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You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.1

Question 1.
Suppose, you have two shorts, one is black and the other one is blue; three shirts which are in white, blue and red. You again wish to make different combinations, but you always want to make sure that the shorts and shirt that you wear are of different colours. List and check how many combinations are possible now.
Solution:
We have given two shorts which are black and blue in colour. Take it as T black and T blue.
Also, we have 3 shirts, coloured white, blue and red denoted by S white, S blue and S red.
Now fix T black and then T blue the different combinations are

Thus we get a total of 6 combinations as
Black short and White shirt
Black short and Blue shirt
Black short and Redshirt
Blue short and White shirt
Blue short and Blue shirt
Blue short and Redshirt.
But it is given short and the shirt is of different colours.
We give up Blue short and Blue shirt combination. So we have 5 different combinations.

Question 2.
You have two red and two blue blocks. How many different towers can you build that are four blocks high using these blocks? List all the possibilities.
Solution:
6 Possibilities,
R-B-R-B
R-R-B-B
B-R-R-B
B-R-B-R
B-B-R-R
R-B-B-R

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Ex 3.2

Miscellaneous Practice Problems

Question 1.
A shopkeeper buys three articles for ₹ 325, ₹ 450 and ₹ 510. He is able to sell them for ₹ 350, ₹ 425 and ₹ 525 respectively. Find the gain or loss to the shopkeeper on the whole.
Solution:
C.P of three articles = 325 + 450 + 510 = ₹ 1285
S.P of three articles = 350 + 425 + 525 = ₹ 1,300
Here S.P > C.P
Profit = S.P – C.P = 1,300 – 1285 = ₹ 15
The shopkeeper gained = ₹ 15

Question 2.
A stationery shop owner bought a scientific calculator for Rs 750. He had put a battery worth Rs 100 in it. He had spent Rs 50 for its outer pouch. He was able to sell it for Rs 850. Find his profit or loss.
Solution:
CP = Rs 750 + Rs 100 + Rs 50
= Rs 900
SP = Rs 850
SP < CP
Loss = CP – SP
= Rs 900 – Rs 850
= Rs 50

Question 3.
Nathan paid ₹ 800 and bought 10 bottles of honey from a village vendor. He sold them in a city for ₹ 100 per bottle. Find his profit or loss.
Solution:
C.P of 10 bottles of honey = ₹ 800
C.P of 1 bottle honey = 800/10 = ₹ 80
S.P of a bottle honey = ₹ 100
Here S.P > C.P
Profit per bottle = ₹ 100 – ₹ 80 = ₹ 20
Profit for 10 bottles = 20 × 10 = ₹ 200
Profit = ₹ 200

Question 4.
A man bought 400 metre of cloth for Rs 60,000 and sold it at the rate of Rs 400 per metre. Find his profit or loss.
Solution:
Cost of 400 m of cloth = Rs 60,000
CP = Rs 60,000
Selling price of 400 m of cloth = 400 × Rs 400
SP = Rs 1,60,000
SP > CP
Profit = SP – CP
= Rs 1,60,000 – Rs 60,000
= Rs. 1,00,000

Challenge Problems

Question 5.
A fruit seller bought 2 dozen bananas at ₹ 20 a dozen and sold them at ₹ 3 per banana. Find his gain or loss.
Solution:
Cost of one dozen banana = ₹ 20
Cost of 2 dozen bananas = ₹ 20 × 2 = ₹ 40
C.P = ₹ 40
S.P per banana = ₹ 3
S.P for 2 dozen banana = ₹ 3 × 24 = ₹ 72
Here S.P > C.P
Profit = S.P – C.P = 72 – 40 = 32
Profit = ₹ 32

Question 6.
A store purchased pens at Rs 216 per dozen. He paid Rs 58 for conveyance and sold the pens at the discount of Rs 2 per pen and made an overall profit of Rs 50. Find the M.P. of each pen.
Solution:
Cost of 1 dozen pens = Rs 216 + Rs 58
CP = Rs 274
Discount for each pen = Rs 2
Overall profit = Rs 50
Total discount for 12 pens = Rs 2 × 12
= Rs 24
Selling price of 12 pens = Mp – Discount
SP = Mp – Rs 24
Profit = SP – CP
Rs 50 = Mp – Rs 24 – Rs 274
MP = Rs 50 + Rs 24 + Rs 274
= Rs 348 (1 dozen)
MP of each pen = Rs 348 / 12
= Rs 174/6
= Rs 29

Question 7.
A vegetable vendor buys 10 kg of tomatoes per day at ₹ 10 per kg, for the first three days of a weak. 1 kg of tomatoes got smashed on every day for those 3 days. For the remaining 4 days of the week, he buys 15 kg of tomatoes daily at ₹ 8 per kg. If for the entire week he sells tomatoes at ₹ 20 per kg, then find his profit or loss for the week.
Solution:
Total tomatoes bought for 3 days = 3 × 10 = 30 kg
Cost of 1 kg = ₹ 10
Cost of 30kg tomatoes = 30 × 10 = ₹ 300
Total tomatoes bought for other 4 days = 4 × 15 = 60 kg
Cost of 1 kg = ₹ 8
Cost of 60 kg tomatoes = 60 × 8 = ₹ 480
Total cost of 90 kg tomatoes = 300 + 480 = ₹ 780
C.P = ₹ 780
Tomatoes smashed = 3 kg
Total kg of Tomatoes for sale = 90 – 3 = 87 kg
S.P of 1 kg tomatoes = ₹ 20
S.P of 87 kg tomatoes = 87 × 20 = ₹ 1740
Here S.P > C.P
Profit = S.P – C.P = 1740 – 780 = ₹ 960
Profit = ₹ 960

Question 8.
An electrician buys a used T.V. for Rs 12,000 and a used Fridge for Rs 11,000. After spending Rs 1,000 on repairing the T.V. and Rs 1500 on painting the Fridge, he fixes up the M.P. of T.V. as Rs 15,000 and that of Fridge as Rs 15,500. If he gives Rs 1000 discount on each find his profit or loss.
Solution:
Total cost price of TV and Fridge = Rs 12,000 + Rs 11,000 + Rs 1,000 + Rs 1,500 = Rs 25,500
Selling price of TV = MP – Discount = Rs 15,000 – Rs 1,000 = Rs 14,000
Selling price of Fridge = MP – Discount = Rs 15,500 – Rs 1,000 = Rs 14,500
Total Selling price of TV and Fridge = Rs 14,000 + Rs 14,500 = Rs 28,500
SP > CP
Profit = SP – CP = Rs 28,500 – Rs 25,500 = Rs 3,000

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Miscellaneous Practice Problems

Question 1.
A piece of wire is 36 cm long. What will be the length of each side if we form
i) a square
ii) an equilateral triangle.
Solution:
Given the length of the wire = 36 cm
i) When a square is formed out of it
The perimeter of the square = 36 cm
4 × side = 36
side = \(\frac{36}{4}\) = 9 cm
Side of the square

ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm
i.e., side + side + side = 36 cm .
3 × side = 36 cm
side = \(\frac{36}{3}\) = 12 cm
One side of an equilateral triangle = 12 cm

Question 2.
From one vertex of an equilateral triangle with side 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion?
Solution:

If an equilateral triangle of side 6 cm is removed the perimeter = (40 + 34 + 6 +34) cm = 114 cm
Perimeter of the remaining portion = 114 cm

Question 3.
Rahim and peter go for a morning walk, Rahim walks around a square path of side 50 m and Peter walks around a rectangular path with length 40 m and breadth 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Solution:
Distance covered by Rahim
= 50 × 4 m
= 200 m
If he walks 2 rounds, distance covered = 2 × 200 m
= 400 m
Distance covered by peter
= 2 (40 + 30) m
= 2(70)m
= 140 m
If he walks 2 rounds, distance covered = 2 × 140 m
= 280 m
∴ Rahim covers more distance by (400 – 280) = 120 m

Question 4.
The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park?
Solution:
Given length of rectangular park is 14m more than its breadth.
Let the breadth be b m .
∴ Length of the park will be l = b + 14 m
Given perimeter = 200 m
2 × (l + b) = 200 m
2 × (b + 14 + b) = 200 m [∵ l = b + 14]
2 × (2b + 14) = 200 m
2b + 14 = \(\frac{200}{2}\) m
2b + 14 = 100 m
2b = 100 – 14 m
2b = 86 m
b = \(\frac{86}{2}\) m
b 43 m
Length Length of the park = 57 m
Area of a rectangle = (length × breadth) unit²
= (57 × 43) m² = 2,451 m²
Area of the park = 2,451 m²

Question 5.
Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at Rs 10 per meter.
Solution:
a = 5 m
Perimeter of the garden
= 4 a units
= 4 × 5 m
= 20 m
For 1 row
Amount needed to fence l m= Rs 10
Amount needed to fence 20 m
= Rs 10 × 20
= Rs 200
For 2 rows
Total amount needed = 2 × Rs 200
= Rs 400

Challenge Problems

Question 6.
A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.
Solution:
Number of equal sides in the shape = 20
One of its side = 3 cm
Perimeter = length of one side × Number of equal sides
∴ Perimeter = (3 × 20) cm = 60 cm
∴ Perimeter = 60 cm

Question 7.
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.
Solution:
l = 40 cm, b = 20 cm
Area of the rectangle = l × b sq units
= 40 × 20 cm²
= 800 cm²
a = 10 cm
Area of the square = a × a sq. units
= 10 × 10 cm²
= 100 m²
No of squares formed = \(\frac{800}{100}\) cm²
= 8

Question 8.
The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.
Solution:
Given perimeter of a rectangle = 64 cm
Also given length is three times its breadth.
Let the breadth of the rectangle = b cm
∴ Length = 3 × b cm
Perimeter = 64 m
i.e., 2 × (l + b) = 64 m
2 × (3b + b) = 64 m
2 × 4b = 64m
4b = \(\frac{64}{2}\) = 32 m
b = \(\frac{32}{4}\) = 8 m
l = 3 × b = 3 × 8 = 24 m
∴ Breadth of the rectangle = 8 m
Length of the rectangle = 24 m

Question 9.
How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.
Solution:
12 rectangles
(1, 23), (2, 22), (3, 21), (4, 20), (5, 19), (6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)

Question 10.
Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Solution:

Perimeter of A = s + s + s + s units = 4 s units
Perimeter of B = (2s + 2s + 2s + 2s) units
= 8s units = 2 (4s) units.
∴ Perimeter of B is twice perimeter of A

Question 11.
What will be the area of a new square formed if the side of a square is made one – fourth?
Solution:
Area of the new square is reduced to \(\frac{1}{16}\) th times to that of the original area.

Question 12.
Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Solution:

Given perimeter of square = perimeter of rectangle
4 × side = 2 (length + breadth)
(4 × 10) m = 2(l + 8)m
\(\frac{4 \times 10}{2}\) = l + 8
20 = l + 8
l = 20 – 8
l = 12 m
∴ length of the rectangle = 12 m
Area of the square plot – side × side = 10 × 10 m² = 100 m²
Area of the rectangular plot = length × breadth = (12 × 8) m² = 96 m²
100 m² > 96 m²
∴ Square plot has greater area by 4m²

Question 13.
Look at the picture of the house given and find the total area of the shaded portion.

Solution:
Total area of the shaded region = Area of a right triangle + Area of a rectangle
= (\(\frac{1}{2}\) × b × h) + (l × b) cm²
= [(\(\frac{1}{2}\) × 3 × 4) + (9 × 6)] cm²
= (6 + 54) cm² = 60 cm²

Question 14.
Find the approximate area of the flower in the given square grid.

Solution:
Approximate area = Number of full squares + Number of more than half squares + \(\frac{1}{2}\) × Number of half squares.
= 10 + 5 + (\(\frac{1}{2}\) × 1) Sq units. = 10 + 5 + \(\frac{1}{2}\) sq. units
= 15 \(\frac{1}{2}\) sq. units = 15.5 sq. units.

Students can Download Tamil Nadu 11th Maths Previous Year Question Paper June 2019 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Previous Year Question Paper June 2019 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The range of the function \(\frac{1}{1-2sin x}\) is………….
(a) (-∞, -1) ∪(\(\frac{1}{3}\), ∞)
(b) (-1, \(\frac{1}{3}\))
(c) [-1, \(\frac{1}{3}\)]
(d) (-∞, -1] ∪(\(\frac{1}{3}\), ∞)
Answer:
(d) (-∞, -1] ∪(\(\frac{1}{3}\), ∞)

Question 2.
If the function f : [-3, 3] → S defined by f(x) = x² is onto, then S is…………
(a) [-9, 9]
(b) R
(c) [-3, 3]
(d) [0, 9]
Answer:
(d) [0, 9]

Question 3.
The number of solutions of x² + |x – 1| = 1 is ………..
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(c) 2

Question 4.
cos 1° + cos 2° + cos 3° + …. + cos 179° =……………..
(a) 0
(b) 1
(c) -1
(d) 89
Answer:
(a) 0

Question 5.
If tan α and tan β are the roots of x² + ax + b = 0, then \(\frac{sin(α+β)}{sin α sin β}\) is equal to…………
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(b) \(\frac{a}{b}\)

Question 6.
The number of sides of a polygon having 44 diagonals is…………
(a) 4
(b) 4
(c) 11
(d) 22
Answer:
(d) 22

Question 7.
The H.M. of two positive numbers whose A.M. and G.M. are 16, 8 respectively is………….
(a) 10
(b) 6
(c) 5
(d) 4
Answer:
(d) 4

Question 8.
The nth term of the sequence \(\frac{1}{2}\), \(\frac{3}{4}\), \(\frac{7}{8}\), \(\frac{15}{16}\) is……………
(a) 2ⁿ – n – 1
(b) 1 – 2^-n
(c) 2^-n+ n – 1
(d) 2^n-1
Answer:
(b) 1 – 2^-n

Question 9.
The intercepts of the perpendicular bisector of the line segment joining (1,2) and (3,4) with coordinate axes are
(a) 5, -5
(b) 5, 5
(c) 5, 3
(d) 5, -4
Answer:
(b) 5, 5

Question 10.
The image of the point (2, 3) in the line y = -x is
(a) (-3, -2)
(b) (-3, 2)
(c) (-2, -3)
(d) (3, 2)
Answer:
(a) (-3, -2)

Question 11.
If A = \(\left[\begin{array}{cc} \lambda & 1 \\ -1 & -\lambda \end{array}\right]\), then for what value of λ, A² = 0?………
(a) 0
(b) ± 1
(c) -1
(d) 1
Answer:
(b) ± 1

Question 12.
If \(\vec{a}\) and \(\vec{b}\) are having same magnitute and angle between them is 60° and their scalar product is \(\frac{1}{2}\), then |\(\vec {a}\)| is
(a) 2
(b) 3
(c) 7
(d) 1
Answer:
(d) 1

Question 13.
\(\lim _{x \rightarrow \infty} \frac{a^{x}-b^{x}}{x}\) = …………..
(a) log ab
(b) log (\(\frac{a}{b}\))
(c) log (\(\frac{b}{a}\))
(d) \(\frac{a}{b}\)
Answer:
(b) log (\(\frac{a}{b}\))

Question 14.
If f(x) = \(\left\{\begin{array}{ccc}x, & x \text { is irrational } \\ 1-x, & x \text { is rational }\end{array}\right.\) then f is………..
(a) discontinuous at x = \(\frac{1}{2}\)
(b) continuous at x = \(\frac{1}{2}\)
(c) continuous everywhere
(d) discontinuous everywhere
Answer:
(b) continuous at x = \(\frac{1}{2}\)

Question 15.
The derivative of f(x) = x |x| at x = -3 is…………..
(a) 6
(b) -6
(c) does not exist
(d) 0
Answer:
(a) 6

Question 16.
If f(x) = x² – 3x, then the points at which f(x) = f'(x) are…………..
(a) both positive integers
(b) both negative integers
(c) both irrational
(d) one rational and another irrational
Answer:
(c) both irrational

Question 17.
\(\int \tan ^{-1}(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}})\) dx ………….
(a) x² + c
(b) 2x² + c
(c) \(\frac{x²}{2}\) + c
(d) –\(\frac{x²}{2}\) + c
Answer:
(c) \(\frac{x^2}{2}\) + c

Question 18.
e^-7x sin 5x dx is…………
(a) \(\frac{e^{-7x}}{74}\) [-7 sin 5x – 5 cos 5x] + c
(b) \(\frac{e^{-7x}}{74}\) [7 sin 5x + 5 cos 5x] + c
(c) \(\frac{e^{-7x}}{74}\) [7 sin 5x – 5 cos 5x] + c
(d) \(\frac{e^{-7x}}{74}\) [-7 sin 5x + 5 cos 5x] + c
Answer:
(a) \(\frac{e^{-7x}}{74}\) [-7 sin 5x – 5 cos 5x] + c

Question 19.
If A and B are any two events then the probability that exactly one of them occur is
(a) P(A ∪\(\bar { B }\)) + P(\(\bar { A }\) ∪B)
(b) P(A ∩\(\bar { B }\)) + P(\(\bar { A }\) ∩B)
(c) P(A) + P(B) – P(A ∩ B)
(d) P(A) + P(B) + 2P(A ∩ B)
Answer:
(a) P(A ∪\(\bar { B }\)) + P(\(\bar { A }\) ∪B)

Question 20.
In a certain college 4% of the boys and 1 % of the girls are taller than 1.8 meter. Further 60% of the students are girls. If a student is selected at random and is taller than 1.8 meters, then the probability that the student is a girl is ………….
(a) \(\frac{2}{11}\)
(b) \(\frac{3}{11}\)
(c) \(\frac{5}{11}\)
(d) \(\frac{7}{11}\)
Answer:
(b) \(\frac{3}{11}\)

PART- II

II. Answer any seven questions. Question No: 30 is compulsory. [7 × 2 = 14]

Question 21.
Resolve \(\frac{3x+1}{(x-2)(x+1)}\) into partial fractions.
Answer:

Equating numerator parts
3x + 1 = A (x + 1) + B (x – 2)
This equation is true for any value of x.
To find A and B
Put x = -1
-3 + 1 = A (0) + B (-1 -2)
-3 B = -2 ⇒ B = 2/3
Put x = 2
3(2) + 1 = A(2 + 1) + B (0)
3A = 7 ⇒ A = 7/3
Hence \(\frac{3x+1}{(x-2)(x+1)}\) = \(\frac{7}{3(x-2)}\) + \(\frac{2}{3(x+1)}\)

Question 22.
Prove that

Answer:
cot (180° + θ) = cot θ
sin (90° – θ) = cos θ
cos (- θ) = cos θ
sin (270 + θ) = – cos θ
tan (-θ) = – tan θ
cosec (360° + θ) = cosec θ

Question 23.
If cos θ = \(\frac{1}{2}\) (a + \(\frac{1}{a}\)), show that cos 3θ = \(\frac{1}{2}\) (a³ + \(\frac{1}{a³}\))
Answer:

Question 24.
If the letters of the word IITJEE are permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE.
Answer:
The lexicographic order of the letters of given word is E, E, I, I, J, T. In the lexicographic order, the strings which begin with E come first. If we fill the first place with E, remaining 5 letters (E, I, I, J, T) can be arranged in \(\frac{5!}{2!}\) ways. On proceeding like this we get,
E – – – – = \(\frac{5!}{2!}\) = 60 ways
IIE – – – = 3! = 6 ways
IIJ – – – = \(\frac{3!}{1!}\) = 3 ways
IITE – – =2! = 2 ways
IITJEE = 1 way
The rank of the word IITJEE = 60 + 6 + 3 + 2 +1 = 72.

Question 25.
Prove that \(\frac{(2n)!}{n!}\) = 2ⁿ (1, 3, 5 ……..(2n – 1)).
Answer:

Question 26.
Write the equation of the line passing through the point (1, -1) and parallel to the line x + 3y – 4 = 0
Answer:
Equation of a line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0
It passes through (1,-1)
⇒ (1) + 3(-1) + k = 0
-2 + k = 0 ⇒ k = 2
So the required equation is x + 3y + 2 = 0

Question 27.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Answer:
Area of Δ with vertices (k, 2) (2, 4) and (3, 2) = \(\frac{1}{2}\) \(\left|\begin{array}{lll} k & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 2 & 1 \end{array}\right|\) = 4 (given)
⇒ \(\left|\begin{array}{lll} k & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 2 & 1 \end{array}\right|\) = 2(4) = 8
(i.e.,) k (4 – 2) -2(2 – 3) + 1 (4 – 12) = ±8
(i.e.,) 2k – 2(-1) + 1(-8) = ± 8
(i.e.,) 2k + 2 – 8 = 8
(i.e.,) 2k = 8 + 8 – 2 = 14
k = 14/2 = 7
∴ k = 7
or
2k + 2 – 8 = -8
⇒ 2k = -8 + 8 – 2
2k = -2
k = -1
So k = 7 (or) k = -1

Question 28.
Find λ, when the projection of \(\vec {a}\) = λ\(\hat{j}\) + \(\hat{j}\) + 4\(\hat{k}\) on \(\vec {b}\) = 2\(\hat{i}\) + 6\(\hat{j}\) + 2\(\hat{k}\) is 4 units.
Answer:
\(\vec {a}\) = λ\(\hat{j}\) + \(\hat{j}\) + 4\(\hat{k}\) and \(\vec {b}\) = 2\(\hat{i}\) + 6\(\hat{j}\) + 2\(\hat{k}\)
Now \(\vec {a}\) – \(\vec {b}\) = (λ) (2) + (1) (6) + (4) (3)
= 2λ + 6 + 12 = 2λ +18
|\(\vec {a}\)| = \(\sqrt{4+36+9}\) = \(\sqrt{49}\) = 7
Here \(\frac{2λ+18}{7}\) = 4
⇒ 2λ+ 18 = 4 × 7 = 28
2λ = 28 – 18 = 10
λ = 10/2 = 5

Question 29.
Find \(\frac{dy}{dx}\) if y = e^x sin x
Answer:
y = e^x sin x
⇒ y’ = uv’ + vu’
Now u = e^x ⇒ u’ = \(\frac{du}{dx}\) = e^x
v = sin x ⇒ v’ = \(\frac{dv}{dx}\) = cos x
y’ = e^x (cos x) + sin x (e^x)
= e^x [sin x + cos x]

Question 30.
Find \(\frac{dy}{dx}\) if x² + y² = 1
Answer:
We differentiate both sides of the equation.
\(\frac{d}{dx}\) (x)² + \(\frac{d}{dx}\) (y)² = \(\frac{d}{dx}\) (1)
2x + 2y \(\frac{dy}{dx}\) = 0
Solving for the derivative yields
\(\frac{dy}{dx}\) = –\(\frac{x}{y}\)

PART – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
From the curve y = sin x, draw y = sin |x|
Answer:

Question 32.
If one root of k (x – 1)² = 5x – 7 is double the other root, show that k = 2 or -25
Answer:
k(x – 1)² = 5x – 7
(i.e.,) k (x² – 2x + 1) – 5x + 7 = 0
x² (k) + x(-2k – 5) + k + 7 = 0
kx² – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α
Sum of the roots = α + 2α = 3α = \(\frac{2k+5}{k}\) α \(\frac{2k+5}{3k}\) ……..(1)
Product of the roots = α(2α) = 2α² = \(\frac{k+7}{k}\)
⇒ α² = \(\frac{k+7}{2k}\) = ……..(2)
Substituting a value from (1) in (2)

2(4k² + 25 + 20k) = 9k (k + 7)
2(4k² + 25 + 20k) = 9k² + 63k
8k² + 50 + 40k – 9k² – 65k = 0
-k² – 25k + 50 = 0
k² + 23k – 50 = 0
(k + 25) (k – 2) = 0
k = -25 or 2

Question 33.
If θ + ∅ = α and tan θ = k tan ∅, then prove that sin(θ – ∅) =\(\frac{k-1}{k+1}\)
Answer:

Question 34.
If a, b, c are in geometric progression, and if a^{\(\frac{1}{x}\)} = b^{\(\frac{1}{y}\)} = c^{\(\frac{1}{z}\)} then prove that x, y, z are in arithmetic progression.
Answer:
Given a, b, c are in G.P.
⇒ b² = ac
⇒ log b² = log ac
(i.e.) 2 log b = log a + log c …(1)
We are given a^{\(\frac{1}{x}\)} = b^{\(\frac{1}{y}\)} = c^{\(\frac{1}{z}\)} = k (say)
⇒ log a^k = \(\frac{1}{x}\) = b^k \(\frac{1}{y}\) = c^k \(\frac{1}{z}\)
⇒ a^k = \(\frac{1}{x}\) ⇒ x = log k^a
Similarly y = log k^b
z = log k^c
Substituting these values in equation (1) we get 2y = x + z ⇒ x, y, z are in A.P.

Question 35.
Show that the points (1,3), (2,1) and (\(\frac{1}{2}\), 4) are collinear.
Answer:
Let the given points be A (1, 3), B (2, 1), and C (\(\frac{1}{2}\), 4)
Slope of AB = \(\frac{1-3}{2-1}\) = \(\frac{-2}{1}\) = -2 = m₁
Slope of BC = \(\frac{4-1}{1/2-1}\) = \(\frac{3}{-3/2}\) = -2 = m₂
Slope of AB = Slope of BC ⇒ AB parallel to BC but B is a common point.
⇒ The points A, B, C are collinear.

Question 36.
If A and B are symmetric matrices of same order, prove that AB + BA is a symmetric matrix.
Answer:
Given A and B are symmetric matrices
⇒ A^T = A and B^T = B
To prove AB + BA is a symmetric matrix.
Proof: Now(AB + BA)^T = (AB)^T + (BA)^T = B^TA^T + A^TB^T
= BA + AB = AB + BA
i.e. (AB + BA)^T = AB + BA
⇒ (AB + BA) is a symmetric matrix.

Question 37.
Let \(\vec {a}\), \(\vec {b}\), \(\vec {c}\) be three vectors such that |\(\vec {a}\)| = 3, |\(\vec {b}\)| = 4, |\(\vec {c}\)| = 5 and each one of them being perpendicular to the sum of the other two, find |\(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\)|.
Answer:
Given |\(\vec {a}\)| = 3; |\(\vec {b}\)| = 4; |\(\vec {c}\)| = 5
Now, (\(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\))² = a^-2 +b ^-2+c^-2 + 2(\(\vec {a}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {a}\).\(\vec {c}\))
= 3² + 4² + 5² + 2 \(\vec {a}\).\(\vec {b}\) + 2\(\vec {b}\).\(\vec {c}\) + 2\(\vec {a}\).\(\vec {c}\)
= 9 + 16 + 25 + \(\vec {a}\).\(\vec {b}\) + \(\vec {a}\).\(\vec {b}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {b}\).\(\vec {c}\) + \(\vec {a}\).\(\vec {c}\) + \(\vec {a}\).\(\vec {c}\)
= 50 + \(\vec {a}\).(\(\vec {b}\) + \(\vec {c}\)) + \(\vec {b}\).(\(\vec {c}\) + \(\vec {a}\)) + \(\vec {c}\).(\(\vec {a}\) + \(\vec {b}\))
= 50 + 0 + 0 + 0 = 50
(one vector is ⊥^r to the sum of other two vectors)
|\(\vec {a}\) + \(\vec {b}\) + \(\vec {c}\)| = \(\sqrt {50}\) = \(\sqrt {25×2}\) = 5√2

Question 38.
Evaluate \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x^{2}}-1}{x}\)
Answer:

Question 39.
Find the derivatives of the following functions y = x^{cos x}
Answer:
y = x^{cos x}
Taking log on both sides, we get
log y = log x^{cos x} = cos x log x
Differentiating w.r.to x we get

Question 40.
Evaluate

Answer:

PART – IV

IV. Answer all the questions. [7 x 5 = 35]

Question 41 (a).
Let f, g: R → R be defined as f(x) = 2x – |x| and g(x) = 2x + |x|. Find fog.
Answer:

fog (x) = f(g(x))
= f(x) = 3x
For x > 0
fog (x) = f(g(x))
= f(3x) = 3x

[OR]

(b) Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}\) = 1
Answer:
Let θ → 0, thought positive values
∴ Let 0 < θ < \(\frac{π}{2}\)
Draw a circule center O and radius unity and let
∠AOB = θ radiAnswer:
Let the tagent at A meet OB product in the piont P.
Jion AB. From the figure it is clear that
Area of ΔAOB < Area of sector AOB < Area of ΔOAP

i.e., \(\frac{1}{2}\) OA. OB sin θ < 1 (radius)² θ < \(\frac{1}{2}\) OA. AP
or sin θ < θ < tan θ
[ ∵ \(\frac{AP}{OA}\) = tan θ or AP = tan θ as OA = 1
Dividing by sin θ, which is positive

Question 42 (a).
(i) If A × A has 16 elements, S = {(a, b) ∈ A × A: a < b} ; (-1, 2) and (0,1) are two elements of S, then find the remaining elements of S.
(ii) Find the range of the function \(\frac{1}{2 cos x – 1}\)
Answer:
(i) n(A × A) = 16
⇒ n(A) = 4
S = {(-1, 0), (-1, 1), (0, 2), (1, 2)}

(ii) The range of cos x is – 1 to 1
-1 ≤ cos x ≤ 1
(× by 2) -2 ≤ 2 cos x ≤ 2
adding -1 throughout
-2 -1 ≤ 2 cos x – 1 ≤ 2 – 1
(i.e.,) -3 ≤ 2 cos x – 1 ≤ 1
so 1 ≤ \(\frac{1}{2 cos x – 1}\) ≤ \(\frac{-1}{3}\)
The range is outside \(\frac{-1}{3}\) and 1
i.e., range is (-∞,\(\frac{-1}{3}\)] ∪ [1, ∞)

[OR]

(b) Evaluate: \(\int \frac{3 x+5}{x^{2}+4 x+7}\) dx
Answer:
Let I = \(\int \frac{3 x+5}{x^{2}+4 x+7}\) dx
3x + 5 = A \(\frac{d}{dx}\) (x² + 4x + 7) + B
3x + 5 = A(2x + 4) + B
Comparing the coefficients of like terms we got
2A = 3 ⇒ A = \(\frac{3}{2}\); 4A + B = 5 ⇒ B = -1

Question 43 (a).
Derive cosine formula using the law of sines in a ΔABC.
Answer:

[OR]

(b) Find \(\sqrt[3]{65}\) using binomial expansion upto two decimal places.
Answer:
We know that for |x| < 1,

≈ 4 + 0.02 ( Since \(\frac{1}{36864}\) + …….. is very small)
\(\sqrt[3]{65}\) = 4.02 (approximately)

Question 44 (a).
(i) Do the limit of the function \(\frac{sin |x|}{x}\) Li exist as x → 0? State reason for the answer.
Answer:

Hence the limit does ndt exist. Since that f(0^–) ≠ f(0⁺)

[OR]

(b)
Answer:

Question 45.
(i) Evaluate \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\) dx
Answer:
Let sin^-1 x = t
⇒ \(\frac{1}{\sqrt{1-x^{2}}}\) dx = dt
Now sin^-1 x = t ⇒ x = sin t
So I = ∫(sin t) (t) dt = ∫ t sin t dt
Now ∫t sin t dt = ∫ t d (-cos t)
= t(- cos t) – ∫(-cos t) dt
= -t cos t + ∫cos t dt
= – t cos t + sin t

[OR]

(b) Show that the points whose position vectors 4\(\hat{i}\) + 5\(\hat{j}\) + \(\hat{k}\),- \(\hat{j}\) – \(\hat{k}\), 3\(\hat{i}\) +9\(\hat{j}\) + 4\(\hat{k}\) and -4\(\hat{i}\) + 4\(\hat{j}\) + 4\(\hat{k}\) are coplanar.
Answer:
Let the given points be A, B, C and D. To prove that the point A, B, C, D are coplanar, we have to prove that the vectors \(\overrightarrow{\mathrm{AB}}\), \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AD}}\) are coplanar.

Equating \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) components we get
-4 = -m – 8n
⇒ m + 8n = 4 ….(i)
-6 = 4m – n
⇒ 4m – n = -6 ……. (ii)
-2 = 3m + 3n
⇒ 3m + 3n = -2 …….. (iii)
Solving (i) and (ii)

Substituting m = –\(\frac{4}{3}\) in (i) we get,
–\(\frac{4}{3}\) + 8n = 4

we are able to write one vector as a linear combination of the other two vectors ⇒ the given vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
(i.e.,) The given points A, B, C, D are coplanar.

Question 46 (a).
If Q is a point on the locus of x² + y² + 4x – 3x + 7 = 0, then find the equation of locus of P which divides segment OQ externally in the ratio 3 : 4, where O is the origin.
Answer:
Let (h, k) be the moving-point O = (0, 0), Let PQ = (a, b) on x² + y² + 4x – 3y + 7 = 0
P divides OQ externally in the ratio 3 : 4

h² + k² – 12h + 9k + 63 = 0, Locus of (h, k) is x² + y² – 12x + 9y + 63 = 0

[OR]

(b) A consulting firm rents car from three agencies such that 50% from agency L, 30% from agency M and 20% from agency N. If 90% of the cars from L, 70% of cars from M and 60% of cars from N are in good conditions (i) what is the probability that the firm will get a car in good condition? (ii) if a car is in good condition, what is probability that it has come from agency N?
Answer:
Let A₁, A₂ and A₃ be the events that the cars are rented from the agencies X, Y and Z respectively.
Let G be the event of getting a car in good condition.
We have to find

(i) The total probability of event G that is, P(G).
(ii) Find the conditional probability A3, given G that is, P(A₃/G) We have
P(A₁) = 0.50, P(G/A₁) = 0.90
P(A₂) = 0.30, P(G/A₂) = 0.70
P(A₃) = 0.20, P(G/A₃) = 0.60.

(i) Since A₁, A₂ and A₃ are mutually exclusive and exhaustive events and G is an event in S, then the total probability of event G is P(G).
P(G) = P(A₁) P(G/A₁) + P(A₂) P(G/A₂) + P(A₃) P(G/A₃)
P(G) = (0.50) (0.90) + (0.30) (0.70) + (0.20) (0.60)
P(G) = 0.78.

(ii) The conditional probability A₃ given G is P(A₃/G)
By Bayes’ theorem,

Question 47 (a).
Using the Mathematical induction, show that for any natural number n,

Answer:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by principle of mathematical induction,
P(n) is true for all n ∈ Z.

(b) If 7 = (cos^-1 x)², prove that (1 – x²) \(\frac{d²y}{dx²}\) -x\(\frac{dx}{dy}\) -2 = 0. Hence find y₂, when x = 0.
Answer:

Squaring on both sides (1 – x²) (\(y_{1}^{2}\)) = 4(cos^-1 x)² = 4y
⇒ (1 – x²)(\(y_{1}^{2}\)) = 4y
Differentiating again w.r.to x we get
(1 – x²) (2y₁, y₂) + (\(y_{1}^{2}\)) (-2x) = 4y₁
⇒ (1 – x²) (2y₁, y₂) = 4y₁ + 2xy\(y_{1}^{2}\)
(i.e.,) (1 – x²) (2y₁, y₂) = 2y₁ (2 +xy₁)
(÷ by 2y₁)(1 – x²)y² = 2 + xy₁
So (1 – x²)y₂ – xy₁ – 2 = 0
When x = 0
(1 – 0) y₂ – 0y₁ – 2 = 0
y₂ – 2 = 0
y₂ = 2

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Additional Questions

Question 1.
Let A = {1, 2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3,10), (4, 9)} ⊂ A × B be a relation. Show that R is a function and find its domain, co-domain and the range of R.
Answer:
Domain of R = {1, 2, 3, 4}
Co-domain of R = B = {-1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12}
Range of R= {3, 6, 10, 9}

Question 2.
Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f: A → B be a function given by f(x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow and (iv) a graph.
Solution:
A = {0, 1, 2, 3}, B = {1, 3, 5, 7, 9}
f(x) = 2x + 1
f(0) = 2(0) + 1 = 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

(iii) An arrow diagram

Question 3.
State whether the graph represent a function. Use vertical line test.

Solution:
It is not a function as the vertical line PQ cuts the graph at two points.

Question 4.
Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3)} be a function from A = {-1, 0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Is this (i) an one-one function (ii) an onto function, (iii) both one- one and onto function?
Solution:
It is both one-one and onto function.

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Question 5.
A function f: (-7,6) → R is defined as follows.

Find (i) 2f(-4) + 3 f(2)
(ii) f(-7) – f(-3)
Solution:

(i) 2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 × 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

(ii) f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 – 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) – f(-3) = 36 – 2 = 34

Question 6.
If A = {2,3, 5} and B = {1, 4} then find
(i) A × B
(ii) B × A
Answer:
A = {2, 3, 5}
B = {1, 4}

(i) A × B = {2,3,5} × {1,4}
= {(2, 1) (2, 4) (3, 1) (3, 4) (5,1) (5, 4)}.

(ii) B × A = {1,4} × {2,3,5}
= {(1,2) (1,3) (1,5) (4, 2) (4, 3) (4, 5)}

Question 7.
Let A = {5, 6, 7, 8};
B = {- 11, 4, 7, -10, -7, – 9, -13} and
f = {(x,y): y = 3 – 2x, x ∈ A, y ∈ B}.
(i) Write down the elements of f.
(ii) What is the co-domain?
(iii) What is the range?
(iv) Identify the type of function.
Answer:
Given, A = {5, 6, 7, 8},
B = {- 11,4, 7,-10,-7,-9,-13}
y = 3 – 2x
ie; f(x) = 3 – 2x
f(5) = 3 – 2 (5) = 3 – 10 = – 7
f(6) = 3 – 2 (6) = 3 – 12 = – 9
f(7) = 3 – 2(7) = 3 – 14 = – 11
f(8) = 3 – 2 (8) = 3 – 16 = – 13
(i) f = {(5, – 7), (6, – 9), (7, – 11), (8, – 13)}
(ii) Co-domain (B)
= {-11,4, 7,-10,-7,-9,-13} i
(iii) Range = {-7, – 9, -11,-13}
(iv) It is one-one function.

Question 8.
A function f: [1, 6] → R is defined as follows:

Find the value of (i) f(5)
(ii) f(3)
(iii) f(2) – f(4).
Solution:

(i) f(5) = 3x² – 10
= 3 (5²) – 10 = 75 – 10 = 65
(ii) f(3) = 2x – 1
= 2(3) – 1 = 6 – 1 = 5
(ii) f(2) – f(4)
f(2) = 2x – 1
= 2(2) – 1 = 3
f(4) = 3x² – 10
= 3(4²) – 10 = 38
∴ f(2) – f(4) = 3 – 38 = 35

Question 9.
The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11}, where f(x) = 2x – 1. Find the values of a and b.
Solution:

A = {5, 6, 8, 10}, B = {19, 15, 9, 11}
f(x) = 2x – 1
f(5) = 2(5) – 1 = 9
f(8) = 2(8) – 1 = 15
∴ a = 9, b = 15

Question 10.
If R = {(a, -2), (-5, 6), (8, c), (d, -1)} represents the identity function, find the values of a,b,c and d.
Solution:
R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.6

Question 1.
If n(A × B) = 6 and A = {1, 3} then n(B) is
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3
Hint:
If n(A × B) = 6
A = {1, 1}, n(A) = 2
n(B) = 3

Question 2.
A = {a, b,p}, B = {2, 3}, C = {p, q, r, s)
then n[(A ∪ C) × B] is ………….
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12
Hint: A ∪ C = [a, b, p] ∪ [p, q, r, s]
= [a, b, p, q, r, s]
n (A ∪ C) = 6
n(B) = 2
∴ n [(A ∪ C)] × B] = 6 × 2 = 12

Question 3.
If A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} then state which of the following statement is true.
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Answer:
(1) (A × C) ⊂ (B × D)]
Hint:
A = {1, 2}, B = {1, 2, 3, 4},
C = {5, 6}, D ={5, 6, 7, 8}
A × C ={(1,5), (1,6), (2, 5), (2, 6)}
B × D = {(1, 5),(1, 6),(1, 7),(1, 8),(2, 5),(2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)}
∴ (A × C) ⊂ B × D it is true

Question 4.
If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is ………………….
(1) 3
(2) 2
(3) 4
(4) 8
Answer:
(2) 2
Hint: n(A) = 5
n(A × B) = 10
(consider 1024 as 10)
n(A) × n(B) = 10
5 × n(B) = 10
n(B) = \(\frac { 10 }{ 5 } \) = 2
n(B) = 2

Question 5.
The range of the relation R = {(x, x²)|x is a prime number less than 13} is
(1) {2, 3, 5, 7}
(2) {2, 3, 5, 7, 11}
(3) {4, 9, 25, 49, 121}
(4) {1, 4, 9, 25, 49, 121}
Answer:
(3) {4, 9, 25, 49, 121}]
Hint:
R = {(x, x²)/x is a prime number < 13}
The squares of 2, 3, 5, 7, 11 are
{4, 9, 25, 49, 121}

Question 6.
If the ordered pairs (a + 2,4) and (5, 2a + 6) are equal then (a, b) is ………
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
Answer:
(4) (3, -2)
Hint:

The value of a = 3 and b = -2

Question 7.
Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is
(1) mⁿ
(2) n^m
(3) 2^mn – 1
(4) 2^mn
Answer:
(4) 2^mn
Hint:
n(A) = m, n(B) = n
n(A × B) = 2^mn

Question 8.
If {(a, 8),(6, b)} represents an identity function, then the value of a and 6 are respectively
(1) (8,6)
(2) (8,8)
(3) (6,8)
(4) (6,6)
Answer:
(1) (8,6)
Hint: f = {{a, 8) (6, 6)}. In an identity function each one is the image of it self.
∴ a = 8, b = 6

Question 9.
Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}. A function f : A → B given by f = {(1, 4),(2, 8),(3, 9),(4, 10)} is a
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
Answer:
(3) One-to one function
Hint:
A = {1, 2, 3, 4), B = {4, 8, 9,10}

Question 10.
If f(x) = 2x² and g (x) = \(\frac{1}{3 x}\), Then fog is

Answer:
(3) \(\frac{2}{9 x^{2}}\)
Hint:
f(x) = 2x²
g(x) = \(\frac{1}{3 x}\)
fog = f(g(x)) = \(f\left(\frac{1}{3 x}\right)=2\left(\frac{1}{3 x}\right)^{2}\)
= 2 × \(\frac{1}{9 x^{2}}=\frac{2}{9 x^{2}}\)

Queston 11.
If f: A → B is a bijective function and if n(B) = 7 , then n(A) is equal to …………..
(1) 7
(2) 49
(3) 1
(4) 14
Answer:
(1) 7
Hint:
n(B) = 7
Since it is a bijective function, the function is one – one and also it is onto.
n(A) = n(B)
∴ n(A) = 7

Question 12.
Let f and g be two functions given by f = {(0, 1), (2, 0), (3, -4), (4, 2), (5, 7)} g = {(0, 2), (1, 0), (2, 4), (-4, 2), (7, 0)} then the range of fog is
(1) {0, 2, 3, 4, 5}
(2) {-4, 1, 0, 2, 7}
(3) {1, 2, 3, 4, 5}
(4) {0, 1, 2}
Answer:
(4) {0, 1, 2}
Hint:
gof = g(f(x))
fog = f(g(x))
= {(0, 2),(1, 0),(2, 4),(-4, 2),(7, 0)}
Range of fog = {0, 1, 2}

Question 13.
Let f (x) = \(\sqrt{1+x^{2}}\) then ………………..
(1) f(xy) = f(x) f(y)
(2) f(xy) > f(x).f(y)
(3) f(xy) < f(x). f(y)
(4) None of these
Answer:
(3) f(xy) < f(x) . f(y)

Question 14.
If g = {(1, 1),(2, 3),(3, 5),(4, 7)} is a function given by g(x) = αx + β then the values of α and β are
(1) (-1, 2)
(2) (2, -1)
(3) (-1, -2)
(4) (1, 2)
Answer:
(2) (2,-1)
Hint:
g(x) = αx + β
α = 2
β = -1
g(x) = 2x – 1
g(1) = 2(1) – 1 = 1
g(2) = 2(2) – 1 = 3
g(3) = 2(3) – 1 = 5
g(4) = 2(4) – 1 = 7

Question 15.
f(x) = (x + 1)³ – (x – 1)³ represents a function which is …………….
(1) linear
(2) cubic
(3) reciprocal
(4) quadratic
Answer:
(4) quadratic
Hint: f(x) = (x + 1)³ – (x – 1)³
[using a³ – b³ = (a – b)³ + 3 ab (a – b)]
= (x + 1 – x + 1)³ + 3(x + 1) (x – 1)
(x + 1 – x + 1)
= 8 + 3 (x² – 1)²
= 8 + 6 (x² – 1)
= 8 + 6x² – 6
= 6x² + 2
It is quadratic polynomial

Students can Download Tamil Nadu 11th Maths Previous Year Question Paper March 2019 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Maths Previous Year Question Paper March 2019 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
The value of x, for which the matrix A = \(\left[\begin{array}{cc} e^{x-2} & e^{7+x} \\ e^{2+x} & e^{2 x+3} \end{array}\right]\) is singular, is…………..
(a) 7
(b) 6
(c) 9
(d) 8
Answer:
(d) 8

Question 2.
The nth term of the sequence 2, 7, 14, 23 …. is …………..
(a) n² + 2n + 1
(b) n² + 2n – 1
(c) n² – 2n – 1
(d) n² – 2n + 1
Answer:
(b) n² + 2n – 1

Question 3.
\(\int \frac{\sec x}{\sqrt{\cos 2 x}} d x=\)
(a) tan^-1 (tan x) + c
(b) sin^-1 (tan x) + c
(c) tan^-1 (sin x) + c
(d) 2 sin^-1 (tan x) + c
Answer:
(b) sin^-1 (tan x) + c

Question 4.
The line \(\frac{x}{a}\) – \(\frac{y}{b}\) = 0 has the slope 1, if ………..
(a) a = b
(b) only for a = 1, b = 1
(c) a > b
(d) a < b
Answer:
(a) a = b

Question 5.
The number of five digit numbers in which all digits are even, is
(a) 4 × 5⁴
(b) 4 × 5⁵
(c) 5⁵
(d) 5 × 5
Answer:
(a) 4 × 5⁴

Question 6.

(a) f (x) is continuous for all x in R
(b) f(x) is differentiable for all x > a
(c) f(x) is not differentiable at x = a
(d) f (x) is discontinuous at x = a
Answer:
(c) f(x) is not differentiable at x = a

Question 7.
A number is selected from the set {1, 2, 3, 20}. The probability that the selected number is divisible by 3 or 4 is………..
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{1}{8}\)
Answer:
(a) \(\frac{1}{2}\)

Question 8.
Which of the following is not a periodic function with period 2π?
(a) tan x
(b) cos x
(c) sin x
(d) cosec x
Answer:
(a) tan x

Question 9.
The straight line joining the points (2, 3) and (-1, 4) passes through (α, β) if………..
(a) α + 3β = 11
(b) 3α + β = 11
(c) α + 2β = 7
(d) 3α + β = 9
Answer:
(a) α + 3β = 11

Question 10.
The minimum and the maximum values of |cos x| -2 are respectively………..
(a) 0 and 2
(b) -2 and 0
(c) -2 and -1
(d) -1 and 1
Answer:
(c) -2 and -1

Question 11.
If A = {(x, y) / y – e^x, x ∈ [0, ∞)} and B = {(x, y) / y = sin x, x ∈ [0, ∞)} then n(A ∩ B) is………..
(a) ∞
(b) 1
(c) φ
(d) 0
Answer:
(d) 0

Question 12.

(a) \(\underset { x\rightarrow { 2 }^{ – } }{ lim } \) f(x) = -1
(b) \(\underset { x\rightarrow { 0 } }{ lim } \) f(x) does not exist
(c) \(\underset { x\rightarrow { 0 }^{ – } }{ lim } \) f(x) = -1
(d) \(\underset { x\rightarrow { 0 }^{ + } }{ lim } \) f(x) = 1
Answer:
(b) \(\underset { x\rightarrow { 0 } }{ lim } \) f(x) does not exist

Question 13.
If f(x) = x² – 3x, then the points at which f(x) =f'(x) are…………
(a) both irrational
(b) one rational and another irrational
(c) both positive integers
(d) both negative integers
Answer:
(a) both irrational

Question 14.
The unit vector parallel to the resultant of the vectors \(\hat { i }\) + \(\hat { j }\) – \(\hat { k }\) and \(\hat { i }\) – 2\(\hat { j }\) + \(\hat { k }\) is …………
(a) \(\frac{2 \hat{i}-\hat{j}+\hat{k}}{\sqrt{5}}\)
(b) \(\frac{2 \hat{i}-\hat{j}}{\sqrt{5}}\)
(c) \(\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{5}}\)
(d) \(\frac{2 \hat{i}+\hat{j}}{\sqrt{5}}\)
Answer:
(d) \(\frac{2 \hat{i}+\hat{j}}{\sqrt{5}}\)

Question 15.
It is given that the events A and B are such that P(A) = \(\frac{1}{4}\), P(A/B) = \(\frac{1}{2}\) and P(B/A) = \(\frac{2}{3}\). The P(B) is………..
(a) \(\frac{2}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{3}\)
Answer:
(d) \(\frac{1}{3}\)

Question 16.
If \(\vec { a }, \vec { b }\) are the position vectors of A and B, then which one of the following points whose position vector lies on AB?
(a) \(\frac{2 \vec{a}+\vec{b}}{2}\)
(b) \(\frac{\vec{a}-\vec{b}}{3}\)
(c) \(\vec { a } + \vec { b }\)
(d) \(\frac{2 \vec{a}-\vec{b}}{2}\)
Answer:
(a) \(\frac{2 \vec{a}+\vec{b}}{2}\)

Question 17.
If |x + 2| ≤ 8, then x belongs to…………
(a) (6, 10)
(b) (-10, 6)
(c) [6, 10]
(d) [-10, 6]
Answer:
(d) [-10, 6]

Question 18.
The expansion of (1 – x)^-2 is ………..
(a) 1 – x + x² – ……
(b) 1 + x + x² + ……
(c) 1 – 2x + 3x² – ….
(d) 1 + 2x + 3x² + ……
Answer:
(d) 1 + 2x + 3x² + ……

Question 19.
If f : R → R is defined by f(x) = |x| – 5, then the range of f is………..
(a) (-∞, -5)
(b) (-∞, 5)
(c) [-5, ∞)
(d) (-5, ∞)
Answer:
(c) [-5, ∞)

Question 20.
Which one of the following is not true about the matrix
(a) an upper triangular matrix
(b) lower triangular matrix
(c) a scalar matrix
(d) a diagonal matrix
Answer:
(c) a scalar matrix

PART-II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Write the use of horizontal line test.
Answer:
Horizontal line test is used to check whether a function is one-one, onto or not.

Question 22.
Write the relationship between Permutation and Combination.
Answer:
Permutation means selection followed by arrangement.
Combination means selection only now selcting Y from ‘n’ things can be done in ⁿC_r ways ….(1) (r ≤ n)
Selecting and arranging r from n things can be done in ⁿP_r ways ……(2)
‘r’ things can be arranged in r! ways …… (3)
Now (2) = (1) × (3)
(i.e.,) ⁿP_r = ⁿC_r × r!
⇒ ⁿC_r = \(\frac{^{n}P_r}{r!}\)

Question 23.
Count the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digits are repeated.
Answer:
The required numbers is = 1 × 8 × 7 × 2
= 112

Question 24.
Find the separate equations from a combined equation of a straight line 2x² + xy- 3y² = 0
Answer:
The combined equation is 2x² +xy – 3y² = 0

Factorizing (2) (-3) = -6
2x² + xy – 3y² = 2x² – 2xy + 3xy – 3y²
= 2x(x – y) + by (x – y) = (2x + by) (x – y)
∴ 2x² + xy – by² = 0 ⇒ (2x + 3y) (x – y) = 0
So the separate equations are 2x + 3y = 0; x – y = 0

Question 25.
Define diagonal and scalar matrices.
Answer:
A square matrix A = [a_ij]_{n × n} is called a diagonal matrix
If a_ij = 0 whenever i ≠ j
A square matrix A = [a_ij]_{n × n} is called a scalar matrix

Question 26.
Find a unit vector along the direction of the vector \(5 \hat{i}-3 \hat{j}+4 \hat{k}\)
Answer:
Let \(\hat{a}\) = \(5 \hat{i}-3 \hat{j}+4 \hat{k}\)

Question 27.
Define a continuous function on the closed interval [a, b]
Answer:
A function f : [a, b] → R is said to be continuous on the closed interval [a, b] if it continuous on the open interval (a, b) and
\(\underset { x\rightarrow { a }^{ + } }{ lim } \) f(x) = f(a) and \(\underset { x\rightarrow { b }^{ – } }{ lim } \) f(x) = 1 f(x) = f(b)

Question 28.
Consider the function f(x) = √x, x ≥ 0. Does \(\underset { x\rightarrow { 0 }}{ lim } \) f(x) exist?
Answer:
\(\underset { x\rightarrow { 0 }^{ – } }{ lim } \) √x is does not exist
\(\underset { x\rightarrow { 0 }^{ + } }{ lim } \) √x = 0
\(\underset { x\rightarrow { 0 } }{ lim } \) √x is does not exist

Question 29.
An integer is chosen at random from the first ten positive integers. Find the probability that it is multiple of three.
Answer:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {3, 6, 9}
n(S) = 10; n(A) = 3
P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{10}\)

Question 30.
Is it correct to say A × A = {(a, a): a ∈ A}? Justify your answer.
Answer:
It is not correct
Since A × A = {(a, b); a, b ∈ A} is true

PART-III

Answer any seven questions. Question no. 40 is compulsory [7 × 3]

Question 31.
A football player can kick a football from ground level with an initial velocity (u) of 80 ft/second. Find the maximum horizontal distance the football travels and at what angle
Take R = \(\frac{u² sin 2α}{g}\), and g = 32
Answer:

Maximum distance = 200ft
The required angle is \(\frac{π}{4}\)

Question 32.
Find the coefficient of x3 in the expansion of (2 – 3x)⁷.
Answer:
The general term is (2 – 3x)⁷ is t_{r + 1} = ⁷C_r (2)^{7 – r} (-3x)^r
= ⁷C_r (2)^{7 – r} (-3x)^r (x)^r
So the co-efficient of x^r is ⁷C_r (2)^{7 – r} (-3)^r
To find co-efficient of x³
equate r to 3 (i.e.,) r = 3
∴ co-efficient of x³ is ⁷C_r (2)^{7 – 3} (-3)³
= \(\frac{7×6×5}{3×2×1}\) (2)₄ (-27)
= 7 × 5 × (16) (-27)
= -15120

Question 33.
Find the nearest point on the line x – 2y – 5 from the origin.
Answer:

x = 1, y = -2
Nearest point is (x₁, y₁) = (1, -2)

Question 34.
Prove that square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix.
Answer:
Let A be a square matrix. Then A = \(\frac{1}{2}\)(A + A^T) + \(\frac{1}{2}\)(A – A^T)
We know that, (A + A^T) is a symmetric, (A – A^T) is a skew-symmetric
A can be written as sum of a symmetric skew symmetric matrices.

Question 35.
If \(\vec { a }\) , \(\vec { b }\), \(\vec { c }\) are three vectors such that \(\vec { a }\) + 2\(\vec { b }\) + \(\vec { c }\) = 0, and |\(\vec { a }\)|=3, |\(\vec { b }\)| =4, |\(\vec { c }\)| =7, find the angle between \(\vec { a }\) and \(\vec { b }\).
Answer:
\(\vec { a }\) + 2\(\vec { b }\) + \(\vec { c }\) = 0
∴ \(\vec { a }\) + 2\(\vec { b }\) = \(\vec { c }\)
squaring on both sides
(\(\vec { a }\) + 2\(\vec { b }\))² = c²
a² + 4b² + 4\(\vec { a }\)\(\vec { b }\) = c²
3² + 4(4²) + 4 \(\vec { a }\) \(\vec { b }\) = 7²
9 + 64 + 4\(\vec { a }\) \(\vec { b }\) = 49
4\(\vec { a }\) \(\vec { b }\) = 49 – 73 = -24
\(\vec { a }\) \(\vec { b }\) = –\(\frac {24}{4}\) = -6
|\(\vec { a }\)| |\(\vec { b }\)| cos θ = -6 ⇒ (3) (4) cos θ = -6
⇒ cos θ = –\(\frac {6}{12}\) = –\(\frac {1}{2}\) ⇒ θ = π – \(\frac {π}{3}\) = \(\frac {2π}{3}\)

Question 36.
Examine the continuity of the function cot x + tan x.
Answer:
Let f (x) = cot x + tan x
cot x is continuous in R – nπ
tan x is continuous in R – (n + 1) \(\frac {π}{2}\)
∴ f(x) is continuous in R – \(\frac {nπ}{2}\), (n∈Z)

Question 37.
Differentiate y = sin^-1 (\(\frac {1-x²}{1+x²}\))
Answer:

Question 38.
find \(\frac {dy}{dx}\) if x = a (t – sin t), y = a(1 – cos t)
Answer:
x = a (t – sin t) ⇒ \(\frac {dy}{dx}\) = a(1 – cos t)
y = a(1 – cos t) ⇒ \(\frac {dy}{dt}\) = a(sin t)

Question 39.
Evaluate \(\int { (x+3) } \sqrt { x+2 }\) dx
Answer:

Question 40.
Construct a suitable domain X such that f : X → N defined by/(n) = n + 3 to be one to one and onto.
Answer:
f : X → N is defined by f(x) = n + 3
Since f is one-one and onto function
suitable Domain is (-2, 1,0} ∪ N (or) {-2, -1, 0, 1, 2}

PART- IV

IV. Answer all the questions. [7 x 5 = 35]

Question 41 (a).
For the given base curve y = sin x, draw y = \(\frac {1}{2}\) sin 2x
Answer:

[OR]

(b) Write any five different forms of an equation of a straight line.
Answer:
(i) Slope intercept form : y = mx + c, c ≠ 0
(ii) Point and slope form : y – y₁ = m (x – x₁)
(iii) Two Point form : \(\frac {y – y_1}{y_{2}-y_1}\) \(\frac {x – x_1}{x_{2}-x_1}\)
(iv) Intercept form : \(\frac {x}{a}\) + \(\frac {y}{b}\) = 1
(v) Normal form : x cos α + y sin α

Question 42 (a).
Solve the equation \(\sqrt {6-4x-x²}\) = x + 4
Answer:
Squaring both sides
6 – 4x – x² = (x + 4)²
(i.e.,) x² + 6x + 5 = 0 ⇒ (x + 1) (x + 5) = 0
⇒ x = -1, x = -5
But x ≥ – 4
⇒ x = -1

[OR]

(b) Prove that in any ΔABC, Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\), where s is the semi-perimeter of ΔABC.
Answer:
Δ = \(\frac {1}{2}\) ab sin C = \(\frac {1}{2}\) ab (2sin \(\frac {C}{2}\) cos \(\frac {C}{2}\))

Question 43 (a).
State and prove any one of the Napier’s formulae.
Answer:

[OR]

(b) Do the limit of the function \(\frac {sin(x-\left\lfloor x \right\rfloor)}{x – \left\lfloor x \right\rfloor}\) exist as x → 0? State the reasons for your answer.
Answer:

So, the limit does not exist.

Question 44. (a)
Prove that for any natural number n, aⁿ – bⁿ is divisible by a – b, where a > b.
Answer:
a = a – b + b
So, aⁿ =[a- b + b]ⁿ = [(a -b) + b]ⁿ
= ⁿC₀ (a – b)ⁿ + ⁿC₁ (a – b)^n-1b¹ + ⁿC₂ (a – b)^n-2b² + ……. + ⁿC_n-1, (a – b) b^n-1+ ⁿC_n(bⁿ)
⇒ aⁿ – bⁿ = (a – b)ⁿ + ⁿC₁ (a – b)^n-1b + ⁿC₂ (a – b)^n-2b²) + …….. + ⁿC_n-1 (a – b) b^n-1
= (a – b) [(a – b)^n-1 + ⁿC₁ (a – b)^n-2b + ⁿC₂ (a – b)^n-3b² + …….. + ⁿC_n-1 b^n-1]
= (a – b) [aⁿ integer]
⇒ aⁿ – bⁿ If is divisible by(a – b).

[OR]

(b) Evaluate: \(\int \frac{2 x+4}{x^{2}+4 x+6}\)
Answer:
Let 2x + 4 = A \(\frac {d}{dx}\) (x² + 4x + 6) + B
(i.e.,) 2x + 4 = A(2x + 4) + B
Equating the coefficient of x and constant term we get
A = 1, B = 0

Question 45 (a).
Prove that \(\sqrt[3]{x^{3}+7}-\sqrt[3]{x^{3}+4}\) is approximately equal to \(\frac {1}{x²}\)when x is large.
Answer:

[OR]

(b) Find the unit vectors perpendicular to each of the vectors \(\vec { a }\) + \(\vec { b }\) and \(\vec { a }\) – \(\vec { b }\) where \(\vec { a }\) = \(\hat { i }\) + \(\hat { j }\) + \(\hat { j }\) and \(\vec { b }\) = \(\hat { i }\) + 2\(\hat { j }\) + 3\(\hat { k }\).
Answer:

Question 46 (a).
Find \(\frac {d²y}{dx}\) if x² + y² = 4
Answer:
we have x² + y² = 4

[OR]

(b) The chances of X, Y and Z becoming managers of a certain company are 4 : 2 : 3. The probabilities that bonus scheme will be introduced if X, Y and Z become managers are 0.3, 0.5 and 0.4 respectively. If the bonus scheme has been introduced, what is the probability that Z was appointed as the manager?
Answer:
Given X : Y : Z = 4 : 2 : 3

Question 47 (a).
Prove that

Answer:

∴ (x – y) is a factor
Similarly (y – z)(z – x) are also factors.
The other factor is k(x² + y² + z²) + l(xy + yz + zx)

Putting x = 0, y = 1, z = 2 ⇒ 5k + 2l = 2 …. (1)
Putting x = 0, y = – 1, z = 1 ⇒ 2k – l = -1 …. (2)
from (1) & (2) k = 0, l = -1

[OR]

(b) Evaluate \(\int \sqrt{x^{2}+x+1}\) dx
Answer:

Students can Download Tamil Nadu 12th Computer Science Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Computer Science Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Computer Science Model Question Paper 5 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

All questions are compulsory. [15 × 1 = 15]

Choose the most appropriate answer from the given four ‘alternatives and write the option code with the corresponding answer.

Question 1.
Bundling two values together into one can be considered as
(a) Pair
(b) Triplet
(c) Single
(d) Quadrat
Answer:
(a) Pair

Question 2.
The kind of scope of the variable ‘a’ used in the pseudo code given below
(i) Disp():
(ii) a: = 7
(iii) print a
(iv) Disp()
(a) Local
(b) Global
(c) Enclosed
(d) Built-in
Answer:
(b) Global

Question 3.
Big Q is the reverse of…………
(a) Big O
(b) Big 0
(c) Big A
(d) Big S
Answer:
(a) Big O

Question 4.
Extension of Python files is………..
(a) .Pyt
(b) .txt
(c) .Pdm
(d) .Py
Answer:
(d) .Py

Question 5.
The output of the Segment
for i in range (10, 0, 2)
print (i)
(a) 10 8 6 420
(b) 10 8 6 4 2
(c) 0 2 4 6 8 10
(d) Error
Answer:
(d) Error

Question 6.
The bin() function returns a binary string prefixed with:
(a) 0
(b) 1
(c) 0b
(d) lb
Answer:
(c) 0b

Question 7.
The positive and negative index value of’P’ in the string Strl = ‘COMPUTER’ are
(a) 3, -4
(b) 4, -4
(c) 3, -5
(d) 4, -5
Answer:
(c) 3, -5

Question 8.
Which of the following set operation includes all the elements that are in two sets but not the one that are common to two sets?
(a) Symmetric difference
(b) Difference
(c) Intersection
(d) Union
Answer:
(a) Symmetric difference

Question 9.
A variable prefixed with double underscore is
(a) private
(b) public
(c) protected
(d) static
Answer:
(a) private

Question 10.
The data model developed by IBM is
(a) hierarchical
(b) relational
(c) network
(d) ER
Answer:
(a) hierarchical

Question 11.
The SQL command to make a database as current active database is
(a) CURRENT
(b) USE
(c) DATABASE
(d) NEW
Answer:
(b) USE

Question 12.
The expansion of CRLF is…………..
(a) Control Return and Line Feed
(b) Carriage Return and Form Feed
(c) Control Router and Line Feed
(d) Carriage Return and Line Feed
Answer:
(d) Carriage Return and Line Feed

Question 13.
The function call statement of the segment………….
if_name_ == ‘_main_’:
main(sys.argv[1:])
is
(a) main(sys.argv[l:])
(b) _name_
(c) _main_
(d) argv
Answer:
(b) _name_

Question 14.
Which is not a SQL clause?
(a) GROUP BY
(b) ORDER BY
(c) HAVING
(d) CONDITION
Answer:
(d) CONDITION

Question 15.
To make a bar chart with Matplotlib, which function should be used?
(a) plt.bar()
(b) plt.chart()
(c) pip.bar()
(d) pip.chart()
Answer:
(a) plt.bar()

PART – II

Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
What do you mean by Namespaces?
Answer:
Namespaces are containers for mapping names of variables to objects.

Question 17.
What is searching? Write its types.
Answer:
A search algorithm is the step-by-step procedure used to locate specific data among a collection of data. Types of searching algorithms are:

1. Linear search
2. Binary search
3. Hash search
4. Binary Tree search

Question 18.
Define Operator and Operand.
Answer:
In computer programming languages operators are special symbols which represent computations, conditional matching etc. The value of an operator used is called operands. Operators are categorized as Arithmetic, Relational, Logical, Assignment etc. Value and variables when used with operator are known as operands.

Question 19.
What are the types of looping supported by Python?
Answer:
Python provides two types of looping constructs:
while loop:
In the ‘while loop’, the condition is any valid Boolean expression returning True or False. The else part of while is optional part of while.

for loop:
‘for loop’ is the most comfortable loop. It is also an entry check loop. The condition is checked in the beginning and the body of the loop(statements-block 1) is executed if it is only True otherwise the loop is not executed.

Question 20.
What is the use of the operator += in python string operation?
Answer:
Adding more strings at the end of an existing string is known as append (+=). The operator += is used to append a new string with an existing string.
Example:
>>> strl = “Welcome to”
>>> strl += “Leam Python”
>>> print (strl)
Welcome to Learn Python

Question 21.
What will be the output of the following snippet?
alpha = list(range(65, 70))
for x in alpha:
print(chr(x), end=’\t’)
Answer:
Output:
A B C D E

Question 22.
What is the use of WHERE clause in SQL?
Answer:
The WHERE clause in the SELECT command specifies the criteria for getting the desired result. The general form of SELECT command with WHERE Clause is:
SELECT[,,….] FROM WHERE condition>;
The relational operators like =, <, <=, >, >=, <> can be used to compare two values in the SELECT command used with WHERE clause. The logical operaors OR, AND and NOT can also be used to connect search conditions in the WHERE clause. For example:
SELECT Admno, Name, Age, Place FROM Student WHERE (Age>=18 AND Place = “Delhi”);

Question 23.
What are the steps involved in file operation of Python?
Answer:
When you want to read from or write to a file ,you need to open it. Once the reading is over it needs to be closed. So that, resources that are tied with the file are freed. Hence, in Python, a file operation takes place in the following order.
Step 1 → Open a file
Step 2 → Perform Read or write operation
Step 3 → Close the file

Question 24.
Distinguish compiler and interpreter.
Answer:

PART – III

Answer any six questions. Question No. 29 is compulsory. [6 x 3 = 18]

Question 25.
Why strlen is called pure function?
Answer:
strlen (s) is called each time and strlen needs to iterate over the whole of ‘s’. If the compiler is smart enough to work out that strlen is a pure function and that ‘s’ is not updated in the loop, then it can remove the redundant extra calls to strlen and make the loop to execute only one time. This function reads external memory but does not change it, and the value returned derives from the external memory accessed.

Question 26.
Which strategy is used for program designing? Define the strategy.
Answer:
We are using here a powerful strategy for designing programs: ‘wishful thinking’.
Wishful Thinking is the formation of beliefs and making decisions according to what might be pleasing to imagine instead of by appealing to reality.

Question 27.
Which jump statement is used as placeholder? Why?
Answer:
pass statement is generally used as a placeholder. When we have a loop or function that is to be implemented in the future and not now, we cannot develop such functions or loops with empty body segment because the interpreter would raise an error. So, to avoid this we can use pass statement to construct a body that does nothing.

Question 28.
What are the points to be noted while defining a function?
Answer:

1. Function blocks begin with the keyword “def ” followed by function name and parenthesis ().
2. Any input parameters or arguments should be placed within these parentheses when you define a function.
3. The code block always comes after a colon (:) and is indented.
4. The statement “return [expression]” exits a function, optionally passing back an expression to the caller. A “return” with no arguments is the same as return None.

Question 29.
Write a Python program to display the given pattern
COMPUTER
COMPUTE
COM PUT
COMPU
COMP
COM
CO
C
Answer:
Program:
strl = “COMPUTER”
index = len(strl)
for i in strl:
print(strl[: index])
index – = 1

Question 30.
What is the output of the following program?
class Greeting:
def_init_(self, name):
self._name = name
def displayf self):
print (“Good Morning”, self._name)
obj = Greeting(‘Tamil Nadu’)
obj.displayO
Answer:
Output: Tamil Nadu Good Morning

Question 31.
Explain Cartesian product with a suitable example.
Answer:
PRODUCT OR CARTESIAN PRODUCT (Symbol: X)
Cross -product is a way of combining two relations. The resulting relation contains, both relations being combined.
A × B means A times B, where the relation A and B have different attributes.
This type of operation is helpful to merge columns from two relations.

Question 32.
Write a short note on (i) fetchallf) (ii) fetchonef) (iii) fetchmany
Answer:
cursor.fetchall() -fetchall ()method is to fetch all rows from the database table.
cursor. fetchoneQ – The fetchone () method returns the next row of a query result set or None in case there is no row left.
cursor, fetchmany() method that returns the next number of rows (n) of the result set

Question 33.
Write a Python code to display the following chart.

Answer:
Import matplotlib.pyplot as pit
x = [1, 2, 3,4, 5, 7,8]
y = [1,2.5, 4, 5, 6, 7, 7]
plt.plot([1, 2, 3, 4])
plt.show()

PART – IV

Answer all the following questions. [5 x 5 = 25]

Question 34 (a).
Write any five benefits in using modular programming.
Answer:
The benefits of using modular programming include

1. Less code to be written.
2. A single procedure can be developed for reuse, eliminating the need to retype the code many times.
3. Programs can be designed more easily because a small team deals with only a small part of the entire code.
4. Modular programming allows many programmers to collaborate on the same application.
5. The code is stored across multiple files.
6. Code is short, simple and easy to understand.
7. Errors can easily be identified, as they are localized to a subroutine or function.
8. The same code can be used in many applications.
9. The scoping of variables can easily be controlled.

(OR)

(b) Explain input() and print() functions of Python with example.
Answer:
Input() function:
In Python, input() function is used to accept data as input at run time. The syntax for input() function is,
Variable = input (“prompt string”)
Where, prompt string in the syntax is a statement or message to the user, to know what input can be given.
If a prompt string is used, it is displayed on the monitor; the user can provide expected data from the input device. The input() takes whatever is typed from the keyboard and stores the entered data in the given variable. If prompt string is not given in input() no message is displayed on the screen, thus, the user will not know what is to be typed as input.
Example 1:
input() with prompt string
>>> city = input (“Enter Your City: ”)
Enter Your City: Madurai
>>> print (“I am from“, city)
I am from Madurai

Example 2:
input() without prompt string
>>> city = input()
Rajarajan
>>> print (I am from”, city)
I am from Rajarajan

The print() function:
In Python, the print() function is used to display result on the screen. The syntax for print() is as follows:
Example:
print (“string to be displayed as output ”)
print (variable)
print (“String to be displayed as output ”, variable)
print (“String 1 ”, variable, “String 2”, variable, “String 3”………. )

Example:
>>> print (“Welcome to Python Programming”)
Welcome to Python Programming
>>> x = 5
>>> y = 6
>>> z = x + y
>>> print (z)
11
>>> print (“The sum = ”, z)
The sum = 11
>>> print (“The sum of ”, x, “ and ”, y, “ is ”, z)
The sum of 5 and 6 is 11

The print () evaluates the expression before printing it on the monitor. The print () displays an entire statement which is specified within print (). Comma (,) is used as a separator in print () to print more than one item.

Question 35 (a).
Write a detail note on for loop in Python.
Answer:
for loop
for loop is the most comfortable loop. It is also an entry check loop. The condition is checked in the beginning and the body of the loop(statements-block 1) is executed if it is only True otherwise the loop is not executed.
Syntax:
for counter_variable in sequence:
statements block 1
# optional block
[else:
statements block 2]
The counter_variable mentioned in the syntax is similar to the control variable that we used in the for loop of C++ and the sequence refers to the initial, final and increment value. Usually in Python, for loop uses the range() function in the sequence to specify the initial, final and increment values. range() generates a list of values starting from start till stop-1.

The syntax of range() is as follows:
range (start, stop, [step])
Where,
start – refers to the initial value
stop – refers to the final value
step – refers to increment value, this is optional part.

Examples for range()
range (1, 30, 1) will start the range of values from 1 and end at 29
range (2, 30, 2) will.start the range of values from 2 and end at 28
range (30, 3, -3) will start the range of values from 30 and end at 6
range (20) will consider this value 20 as the end value(or upper limit) and starts the range count from 0 to 19 (remember always range() will work till stop -1 value only)

#program to illustrate the use of for loop – to print single digit even number
for i in range (2, 10, 2):
print (i, end = ‘ ‘)
Output:
2 4 6 8

[OR]

(b) Explain the different types of functions in Python with example.
Answer:
Types of Functions
Basically, we can divide functions into the following types:

1. User-defined Functions
2. Built-in Functions
3. Lambda Functions
4. Recursion Functions

1. Syntax for User defined function
def < function_name ([parameter 1, parameter2 …… ])> :

return
Example:
def hello():
print (“hello – Python”)
return

Advantages of User-defined Functions:

1. Functions help us to divide a program into modules. This makes the code easier to manage.
2. It implements code reuse. Every time you need to execute a sequence of statements, all you need to do is to call the function.
3. Functions, allows us to change functionality easily, and different programmers can work on different functions.

2. Anonymous Functions
In Python, anonymous function is a function that is defined without a name. While normal functions are defined using the def keyword, in Python anonymous functions are defined using the lambda keyword. Hence, anonymous functions are also called as lambda functions.
The use of lambda or anonymous function:

1. Lambda function is mostly used for creating small and one-time anonymous function.
2. Lambda functions are mainly used in combination with the functions like filter(), map() and reduce().

Syntax of Anonymous Functions:
The syntax for anonymous functions is as follows:
lambda [argument(s)]:expression
Example:
sum = lambda argl, arg2: argl + arg2
print (‘The Sum is :’, sum(30,40))
print (‘The Sum is :’, sum(-30,40))
Output:
The Sum is : 70
The Sum is : 10
The above lambda function that adds argument argl with argument arg2 and stores the result in the variable sum. The result is displayed using the print().

3. Functions using libraries
Built-in and Mathematical functions

4. Recursive functions
When a function calls itself is known as recursion. Recursion works like loop but sometimes it makes more sense to use recursion than loop. You can convert any loop to recursion.
Example:
def fact(n):
if n = = 0:
return 1
else:
return n * fact (n – 1)
print (fact (0))
print (fact (5))
Output:
1
120

Question 36 (a).
Explain about the find() function in Python with example.
Answer:

(OR)

(b) Compare remove(), pop() and clear() function in Python.
Answer:
The remove( ) function can also be used to delete one or more elements if the index value is not known. Apart from remove() function, pop() function can also be used to delete an element using the given index value. pop() function deletes and returns the last element of a list if the index is not given.
The function clear() is used to delete all the elements in list, it deletes only the elements and retains the list. Remember that, the del statement deletes entire list.
Syntax:
List.remove(element)
# to delete a particular element
List.pop(index of an element)
List, clear ()
Example:
>>> MyList = [12, 89, 34,’Kannan’, ‘Gowrisankar’, ‘Lenin’]
>>> print(MyList)
[12, 89, 34, ‘Kannan’, ‘Gowrisankar’, ‘Lenin’]
>>> MyList.remove(89)
>>> print(MyList)
[12, 34, ‘Kannan’, ‘Gowrisankar’, ‘Lenin’]
In the above example, MyList has been created with three integer and three string elements, the following print statement shows all the elements available in the list. In the statement.
>>> MyList.remove(89), deletes the element 89 from the list and the print statement shows the remaining elements.
Example:
>>> MyList.pop(l)
34
>>> print(MyList)
[12, ‘Kannan’, ‘Gowrisankar’, ‘Lenin’]
In the above code, pop() function is used to delete a particular element using its index value, as soon as the element is deleted, the pop() function shows the element which is deleted. pop() function is used to delete only one element from a list. Remember that, del statement deletes multiple elements.
Example:
>>> MyList.clear()
>>> print(MyList)
[]
In the above code, clear() function removes only the elements and retains the list. When you try to print the list which is already cleared, an empty square bracket is displayed without any elements, which means the list is empty.

Question 37 (a).
Explain the components of DBMS.
Answer:
Components of DBMS:

The Database Management System can be divided into five major components as follows:

1. Hardware
2. Software
3. Data
4. Procedures/Methods
5. Database Access Languages

1. Hardware: The computer, hard disk, I/O channels for data, and any other physical component involved in storage of data.
2. Software: This main component is a program that controls everything. The DBMS software is capable of understanding the Database Access Languages and interprets into database commands for execution.
3. Data: It is that resource for which DBMS is designed. DBMS creation is to store and utilize data.
4. Procedures/Methods: They are general instructions to use a database management system such as installation of DBMS, manage databases to take backups, report generation, etc.
5. DataBase Access Languages: They are the languages used to write commands to access, insert, update and delete data stored in any database.

[OR]

(b) What are the components of SOL? Write the commands in each.
Answer:
Components of SQL
SQL commands are divided into five categories:

a. Data Definition Language
The Data Definition Language (DDL) consist of SQL statements used to define the database structure or schema. It simply deals with descriptions of the database schema and is used to create and modify the structure of database objects in databases.
SQL commands which comes under Data Definition Language are:

b. Data Manipulation Language
A Data Manipulation Language (DML) is a computer programming language used for adding (inserting), removing (deleting), and modifying (updating) data in a database.
SQL commands which comes under Data Manipulation Language are :

c. Data Control Language:
A Data Control Language (DCL) is used for controlling privileges in the database SQL commands: GRANT, REVOKE

d. Transactional Control Language;
Transactional control language (TCL) is used to manage transactions i.e. changes made to the data in the database.
SQL commands: COMMIT, ROLLBACK, SAVEPOINT.

e. Data Query Language
The Data Query Language (DQL) have commands to query or retrieve data from the database. SQL commands: SELECT.

Question 38 (a).
Explain the following operators in Relational Algebra with suitable example
1. Union, (∪) 20 Intersection (∩)
UNION (Symbol :∪)
It includes all tuples that are in tables A or in B. It also eliminates duplicates. Set A Union Set B would be expressed as A ∪ B
Example 2
Consider the following tables

INTERSECTION (symbol: ∩) A ∩ B
Defines a relation consisting of a set of all tuple that are in both in A and B. However, A and B must be union-compatible.
Example 5 (using Table B)

[OR]

(b) Draw the output for the following data visualization plot.
import matplotlib.pyplot as pit
plt.bar([1, 3, 5, 7, 9],[5, 2, 7, 8, 2], label=”Example one”)
plt.bar([2, 4, 6, 8, 10],[8, 6, 2, 5, 6], label=”Example two”, color=’g’)
plt.legendO
plt.xlabel(‘bar number’)
plt.ylabel(‘bar height’)
plt.title(‘Epic Graph\nAnother Line! Whoa’)
plt.show()
Answer:

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Additional Questions

Question 1.
Use Euclid’s algorithm to find the HCF of 4052 and 12756.
Solution:
Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get HCF
12576 = 4052 × 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 × 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 × 1 + 148, 148 ≠ 0.
∴ Again by division lemma
272 = 148 × 1 + 124, here 124 ≠ 0.
∴ Again by division lemma
148 = 124 × 1 + 24, Here 24 ≠ 0.
∴ Again by division lemma
124 = 24 × 5 + 4, Here 4 ≠ 0.
∴ Again by division lemma
24 = 4 × 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.

Question 2.
If the HCF of 65 and 117 is in the form (65m – 117) then find the value of m.
Answer:
By Euclid’s algorithm 117 > 65
117 = 65 × 1 + 52
52 = 13 × 4 × 0
65 = 52 × 1 + 13
H.C.F. of 65 and 117 is 13
65m – 117 = 13
65 m = 130
m = \(\frac { 130 }{ 65 } \) = 2
The value of m = 2

Question 3.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Solution:
We have 6 = 2¹ × 3¹ and
20 = 2 × 2 × 5 = 2² × 5¹
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 × 2 × 3 × 5 = 60. As done in your earlier classes. Note that HCF (6, 20) = 2¹ = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 2² × 3¹ × 5¹ = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

Question 4.
Prove that \(\sqrt { 3 }\) is irrational.
Answer:
Let us assume the opposite, (1) \(\sqrt { 3 }\) is irrational.
Hence \(\sqrt { 3 }\) = \(\frac { p }{ q } \)
Where p and q(q ≠ 0) are co-prime (no common factor other than 1)

Hence, 3 divides p²
So 3 divides p also …………….. (1)
Hence we can say
\(\frac { p }{ 3 } \) = c where c is some integer
p = 3c
Now we know that
3q² = p²
Putting = 3c
3q² = (3c)²
3q² = 9c²
q² = \(\frac { 1 }{ 3 } \) × 9c²
q² = 3c²
\(\frac{q^{2}}{3}\) = C²
Hence 3 divides q²
So, 3 divides q also ……………. (2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 }\) is irrational.

Question 5.
Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
(i) 4, 10, 16, 22, …
(ii) 1, -1,-3, -5,…
(iii) -2, 2, -2, 2, -2, …
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3,…
Solution:
(i) 4, 10, 16, 22, …….
We have a₂ – a₁ = 10 – 4 = 6
a₃ – a₂ = 16 – 10 = 6
a₄ – a₃ = 22 – 16 = 6
∴ It is an A.P. with common difference 6.
∴ The next two terms are, 28, 34

(ii) 1, -1, -3, -5
t₂ – t₁ = -1 – 1 = -2
t₃ – t₂ = -3 – (-1) = -2
t₄ – t₃ = -5 – (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

(iii) -2, 2,-2, 2,-2
t₂ – t₁ = 2-(-2) = 4
t₃ – t₂ = -2 -2 = -4
t₄ – t₃ = 2 – (-2) = 4
It is not an A.P.

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3
t₂ – t₁ = 1 – 1 = 0
t₃ – t₂ = 1 – 1 = 0
t₄ – t₃ = 2 – 1 = 1
Here t₂ – t₁ ≠ t₃ – t₂
∴ It is not an A.P.

Question 6.
Find n so that the n^th terms of the following two A.P.’s are the same.
1, 7,13,19,… and 100, 95,90,…
Answer:
The given A.P. is 1, 7, 13, 19,….
a = 1, d = 7 – 1 = 6
t_n1 = a + (n – 1)d
t_n1 = 1 + (n – 1) 6
= 1 + 6n – 6 = 6n – 5 … (1)
The given A.P. is 100, 95, 90,….
a = 100, d = 95 – 100 = – 5
tn₂ = 100 + (n – 1) (-5)
= 100 – 5n + 5
= 105 – 5n …..(2)
Given that, t_n1 = t_n2
6n – 5 = 105 – 5n
6n + 5n = 105 + 5
11 n = 110
n = 10
∴ 10^th term are same for both the A.P’s.

Question 7.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 is the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?
Answer:
The number of rose plants in the 1st, 2nd, 3rd,… rows are
23, 21, 19,………….. 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = -2, l = 5.
As, a_n = a + (n – 1)d i.e. t_n = a + (n – 1)d
We have 5 = 23 + (n – 1)(-2)
i.e. -18 = (n – 1)(-2)
n = 10
∴ There are 10 rows in the flower bed.

Question 8.
Find the sum of the first 30 terms of an A.P. whose n^th term is 3 + 2n.
Answer:
Given,
t_n = 3 + 2n
t₁ = 3 + 2 (1) = 3 + 2 = 5
t₂ = 3 + 2 (2) = 3 + 4 = 7
t₃ = 3 + 2 (3) = 3 + 6 = 9
Here a = 5,d = 7 – 5 = 2, n = 30
S_n = \(\frac { n }{ 2 } \) [2a + (n – 1)d]
S₃₀ = \(\frac { 30 }{ 2 } \) [10 + 29(2)]
= 15 [10 + 58] = 15 × 68 = 1020
∴ Sum of first 30 terms = 1020

Question 9.
How many terms of the AP: 24, 21, 18, . must be taken so that their sum is 78?
Solution:
Here a = 24, d = 21 – 24 = -3, S_n = 78. We need to find n.
We know that,
S_n = \(\frac { n }{ 2 } \) (2a + (n – 1)d)
78 = \(\frac { n}{ 2 } \) (48 + 13(-3))
78 = \(\frac { n}{ 2 } \) (51 – 3n)
or 3n² – 51n + 156 = 0
n² – 17n + 52 = 0
(n – 4) (n – 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

Question 10.
The sum of first n terms of a certain series is given as 3n² – 2n. Show that the series is an arithmetic series.
Solution:
Given, S_n = 3n² – 2n
S₁ = 3 (1)² – 2(1)
= 3 – 2 = 1
ie; t₁ = 1 (∴ S₁ = t₁)
S₂ = 3(2)² – 2(2) = 12 – 4 = 8
ie; t₁ + t₂ = 8 (∴ S₂ = t₁ + t₂)
∴ t₂ = 8 – 1 = 7
S₃ = 3(3)² – 2(3) = 27 – 6 = 21
t₁ + t₂ + t₃ = 21 (∴ S₃ = t₁ + t₂ + t₃)
8 + t₃ = 21 (Substitute t₁ + t₂ = 8)
t₃ = 21 – 8 ⇒ t₃ = 13
∴ The series is 1,7,13, …………. and this series is an A.P. with common difference 6.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Unit Exercise 2

Question 1.
Prove that n² – n divisible by 2 for every positive integer n.
Answer:
To prove n² – n divisible by 2 for every positive integer n.
We know that any positive integer is of the form 2q or 2q + 1, for some integer q.
So, following cases arise:
Case I. When n = 2q.
In this case, we have
n² – n = (2q)² – 2q = 4q² – 2q = 2q(2q – 1)
⇒ n² – n = 2r where r = q(2q – 1)
⇒ n² – n is divisible by 2.

Case II. When n = 2q + 1.
In this case, we have
n² – n = (2q + 1)² – (2q + 1)
= (2q + 1)(2q + 1 – 1) = (2q + 1)2q
⇒ n² – n = 2r where r = q (2q + 1)
⇒ n² – n is divisible by 2.
Hence n² – n is divisible by 2 for every positive integer n.

Question 2.
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following (i) Capacity of a can
(ii) Number of cans of cow’s milk
(iii) Number of cans of buffalow’s milk.
Answer:
Cow’s milk = 175 litres
Buffalow’s milk = 105 litres
Find the H.C.F. of 175 and 105 using Euclid’s division method of factorisation method.

175 = 5 × 5 × 7
105 = 3 × 5 × 7
H.C.F. of 175 and 105 = 5 × 7 = 35
(i) The capacity of the milk can’s is 35 litres

(ii) Cows milk = 175 litres
Number of cans = \(\frac { 175 }{ 35 } \) = 5

(iii) Buffalow’s milk = 105 litres
Number of cans = \(\frac { 105 }{ 35 } \) = 3
(i) Capacity of one can = 35 litres
(ii) Number of can’s for cow’s milk= 5 litres
(iii) Number of can’s for Buffalow’s milk = 3 litres

Question 3.
When the positive integers a, b and c are divided by 13 the respective remainders are 9, 7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Answer:
Let the positive integers be a, b, and c.
a = 13q + 9
b = 13q + 7
c = 13q + 10
a + 2b + 3c = 13 q + 9 + 2(13q + 7) + 3(13q + 10)
= 13q + 9 + 269 + 14 + 39q + 30
= 78q + 53 = (13 × 6)q + 53
The remainder is 53.
But 53 = 13 × 4 + 1
∴ The remainder is 1

Question 4.
Show that 107 is of the form 4q + 3 for any integer q.
Answer:

107 = 4 x 26 + 3
This is in the form of a = bq + r
Hence it is proved.

Question 5.
If (m + 1)^th term of an A.P. is twice the (n + 1)^th term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Solution:
t_n = a + (n – 1)d
t_m+1 = a + (m + 1 – 1)d
= a + md
t_n+1 = a + (n + 1 – 1)d
= a + nd
2(t_n+1) = 2(a + nd)
t_m+1 = 2t_n+1 …………… (1)
⇒ a + md = 2(a + nd)
2a + 2nd – a – md = 0
a + (2n – m)d = 0
t_(3m+1) = a + (3m + 1 – 1)d
= a + 3md
t_(m+n+1) = a + (m + n + 1 – 1)d
= a + (m + n)d
2(t_(m+n+1)) = 2(a + (m + n)d)
= 2a + 2md + 2nd
t_(3m+1) = 2t_(m+n+1) ………….. (2)
a + 3md = 2a + 2md + 2nd
2a + 2md + 2nd – a – 3md = 0
a – md + 2nd = 0
a + (2n – m)d = 0
∴ It is proved that t_(3m+1) = 2t_(m+n+1)

Question 6.
Find the 12^th term from the last term of the A.P -2, -4, -6, … -100.
Solution:

12^th term from the last = 39^th term from the beginning
∴ t₃₉ = a + 38d
= -2 + 38(-2)
= – 2 – 76
= – 78

Question 7.
Two A.P.’s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. Show that the difference between their 10^th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
Answer:
Let the common difference for the 2 A.P be “d”
For the first A.P
a = 2, d = d, n = 10
t_n = a + (n – 1) d
t₁₀ = 2 + 9 d ….(1)
For the 2^nd A.P
a = 7, d = d n = 10
t₁₀ = 7 + (9)d
= 7 + 9d ….(2)
Difference between their 10th term ⇒ (1) – (2)
= 2 + 9d – (7 + 9d)
= 2 + 9d – 7 – 9d
= -5
For first A.P when n = 21, a = 2, d = d
t₂₁ = 2 + 20d …….(3)
For second A.P when n = 21, a = 7, d = d
t₂₁ = 7 + 20d …….(4)
Difference between the 21^st term ⇒ (3) – (4)
= 2 + 20d – (7 + 20d)
= 2 + 20d – 7 – 20d
= -5
Difference between their 10th term and 21^st term = -5
Hence it is proved.

Question 8.
A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?
Solution:
S₁₀ = ₹16500
a, a + d, a + 2d…
d = 100
n = 10
S_n = \(\frac { n }{ 2 } \)(2a+(n-1)d)
S₁₀ = 16500
S₁₀ = \(\frac { 10 }{ 2 } \)(2×a+9×100)
16500 = 5(2a+900)
16500 = 10a + 4500
10a = 16500 – 4500
10a = 12000
a = \(\frac { 12000 }{ 10 } \) = ₹1200
∴ He saved ₹1200 in the first year

Question 9.
Find the G.P. in which the 2nd term is \(\sqrt { 6 }\) and the 6th term is 9\(\sqrt { 6 }\).
Solution:

Question 10.
The value of a motorcycle depreciates at the rate of 15% per year. What will be the value of the motor cycle 3 year hence, which is now purchased for ₹45,000?
Answer:
Value of the motor cycyle = ₹ 45000
a = 45000
Depreciation = 15% of the cost value
= \(\frac { 15 }{ 100 } \) × 45000
= 15 × 450
= 6750
d = – 6750 (decrease it is depreciation
Value of the motor cycle lightning of the 2nd year = 45000 – 6750
= ₹ 38250
Depreciation for the 2nd year = \(\frac { 15 }{ 100 } \) × 38250
= ₹ 57370.50