Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.4

Question 1.
A factory has two Machines – I and II. Machines – I produce 60 % of items and Machine – II produces 40 % of the items of the total output. Further 2 % of the items produced by Machine – I are defective whereas 4 % produced by Machine -II are defective. If an item is drawn at random what is the probability that it is defective?
Answer:
Let A₁ be the event that items are produced by machine – I, A₂ be the event that items are produced by machine – II.
Let B be the event of drawing a defective item. We have to find the total probability of event B. That is P(B) clearly A₁ and A₂ are mutually exclusive and exhaustive events.
∴ P(B) = P(A₁) . P(B/A₁) + P(A₂) . P(B/A₂)

P(B) = P(A₁) . P(B/A₁) + P(A₂) . P(B/A₂)
= 0.60 × 0.02 + 0.40 × 0.04
= 0.012 + 0.016
= 0.028

Question 2.
There are two identical urns containing respectively 6 black and 4 red balls, 2 black, and 2 red balls. An urn is chosen at random and a ball is drawn from it.
(i) Find the probability that the ball is black
(ii) if the ball is black, what is the probability that it is from the first urn?
Answer:

(i) Let A₁ be the event of selecting Urn – I and A₂ be the event of selecting Urn – II.
Let B be the event of selecting one black ball.
We have to find the total probability of event B. That is P(B).
Clearly, A₁ and A₂ are mutually exclusive and exhaustive events.
Probability of selecting Urn – I
P(A₁) = \(\frac{1}{2}\)
Conditional Probability of B, given A₁

Probability of selecting Urn – II
P(A₂) = \(\frac{1}{2}\)
Conditional Probability of B, given A₂

(ii) The conditional Probability of A₁ given B is P(A₁/B)
By Bayes’ theorem

Question 3.
A firm manufactures PVC pipes in the three plants viz, X, Y, and Z. The daily production volumes from the three firms X, Y, and Z are respectively 2000 units, 3000 units, and 5000 units. It is known from past experience that 3 % of the output from plant X, 4 % from plant Y, and 2 % from plant Z are defective. A pipe is selected at random from a day’s total production
(i) find the probability that the selected pipe is a defective one?
(ii) if the selected pipe is defective, then what is the probability that it was produced by plant Y?
Answer:
Let A₁ be the daily volume of production by plant X, A₂ be the daily volume of production by plant Y, A₃ be the daily volume of production by plant Z.
Let B be the defective output we have to find P (B).

(i) Find the probability that the selected pipe is a defective one:
Clearly, A₁, A_2, and A₃ are mutually exclusive and exhaustive events.

Probability that the selected pipe is a defective one = \(\frac{7}{250}\)

(ii) If the selected pipe is defective, then what is the probability that it was produced by plant Y?


Probability that the defective pipe produced by plant Y = \(\frac{3}{7}\)

Question 4.
The chances of A, B, and C becoming manager of a certain company are 5 : 3 : 2. The probabilities that the office canteen will be improved if A, B, and C become managers are 0.4, 0.5 and 0.3 respectively. If the office canteen has been improved, what is the probability that B was appointed as the manager?
Answer:
Let A₁, A_2, and A₃ be the events of A, B, and C becoming managers of the company respectively.
Let B be the event that the office canteen will be improved.
We have to find the conditional probability P (A₂/B).
Since A₁, A₂ and A₃ are mutually exclusive and exhaustive events, applying Bayes theorem.

If the office canteen is improved than the probability of that B was appointed as the manager is \(\frac{15}{41}\)

Question 5.
An advertising executive is studying television viewing habits of married men and women during prime time hours. Based on the past viewing records he has determined that during prime time wives are watching television 60 % of the time. It has also been determined that when the wife is watching television, 40 % of the time the husband is also watching. When the wife is not watching the television, 30 % of the time the husband is watching the television. Find the probability that
(i) the husband is watching the television during the prime time of television
(ii) if the husband is watching the television, the wife is also watching the television.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.3

Question 1.
Can two events be mutually exclusive and independent simultaneously?
Answer:
When A and B are independent
P(A ∩ B) = P(A) P(B)
But when A and B are mutually
Exclusive P(A ∩ B) = 0

Question 2.
If A and B are two events such that
P ( A ∪ B ) = 0.7 , P (A ∩ B) = 0.2 , and P (B) = 0.5 , then show that A and B are independent.
Answer:
Given A and B are twp events such that
P(A ∪ B) = 0.7, P(A ∩ B) = 0.2 and P(B) = 0.5
To prove A and B are independent it is enough to prove
P(A ∩ B) = P(A) . P(B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = P (A) + 0.5 – 0.2
0.7 = P(A) + 0.3
P(A) = 0.7 – 0.3 = 0.4
P(A) . P(B) = 0.4 × 0.5 = 0.20
= P(A ∩ B)
∴ P(A∩B) = P(A) . P(B)
∴ A and B are independent.

Question 3.
If A and B are two independent events such that P(A ∪ B) = 0.6, P(A) = 0.2, find p(B).
Answer:
Given A and B are independent.
⇒ P(A ∪ B) = P(A).P(B)
Here P(A ∪ B) = 0.6 and P(A) = 0.2
To find P(B):
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
(i.e.,) 0.6 = 0.2 + P(B) (1 – 0.2)
P(B) (0.8) = 0.4
⇒ P(B) = \(\frac{0.4}{0.8}=\frac{4}{8}=\frac{1}{2}\) = 0.5

Question 4.
If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find P(A/B) and P(A ∪ B).
Answer:
Given P(A) = 0.5, P(B) = 0.8
and P(B/A) = 0.8

P(A ∩ B) = P(B/A) P(A)
Substituting in equation (1) we get

P(A/B) = 0.5
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ………. (2)

P(A ∩ B) = P(A/B) . P(B)
= 0.5 × 0.8
P(A ∩ B) = 0.40
(2) ⇒ P(A ∪ B) = 0.5 + 0.8 – 0.40
= 1.3 – 0.40
P(A ∪ B) = 0.90

Question 5.
If for two events A and B, P(A) = \(\frac{3}{4}\) P(B) = \(\frac{2}{5}\) and A ∪ B = S (sample space), find the conditional probability p (A/B).
Answer:

Question 6.
A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{1}{5}\).
(i) What is the probability that the problem is solved?
Answer:
(i) What is the probability that the problem is solved?
Let A₁ denote the event of that first student solves the problem.
A₂ denote the event that second student solves the problem.
A₃ denote the event that third student solves the problem.
Given P(A₁) = \(\frac{1}{3}\), P(A₂) = \(\frac{1}{4}\) P(A₃) = \(\frac{1}{5}\)
We note that A₁, A₂, A₃ are independent events.
The problem will be solved if atleast one of them
solves it we have to find P(A₁ ∪ A₂ ∪ A₃)
Probability of at least one solves the problem = 1 – Probability of no one solving it
P(A₁ ∪ A₂ ∪ A₃) = 1 – P(A̅₁ ∪ A̅₂ ∪ A̅₃)
= 1 – P(A̅₁) . P(A₂) . P(A₃)
A₁, A₂, A₃ are independent then A̅₁, A̅₂, A̅₃ are also independent.
= 1 – [1 – p(A₁)] [1 – P(A₂)] [1 – P(A₃)]

(ii) What is the probability that exactly one of them will solve it?
Answer:
p(A̅₁) = 1 – P(A₁) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
p(A̅₁) = 1 – P(A₂) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
p(A̅₁) = 1 – P(A₃) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Probability of Exactly one student solving the problem

[Probability of exactly one student solving the problem = Probability [(A₁ solving the problem and A₂, A₃ non solving the problem) or (A₁, A₃ non solving and A₂ solving) or (A₁, A₂ solving and A₃ non solving)]

Question 7.
The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that the both the oil and filter need changing is 0.15.
(i) If the oil had to be changed, what is the probability that a new oil filter is needed?
(ii) If a new oil filter is needed, what is the probability that the oil has to be changed?
Answer:
Let A be the event of changing oil, B be the event of changing oil filter.
Given P(A) = 0.30, P(B) = 0.40, P(A ∩ B) = 0.15

(i) Probability of new oil filter B needed when the oil A changed is

(ii) Probability of oil A changed when new oil filter B is changed is

Question 8.
One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that
(i) both are white
(ii) both are black
(iii) one white and one black
Answer:
First Bag contains 5 white and 3 black balls Total number of balls in the first bag 8 Second Bag contains 4 white and 6 black halls Total number of balls in the second bag = 10
One ball is drawn from each bag.

(i) Probability of both are white:
P (getting both are white) = P (getting white ball from the first bag) × P (getting the white ball from the second bag)

(ii) Probability of both are black:
P (getting both are black) = P (getting black ball from the first bag) × P (getting the ball from the second bag)

(iii) Probability of one white and one black.
P (getting one white and one black) = P ( getting one white from the first bag or one white from the second bag) + P (getting one black from the first bag or one black from the second bag)

Question 9.
Two-thirds of students in a class are boys and the rest girls. It is known that the probability of a girl getting the first grade is 0.85 and that of boys is 0.70. Find the probability that a student chosen at random will get first-grade marks.
Answer:
Let G be the event of choosing a boy and G be the event of choosing a Girl.
Given P(B) = \(\frac{2}{3}\), P(G) = \(\frac{1}{3}\)
Let B₁ be the event of a boy getting first grade
P(B₁) = 0.70
Let G₁ be the event of a girl getting first grade
P(G₁) = 0.85
Probability of a. student getting a first grade = Probability of a boy getting first grade or Probability
of a Girl getting first grade
= P(B) × P(B₁) + P(G) × P(G₁)
= \(\frac{2}{3}\) × 0.70 + \(\frac{1}{3}\) × 0.85
= \(\frac{1.4+0.85}{3}\)
= \(\frac{2.25}{3}\) = 0.75

Question 10.
Given P(A) = 0.4 and P(A ∪ B) = 0.7 Find P(B) if
(i) A and B are mutually exclusive
(ii) A and B are independent events
(iii) P(A/B) = 0.4
(iv) P(B/A) = 0.5
Answer:
P(A) = 0.4, P(A ∪ B) = 0.7
(i) When A and B are mutually exclusive
P(A ∪ B) = P(A) P(B)
(i.e.,) 0.7 = 0.4 + P(B)
0.7 – 0.4 = P(B)
(i.e.,) P(B) = 0.3

(ii) Given A and B are independent
⇒ P(A ∩ B) = P(A). P(B)
Now, P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
(i.e.,) 0.7 = 0.4 + P(B) – (0.4) (P(B))
(i.e.,) 0.7 – 0.4 = P(B) (1 – 0.4)
0.3 = P (B) 0.6
⇒ P(B) = \(\frac{0 \cdot 3}{0 \cdot 6}=\frac{3}{6}\) = 0.5

(iii) P(A/B) = 0.4
(i.e.,) \(\frac{P(A \cap B)}{P(B)}\) = 0.4
⇒ P(A ∩ B) = 0.4 [P(B)] …………. (i)
But We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A ∩ B) = 0.4 + P(B) – 0.7
= P(B) – 0.3 …………. (ii)
from (i) and (ii) (equating R.H.S) We get
0.4 [P(B)] = P(B) – 0.3
0.3 = P(B) (1 – 0.4)
0.6 (P(B)) = 0.3 ⇒ P(B) = \(\frac{0.3}{06}=\frac{3}{6}\) = 0.5

(iv) P(B/A) = 0.5
(i.e.,) \(\frac{P(A \cap B)}{P(A)}\) = 0.5
(i.e.,) P(A ∩ B) = 0.5 × P(A)
= 0.5 × 0.4 = 0.2
Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.7 = 0.4 + P(B) – 0.2
⇒ 0.7 = P(B) + 0.2
⇒ P(B) = 0.7 – 0.2 = 0.5

Question 11.
A year is selected at random. What is the probability that
(i) it contains 53 Sundays
(ii) it is a leap year which contains 53 Sundays.
Answer:
Probability of year being a leap year = \(\frac{1}{4}\)
Probability of year being non – leap year = \(\frac{3}{4}\)

(i) It contains 53 Sundays:
A non – leap year has 365 days.
365 days = 52 weeks + 1 day.
52 weeks contain 52 Sundays.
In order to get 53 Sundays in a non – leap year the remaining I day must be a Sunday.
Remaining one day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.
Probability of getting Sunday from the remaining one day = \(\frac{1}{7}\)
A leap year has 366 days.
366 days = 52 weeks + 2 odd days
52 weeks contain 52 Sundays.

In order to get 53 Sundays in a leap year the remaining 2 days must contain a Sunday. Remaining Two days may be

S = { (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday ), ( Friday, Saturday), (Saturday, Sunday) }
n(S) = 7

Let A be the event set of getting a Sunday then
A = { (Sunday, Monday), ( Saturday , Sunday) }
n(A) = 2

P (getting a Sunday from the remaining 2 days)

P(getting 53 Sundays in a year) = P(getting a leap year) × P(getting a Sunday from the remaining 2 days) + P(getting a non-leap year) × P(getting a Sunday from the remaining 1 day)

∴ Probability of getting 53 Sundays in a year = \(\frac{5}{28}\)

(ii) A leap year has 366 days
\(\frac{366}{7}\) = 52 weeks + 2 days
In 52 weeks, we get 52 Sundays.
From the remaining two days we should get one Sunday, the remaining two days can be any one of the following combinations.
Saturday and Sunday, Sunday and Monday, Monday and Tuesday, Tuesday and Wednes¬day, Wednesday and Thursday, Thursday and Friday, Friday and Saturday of the seven combinations two have Sundays.
∴ (Probability of getting a Sunday = \(\frac{2}{7}\)
Selecting a leap year = \(\frac{1}{4}\)
{∴ In every four consecutive years we get one leap year}
∴ Probability of getting 53 Sundays = \(\frac{2}{7} \times \frac{1}{4} \times \frac{1}{14}\)

Question 12.
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?
Answer:
Given
Probability X hitting the targent P (X) = \(\frac{3}{4}\)
Probability Y hitting the targent P (Y) = \(\frac{4}{5}\)
Probability Z hitting the targent P (Z) = \(\frac{2}{3}\)
P(X̅) = 1 – P(X) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
P(Y̅) = 1 – P(Y) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
p(Z̅) = 1 – P(Z) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Probability hitting the target exactly by 2 hits

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2

Question 1.
If A and B are mutually exclusive events P(A) = \(\frac{3}{8}\) and P(B) = \(\frac{1}{8}\), then find
(i) P(A̅)
(ii) P(A ∪ B)
(iii) P(A̅ ∩ B)
(iv) P(A̅ ∪ B̅)
Answer:
(i) P(A̅)
P(A̅) = 1 – P(A)
= 1 – \(\frac{3}{8}\)
P(A̅) = \(\frac{8-3}{8}\) = \(\frac{5}{8}\)

(ii) P(A ∪ B)
P(A ∪ B) = P(A) + P(B)
since A and B are mutually exclusive events.

(iii) P(A̅ ∩ B)
P(A̅ ∩ B) = P(B) – P(A ∩ B)
Since A and B are mutually exclusive we have A ∩ B = Φ
∴ P(A ∩ B) = 0
∴ P(A̅ ∩ B) = p(B)
= \(\frac{1}{8}\)

(iv) P(A̅ ∪ B̅)
P(A̅ ∪ B̅) = P\((\overline{A \cup B})\)
= 1 – P(A ∩ B)
Since A and B are mutually exclusive we have A ∩ B = Φ
∴ P(A ∩ B) = 0
P(A̅ ∪ B̅) = 1 – 0 = 1

Question 2.
If A and B are two events associated with a random experiment for which P (A) = 0.35, P (A or B ) = 0.85 , and P (A and B) = 0.15 find (i) P (only B) (ii) P (B) (iii) P (only A)
Answer:
Given P (A) = 0.35 ,
P (A and B) = P(A ∩ B) = 0.15
P(A or B) = P(A ∪ B) = 0.85
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.85 = 0.35 + P (B) – 0.15
0.85 + 0.15 – 0.35 = P(B)
P(B) = 1 – 0.35
P(B) = 0.65

(i) P (only B)
P (only B) = P(A̅ ∩ B)
= P(B) – P(A ∩ B)
= 0.65 – 0.15
P (only B) = 0.50

(ii) P (B̅)
P (B̅) = 1 – P(B)
= 1 – 0.65
P (B̅) = 0.35

(iii) P (only A)
P (only A) = P(A ∩ B̅)
= P(A) – P(A ∩ B)
= 0.35 – 0.15
P (only A) = 0.20

Question 3.
A die is thrown twice. Let A be the event, ‘First die shows 5’ and B be the event , ‘second
die shows 5’. Find P(A ∪ B).
Answer:
A die is thrown twice. Let S be the sample space
s = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3 ,2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2) ,(5, 3 ) ,(5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
n(S) = 36
Let A be the event ‘First die shows 5’
A = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,0)}
n(A) = 6

Let B be the event ‘Second die shows 5’
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
n(B) = 6

Question 4.
The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of
(i) P(A∪B)
(ii) P(A ∩ B̅)
(iii) P(A̅ ∩ B)
Answer:
P(A) = 0.5, P(B) = 0.3
Here A and B are mutually exclusive.
(i) P(A ∪ B) = P(A) + P(B)
= 0.5 + 0.3 = 0.8
(ii) P(A ∩ B) = P(A) + P(B) – P(A ∪ B) = 0.5 + 0.3 – 0.8
P(A ∩ B) = 0
P(A ∩ \(\overline{B}\)) = P(A) – P(A ∩ B) = 0.5 – 0 = 0.5
(iii) P(\(\overline{A}\) ∩ B) = P(B) – P(A ∩ B) = 0.3 – 0 = 0.3

Question 5.
A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96.
(i) What is the probability that a fire engine is available when needed?
(ii) What is the probability that neither is available when needed?
Answer:
A be the event of availability of a fire B be the event of a fire engine when needed. Availability of a second fire engine when needed.
Given P(A) = 0.96, P(B) = 0.96
Then A̅ is the event of non-availability of the first fire engine and B̅ is the event of non-availability of second fire engine when needed.
P (A̅) = 1 – P(A)
= 1 – 0.96
= 0.04
Also P(B) = 0.04

(i) P(atleast one engine is available) = (1 – probability of no engine available)
= 1 – P(A’ ∩ B’)
= 1 – P (A’) P(B’)
= 1 – (0.04) (0.04)
= 1 – 0.0016
= 0.9984

(ii) P (A’ ∩ B’) = P (A’) P(B’)
= 0.04 × 0.04
= 0.0016

Question 6.
The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that
(i) it will get at least one of the two awards
(ii) it will get only one of the awards.
Answer:
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.48 + 0.36 – 0.2
= 0.64.

(ii) P (Getting only one award)
= P(A) – P(A ∩ B) + P(B) – P(A ∩ B)
= (0.48 – 0.2) + (0.36 – 0.2)
= 0.28 + 0.16
= 0.44.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 12 Introduction to Probability Theory Ex 12.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.1

Question 1.
An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible.
(i) P (A) = 0.15, P (B) = 0.30,
P (C) = 0.43, P (D) = 0.12
Answer:
P(A) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D) = 0.12
Now P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 = 1
0.15 + 0.30 + 0.43 + 0.12 = 1
∴ The assignment of probability is permissible.

(ii) P (A) = 0.22, P (B) = 0.38,
P (C) = 0.16, P (D) = 0.34
Answer:
Given that
P (A) = 0.22 ≥ 0,
P (B) = 0.38 ≥ 0,
P(C) = 0.16 ≥ 0 ,
P (D) = 0.34 ≥ 0
P(S) = P (A) + P(B) + P(C) + P(D)
= 0.22 + 0.38 + 0.16 + 0.34
= 1.1 > 1
Therefore the assignment of probability isn’t permissible

(iii) P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\),
P(C) = – \(\frac{1}{5}\), P(D) = \(\frac{1}{5}\)
Answer:
P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\), P(C) = \(-\frac{1}{5}\), P(D) = \(\frac{1}{5}\)
P(C) = \(-\frac{1}{5}\) which is not possible
(i.e.) for any event A, (0 ≤ P(A) ≤ 1)
∴ The assignment of probability is not permissible.

Question 2.
If two coins are tossed simultaneously, then find the probability of getting
(i) one head and one tail
(ii) at most two tails
Answer:
Two coins are tossed simultaneously = one coin is tossed two times.
The sample space S = { H , T } × { H , T }
S = {HH, HT, TH, TT }
n(S) = 4
Let A be the event of getting one head and one tail, B be the event of getting atmost two tails.
A = {HT, TH}
n(A) = 2
B = {HH, HT , TH, TT}
n (B) = 4

(i) P (getting one head and one tail)

(ii) P (getting atmost two tails)

Question 3.
Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that
(i) one is a mango and the other is an apple
(ii) both are of the same variety.
Answer:
(i) One is a mango and the other is an apple:
Let S be the sample space, A be the event of taking one mango and one apple.
n(S) = 9C₂
n (A) = 5C₁ × 4C₁
= 5 × 4 = 20

(ii) Both are of the same variety:
Let S be the sample space, A be the event of taking 2 mangoes and B be the event of taking 2 apples
∴ n(s) = 9C₂

P(taking 2 fruits are of the same colour)
= P(A or B)
= P(A ∪ B)
= P(A) + P(B)

Question 4.
What is the chance that
(i) Non – leap year
Answer:
Non-leap year
No of days = 365
= \(\frac{365}{7}\) weeks = 52 weeks + 1 day
In 52 weeks we have 52 Sundays. So we have to find the probability of getting the remaining one day as Sunday. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
(i.e.,) n(S) = 7
In the n (Sunday) = A {Saturday to Sunday or Sunday to Monday}
(i.e.,) n(A) = 1
So, P(A) = 1.
∴ Probability of getting 53 Sundays \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{1}{7}\)

(ii) Leap year should have fifty-three Sundays?
Answer:
In 52 weeks we have 52 Sundays. We have to find the probability of getting one Sunday to form the remaining 2 days can be a combination of the following S = {Saturday to Sunday, Sunday to Monday, Monday, to Tuesday, Tuesday to wednes¬day, Wednesday to Thursday, Thursday to Friday, Friday and Saturday}.
(i.e) n(s) =7
In this = A {Saturday to Sunday, Sunday to Monday}
(i.e) n(A) = 2
So, P(A) = \(\frac{2}{7}\)

Question 5.
Eight coins are tossed once, find the probability of getting
(i) exactly two tails
(ii) at least two tails
(iii) at most two tails
Answer:
Eight coins are tossed simultaneously one time = one coin is tossed eight times.
Let S be the sample space.
S = {H, T} × {H, T} × ………….. × {H, T} 8 times
Let A be the event of getting exactly two heads,
B be the event of getting atleast two tails and
C be the event of getting atmost two tails.
When eight coins are tossed, the number of elements in the sample space
n(S) = 2⁸ = 256
n(A) = 8C₂
= \(\frac{8 \times 7}{1 \times 2}\) = 28
n (B) = 8C₂ + 8C₃ + 8C₄ + 8C₅ + 8C₆ + 8C₇ + 8C₈
= n (S) – (8C₈ + 8C₁)
= n (S) – {n (Event of getting all heads) + n (Event of getting one head)}
= n(S) – (1 + 8)
= 256 – 9 = 247
n (C) = 8C₀ + 8 C₁ + 8 C₂
= 1 + 8 + \(\frac{8 \times 7}{1 \times 2}\)
= 1 + 8 + 28 = 37

(i) P {getting exactly two tails) =

(ii) P (getting atleast two tails ) =

(iii) P (getting atmost two tails) =

Question 6.
An integer is chosen at random from the first 100 positive integers. What is the probability that the integer chosen is a prime or multiple of 8 ?
Answer:
Let S be the sample space
S = { 1, 2, 3, …………., 100 }
Let A be the event of choosing a prime number
A = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
n(A) = 25
Let B be the event of choosing an integer a multiple of 8.
B = { 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
n(B) = 12
P(Choosing a prime number) = P(A)

P (Choosing an integer a multiple of 8) = P (B)

P ( Choosing an integer a prime or multiple of 8)
= P(A or B)
= P(A ∪ B)
P(A) + P(B)
(since A and B are mutually exclusive that is A ∩ B = Φ)

Question 7.
A bag contains 7 red 4 black balls, 3 balls are drawn at random. Find the probability that
(i) all are red
(ii) one red and 2 black.
Answer:
Let S be the sample space, A be the event of taking 3 red balls and B be the event of taking one red and
2 black balls.
A bag contains 7 red balls and 3 black balls.
3 balls are drawn at random.
∴ The number of outcomes n(S) = 11C₃
= \(\frac{11 \times 10 \times 9}{1 \times 2 \times 3}\)
= 11 × 5 × 3 = 165

n (A) = 7C₃

n(B) = 7C₁ × 4C₂
= 7 × \(\frac{4 \times 3}{1 \times 2}\) = 7 × 6 = 42

(i) p (getting 3 red balls) = p(A)

(ii) p (getting one red and 2 blacks) = p(B)

Question 8.
A single card is drawn from a pack of 52 cards. What is the probability that
(i) the card is an ace or a king
(ii) the card w11 be 6 or smaller
(iii) the card is either a queen or 9?
Answer:
S be the sample space one card is drawn from a pack of 52 cards.
∴ n(S) = 52C₁
n(S) = 52

(i) The card is an ace or a king
Let A be the event of getting an ace.
n(A) = 4C₁ = 4
Let B be the event of getting a king.
n(B) = 4C₁ = 4
P (getting an ace or a king)
= P(A or B)
= P(A ∪ B)
= P(A) + P(B)
(since A and B are mutually exclusive events, A ∩ B = Φ)

(ii) The card will be 6 or smaller:
Let A be the event of getting a number 6.
∴ n(A) = 4C₁ = 4
Let B be the event of getting numbers less than 6.
n(B) = 16C₁ = 16
P (the card will be 6 or less than 6)
= P(A or B)
= P(A ∪ B)
= P(A) + P(B)
(since A and B are mutually exclusive events A ∩ B = Φ)

(iii) The card is either a queen or 9?
Let A be the event of getting a Queen.
∴ n(A) = 4C₁ = 4
Let B be the event of getting a number 9.
n(B) = 4C₁ = 4
P (the card is either a Queen or 9)
= P(A or B)
= P(A ∪ B
= P(A) + P(B)
(since A and B are mutually exclusive events)

Question 9.
A cricket club has 16 members, of whom only 5 can bowl. What is the probability that in a team of 11 members at least 3 bowlers are selected?
Answer:
Number of members in the cricket club = 16
Number of bowlers = 5
Number of batters = 16 – 5 = 11
The probability that a team of 11 members consisting of atleast 3 bowlers = (Probability of selecting 3 bowlers and 8 batters) + ( Probability of selecting 4 bowlers and 7 batters) + (Probability of selecting 5 bowlers and 6 batters)
The probability that a team of 11 members consisting of atleast 3 bowlers

[Selection procedure:
First out of total 16 members selecting 11 members in 16C₁₁ ways.
Selection of 11 members consisting minimum of 3 bowlers.
∴ Selection of 11 members as follows
(1) 3 bowlers from 5 bowlers and 8 batters from 11 batters.
(2) 4 bowlers from 5 bowlers and 7 batters from 11 batters.
(3) 5 bowlers from 5 bowlers and 6 batters from 11 batters.]

Question 10.
(i) The odds that event A occurs is 5 to 7, find P(A).
Answer:
Given the odds, that event A occurs is 5 to 7.

(ii) Suppose p (B) = \(\frac{2}{5}\). Express the odds that the event B occurs.
Answer:

∴ The odds that the event B occurs is 2 to 3.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.13 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.13

Choose the correct or the most suitable answer from given four alternatives.

Question 1.
If ∫ f(x) dx = g(x) + c. then ∫ f(x) g (x)dx
(1) ∫ (f(x)² dx
(2) ∫ f(x) g(x) dx
(3) ∫ f'(x) g(x) dx
(4) ∫ (g(x))² dx
Answer:
(1) ∫ (f(x)² dx

Explaination:
Given ∫ f (x) dx = g (x) + c
\(\frac{\mathrm{d}}{\mathrm{d} x}\) ∫ f(x)dx = \(\frac{\mathrm{d}}{\mathrm{d} x}\) (g(x) + c)
∫ \(\frac{\mathrm{d}}{\mathrm{d} x}\) (f(x)) dx = g'(x)
∫ d(f(x)) = g'(x)
f(x) = g(x)
∴ ∫ f(x) g'(x) dx = ∫ f(x) f(x) dx
= ∫ [f(x)]² dx

Question 2.
If \(\int \frac{3^{\frac{1}{x}}}{x^{2}}\) dx = k\(\left(3^{\frac{1}{x}}\right)\) + c, then the value of k is
(1) log 3
(2) – log 3
(3) \(-\frac{1}{\log 3}\)
(4) \(\frac{1}{\log 3}\)
Answer:
(3) \(-\frac{1}{\log 3}\)

Explaination:

Question 3.
If ∫ f'(x) e^x2 dx = (x – 1)e^x2 + c, then f(x) is
(1) 2x³ – \(\frac{x^{2}}{2}\) + x + c
(2) \(\frac{x^{3}}{2}\) + 3x² + 4x + c
(3) x³ + 4x² + 6x + c
(4) \(\frac{2 x^{3}}{3}\) – x² + x + c
Answer:
(4) \(\frac{2 x^{3}}{3}\) – x² + x + c

Explaination:
Given ∫ f'(x) e^x2 dx = (x – 1)e^x2 + c
Differentiating both sides with respect to x we have

Question 4.
The gradient (slope) of a curve at any point (x, y) is \(\frac{x^{2}-4}{x^{2}}\). If the curve passes through the point (2, 7), then the equation of the curve is
(1) y = x + \(\frac{4}{x}\) + 3
(2) y = x + \(\frac{4}{x}\) + 4
(3)y = x² + 3x + 4
(4) y = x² – 3x + 6
Answer:
(1) y = x + \(\frac{4}{x}\) + 3

Explaination:


Given, this curve passes through the point (2, 7)
∴ 7 = 2 + \(\frac{4}{2}\) + c
7 = 2 + 2 + c
c = 7 – 4 = 3
∴ The required equation is
y = x + \(\frac{4}{x}\) + 3

Question 5.

(1) cot (xe^x) + c
(2) sec (xe^x) + c
(3) tan (xe^x) + c
(4) cos (xe^x) + c
Answer:
(3) tan (xe^x) + c

Explaination:

= ∫sec² t
= tan t + c
= tan (x e^x) + c

Question 6.

(1) \(\sqrt{\tan x}+\mathbf{c}\)
(2) \(2 \sqrt{\tan x}+\mathbf{c}\)
(3) \(\frac{1}{2} \sqrt{\tan x}+c\)
(4) \(\frac{1}{4} \sqrt{\tan x}+c\)
Answer:
(1) \(\sqrt{\tan x}+\mathbf{c}\)

Explaination:

Question 7.
∫sin³ dx is
(1)
(2)
(3)
(4)
Answer:
(3)

Explaination:
∫sin³ dx
sin 3x = 3 sin x – 4 sin³ x
4 sin³ x = 3 sin x – sin 3x
sin³ x = \(\frac{1}{4}\)(3 sin x – sin 3x)

Question 8.

(1) x + c
(2) \(\frac{x^{3}}{3}\) + c
(3) \(\frac{3}{x^{3}}\) + c
(4) \(\frac{1}{x^{2}}\) + c
Answer:
(2) \(\frac{x^{3}}{3}\) + c

Explaination:

Question 9.

(1) tan^-1 (sin x) + c
(2) 2 sin^-1 (tan x) + c
(3) tan^-1 (cos x) + c
(4) sin^-1 (tan x) + c
Answer:
(4) sin^-1 (tan x) + c

Explaination:

Question 10.

(1) x² + c
(2) 2x² + c
(3) \(\frac{x^{2}}{2}\) + c
(4) – \(\frac{x^{2}}{2}\) + c
Answer:
(3) \(\frac{x^{2}}{2}\) + c

Explaination:

Question 11.
∫2^3x+5 dx is
(1)
(2)
(3)
(4)
Answer:
(4)

Explaination:
∫2^3x+5 dx
Put 3x + 5 = t
3 dx = dt
dx = \(\frac{1}{3}\) dt

Question 12.

(1) \(\frac{1}{2}\) sin 2x + c
(2) –\(\frac{1}{2}\) sin 2x + c
(3) \(\frac{1}{2}\) cos 2x + c
(4) –\(\frac{1}{2}\) cos 2x + c
Answer:
(2) –\(\frac{1}{2}\) sin 2x + c

Explaination:

Question 13.

(1) e^x tan^-1 (x + 1) + c
(2) tan^-1 (e^x) + c
(3) e^x \(\frac{\left(\tan ^{-1} x\right)^{2}}{2}\) + c
(4) e^x tan^-1 x + c
Answer:
(4) e^x tan^-1 x + c

Explaination:

Question 14.

(1) cot x + sin^-1 x + c
(2) – cot x + tan^-1x + c
(3) – tan x + cot^-1 x + c
(4) – cot x – tan^-1x + c
Answer:
(4) – cot x – tan^-1x + c

Explaination:

Question 15.
∫x² cos x dx is
(1) x² sin x + 2x cos x – 2 sin x + c
(2) x² sin x – 2x cos x – 2 sin x + c
(3) – x² sin x + 2x cos x + 2 sin x + c
(4) – x² sin x – 2x cos x + 2 sin x + c
Answer:
(1) x² sin x + 2x cos x – 2 sin x + c

Explaination:
\(\int x^{2} \cos x d x\)
By Bernoullis formula dv = cosxdx
u = x² v = sinx
u’ = 2x v₁ = -cos x
u” = 2 v₂ = -sinx
= uv – u’v₁ + u”v₂
= x²sin x + 2x cos x – 2 sin x + c

Question 16.

(1)
(2)
(3)
(4)
Answer:
(1)

Explaination:

Question 17.
\(\int \frac{d x}{e^{x}-1}\) is
(1) log |e^x| – log |e^x – 1| + c
(2) log |e^x| + log |e^x – 1| + c
(3) log |e^x – 1| – log |e^x| + c
(4) log |e^x + 1| – log |e^x| + c
Answer:
(3) log |e^x – 1| – log |e^x| + c

Explaination:

Question 18.
∫e^-4x cos x dx is
(1) \(\frac{e^{-4 x}}{17}\) [4 cos x – sin x] + c
(2) \(\frac{e^{-4 x}}{17}\) [- 4 cos x – sin x] + c
(3) \(\frac{e^{-4 x}}{17}\) [4 cos x + sin x] + c
(4) \(\frac{e^{-4 x}}{17}\) [- 4 cos x – sin x] + c
Answer:
(2) \(\frac{e^{-4 x}}{17}\) [- 4 cos x – sin x] + c

Explaination:

Question 19.

(1)
(2)
(3)
(4)
Answer:
(4)

Explaination:

Question 20.
∫e^-7x sin 5x dx is
(1) \(\frac{e^{-7 x}}{74}\) [- 7 sin 5x – 5 cos 5x] + c
(2) \(\frac{e^{-7 x}}{74}\) [7 sin 5x + 5 cos 5x] + c
(3) \(\frac{e^{-7 x}}{74}\) [7 sin 5x – 5 cos 5x] + c
(4) \(\frac{e^{-7 x}}{74}\) [- 7 sin 5x + 5 cos 5x] + c
Answer:
(1) \(\frac{e^{-7 x}}{74}\) [- 7 sin 5x – 5 cos 5x] + c

Explaination:

Question 21.
∫x² e^x/2 dx is
(1)
(2)
(3)
(4)
Answer:
(3)

Explaination:

Question 22.

(1)
(2)
(3)
(4)
Answer:
(4)

Explaination:

Question 23.

(1)
(2)
(3)
(4)
Answer:
(3)

Explaination:

Question 24.
∫sin √x dx is
(1) 2(- √x cos √x + sin √x) + c
(2) 2(- √x cos √x + sin √x) + c
(3) 2(- √x sin √x – cos √x) + c
(4) 2(- √x sin √x + cos √x) + c
Answer:
(1) 2(- √x cos √x + sin √x) + c

Explaination:

Question 25.

(1)
(2)
(3)
(4)
Answer:
(4)

Explaination:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.12 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.12

Integrate the following with respect to x.

Question 1.
(i) \(\sqrt{x^{2}+2 x+10}\)
Answer:

(ii) \(\sqrt{x^{2}-2 x-3}\)
Answer:

(iii) \(\sqrt{(6-x)(x-4)}\)
Answer:

Question 2.
(i) \(\sqrt{9-(2 x+5)^{2}}\)
Answer:

(ii) \(\sqrt{81+(2 x+1)^{2}}\)
Answer:

(iii) \(\sqrt{(x+1)^{2}-4}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.6 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.6

Question 1.
\(\frac{x}{\sqrt{1+x^{2}}}\)
Answer:
Put 1 + x² = u
2x dx = du
x dx = \(\frac{1}{2}\) du

Question 2.
\(\frac{x^{2}}{1+x^{6}}\)
Answer:

Put x³ = u
3x² dx = du
x² dx = \(\frac{1}{3}\) du

Question 3.
\(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)
Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.
\(\frac{\cot x}{\log (\sin x)}\)
Answer:

Question 7.

Answer:

Question 8.

Answer:

Put a² + b² sin²x = u
(0 + b² × 2 sin x cos x) dx = du
b² sin 2x dx = du
sin 2x dx = \(\frac{1}{b^{2}}\) du

Question 9.
\(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\)
Answer:

Question 10.
\(\frac{\sqrt{x}}{1+\sqrt{x}}\)
Answer:

Question 11.

Answer:

Question 12.
α β x^α-1 e^{-β xα}
Answer:
∫α β x^α-1 e^{-β xα}
Put β x^α = u
α β x^α-1 dx = du

Question 13.
\(\tan x \sqrt{\sec x}\)
Answer:

put cos x = u
– sin x dx = du
sin x dx = – du

Question 14.
x (1 – x)¹⁷
Answer:
∫x (1 – x)¹⁷ dx
put 1 – x = u
-dx = du

Question 15.
sin⁵ x cos³ x
Answer:
∫ sin⁵ x cos³ x dx = ∫ sin⁵ x cos² x . cos x dx
= ∫ sin⁵ x (1 – sin² x) . cos x dx
= ∫ (sin⁵ x – sin⁷ x) . cos x dx
= ∫ sin⁵ x cos x dx – ∫ sin⁷ x cos x dx
put u = sin x
du = cos x dx

Question 16.

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.11 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.11

Integrate the following with respect to x:

Question 1.
(i) \(\frac{2 x-3}{x^{2}+4 x-12}\)
Answer:
Let 2x – 3 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 4x – 12) + B
2x – 3 = A (2x + 4) + B
2x – 3 = 2Ax + 4A + B
2A = 2 ⇒ A = 1
4A + B = – 3
4 × 1 + B – 3 ⇒ B = – 3 – 4 = – 7
2x – 3 = 1 (2x + 4) – 7

(ii) \(\frac{5 x-2}{2+2 x+x^{2}}\)
Answer:
Let 5x – 2 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 2x + 2) + B
5x – 2 = A (2x + 2) + B
5x – 2 = 2Ax + 2A + B
2A = 5 ⇒ A = \(\frac{5}{2}\)
2A + B = -2
2 × \(\frac{5}{2}\) + B = -2 ⇒ B = -2 – 5 = -7
5x – 2 = \(\frac{5}{2}\) (2x + 2) – 7

Put x² + 2x + 12 = t
(2x + 2) dx = dt

Put x + 1 = u
dx = du

(iii) \(\frac{3 x+1}{2 x^{2}-2 x+3}\)
Answer:
Let 3x + 1 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (2x² – 2x + 3) + B
3x + 1 = A (4x – 2) + B
3x + 1 = 4Ax – 2A + B

Question 2.
(i) \(\frac{2 x+1}{\sqrt{9+4 x-x^{2}}}\)
Answer:
Let 2x + 1 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (9 + 4x – x²) + B
2x + 1 = A(4 – 2x) + B
2x + 1 = 4A – 2Ax + B
-2A = 2 ⇒ A = -1
4A + B = 1 ⇒ 4 (-1) + B = 1
B = 1 + 4 = 5
B = 5
2x + 1 = – 1 (4 – 2x) + 5

Put 9 + 4x – x² = t
(4x – 2) dx = dt

Put x – 2 = u
dx = du

(ii) \(\frac{x+2}{\sqrt{x^{2}-1}}\)
Answer:
Let x + 2 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² – 1) + B
x + 2 = A (2x) + B
2A = 1 ⇒ A = \(\frac{1}{2}\)
B = 2

(iii) \(\frac{2 x+3}{\sqrt{x^{2}+4 x+1}}\)
Answer:
Let 2x + 3 = A \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x² + 4x + 1) + B
2x + 3 = A (2x + 3) + B
2A = 2 ⇒ A = 1
4A + B = 3
4 × 1 + B = 3 ⇒ B = 3 – 4
B = -1
(2x + 3) = (2x + 4) – 1

Put x² + 4x + 1 = t
(2x + 4) dx = dt

Put x + 2 = u
dx = dt

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.10

Find the integrals of the following:

Question 1.
(i) \(\frac{1}{4-x^{2}}\)
Answer:

(ii) \(\frac{1}{25-4 x^{2}}\)
Answer:

(iii) \(\frac{1}{9 x^{2}-4}\)
Answer:

Question 2.
(i) \(\frac{1}{6 x-7-x^{2}}\)
Answer:

(ii) \(\frac{1}{(x+1)^{2}-25}\)
Answer:

(iii) \(\frac{1}{\sqrt{x^{2}+4 x+2}}\)
Answer:

Question 3.
(i) \(\frac{1}{\sqrt{(2+x)^{2}-1}}\)
Answer:

(ii) \(\frac{1}{\sqrt{x^{2}-4 x+5}}\)
Answer:

(iii) \(\frac{1}{\sqrt{9+8 x-x^{2}}}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.9

Integrate the following with respect to x:

Question 1.
e^x (tan x + log sec x)
Answer:
Let I = ∫e^x (tan x + log sec x) dx
Take f(x) = log sec x
f'(x) = \(\frac{1}{\sec x}\) sec x tan x
f'(x) = tan x
[∫e^x [f (x) + f (x)] dx = e^x f(x) + c]
∴ I = e^x log |sec x| + c

Question 2.

Answer:

[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]

Question 3.
e^x sec x (1 + tan x)
Answer:
Let I = \(\mathrm{I}=\int e^{x}(\sec x+\sec x \tan x) d x\)
Take f(x) = sec x
f ‘ (x) = sec x tan x
This is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\) = e^x f(x) + c
= e^x sec x + c

Question 4.

Answer:

Take f(x) = tan x
f'(x) = sec² x
[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]
I = e^x tan x + c

Question 5.

Answer:

Put tan^-1 x = t
\(\frac{1}{1+x^{2}}\)dx = dt
x = tan t
∴ I = ∫e^t [1 + tan t + tan²t] dt
= ∫e^t [1 + tan² t + tan t] dt
= ∫e^t [sec²t + tan t] dt
= ∫e^t [tan t + sec²t] dt
f(x) = tan t
f'(x) = sec²t
[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]
∴ I = e^t tan t + c
I = e^tan-1(x) . x + c
I = x e^tan-1(x) + c

Question 6.

Answer: