i⁴ + i⁵ + i⁶ + i⁷.
Answer:
Let z = (1 + √3i) (cos θ + isin θ)
= cos θ + √3sin θi² + isin# + √3i cos θ
= (cos θ – sflsin θ) + i(sin θ + √3cos θ)
Re(z) = cos θ – √3sin θ
and Im(z) = (sin θ + √3cos θ)

Now, argument of z is Im(z)

α = (\(\frac{\pi}{3}\) + θ) is the required argument.

Commonly Made Error
Some students do not know how to simplify the real and imaginary part to find the argument.

Answering Tip
Revise all the formulae and practice as many i questions as possible.

OR
i⁴ + i⁵ + i⁶ + i⁷ = (i²)² + i⁴i + i⁴.i² + i⁴.i³
Since, i = ~P i² = -1, i³ = -i, i⁴ = 1,
So, i⁴ + i⁵ + i⁶ + i⁷ = (-1)² + 1.i + 1(i)² + 1.(-i)
= 1 + i + 1 × -1 – i
= 1 + i – 1- i
= 0

Commonly Made Error
Students make errors while putting the values of i.

Answering Tip
Always remember, for any integer k, i^4k = 1, i^4k+1 = i, i^4k+2 = -1, i^4k+3 = -i

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

Question 21.
Solve \(\frac{2 x+3}{4}\) – 3 < \(\frac{x-4}{3}\) – 2, x R
OR
Solve \(\frac{2 x-1}{3} \geq\left(\frac{3 x-2}{4}\right)-\left(\frac{2-x}{5}\right) .\)
Answer:

Commonly Made Error
Students make errors in putting correct sign < while transposing the terms in an inequality. i

Answering Tip
Students should always remember that two real number or two algebraic expressions related to symbolic <, >, < or > form an inequality.

OR

⇒ 20(2s – 1) ≥ 3(19s – 18)
⇒ 40s – 20 ≥ 57s – 54
Transposing the term 57s to LHS and the term -20 to RHS, 40s – 57s ≥ -54 + 20
-17s ≥ -34
Divide both sides by -17,
\(\frac{-17 x}{-17} \leq \frac{-34}{-17}\)
x ≤ \(\frac{-34}{-17}\)
x ≤ 2
∴ Solution set = (-∞, 2]

Question 22.
If a = cos θ + i sin θ, then find the value of \(\frac{1+a}{1-a}\)
Answer:
a = cos θ + i sin θ

Question 23.
If a, b, c are in G.P, then show that a² + b², ab + bc, b² + c² are also in G.P
OR
The 5th, 8th and 11th terms of G.E are p, q and s respectively. Show that q² = ps.
Answer:
Given that, a, b, c are in G.P
⇒ b² = ac
⇒ b² – ac = 0
⇒ (b² – ac)² = 0
⇒ b⁴ + a²c² – 2b²ac = 0
⇒ a²b² + b²c² + 2b²ac = a²b² + b²c² + a²c² + b⁴

(Adding a²b² + b²c² both sides)
⇒ (ab + bc)² = (a² + b²)(b² + c²)
⇒ a² + b², ab + bc,
b² + c² are in G.P
Hence Proved.

OR

Let a and rbe the first term and common ratio of the G.P
Given, T₅ = p ⇒ ar⁴ = p ..(i)
T₈ = q ⇒ ar⁷ = q …(ii)
and T₁₁ = s ⇒ ar¹⁰ = s

Now, multiplying Eqs. (i) and (iii),
ar⁴ x ar¹⁰ = ps
a²r¹⁴ = ps
⇒ [(ar)⁷]² = ps [From Eq. (ii)]
⇒ q² = ps
Hence Proved.

Question 24.
Evaluate \(\lim _{x \rightarrow \pi / 6} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}\)
Answer:

Question 25.
Let A = {1, 2,4,5},B = {2,3,5,6}, C = {4,5, 6, 7} verify the following identity
A ∪ (B ∩ C) = [(A ∪ B) ∩ (A ∩ C)]
Answer:
L.H.S. = A ∪ (B ∩ C)
= (1, 2,4,5} ∪ [{2, 3,5,6} ∩ {4, 5, 6, 7}]
= {1, 2,4,5} ∪ {5, 6}
= {1,2,4,5,6}
R. H. S. = (A ∪ B) ∩ (A ∪ C)
= [{1, 2,4, 5} ∪ {2, 3, 5,6}] ∩ [{1,2,4,5} ∪ {4, 5,6, 7}]
= (1,2, 3,4,5, 6} ∩ {1,2,4,5,6, 7}
= {1,2, 4,5,6}
∴ L. H. S. = R. H. S.
Hence verified.

Question 26.
If A and B be the point (3,4,5) and (-1,3, -7), respectively, find the equation of the set of points P such that PA² + PB² = k², where k is constant.
Answer:
The coordinates of points A and B are given as (3, 4, 5) and (-1, 3, -7) respectively. Let the coordinates of point p be (x, y, z).
On using distance formula, we obtain
PA² = (x – 3)² + (y – 4)² + (z – 5)²
= x² + 9 – 6x + y² + 16 – 8y + z² + 25 – 10z
= x² + y² + z² – 6x – 8 y – 10z + 50
PB² = (x + 1)² + (y – 3)² + (z + 7)²
= x² + 1 + 2x + y² + 9 – 6y + z² + 49 + 14z
= x² + y² + z² + 2x – 6y + 14z + 59

Now, if PA² + PB² = K², then
(x² + y² + z² – 6x – 8y – 10z + 50) + (x² + y² + z² + 2x – 6y + 14z + 59)
⇒ 2x² +2y² + 2z² – 4x- 14y + 4z + 109 = K²
⇒ 2(x² + y² + z² – 2x – 7y + 2z) + 109 = K²
⇒ 2(x² + y²² + z² – 2x – 7y + 2z) = K² – 109
⇒ x² + y² + z² – 2x – 7y + 2z = \(\frac{k^2-109}{2}\)
Thus, the required equation is x² + y² + z²– 2x – 7y + 2z = \(\frac{k^2-109}{2}\)

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

Question 27.
Prove that: tan α. tan (60° – α) tan (60° + α) = tan 3α
OR
Prove that: \(\sqrt{2+\sqrt{2+2 \cos 4 \theta}}\) = 2cos θ.
Answer:
L.H.S = tan α. tan (60° – α) tan (60° + α)

OR

Hence Proved.

Question 28.
If z = 2 – 3i show that z² – 4z + 13 = 0, hence find the value of 4z³ – 3z² +169
OR
Convert complex number -√3 + i in polar form.
Answer:
We have,
z = 2 – 3i
z – 2 = -3i
(z – 2)² = (-3i)²
z² – 4z + 4 = 9i²

4z³ – 3z² + 169 = (z² – 4z + 13)(4z + 13)
= (0) (4z + 13)

4z³ – 3z² + 169 = 0 [∵ z² – 4z + 13 = 0]
OR
Let z = -√3 + i, then
r(cos θ + isin θ) = -√3 + i

Then on comparing real & imaginary parts we get
r cos θ = – √3 ……(i)
r sin θ = 1 ………..(ii)

Squaring & adding both eq. we get
⇒ r²cos²θ + r²sin²θ = (-√3)² + (i)² = 3+1
⇒ r²(cos²θ + sin²θ) = 4
r² × 1=4
r = 2
On dividing eq. (ii) from eq.(i)

Commonly Mode Error
Students don’t remember the correct expression of polar form which leads to deduction of marks.

Answering Tip
Revise the method of polar form and practice as many questions as possible.

Question 29.
An arc is in the form of a semi ellipse, It is 8m wide and 2m high at the centre. Find the height of arc at a point 1.5m from one end.
Answer:
Let ABA’ be the given arc such that AA’ = 8m and OB = 2m

The arc is a part of the ellipse.
AA’ = 8m ⇒ 2a = 8
⇒ a = 4
and OB = 2m
⇒ b = 2

The equation of the ellipse is
\(\frac{x^2}{16}+\frac{y^2}{4}\) = 1 ………(1)

To find PM.
Putting x = 2 . 5 = \(\frac{5}{2}\) and y = PM
From eqn. (1) ⇒ \(\frac{25}{4 \times 16}+\frac{P M^2}{4}\) = 1
\(\frac{P M^2}{4}=\frac{39}{64}\)
PM² = \(\frac{39}{16}\)
⇒ PM = \(\frac{\sqrt{39}}{4}\)
Height of the arc at a point 1.5 from one end is \(\frac{\sqrt{39}}{4}\) m.

Commonly Made Error:
Some students have difficulty in finding the value of major and minor axis from the information is given in the question.

Answering Tip:
To avoid such errors, a thorough revision on i concept of the ellipse is must and practice more such problems based on it.

Question 30.
Evaluate \(\lim _{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2}\)
Answer:

Question 31.
Find the mean and variance for the following frequency distribution.

Answer:


= 30[76 – 0.13]
Variance = 2276
∴ Mean = 107 and Variance = 2276

Question 32.
Find the coefficient of x⁴ in (1 – x)²(2 + x)⁵ using binomial theorem.
Answer:
Given, (1 – x)² (2 + x)⁵
= (1 + x² – 2x)2⁵ + ⁵C₁ 2^5-1 x¹ + ⁵C₂ 2^5-2 x² + ⁵C₃ 2^5-3 x³ + ⁵C₄ 2^5-4 x⁴ + ⁵C₅ 2^5-5 x⁵)
= (1 + x² -2x)2⁵ + 5.2⁴ – x + 10.2³ x² + 10.2² x³ + 5.2 x⁴ +x⁵
= (1 + x² – 2x)(32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵)
∴ Coefficient of x⁴ = 80 – 80 + 10
= 10

Commonly Made Error
Students had difficulty in using the binomial theorem formulae to solve for power and often substituted the coefficients when equating to the power of a given term.

Answering Tip
Revise the concept of the binomial theorem thoroughly and practice more such problems based on it.

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

Question 33.
In a group of 500 persons, 300 take tea, 150 take coffee, 250 take a cold drink, 90 take tea and coffee, 110 take tea and a cold drink, 80 take coffee and a cold drink and 50 take all the three drinks.
(i) Find the number of persons who take none of the three drinks.
(ii) Find the number of persons who take only tea.
(iii) Find the number of persons who take coffee and cold drink but not tea.
Answer:
Given,
n(U) = 500, n(T) = 300, n(Co) = 150, n(Cd) = 250,
n(T ∩ Co) = 90, n(T ∪ Cd) = 110,
n(Co ∩ Cd) = 80, n(T ∩ Cd ∩ Co) = 50

Here, n(T) = like tea,
n(Co) = like coffee,
n(Cd) = like cold drink

(i) Number of persons who take none of three drinks n(U) – n(T ∪ Co ∪ Cd)
n(T ∪ Co ∪ Cd) = n(T) + n(Co) + n(Cd) – n(T ∩ Co) – n(T ∩ Cd) – n(Co ∩ Cd) + n(T ∩ Cd ∩ Co)
= 300 + 150 + 250-90-110-80 + 50
= 750-280
= 470

n(U) – n(T ∩ Co ∩ Cd) = 500 – 470 = 30
∴ Number of persons who take none of three drinks = 30.

(ii) Number of person who take only tea
= n(T) – n(T ∩ Co) – n(T ∩ Cd) + n(T ∩Cd ∩ Co)
= 300-90-110 +50
= 350-200
= 150
∴ Number of person who take only tea = 150

(iii) Number of persons who take coffee and cold drink but not tea
= n(Co ∩ Cd)- n(T ∩ Co ∩ Cd)
= 80 -50 = 30
∴ Number of persons who take coffee and cold drink but not tea = 30.

Question 34.
A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2,0), (0, 2) and (1,1) on the line is zero. Find the coordinates of the point P.
OR
If the sum of the distance of a moving point in a plane from the axes is 1, then find the locus of the point.
Answer:
Let slope of the line be m and the coordinates of fixed point p are (x₁, y₁)
Equation of line y – y₁ = m (x – x₁)
Since, the given points are A(2, 0), B(0, 2) and C(1, 1).
Now, perpendicular distance from A, is

⇒ -3y₁ – 3m + 3 mx₁ + 3 = 0
⇒ -y₁ – m + mx₁ + 1 = 0
Since, (1,1) lies on this, So, the point P is (1,1).

OR

Let the coordinates of moving point B be (x, y). Given that, the sum of distances of this point in a plane from the axes is 1.
|x| + |y| = 1
⇒ ±x ± y = 1
⇒ x + y = 1
⇒ -x – y = 1
⇒ -x + y = 1

So, these equation give us locus of the point which is a square.

Question 35.
Find the linear inequalities for which the shaded region in the given figure is the solution set.

Answer:
Consider the line x + y = 4.
We observe that the shaded region and the origin lie on the opposite side of this line and (0,0) satisfies x + y ≥ 4. Therefore, we must have x + y ≥ 4 as the linear inequation corresponding to he line x + y = 4. Consider the line x + y = 8, clearly the shaded * region and origin lie on the same side of this line and (0, 0) satisfies the constraints x + y ≤ 8. Therefore, we must have x + y ≤ 8, as the linear inequation corresponding to the line x + y = 8. Consider the line x = 5. It is clear from the graph that the shaded region and origin are on the left of this line and (0, 0) satisfy the constraint x ≤ 5. Hence, x ≤ 5 is the linear inequation corresponding to x = 5.

Consider the line y = 5, clearly the shaded region and origin are on the same side (below) of the line , and (0, 0) satisfy the constrain y ≤ 5.
Therefore, y ≤ 5 is an inequation corresponding to the line y = 5.
We also notice that the shaded region is above the X-axis and on the right of the Y-axis i.e., shaded region is in first quadrant. So, we must have x ≥ 0, y ≥ 0.
Thus, the linear inequations comprising the given solution set are x + y ≥ 4; x + y ≤ 8; x ≤ 5; y ≤ 5; x ≥ 0 and y ≥ 0.

Question 36.
The sum of an infinite G.E is 57 and the sum of the cubes of its terms is 9747, find the G.E
OR
Find the sum of first n terms of the series \(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}\) + …+n terms.
Answer:
Let the first term of G.P be ‘a’ and the common ratio be ‘r’ where – 1 < r < 1 The G.P is a, ar, ar2,…
Therefore the sum of the infinite terms of the G.P. is \(\frac{a}{1-r}\) = 57 ……..(i)

If taking the cube of each terms the new G.P is
a³, a³r², a⁴r⁶,…

Therefore the sum of their cube is
\(\frac{a^3}{1-r^3}\) = 9747 …………(ii)

Taking the cube of the (i)
\(\frac{a^3}{1-r^3}\) = (57)³ ⇒ a³ = (57)(1 – r)³ ………(iii)

Substituting the value of ‘a’ in terms of r in

19(1 + r² – 2r) = 1 + r² + r
18r² – 39r + 18 = 0
6r² – 13r + 6 = 0
6r² – 9r – 4r + 6 = 0
(2r – 3)(3r – 2) = 0

Therefore, r = \(\frac{3}{2}\) or r = \(\frac{2}{3}\) since r < 1.
r = \(\frac{2}{3}\)

Substitute in eq. (i)
\(\frac{a}{1-r}\) = 57
\(\frac{a}{1-\frac{2}{3}}\) = 57
⇒ 3a = 57
⇒ a = \(\frac{57}{3}\) = 19
Thus, the first term of the G.P is 19 and the common ratio is \(\frac{2}{3}\).

The G.P is 19, \(\frac{38}{3}, \frac{76}{9}\) and so on.

OR

The given sequence is
\(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}\) + ……. + n terms

We can write each individual as,