Plus One

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 9 Sequences and Series.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 9 Sequences and Series

Plus One Maths Sequences and Series 3 Marks Important Questions

Question 1.
Consider the GP 3,3²,3³, _______. (IMP-2014)
i) Find the sum to n terms of this GP.
ii) Find the value of n so that the sum to n terms of this GP is 120.
Answer:
i)

ii)

Plus One Maths Sequences and Series 4 Marks Important Questions

Question 1.
Given sum of three consecutive terms in an AP is 21 and their product is 280  (IMP-2011)
i) Find the middle term of the above terms.
ii) Find the remaining two terms of the above AP.
Answer:
i) Let the three consecutive terms be
a-d, a, a + d
a-d + a + a + d = 21
=>3a = 21
=>a = 7
ii) Then the AP becomes 7 – d,7, 7 + d
Given product is 280;
(7 – d)(7)(7 + d) = 280
=> (7 – d)(7 + d) = 40
==> 49 – d² = 40
=> <d² = 9 => d= 3,- 3
Therefore the AP is 4,7,10 or 10,7,4.

Question 2.
Consider the GP 3,6,12  (IMP-2011)
i) Which term of this GP is 96?
ii) Find the value of n so that sum to n terms of this GP is 381.
Answer:

Question 3.
i) What is the sum of the first ‘n’ natural numbers?  (IMP-2012)
ii) Find the sum to ‘n’ terms of the series
3 x 8 + 6 x 11 + 9 x 14 + ______.
Answer:

Question 4.
If the sum of the first n terms of an Arithmetic progression is ——,where X and Y are constants, find  (IMP-2012)
i) S₁ and S₂
ii) The first term and common difference.
iii) The n^th term.
Answer:
i) S₁ = X
S₂ =2X + 1/2(2 – 1)Y=2X + Y
ii) First term = a, = Sx = X
S₂ =2 X + Y
=> a₁ +a₂ =2 X + Y
=> a₂ =2X + Y
=>a₂ = X+ Y
Common difference =
a₂ – a₁ =X + Y – X = Y
iii) n^thterm = a_n = a + (n-1)d = X + (n – 1)Y

Question 5.
Find the sum to n terms of the series;  (IMP-2012)
2² + 5² + 8² +_______
Answer:

Question 6.
i) Write the first four terms of the sequence whose nth term \(a_{n}=\frac{n}{n+1}\) (MARCH-2013)
ii) The sum of the first three terms of a GP is \(\frac {12}{13}\) and their product is -1. Find the common ratio and the terms.
Answer:

Question 7.
If the numbers \(\frac { 5 }{ 2 }\) x \(\frac { 5 }{ 8 }\) are three consecutive terms of a GP, then find x. (MARCH-2014)
Find the sum of the first n-terms of the series. 2 +22+222 + _____
Answer:

Question 8.
i) Find the 5^th term of the sequence whose nth term is \(a_{n}=\frac{n(n-2)}{(n+3)}\) (MARCH-2014)
ii) Write the sum of first n natural numbers.
iii) The 5^th, 8^th and 11^th terms of a GP are p, q and s respectively. Prove that q² – ps
Answer:

Question 9.
i) A man starts repaying a loan as a first instalment of Rs. 1,000. If he increases the instalment by Rs. 150 every month, what amount will he pay in the 30^th instalment?  (IMP-2014)
ii) Find the sum to n terms of the sequence:
7,77,777,7777 ______.
Answer:

Question 10.
i) Consider the AP 4,10,16,22…….. Find its common difference and the 7^th terms.  (IMP-2014)
ii) If the m^th term of an AP is \(\frac { 1 }{ n }\) and the nth term is \(\frac { 1 }{ m }\) , prove that the sum of the first ‘mn’ terms is \(\frac { 1 }{ 2 }\)(mn +1)
Answer:

Question 11.
The 6^th term of the sequence whose n^th term is \(t_{n}=\frac{2 n-3}{6}\) is _____. (MARCH-2015)
a) 3
b) \(\frac { 1 }{ 2 }\)
c) \(\frac { 3 }{ 2 }\)
d) \(\frac { 1 }{ 3 }\)
ii) Find the sum to infinity of the sequence 1,\(\frac { 1 }{ 3 }\) ,\(\frac { 1 }{ 9 }\), ………
iii) If a, b, c are in AP and \(a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}\), prove that x, y, z are in AP.
Answer:

Plus One Maths Sequences and Series 6 Marks Important Questions

Question 1.
i) In an AP, the first term is 2 and the sum of the first five terms is one fourth the sum of the next five terms. (MARCH-2010)
a) Find the common difference.
b) Find the 20^th term.
ii) If AM and GM of two numbers are 10 and 8 respectively, find the numbers.
Answer:

Question 2.
i) In an AP if m^th term is ‘n’ and nth term is ‘m’ .find the (m + n)^th term.  (IMP-2010)
ii) If 3^rd, 8^th and 13^th terms of a GP are x,y,z respectively, prove that x,y,z are in GP.
iii) Prove that x,y,z in the above satisfies the equation \(\frac{y^{10}}{(x z)^{5}}=1\)
Answer:

Question 3.
Which of the following is the n^th term of an AP? (MARCH-2011)
a) 3 – 2n
b)n² – 3
c) 3ⁿ – 2
d) 2 – 3n²
ii) Find the 10th term of the sequence
– 6,- \(\frac { 11 }{ 2 }\), – 5,….
iii) The sum of the first three terms of a GP is \(\frac { 39 }{ 10 }\) and their product is 1. Find the common ratio and the terms.
Answer:

Question 4.
Find the 10^th term of an AP whose n^th \(\frac{2 n-3}{6}\) term is (MARCH-2012)
ii) Find the sum of the first 10 terms of the above AP.
iii) Find the sum of the first 10 terms of a GP, whose 3^rd term is 12 and 8^th term is 384.
Answer:

Question 5.
i) Find the 5^th term of the sequence whose n^th term, \(a_{n}=\frac{n^{2}-5}{4}\) (MARCH-2013)
ii) Find 7 + 77 + 777 +……. to n terms.
iii) Find the sum to n terms of the series.
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ………
Answer:

Question 6.
i) Find the sum of multiple of 7 between 200 and 400.  (IMP-2013)
ii) The sum of first 3 terms of a GP is \(\frac { 39 }{ 10 }\) and their product is 1. Find the terms.
Answer:

Question 7.
If ‘a’ is the first term and ‘cf is the common difference of an AP, then the nth term of the AP, a_n = ……. (MARCH-2014)
ii) In an AP, if the m^th‘ term is ‘n’ and the n^th term is ‘m’, where , prove that its p^th term is n + m – p.
iii) Find the sum to ‘n’ terms of the series:
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + _______.
Answer:

Question 8.
i) If the sum of certain number of terms of the AP 25,22,19 is 116, then find the last term.  (IMP-2014)
ii) Find the sum to n terms of the series
1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ………
(Imp (Science) – 2014)
Answer:

Question 9.
i) The 3^rd term of the sequence whose n^th term is (MARCH-2015)
ii) Insert three numbers between 1 and 256 so that the resulting sequence is a GP.
iii) If p^th term of an AP is q and q^th term is ‘p’, where p ≠ qfind r^th term.
Answer:

Question 10.
i) Geometric mean of 16 and 4 is ______.  (IMP-2015)
(a) 20
(b) 4
(c) 10
(d) 8
ii) Find the sum to n terms of the series: 5 + 55 + 555 + ________.
iii) Find the sum to n terms of the AP,
whose K^th term is a_k = 5K +1
Answer:

Question 11.
i) If the first three terms of an AP is x – 1,x + 1, 2x + 3, then x is  (IMP-2015)
(a)- 2
(b) 2
(c) 0
(d) 4
ii) Find the sum to n terms of the sequence.
1 x 2 + 2 x 3 + 3 x 4 + _______
iii) The nth term of the GP 5,- \(\frac { 5 }{ 2 }\),\(\frac { 5 }{ 4 }\),\(\frac { 5 }{ 8 }\),….. is \(\frac { 5 }{ 1024 }\) find ‘n’.
Answer:

Question 12.
The n^th term of the GP 5,25,125 (MARCH-2016)
is
(a) n⁵
(b) 5ⁿ
(c) (2n)⁵
(d) (5)²ⁿ
ii) Find the sum of .all natural numbers between 200 and 1000 which are multiples of 10.
iii) Calculate the sum of n-terms of the series whose n81 term is a_n = n(n + 3)
Answer:

Question 13.
i) Which among the following represents the sequence whose n^th terms is \(\frac { n}{ n+1 }\) ? (MAY-2017)
a) 1,2,3,4,5,6
b) 2,3,4,5,6
c) 2,\(\frac { 3 }{ 2 }\),\(\frac { 4 }{ 3 }\),\(\frac { 5 }{ 4 }\),\(\frac { 6 }{ 5 }\)
d) \(\frac { 1 }{ 2 }\),\(\frac { 2 }{ 3 }\),\(\frac { 3 }{ 4 }\),\(\frac { 4 }{ 5 }\),\(\frac { 5 }{ 6 }\)
ii) Using progression, find the sum of first five terms of the series 1 + \(\frac { 2 }{ 3 }\) + \(\frac { 4 }{ 9 }\) + …..
iii) Calculate: 0.6 + 0.66 + 0.666 + ………. n terms.
Answer:

Question 14.
The sum of the infinite series is 1, \(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 9 }\) ………is ________. (MARCH-2017)
(a) \(\frac { 3 }{ 2 }\)
(b) \(\frac { 5 }{ 2 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { 7 }{ 2 }\)
ii) Find the sum of all natural numbers between 100 and 1000 which is a multiple of 5.
iii) Find the sum to n terms of the series 8,88,888 ………
Answer:

Question 15.
The 6^th term of the GP \(\frac { 1 }{ 2 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 8 }\), ………. (MARCH-2017)
a) \(\frac { 1 }{ 32 }\)
b) \(\frac { 1 }{ 64 }\)
c) \(\frac { 1 }{ 16 }\)
d) \(\frac { 1 }{ 128 }\)
ii) The sum of 1st 3 terms of a G.P is \(\frac { 13 }{ 12 }\) and their product is – 1. Find the common ratio and terms.
iii) Find the sum to n terms of the series \(3 \times 1^{2}+5 \times 2^{2}+7 \times 3^{2}\) + ………
Answer:

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 7 Permutation and Combinations.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 7 Permutation and Combinations

Plus One Maths Permutation and Combinations 3 Marks Important Questions

Question 1.
i) if \({ }^{n} C_{9}={ }^{n} C_{8}\) find ‘n’ and \({ }^{n} C_{17}\) (IMP-2014)
ii) How many chords can be drawn, through 23 points on a circle?
Answer:

Plus One Maths Permutation and Combinations 4 Marks Important Questions

Question 1.
i) Simplify (MARCH-2011)

ii) In how many different ways can the letters of the word HEXAGON be permuted?
iii) In how many different ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
(March (Science) – 2011)
Answer:

Total number of ways = 10 x 4 = 40

Question 2.
i) If \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}\) (MARCH-2011)
then find x.
ii) How many 4 digit numbers are there with no digit repeated?
iii) If nC₈ = nC₂, then find nC₃ ?
Answer:

Question 3.
Consider all the letters of the word ‘FALIURE’.  (IMP-2011)
i) How many words can be formed using these letters?
ii) How many words can be formed so that the vowels being together?
iii) How many words begin with A and end with E?
(Imp (Commerce) – 2011)
Answer:
i) ‘FALIURE’ word has 7 letters in it, can be arranged in 7! Ways = 7.6.5.4.3.2.1 = 5040
ii)

A,E,I,U are vowels in the word, should be kept together, so should be treated as on block. Hence there are 4 such blocks can be arranged in 4! ways. These 4 vowels can be arranged in 4! Ways.
Hence the total words = 4! x 4! = 24 x 24 = 576
iii)

The only possible arrangement is for 5 blocks; hence total number of ways is 5! = 120.

Question 4.
i) Find the value of n if (IMP-2012)

ii) How many words, with or without meaning,
can be formed using all the letters of the word CHEMISTRY, using each letter exactly once? How many of them start with C and end with Y?
Answer:
i)

n —10,- 3
n = – 3 is not possible since negative son = 10
ii) Total number of words = 9!
Total number of words starting by C and ending by Y= 7!

Question 5.
i) If 2nC₃ : nC₃ = 12 : 1 find n. (IMP-2012)
ii) What is the total number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these four cards of the same suit?
Answer:

Question 6.
i) If nC₉ = nC₈, find nC₁₇ 
ii) A committee of 5 person is to be selected from a group of 4 men and 5 women. In how many ways can this be done? How many of these committees would consist of 2 men and 3 women? (IMP-2012)
Answer:

Number of different selection = 6 x 10 = 60

Question 7.
i) If nC₉ = nC₈, find nC₁₇ 
ii) How many three digit number can be formed using the digits 1,2,3,4,5 if repetition is not allowed? (MARCH-2013)
iii) In How many ways can a team of 4 boys and 3 girls be selected from 6 boys and 4 girls?
Answer:

Question 8.
i) If nC₅ = nC₄, find nC₈
ii) How many chords can be drawn through 20 points on a circle? (MARCH-2014)
iii) A bag contains 6 red and 5 blue balls. In how many ways can one choose 3 red and 2 blue balls from this bag?
Answer:

Question 9.
i) Find the number of permutation of the letters of the word ALLAHABAD.
ii) Find r,if \({ }^{5} P_{r}=2 \times{ }^{6} P_{r-1}\) (IMP-2014)
Answer:
i) Total number of letters is 9.
A: 4 times; L: 2 times.

Therefore r = 3.

Plus One Maths Permutation and Combinations 6 Marks Important Questions

Question 1.
i) Find the value of n such that (MARCH-2010)
nP₅ = 42 x nP₃ for n > 4
ii) A committee of 3 persons is to be constituted from a group of 2 men and 3 women.
a) In how many ways can this be done?
b) How many of these committees would consist of 1 man and 2 women? (March (Science) – 2010)
Answer:
i) nP₅ = 42 x nP₂
=> n(n – 1)(n – 2 )(n – 3)(n – 4) = 42 x n(n -1)(n- 2)
=> (n – 3)(n – 4) = 42
=>n² – 7n + 12 = 42
=> n² – 7n – 30 = 0
=> (n -10)(n + 3) = 0
=> n = 10; n = – 3
n can’t be negative, so the acceptable value is n = 10
ii) a) 3 person can be selected from 5 in 5C₃ = 10
(b) 1 man can be selected from 2 in 2C₁ = 2 ways.
2 women can be selected from 3 in 3 C₂ = 3 ways.
Total ways = 2×3 = 6

Question 2.
i) lf \({ }^{n} C_{2}:{ }^{2 n} C_{1}=3: 2\), find n. (MARCH-2010)
ii) a) Find the number of words that can be
formed from the letters of the word MALAYALAM.
b) How many of these arrangements start with Y?
(March (Science) – 2010)
Answer:
i)

ii) a) MALAYALAM, this word has 9 letters
M- repeated 2 times.
A- repeated 4 times.
L- repeated 2 times.
Number of words formed by these 9 letters
= \(\frac{9 !}{2 ! \times 4 ! \times 2 !}\)
b) If the word starts with Y, then total number of letters that can be arranged become 8. Number of words formed which begin with Y
= \(\frac{8 !}{2 ! \times 4 ! \times 2 !}\)

Question 3.
i) if \(5 \times 4 P_{r}=6 \times 5 P_{r-1}\) find ‘r’, (IMP-2014)
ii) How many 3 digit number can be formed with the digits 0,1,2,3 and 4?
iii) In a Panchayath there are 10 Panchayath members. Ladies contested only in the 50 % reserved constituency. If the post president and vice president are reserved for ladies, in how many ways both the president and vice president can be selected?
(Imp (Science) – 2010)
Answer:
i)

ii)

Total 3 digit numbers = 4x5x5 = 100 (IMP-2010)
iii) The 10 member Panchayath has 5 men and 5 ladies. The president and vice president are to be selected from these ladies in 5C₁₂ = 10 ways.

Question 4.
i) Prove that nC_r = nC_n-r
ii) Twenty eight matches were played in a volley ball tournament. Each team playing one against each of others. How many teams were there? (IMP-2010)
iii) If the letters of the word ‘TUTOR’ be . permuted among themselves and arranged as in a dictionary, then find the position of the word ‘TUTOR’.
Answer:

Question 5.
A student is instructed to answer any 8 out of 12 questions. (IMP-2011)
i) How many different ways he can choose the questions?
ii) How many different ways he can choose the questions so that question no.1 will be included?
iii) How many different ways, he can choose the questions so that question no.1 will be included and question no.10 will be excluded?
Answer:
i) 8 out of 12 questions can be selected

ii) Since question no.1 is included, the possible is selection is from 11 questions and the number of questions to be selected becomes 7.
Hence the total selection

iii) Question no.10 will be excluded so total questions become 11. Question no.1 is included again total questions reduced to 10. Now we have to select 7 questions out of 10, can be done in 10C₇ = 10C₃ = \(\frac{10 \times 9 \times 8}{1 \times 2 \times 3}\) = 120.

Question 6.
Solve for the natural n; (MARCH-2012)
12.(n-1)P10C₃ =5.(n + 1)P10C₃
In how many ways seven althlets can be chosen out of 12?
iii) The English alphabets has 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed without repetition of letters?
Answer:
i)

ii) 7 athletes be chosen out of 12 in 12C₇ =12C₅ ways
iii) Two different vowels can be selected in 5C₂. Two different consonants can be selected in 21C₂.
Therefore total numbers of words

Question 7.
i) Find r if 5P_r = 6P_r-1.
ii) If there are 12 persons in a party and each of them shake hands with all others, what is the total number of handshakes? (MARCH-2012)
iii) In How many ways can a committee of 3men and 2 women be selected out of 7 men and 5 women?
Answer:

Question 8.
i) Find the value of n such that (MARCH-2013)
3.nP₄ = 5.(n -1 )P₄,n > 4
ii) In how many ways can 5 students be seated on a bench?
iii) Find the number of different 8-letter arrangements that can be made from the letters of the word, ‘DAUGHTER’ so that:
a) All vowels are occur together.
b) All vowels do not occur together.
Answer:

Question 9.
i) Determine n if 2nC₃ = 11.nC₃ 
ii) In how many ways can a cricket team of 11 of players be selected from 15 players? (MARCH-2013)
iii) A bag contains 5 white, 6 red and 4 blue balls. Determine the number of ways in which 2 white, 3red and 2 blue balls can be selected.
Answer:

Question 10.
i) The number of 3 digit numbers can be formed from the digits 1,2,3,4,5 assuming that repetition of the digits is not allowed is _______.
ii) If \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\), find x. (IMP-2013)
iii) How many words, with or without meaning, can be formed using all the letters of the word ‘FRIDAY’, using each letter exactly once? How many of them have first letter is a vowel?
Answer:

Question 11.
i) nC₇=nC₅,n =
ii) A bag contains 5 blue and 6 white balls. Determine the number of ways in which 3 blue and 4 white balls can be selected. (IMP-2013)
iii) What is number of choosing 3 cards from a pack of 52 playing cards? In how many of these 3 cards of the same colour?
Answer:

Question 12.
i) if \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}\) find x? (MARCH-2014)
ii) How many four digit numbers can be formed using the digits 4,5,6,7,8 if repetition of digits is not allowed?
iii) Find the number of arrangements that can be made from the letters of the word ‘MOTHER’ so that all vowels occur together.
Answer:

Question 13.
i) In how many ways can the letters of the word PERMUTATIONS be arranged if; (MARCH-2014)
a) the word starts with P and ends with S?
b) there are always 4 letters between P and S?
ii) In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
iii) How many chords can be drawn through 21 points?
Answer:
i) a) In the word T is repeated twice

b) P should move from 1 to 7th position and S should move from 6th to 12th position. Hence the arrangements

ii) First arrange 5 girls in 5P₅ ways. Arrange 3 boys in 6P₃.
Hence the total arrangements
= 5P₅ x 6P₃ = 14400
iii) Chord is the join of two points. Hence selection 2 points from 21, which can be done in 21C₂ =210

Question 14.
i) What is the minimum number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
a) are 4 cards of the same suit? (MARCH-2014)
b) do 4 cards belong to 4 different suits?
ii) Find the number of permutation of the letters of the word, ALLAHABAD.
iii) How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer:

Question 15.
i) Find ‘n’ if \(9 \times^{(n-1)} P_{3}={ }^{n} P_{4}\) (IMP-2014)
ii) Find the number of words that can be formed from the letters of the word, COMMERCE’.
iii) In how many ways can a group of 12 students be selected from 15 students? How many of these groups would include one particular student?
Answer:

Question 16.
i) \(\frac{0 !}{1 !}\) = _____ (MARCH-2015)
a) 0
b) 1
c) 2
d) 3
ii) Find r, if \(5 \times^{4} P_{r}=6 \times^{5} P_{r-1}\)
iii) Find the number of 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together.
Answer:

Question 17.
nC_{n – 1} = _____ (MARCH-2015)
(a) n-1
(b) n
(c) 0
(d) 1
If nC₉=nC₈,find nC₂
How many ways can a team of 5 persons be selected out of a group of 4 men and 7 women, if the team has at least one man and one women?
Answer:

Question 18.

i) \({ }^{7} P_{7}\) = _____ (IMP-2014)
a) 7
b) 7!
c) 1
d) 77
ii) Find the number of words that can be formed from the letters of the word “MALAYALAM”. How many of them start with Y?
iii) \({ }^{2 n} C_{3}=11 \times{ }^{n} C_{3}\) Find ’n’.
Answer:

iii)

Question 19.
i) \({ }^{29} C_{29}\) = ______ (IMP-2015)
(a) 0
(b) 1
(c) 2
(d) 3
ii) Prove that \({ }^{61} C_{57}-{ }^{60} C_{56}={ }^{60} C_{3}\)
iii) In how many ways can the letters of the word ‘ARRANGE’ be arranged such that two A’s do not occur together?
Answer:

Question 20.
Write the value of \({ }^{7} C_{5}\) (MARCH-2016)
Find the vale of n, if \(3 \times^{n} P_{4}=5 \times^{n-1} P_{4}\)
What is the number of ways of choosing 4 cards from a pack of 52 cards, provided all 4 cards belong to 4 different suits? (March (Science) – 2016)
Answer:

Question 21.
i) \({ }^{29} C_{29}\) = ______ (MARCH-2016)
a) 0
b) 1
c )2
d )3
ii) Find the value of n,
if \(12 \times^{n-1} P_{3}=5 \times^{n+1} P_{3}\)
iii) A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has at least one boy and one girl?
Answer:

Question 22.
i) How many 4 digit numbers can be formed using the digits 9, 8, 7, 6, 5, 4, if no digits are repeated? (MAY-2017)
(a) 630
(b) 603
(c) 306
(d) 360
ii) In how many ways a committee of 3 persons can be formed from a group of 2 men and 3 women?
iii) Find the value of n,
if \({ }^{2 n} C_{3}=11 \times{ }^{n} C_{3}\)
Answer:

Question 23.
i) \({ }^{569} \mathrm{C}_{569}\) = ______. (MAY-2017)
ii) \({ }^{2 n} C_{3}:{ }^{n} C_{3}=12: 1\) Find n.
iii) If the letters of the word EQUATION are arranged, find the number of arrangements in which no two consonants occur together?
Answer:
i) 1

Question 24.
i) if \(\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}\) , then x is ______.
(a) 32
(b) 16
(c) 64
(d) 8
ii) Given 5 flags of different colour, how many different signals can be generated if each other.
iii) Find r, if \({ }^{5} P_{r}=2 \times{ }^{6} P_{r-1}\)
Answer:
i) c) 64
ii) Number of ways = 5 x 4 = 20

Question 25.
i) lf nC₉=nC₈, then n = ______.
a) 9
b) 8
c) 17
d) 1
ii) How many chords can be drawn through 12 point on a circle?
iii) What is the number of way of choosing 4 cards from a pack 52 playing cards? In how many of these:
a) Four cards are of the same suit.
b) Cards are of the same colour.
Answer:

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 6 Linear Inequalities.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 6 Linear Inequalities

Plus One Maths Linear Inequalities 4 Marks Important Questions

Question 1.
i) Draw the graphs of 2x + 3y = 24 and x + y = 9 (IMP-2011)
ii) Solve the following system of inequalities graphically;
2x + 3y ≤ 24,
x + y ≤ 9, x, y ≥ 0
Answer:

Question 2.
i) Solve 4x – 5 < 7, when x is a real number. (IMP-2012)
ii) Solve the following system of inequalities
graphically. 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Answer:
i) 4x – 5 < 7 => 4x < 12 => x < 3
ii) 3x + 4y ≤ 12, x ≥ 0, y ≥ 0

Question 3.
Solve: 4x + 3 < 3x + 7 represent the solution on the real line. (MARCH-2013)
ii) Solve the following system of inequalities graphically.
3x + 2y ≤ 12;
x,y ≥ 0
Answer:
i) 4x + 3 < 3x + 7 => 4x – 3x < 7 – 3 => x < 4
ii)

Question 4.
i) Represent the inequality x ≥ – 3 on a number line. (IMP-2014)
ii) Solve the following inequalities graphically:
x + y≥5;
x – y≤ 3
Answer:

Question 5.
The interval representing the solution of the inequality 3x-1 ≥ 5, x∈R is (MARCH-2015)
a) [5,∞) b) [2, ∞)
c) [3,∞) d) (— ∞, ∞)
ii) Solve the system of inequality graphically
x + 2y ≤ 8,2x + y ≤ 8,
x ≥ 0,y ≥ 0
Answer:

Question 6.
i) Which among the following is the interval corresponding to the inequality – 2 < x ≤ 3 . (MARCH-2016)
(a) [- 2,3]
(b) [- 2,3)
(c) (- 2,3]
(d) (- 2,3)
ii) Solve the following equation.
2x + y ≥ 4;
x + y ≤ 3;
2x – 3y ≤ 6.
Answer:

Question 7.
i) Which among the following inequality represents the intervals [2,∞)
a) x – 3 ≥ 5, x∈R
b) 3x – 3 ≥5, x∈R
c) 3x – 1≥ 3, x∈R
d) 3x – 1 ≥ 5, x∈R
ii) Solve the following system of inequalities graphically. 3x + 2y ≤ 12; x ≥ 1; y ≥ 2
Answer:

Plus One Maths Linear Inequalities 6 Marks Important Questions

Question 1.
i) Solve the inequality 3(x – 1) ≤ 2(x – 3)  (MARCH-2010)
ii) Solve the following system of inequalities graphically. 5x + 4y ≤ 20; x ≥ 1, y ≥ 2
Answer:
i) 3(x – 1) ≤ 2(x – 3) =>3x – 3 ≤2x – 6
=>3x – 2x ≤ 3 – 6
=> x ≤ – 3
ii) 5x + 4y = 20

Question 2.
i) Arathi took 3 examinations in an year. The marks obtained by her in the second and third examinations are more than 5 and 10 respectively than in the first examination. If her average mark is at least 80 find the minimum mark that she should get in the final examinations? (IMP-2010)
ii) Solve the following system of inequalities graphically 2x + y ≥6; 3x + 4y ≤12
Answer:
Let x denote the marks of arathi in first examination. then mark in second exam and third exam are x +5 and x + 10 respectively. Given average in three examinations is atleast 80.

Question 3.
i) Solve the inequality
2(2x + 3) – 10 < 6(x – 2) (MARCH-2011)
ii) Solve the following inequalities graphically. system of
x – 2y < 3;
3x + 4y ≥ 12; x,y ≥ 0
Answer:
i) 2(2x + 3) – 10 < 6(x – 2)
=> 4x + 6 – 10 ≤ 6x – 12
=> – 2x ≤ -12 + 4
=> – 2x ≤ – 8
=> x ≥ 4
ii)

Question 4.
i) Find all pairs of consecutive odd natural numbers, both of which are smaller than 10, such that their sum is more than 11. (IMP-2012)
ii) Solve 2x + y ≤ 6 graphically.
Answer:
i) Consecutive odd natural numbers be x and x+2. Then,
x + x + 2 > 11;
x + 2 < 10
=> 2x > 11 – 2;
x < 10 – 2
=> x > 9/5 = 4.5;
x < 8
5 ≤ x < 8 Therefore x can take values 5,7.
Hence the pairs are (5,7),(7,9)
ii)

Question 5.
i) Solve the inequality: 3(2 – x) ≥ 2(1 – x)  (MARCH-2013)
ii) Solve the following system of inequalities graphically.
2x + y ≥ 4;
x + y ≤ 3;
2x – 3 ≤ 6
Answer:
i) 3(2 – x) ≥ 2(1 – x)
=> 6 – 3x ≥ 2 — 2x
– 3x + 2x ≥ 2 – 6
=>- x ≥ – 4
=> x ≤ 4
ii)

Question 6.
i) Solve: 5x – 3 < 3x + l (MARCH-2014)
ii) Solve the following inequalities graphically.
x + 2y ≤ 8;
2x + y ≤ 8;
x,y ≥ 0
Answer:

Question 7.
i) Raju obtained 70 and 60 marks in first two examinations. Find the minimum mark he should get in the third examination to have an average of atleast 50 marks. (IMP-2013)
ii) Solve the following system of inequalities graphically.
3x + 2y ≤ 12;
x ≥ 1;
y ≥ 2
Answer:
i) Let x be the mark obtained by Raju in third exam. Then,

Question 8.
i) Solve: 5x + 3 < 2x + 7 represent the solution on the real line. (MARCH-2014)
ii) Solve the following system of inequalities graphically.
x + 2y ≤ 8;
2x + y ≤ 8;
x, y ≥ 0
Answer:

Question 9.
i) Solve: 7x + 3 < 5x + 9 represent the solution on the real line. (MARCH-2014)
ii) Solve the following system of inequalities graphically.
x + 2y ≤ 8;
2x + y ≤ 8;
x, y ≥ 0
Answer:
i)

ii)
15

Question 10.
i) Solve 10x – 23 < 3x + 5
ii) Solve the following system of inequalities graphically: 3x + 5y ≤ 15; 5x + 2y ≤ 10; x,y ≥ 0 (IMP-2014)
Answer:

Question 11.
i) Solve; 7x + 3 ≤ 5x + 9; x∈R . Express the solution on a number line.  (IMP-2015)
ii) Solve graphically; 3x + 4y ≤ 60;
x + 3y ≤ 30;
x,y ≥ 0.
Answer:

Question 12.
Solve the inequality \(\frac{x}{3}>\frac{x}{2}+1\) (MARCH-2017)
Solve the system of inequality graphically
2x + y > 6,3x + 4y ≤ 12
Answer:
i) 2x > 3x + 6
=> – x > 6
=> x < – 6
ii)

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 5 Complex Numbers and Quadratic Equations.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 5 Complex Numbers and Quadratic Equations

Plus One Maths Principle of Complex Numbers and Quadratic Equations 3 Marks Important Questions

Question 1.
Find the modulus and argument of the complex number \(\frac{1+i}{1-i}\) . Find its multiplicative inverse in the form a + ib (IMP-2012)
Answer:
Convert into a + ib form

Hence Modulus is 1 and argument is \(\frac{\pi}{2}\)
Multiplicative inverse of i is \(\frac{1}{i}=\frac{1}{i} \times \frac{-i}{-i}=-i\)

Plus One Maths Principle of Complex Numbers and Quadratic Equations 4 Marks Important Questions

Question 1.
i) Express the complex number \(z=\frac{5 + i}{2 + 3i}\) in the form of a + ib. (MARCH-2010)
ii) Represent z in the polar form.
Answer:

Question 2.
consider the complex number (MARCH-2011)

i) Express z in the form of a + ib
ii) Represent z in the polar form.
Answer:

The complex number lies in the first quadrant;

Question 3.
i) Express \(\frac{1}{1-i}\) in the form of a + ib  (IMP-2011)
ii) Express \(\frac{1}{1-i}\) in the polar form.
Answer:

Question 4.
Represent the complex number 1 + i√3 in the polar form.  (IMP-2012)
Express \(\frac{2 + i}{2 – i}\) in the form of a + ib.
Answer:

Question 5.
Consider the complex number  (MARCH-2012)

i) Express complex number in the form of a + ib.
ii) Express complex number in the polar form
Answer:

Question 6.
i) Express the following complex number in the form a + ib  (MARCH-2013)
(1 +i) – (1 – 6i) + (2 + i)
ii) Represent the complex number z = 1 + i in the polar form.
Answer:

Question 7.
i) Represent the complex number √3+ i in the polar form.  (MARCH-2013)
ii) Solve : √5x² + x + √5 = 0
(March (Science) – 2013)
Answer:

Question 8.
i) Express \(\frac{1+i}{1-i}\) in the form a + ib.  (IMP-2013)
ii) Represent the \(\frac{1+i}{1-i}\) in the polar form.
Answer:

Question 9.
i) Solve the quadratic Equation – x² + x – 2 = 0  (IMP-2014)
ii) Express ‘i’ in the form r(cosθ+i sinθ )
Answer:

Plus One Maths Principle of Complex Numbers and Quadratic Equations 6 Marks Important Questions

Question 1.

a) a + ib form. (IMP-2010)
b) polar form.
Answer:

Question 2.
i) If z = √3 + i , find the conjugate of Z. (IMP-2010)
ii) Write the polar form of the complex number z = √3 + i
iii) Solve 2x² + 3x + 1 = 0
Answer:

Question 3.
i) Solve: √3x² + x + √3 = 0 (MARCH-2014)
ii) Represent the complex number z = 1 + i √3 in the polar form.
Answer:

Question 4.
The conjugate of 1 – 2i is _______. (IMP-2014)
ii) Express the complex number \(\frac{2 + 3i}{1 – 2i}\) in the form a + ib .
iii) Solve x² + 3x + 5 = 0
Answer:

Question 5.
i) Represent the complex number 1 + √3i in the polar form. (MARCH-2015)
ii) Find the square root of the complex number – 7 – 24i.
Answer:

ii)

Since the product xy is negative, we have x = 3, y = – 4 or, x = – 3, y = 4
Thus, the square roots of – 7 – 24i are 3 – 4i and – 3 + 4i.

Question 6.
i) What is i ^{– 35} ? (IMP-2015)
a) i
b) -i
c) 1
d) -1
ii) Express the complex number √3 + i ’ in the polar form.
iii) Solve: √5x² + x + √5 = 0
Answer:
i) a) i
ii)

The complex number lies in the first quadrant;

iii)

Question 7.
i) Which one of the following is the real part and imaginary part of the complex number: (MARCH-2016)

a) 0 and 1
b) 0 and 2
c) 3 and 2
d) 0 and 4
ii) Express the complex number ‘ i ’ in the polar form.
iii) Solve: √5x² + x + √5 = 0
Answer:
i) b) 0 and 2
ii)

iii)

Question 8.
i) Write the real and imaginary part of the complex number – 3 + √- 7 (MAY-2017)
ii) Find the modulus and argument of the complex number 1 + i√3
iii) Solve: x² – 2x + 3 = 0
Answer:
i) Real part is – 3 and imaginary part is √7
ii)

iii)

Question 9.
i) i¹⁸= ________
a) 1
b) 0
c) – 1
d) i
ii) complex number in polar form √3 + i
iii) Find the square root of the complex number – 8 – 6i.
Answer:
i) c) – 1
ii)

The complex number lies in the first quadrant;

iii)

Since the product xy is negative, we have x = 1, y = – 3 or, x = – 1, y = 3 Thus, the square roots of – 8 – 6i are 1 – 3i and – 1 + 3i.

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 4 Principle of Mathematical Induction.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 4 Principle of Mathematical Induction

Plus One Maths Principle of Mathematical Induction 3 Marks Important Questions

Question 1.
Consider the statement “p(n): 9ⁿ – 1 is a multiple of 8 ”. Where n is a natural number. (MARCH-2011)
i) is p true?
ii) Assuming p(k) is true, Show that p(k+1) is true.
Answer:

Question 2.
Consider the statement “ P(n): xⁿ – yⁿ is divisible by x – y” (MAY-2017)
i) Show that P is true.
ii) Using the principle of Mathematical induction verify that P(n) is true for all natural numbers.
Answer:

Hence divisible by (x – y).
Therefore by using the principle of mathematical induction true for all n ∈ N .

Plus One Maths Principle of Mathematical Induction 4 Marks Important Questions

Question 1.
Consider the statement “ 7ⁿ – 3ⁿ is divisible by 4” (MARCH-2010)
i) Verify the result for n = 2.
ii) Prove the statement using mathematical induction.
Answer:


Therefore p(k +1) is multiple of 4. Hence p(n) is true for all natural number n.

Question 2.
i) Which among the following is the least number that will divide 7²ⁿ-4²ⁿ for every positive integer n? [4,7,11,13] (IMP-2010)
ii) Prove by mathematical induction.
(cos θ + i sin θ)ⁿ = (cosnθ + i sin nθ)
Answer:

Therefore true for p(k +1). Hence p(n) is true for all natural number n.

Question 3.
Given P(n): 3²ⁿ -1 is divisible by 8. (IMP-2011)
i) Check whether P(1) is true.
ii) If P(k) is true then prove P(k+1) is true.
iii) Is the statement P(n) true for all natural numbers? Justify your answer.
Answer:
i) P(1):3²⁽¹⁾ = 9 -1 = 8 divisible by 8, hence true.
ii) P(k) : 3^2k -1 is divisible by 8.
3^2k– 1 = 8M .
M is a positive real
P(k + 1):3^2k+1 – 1= 3^2k+2 – 1
= 3^2k3² – 1
= 3^2k x 9 – 1
=3^2k x 9 – 9 + 8
= 9(3^2k – 1) + 8
= 9(8M) + 8
Hence divisible by 8
iii) By PMI; true for all natural number n.

Question 4.
Prove that by 1.2 +2.3 + 3.4 + + n(n +1) = \(\frac{n(n+1)(n+2)}{3}\) by using the principle of mathematical induction for all n∈N. (IMP-2012)
Answer:

Hence by using the principle of mathematical induction true for all n∈N .

Question 5.
By the principle of Mathematical Induction, Prove that (IMP-2012)

Answer:


Hence by using the principle of mathematical induction true for all n∈N .

Question 6.
Consider the statement P(n): n(n+1)(2n+1) is divisible by 6 (MARCH-2012)
i) Verify the statement for n = 2.
ii) By assume that P(k) is true for a natural number k, verify that P(k+1) is true.
Answer:
P : 2(2 + 1)(2 x 2 + 1) = 2 x 3 x 5 = 30
Which is divisible by 6.
ii) Assuming that true for p(k)
p(k): k(k +1)(2k +1) is divisible by 6.
k(k +1)(2k +1) = 6M, M is a positive real
Let p(k +1): (k +1)(k + 2)(2(k +1) +1)
= (k + 1)(k + 2)(2k + 3)
= {k + 1){2k² +7k + 6)}
= {(k + 1){(2k²+k) + (6k + 6)}
= k(k + 1)(2k +1) + 6(k + 1)(k +1)
= 6M + 6(k + 1)(k +1)
Hence divisible by 6. Therefore by using the principle of mathematical induction true for all n∈N .

Question 7.
Consider the statement (MARCH-2013)

i) Check whether P is true.
ii) By assume that P(k) is true, prove that P(k+1) is true.
iii) Is P(n) true for all natural number n?
justify your answer.
Answer:


iii) Hence by using the principle of mathematical induction true for all n∈ N .

Question 8.
Consider the statement (MARCH-2013)

i) Check whether P is true.
ii) If P(k)is true, prove thatP(k+1) is true.
iii) Is P(n) true for all natural number n?
Justify your answer.
Answer:


iii) By PMI; true for all natural number n.

Question 9.
Consider the statement (IMP-2014)

i) Verify the result for n = 2.
ii) Prove the statement using mathematical induction.
Answer:

Hence by using the principle of mathematical induction true for all n∈ N

Question 10.
Consider the statement (MARCH-2014)
P(n) : 1.2 + 2.3 + 3.4 +……… + n(n +1) = \(\frac{n(n+1)(n+2)}{3}\)
i) Prove that P is true.
ii) Assume that P(k) is true for a natural number k, verify that P(k+1) is true.
Answer:

Therefore p(k+1) is true.

Question 11.
Consider the statement (MARCH-2014)
P{n) = 3²ⁿ⁺² – 8n – 9 is divisible by 8
i) Verify the statement for n = 1.
ii) Prove the statement using the principle of mathematical induction for all natural numbers.
Answer:
i) P(1) = 3²⁺² – 8 – 9 = 64 is divisible by 8 . hence true
ii) Assuming true for P(k) is divisible by 8
P(k) = 3^2k+2 – 8k – 9=8M,
M is a positive real
Let P(k +1) = 3^2(k+1)+2 – 8(k) – 9
3^2k+2+2 – 8k – 8 – 9
3^2k+2 x 3² – 8k – 17
= (8M + 8k + 9)9 – 8k – 17
= 72M + 12k + 81 – 8k – 17
= 72M + 64k + 64 = 8(9M + 8k + 8)
Therefore p(k +1) is divisible by 8. Hence p(n) is true for all natural number n.

Question 12.
Consider the statement:(IMP-2014)

i) Prove that P is true.
ii) If P(k) is true, Prove that P(k+1) is true
iii) Is P(n) true for all natural number n? Why?
Answer:
11 i,ii

Question 13.
Using the principal of mathematical induction, prove that

Answer:

It is true.
Therefore P(k+1) is true when ever P(k) is true. By PMI P(n) is true for all n.

Question 14.
A statement p(n) for a natural number n is given by (MARCH-2015)

i) Verify that p(1) is true.
ii) By assuming that p(k) is true for a natural number k, show that p(k+1) is true.
Answer:

Question 15.
Consider the statement (IMP-2015)
P(n): 7ⁿ-3ⁿ is divisible by 4.
i) Show that P(1) is true.
ii) Verify, by the method of Mathematical induction, that P (n) is true for all natural numbers.
Answer:

Question 16.
Consider the following statement:

i) Prove that P is true.
ii) Hence by using the principle of mathematical induction, prove that P(n) is true for all natural numbers n. (MARCH-2016)
Answer:

Hence by PMI the result is true for all natural numbers.

Question 17.
Consider the statement “10^2n-1+1 is divisible by 11” verify that P is true and prove the statement by using Mathematical induction.
(MARCH-2017)
Answer:
p(1): 10¹ + 1 = 11 divisible by 11, hence true. Assuming that true for p(k)
p(k): 10^{2k – 1} + 1 is divisible by 11.
10^{2k – 1} + 1 = 11M
p(k + 1):10^{2(k+1) – 1} + 1
= 1o^{2k + 2 – 1} +1
=10^{2k – 1} x 10² + 1
= 10^{2k – 1} x 100 + 1
(11M – 1)100 + 1
= 1100M – 100+1
= 1100M – 99
= 11(100M – 9)
Hence p(k+1)divisible by 11. Therefore by using the principle of mathematical induction true for all n ∈ N .

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 3 Trigonometric Functions.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 3 Trigonometric Functions

Plus One Maths Trigonometric Functions 3 Marks Important Questions

Question 1.
i) Find the degree measure corresponding to \(\frac { 11 }{ 14 }\) radians.( use π = \(\frac { 22 }{ 7 }\)) (MARCH-2010)
ii) If cos x = \(\frac { -1 }{ 2 }\), x lies in the third quadrant,find sin x and tan x
Answer:
i) Degree measure corresponding to \(\frac { 11 }{ 14 }\) radians
= \(\frac{11}{14} \times \frac{180}{\pi}=\frac{11}{14} \times \frac{180 \times 7}{22}=\frac{1}{2} \times \frac{180}{2}=45^{\circ}\)
ii) since x lies in the third quadrant

sin x = –\(\frac{\sqrt{3}}{2}\)
tan x = √3

Question 2.
prove that (MARCH-2010)

Answer:

Question 3.
Showthat (cos x + cos y)² +(sin x + sin y)² =\(4 \cos ^{2}\left(\frac{x-y}{2}\right)\) (MARCH-2011)
Answer:
(cos x + cos y)² + (sin x + sin y)²
= cos² x + cos² y + 2cosx cosy + sin² x + sin² y + 2 sin x sin y
= 1+ 1 + 2(cosxcosy + sinxsiny)
= 2 + 2cos(x – y) = 2(1 + cos(x – y))
= \(4 \cos ^{2}\left(\frac{x-y}{2}\right)\)

Question 4.
prove that (MARCH-2013)

Answer:

Question 5.
Consider the trigonometric equation tan x = √3 (IMP-2013)
i) Write the general solution.
ii) Write the principal solution.
Answer:

Question 6.
i) The value of sin(π – x) is _______. (MARCH-2014)
ii) Find the principal and general solution of the equation sin x = \(\frac { √3 }{ 2 }\)
Answer:
i) sin x
ii)

Question 7.
prove that (MARCH-2014)

Answer:

Question 8.
i) 1 + tan² x = _______. (IMP-2014)
ii) If sin x = \(\frac { 3 }{ 5 }\) and x lies in the second quadrant, find the values of cosx, tan x and secx.
Answer:
i) sec² x
ii)

Plus One Maths Trigonometric Functions 4 Marks Important Questions

Question 1.
Expand cos(x + y) and hence prove (IMP-2010)
i) cos 2x = 1 – 2sin² x
ii) Solve the equation tan² θ + cot² θ = 2
Answer:
i) cos(x + y) = cos x cos y — sinx sin y
Put y = x
cos(x + x) = cos x cosx – sin x sin x
=> cos(2x) = cos² x – sin² x
=> cos(2x) = 1 – sin² x – sin² x = 1 – 2sin² x
ii)

Question 2.
Show that (IMP-2010)

Answer:

Question 3.
i) Find the value of sin(\(\frac { 31π }{ 3 }\)) (MARCH-2011)
ii) Find the principle and general solution of the equation cos x = \(\frac {-√3 }{ 2 }\)
Answer:

Question 4.
Solve: sin 2x – sin 4x + sin 6x = 0 (IMP-2012)
Answer:
sin 2x + sin 6x – sin 4x = 0
=> 2sin4xcos2x – sin4x = 0
=> sin4x(2cos2x – 1) = 0
=>sin4x = 0 or (2cos2x – 1) = 0

Question 5.
tan x tan 2x tan 3x = tan 3x – tan 2x – tan x (IMP-2012)
Answer:
We have; 3x= 2x + x
Take tan on both sides;

tan 3x(1 – tan 2x tan x) = tan 2x + tan x
tan 3x- tan 3x tan 2x tan x = tan 2x + tan x
tan x tan 2x tan 3x = tan 3x – tan 2x – tan x

Question 6.
i) Evaluate tan(\(\frac {13π }{ 6 }\)) (MARCH-2012)
ii) If tan x = \(\frac {1 }{ 2 }\) and x is in the third quadrant, find sinx and cosx.
Answer:

Question 7.
Prove that (MARCH-2012)

Answer:

Question 8.
i) \(\frac {tan x + tan y}{ 1 – tan x tan y }\) = _________. (MARCH-2013)
ii) Prove that

Answer:
i) tan(x + y)
ii)

Question 9.
Match the following: (MARCH-2013)

Answer:

Question 10.
i) Prove that (MARCH-2013)

ii) Prove that

Answer:

Question 11.
Show that (IMP-2013)
i) tan 15°=2-√3
ii) tan 15°+cot 15° = 4
Answer:
i)

ii)

Question 12.
i) Prove that (MARCH-2014)

ii)

Answer:
i)

Question 13.
i) If tan x = \(\frac {3 }{ 4 }\); x lies in the third quadrant,
find the value of cos x. (MARCH-2014)
ii) Find the principal and general solution of cos x = \(\frac {1 }{ 2 }\)
Answer:
i) Since x lies in the third quadrant

Plus One Maths Trigonometric Functions 6 Marks Important Questions

Question 1.
i) Write the value of (IMP-2011)
sin 600°; cos 330°; cos 120°; sin 150°
ii) Prove that
sin 600° cos330° +cos120° sin 150° + sin 180° cos 180° = -1
Answer:

Question 2.
i) Find the value of sin 75° (IMP-2012)
ii) Prove that

Answer:

Question 3.
i) \(\frac {2π }{ 3 }\) radians = ______ degree.(IMP-2014)
ii) cos(2π – x) = _______
iii) Find the general solution of sin 2x – sin 4x + sin 6x = 0
Answer:
i) 120°
ii) cos x
iii)

Question 4.
i) sin x cos y + cos x sin y = ______. (IMP-2014)
ii) Find sin 50° cos 10° + cos50° sin10°
iii) Prove that

Answer:
i) sin(x + y)
ii) sin 50° cos 10° + cos 50° cos 10°
= sin(50° +10°) = sin(60°) = \(\frac {√3 }{ 2 }\)
iii)

Question 5.
i) Which one of the following values of sin x is incorrect? (MARCH-2015)
a) 0
b) \(\frac {1 }{ 2 }\)
c) 3
d) 1
ii) Prove that

iii) A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Answer:
i) 3
ii)

iii)

Question 6.
i) sin 225° = _______.(MARCH-2015)

ii) Find the principle and general solutions of sin x = – \(\frac {√3 }{ 2 }\)
iii) Prove that

Answer:
i) \(\frac {- 1 }{ √2 }\)
ii)

iii)

Question 7.
i) Which of the equal to 520° ? (IMP-2015)

ii) Solve Sin 2x – Sin 4x + Sin 6x = 0.
iii) In any triangle ABC, prove that

Answer:
i) \(\frac { 26π }{ 6 }\)
ii)

iii)

Question 8.
i) The degree measure of \(\frac { 7π }{ 6 }\) radian is _____. (MARCH-2016)
(a) 120° (b) 102° (c) 201° (d) 210°
ii) Prove that

iii)A lamp post is situated at the middle point M of the side AC of a triangular plot ABC with BC = 7m, CA = 8m, AB =9m. Lamp post subtends an angle 15° at the point B. Determine the height of the lamp post.
Answer:
i) 210°
ii)

iii)

Question 9.
i) 40°20′ = ____ radians (MAY-2016)
a) \(\frac { 112π }{ 540 }\)
b) \(\frac { 211π }{ 540 }\)
c) \(\frac { 122π }{ 540 }\)
d) \(\frac { 121π }{ 540 }\)
ii) Prove that

iii) Solve sin 2x – sin 4x + sin 6x = 0
Answer:
i)
d) \(\frac { 121π }{ 540 }\)
ii)

iii)

Question 10.
i) sin 405°= _____. (MARCH-2017)

ii)
sin x = \(\frac { 3 }{ 5 }\)
x lies in the second quadrant.Find the values of cosx,secx,tanx,cotx
iii) Solve: sin 2x – sin 4x + sin 6x = 0
Answer:
i) a) \(\frac { 1 }{ 2 }\)
ii)

iii)

Question 11.
i) \(\frac { 7π }{ 6 }\) radian = ______ degree. (MARCH-2017)
a) 200
b) 300
c) 240
d) 120
ii) Find the value of tan 75°
iii) In a triangle ABC, prove that
a sin(B – C)+b sin(C -A)+c sin (A – B) = 0
Answer:
i) a) 210
ii)

iii)

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 2 Relations and Functions.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 2 Relations and Functions

Plus One Maths Relations and Functions 3 Marks Important Questions

Want any assistance in solving the given function? Then, make use of this Function Calculator.

Question 1.
Let A = {1,2,3,4} and B = {1,4,5} be two sets. (IMP-2011)
If R is the relation”<” from A to B then )
i) Write R in Roster form.
ii) Write Domain and Range of R.
iii) Find the number of relations from AtoB.
Answer:
i) R = {(1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (4,5)}
ii) Domain of R = {1,2,3,4}; Range of R = {4,5}
iii) Number of relations from A to B = 2^{4 x 3} =2¹²

Question 2.
\(\left(\frac{2 x}{5}+1, y-\frac{3}{4}\right)=\left(\frac{1}{5}, \frac{1}{4}\right)\) find xand y (IMP-2012)
Answer:

Question 3.
Let A = {1,2,3}; B = {3,4}. Write a relation from A and B having 5 elements. Write its domain, co-domain and range. (IMP-2012)
Answer:
R = {(1,3), (1,4), (2,3), (3,4), (3,3)}
Domain of R = {1,2,3}
Range of R = {3,4}
Co-domain of R = {3,4} = B

Question 4.
The function f is defined by (IMP-2012)

Draw the graph of Find f(x)
Answer:
For x < 0

Question 5.
5. Let A = {1,2,3}, B={4,5} (MARCH- 2013)
i) Find A x B and B x A
ii) Find the number of relations from A to B.
Answer:
i) A x B = {(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)} B x A = {(4,1), (4,2), (4,3), (5,1), (5,2), (5,3)}
ii) n(R) = 2^{2 x 3} = 2⁶ = 64

Question 6.
i) Let A = {7,8} and B = {2,4,5}. Find A x B (MARCH – 2013)
ii) Determine the domain and range of the relation R defined by R={(x,y):y=x+1},
x ∈ {0,1,2,3,4,5,6}
Answer:
i) A x B = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
ii) R= {(0,1),(1,2),(2,3),(3,4),(4,5),(5,6)}
Domain = {0,1,2,3,4,5}
Range = {1,2,3,4,5,6}

Question 7.
i) If A = {2,4}, B = {1,3,5}. Then the number of relations from A to B is …………. (IMP-2013)
ii) If P={-1,1}, form the set P x P x P
Answer:
i) number of relations from A to B 2^{2 x 3} = 2⁶ = 64
ii) P x P = {-1,1} x {-1,1} – {(-1,-1),(-1,1), (1-1), (1,1)}
P x P x P={(-1,-1), (-1,1), (1,-1),(1,1)} x {-1,1}
= {(-1,-1,-1),(-1-1,1),(-1,1,-1),(-1,1,1), (1,-1,-1), (1,-1,1), (1,1-1), (1,1,1)}

Question 8.
Consider the function f :R -> R defined by f(x) = -|x| (IMP-2013)
i) Find the domain and range of f.
ii) Draw the graph of f.
Answer:
i) Domain = R; Range = (-∝,0]
ii)

Plus One Maths Relations and Functions 4 Marks Important Questions

Question 1.
Consider the functions (MARCH- 2010)
\(f(x)=\sqrt{x-2}, \quad g(x)=\frac{x+1}{x^{2}-2 x+1}\)
Find
i) Domain of ‘f.
ii) Domain of ‘g’.
iii) (f + g)(x)
iv) (fg)(x)
Answer:

Question 2.
The Cartesian product P*P has 9 elements among which are found (- a, 0) and (a, 0). A relation from P to P is defined as (IMP- 2010)
R = {(x,y):x + y = 0)
i) Find P.
ii) Depict the relation using an arrow
diagram.
iii) Write down the domain and range of R,
iv) How many relations are possible from P to P?
Answer:
i) (-a,0), (a,0) are elements in P*P and P*P have 9 elements, implies P has 3 elements Therefore; P = {- a,a,0}

ii) Given; R = {(x,.y) : x + y
x = -a=>y = a=> (-a, a)
x = a =>y = -a => (a,-a)
x = 0=>y = 0 => (0,0)
iii) R = {{-a, a), (a-a), (0,0)}
Domain of R = {-a,a,0}; Range of R = {-a,a,0}
iv) Number of relation from P to P = 2₉

Question 3.
Consider the real function (MARCH-2012)
\(f(x)=\frac{x+2}{x-2}\)
i) Find the domain and range of thefunction.
ii) Prove that f(x)f(-x) + f(0) = 0
Answer:

Question 4.
i) Let P = {1,2}. Find P x P x P (MARCH-2014)
ii) Let A = {1,2,3,……..13,14},
R is the relation on A defined by R={(x,y):3x-y=0,x,y∈A}
a) Write R in a tabular form.
b) Find the domain and range of R.
Answer:
i) P x P x P = {(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)}
ii) R = {(1,3), (2,6), (3,9), (4,12)}
Domain = {1,2,3,4}
Range = {3,6,9,12}

Question 5.
i) Write the domain and range of the relation (IMP-2014)
R = {(2,5),(3,10),(4,17),(5,26)}
ii) If f{x) – x² -3x and g(x) = x + 2 find (f + g)(x), (f – g)(x) and (fg)(x)
Answer:
i) Domain = {2,3,4,5}; Range = {5,10,17,26}
ii) (f – g)(x) = f(x) – g(x)
= x² – 3x + x + 2
=x²—2x + 2
(f – g)(x) = f(x) – g(x)
= x² – 3x – x – 2 =x²-4x – 2
(fg)(x) = f(x) x g(x)
= (x² —3x) x (x + 2) = x³ – x² – 6x

Question 6.
i) If P = {m,n}, Q = {n,m} ; state whether the following is TRUE or FALSE (MAY-2016)
P x Q – {(m,n),(n,m)}
ii) Write the relation R= {(x, x³), x is a prime number less than 10}, in roster form.
iii) Let A = {1,2,3,4} B = {1,5,9,11,15,16} and
f = {(1,5),(2,9),(3,1),(4,5),(2,1)}. State with the reason whether f is a relation or a function.
Answer:
i) False
ii) R = {(2,2³)(3,3³),(5,5³),(7,7³)}
= {(2,8)(3,27), (5,125), (7,343)}
iii) F is not a function since the element 2 has two images.

Plus One Maths Relations and Functions 6 Marks Important Questions

Question 1.
Let R be the set of Reals. Define a function (IMP-2011)
f : R→R by f(x) = 2x²-1
i) Find \(\frac{f(-1)+f(1)}{2}\)
ii) Find f[f(x)]
iii) Draw the graph of f(x)
Answer:

Question 2.
i) If A = {-1,1}, Find A x A. (MARCH-2014)
ii) Consider the relation R defined by R = {(x,x + l) : x ∈ {-1,1}} Write R in the roster form. Also find the range.
iii) Draw the graph of the function. y = x, x ∈ R
Answer:
i) A x A ={-1,1} x {-1,1}
= {(-1,-1), (-1,1), (1,-1), (1,1)}
ii) R = {(-1,0),(1,2)}; Range ={0,2}
iii)

Question 3.
Let A = {1,2,3,4,5,6} be a set. Defined a relation R from A to A by R = {(x,y)/y = x+1} (IMP-2014)
i) Express R in the roster form.
ii) Represent the relation R using an arrow diagram.
iii) Write the domain and range of R.
Answer:
i) R = {(1,2),(2,3),(3,4),(4,5),(5,6)}
ii) Given; y = x +1
x = l =>j = 1 + 1 = 2; x = 2=> y = 3
x = 3 =>j = 4; x = 4 =>y = 5
x = 5=> y = 6
R = {(1,2), (2,3), (3,4), (4,5), (5,6)}

iii) Domain of R = {1,2,3,4,5}
Range of R = {2,3,4,5,6}
Codomain of R = {1,2,3,4,5,6}

Question 4.
i) Find the domain of the function. (MARCH-2015)
\(f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\)
ii) Sketch the graph of the function
f(x) = |x + 1|
iii) Consider A = {1,2,3,5} and B = {4,6,9}. Define a relation
R : A → B by R = {(x,y): x – y is odd,x ∈ A,y ∈ B}.
Write R in roster form and find range of R.
Answer:
i) x² – 5x + 4 = 0 => (x-4)(x-l) = 0
Therefore f(x) is defined for all x∈R? , except x = 4 and x = 1. Hence the domain is R – {1,4}.
ii)

iii) R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}
Range = {4,6,9}

Question 5.
i) Match the following  (IMP-2015)

ii) A = {1,2,3 ………. 14}. R is a relation from
A to A defined by R = {(x,y): 3x – y = 0,x,y ∈A}. Write the domain, range, co-domain of ,R.
Answer:
i) 1- Identity function, f :R →R : f(x) = x
2- Modulus function, f :R →R : f(x) = |x|
3 – Signum function f :R → R

Question 6.
i) If (x+1, y-2) = (3,1), write the values of x and y. (MARCH-2016)
ii) Let A = {1,2,3,4,5,} and B = {4,6,9} be two sets. Define a relation R from A to B by R = {(x,y): x – y} is a positive integer}
iii) Define the modulus function. What is its domain? Draw a rough sketch.
Answer:
i) Given; (x + l,y-2) = (3,1)
=> x + l = 3;
y – 2 = 1
=>x = 2; y = 3
ii) A x B= {(1,4),(1,6),(1,9),(2,4),(2,6),(2,9), (3,4),(3,6),(3,9),(4,4),(4,6)(4,9),(5,4),(5,6),(5,9)}
Then R = {(5,4)}

Question 7.
The domain of the function (MARCH-2017)
\(f(x)=\frac{1}{x-1}\)
(a) {1}
(b) R
(c) R – {1}
(d) R – {0}
ii) A relation R on set natural numbers is defined by R = {(x,y):y = x+5,x is a natural number less than 4, x, y ∈ N}
a) Write the relation in roster form.
b) Write the domain and range of the relation.
iii) Draw the graph of the relation
f(x) = |x|,x∈R
Answer:
i) c) R – {1}
ii) a) R = {(1,6), (2,7), (3,8)}
b) Domain = {1,2,3,}
Range of R = {6,7,8}

Kerala State Board New Syllabus Plus One Maths Chapter Wise Previous Questions and Answers Chapter 1 Sets.

Kerala Plus One Maths Chapter Wise Previous Questions Chapter 1 Sets

Plus One Maths Sets 3 Marks Important Questions

Question 1.
Consider the sets A and B given by (MARCH-2010)
A = {x\ x is a natural number, 2 ≤ x ≤ 6 }
B = {x: x is a prime number x ≤ 7 }
C = {x: x² – 5x + 6 = 0 }
i) Write A, B and C in the roster form.
ii) Verify that(A∪B)∪C = A∪(B∪C)
Answer:
i) A = {2,3,4,5,6}; B = {2,3, 5,7};
x² – 5x + 6 = 0
=> (x – 2)(x – 3) = 0
x = 2,3
C = {2,3}

ii) A∪B = {2,3,4,5,6,7};
(A∪B)∪C = {2,3,4,5,6,7}
B∪C = {2,3,5,7};
A∪(B∪C )= {2,3,4,5,6,7}
Hence;
(A∪B)∪C = A∪(B∪C)

Question 2.
Consider the sets A and B given by (MARCH-2011)
A = {x: x is a prime number <10}
B = {y. x is a natural number which divides 12}
i) Write A and B in the roster form.
ii) Find A∪BandB-A.
iii) Verify that(A∪B) – A = B – A.
Answer:
i) A = (2,3,5,7};B = {1,2,3,4,6,12}
ii) A∪B = {1,2,3,4,5,6,7,12};
B-A = {1,4,6,12}
iii) (A∪B)-A = {1,4,6,12} = B – A

Plus One Maths Sets 4 Marks Important Questions

Question 1.
If A = {1},
B = {{1},2}
C = {{1}, 3} and ∪ = {{1}, {2}, {3},1,2,3}, then find (IMP-2010)
i) A∩B
ii) B∩C
iii) n{{A∩B)’∪(B∩C)’)
Answer:
i) A∩B = 𝟇
ii) B∩C={{1}}
iii) (A∩B)’ = ∪ ;
(B∩C)’ = {{2},{3},1,2,3}
(A∩B)’∪(B∩C)’ = U∪{ {2}, {3} ,1,2,3} =U
n((A∩B)'{B∩C)’)= 6

Question 2.
Given U = {1,2,3,4,5,6,7,8,9,10},A= {1,2,3,4,5} and B = {3,4,5,6} (IMP-2011)
i) Write A∪B
ii) Verify whether(A∪B) ‘ = A’∩B’
iii) Verify whether
n(A∪B)=n(A-B)+n(A∩B)+n(B – A)
Answer:
i) A∪B = {1,2,3,4,5,6}
ii) A’ = {6,7,8,9,10} ,B’-= {1,2,7,8,9,10}
A’∩B’ = {7,8,9,10}
(A∪ B)’ = {7,8,9,10}
Hence (A∪B)’ = A’∩B’
iii) (A – B)={1,2};B – A = {6}; A∩B = {3,4,5}
n(A∪B) = 6,n(A – B) = 2,
n{B-A) = 1,n(A∩B) = 3
n(A -B) + n(A ∩ B) + n(B -A)
= 2 + 3 +1 = 6
= n(A ∪ B)
Hence verified.

Question 3.
Let A = {x:x is an integer \(\frac{1}{2}<x<\frac{7}{2}\) } and B = {2,3,4} (MARCH-2012)
i) Write A in the roster form.
ii) Find the power set of (A∪ B).
iii) Verify that(A-B)∪(A∩B) = A
Answer:
i) A = {1,2,3}
ii) A ∪ B= {1,2,3,4}
P(A) = {𝟇, {1} ,{2}, {3}, {4}, {1,2}, {1,3}, {1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4}, {2,3,4}, {1,2,3,4}}
iii) (A – B)∪(A ∩ B)
= {1} ∪ {2,3} = {1,2,3} = A

Question 4.
i) How many elements has P(A) if A = {1,2,3}? (IMP-2012)
ii) U = {1,2,3,4,5,6,7}; A = {1,4,6,7}; B = {1,2,3}. Find
A’,B’,A’∩B’,A∪B and
hence show that A’ ∩ B’ = (A ∪ B)’.
iii) If A and B are two sets such that A⊂B, then what is A ∩ B ?
Answer:
i) n(P(A)) = 2³ = 8
ii) A’ = {2,3,5},B’= {4,5,6,7}
A’ ∩B’ = {5}
A∪B = {1,2,3,4,6,7} =>(A∪B)’={5}
Hence A’∩B’ = (A∪ B)’
iii) A∩B = A

Question 5.
i) Let A = {1,2,3,5,6} and B = {1,3,4,6,7} (MARCH-2013)
a) Find A ∪ B? and A ∩B?
b) Find A -(A ∩B)
ii) If X and Y are two sets such that n(X) = 17, n{Y) = 23 and n(X∪Y)=38, find n(X∩Y).
Answer:
i) a) (A∪B) = {1,2,3,4,5 6,7} ,A∩B = {1,3,6}
b) A – (A∩B) = A – {1,3,6} = {2,5}
ii) We have; n(X∪Y)=n(X) + n(Y) – n(X∩Y)
38 = 17 + 23 – n(X∩Y)
n(X∩Y)= 40 – 38 = 2

Question 6.
Let ∪ = {1,2,3,4,5,6,7,8,9}, A = {1,2,4,7} and B = {1,3,5,7} (MARCH-2013)
i) Find A∪B.
ii) Find A’, B’ and hence show that
(A∪B)’ = A’∩B’
iii) The power set P(A) has …………. elements.
Answer:
i) A∪B={l,2,3,4,5,7}
ii) A’ = {3,5,6,8,9} ,B’ = {2,4,6,8,9}
(A∪B)’ = {6,8,9} = A’ ∩B’
iii) 2⁴= 16

Question 7.
i) If two sets A and B are disjoint, which none among the following is true? (IMP-2014)
a) A∪B = A
b)A∪B = B
c) A∩B = A
d )A∩B = 𝟇
ii) Find the set of the equation x² + x – 2 = 0 in roster form.
iii) In a group of students, 100 students know Hindi, 50 know English and 33 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Answer:
i) d)A∩B = 𝟇
ii) x² + x – 2 = 0
=> (x – 1)(x + 2) = 0 =>
x = 1;-2
Solution set is {1,-2}
iii) Let H-Hindi; E- English
n(H) = 100;
n(E) = 50;
n(H ∩E) = 33
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50-33 = 117

Question 8.
Let A = {x: x ∈ W ,x<5} and B = (x : x is a prime number less than 5} and U = {x: x is an integer O ≤ x ≤ 6} (MARCH-2015)
i) Write A,B in roster form.
ii) Find (A-B)∪(B-A)
iii) Verify (A ∪ B)’ = A’ ∩ B’
Answer:
i) A= {0,1,2,3,4}; B = {2,3}
ii) A – B = {0,1,4}; B – A ={ }
iii) U= {0,1,2,3,4,5,6}
A’ = {5,6}, B’ = {0,1,4,5,6}
(A∪B)’ = {5,6}
A’ ∩ B’ = {5,6}

Question 9.
If U = {1,2,3,4,5,6,7,8}, A = {2,4,6,8} and B = {2,4,8} (MAY-2016)
then:
i) Write A’, B’.
ii) For the above sets A and B prove that (A∪B)’ = A’∩B’
iii) Check whether (A∩B)’ = A’∪B’
Answer:
i) A’ = {1,3,5,7};B’ = {1,3,5,6,7}
ii) A ∪ B = {2,4,6,8}, (A ∪ B)’ = {1,3,5,7}
A’∩B’ = {l,3,5,7} => (A∪B)’ = A’∩B’
iii) A ∩ B = {2,4,8}; (A ∩ B)’ = {1,3,5,6,7}
A’∪B’ = {1,3,5,6,7} => (A∩B)’ = A’∪B’

Plus One Maths Sets 6 Marks Important Questions

Question 1.
i) What is A∪A’? (IMP-2012)
ii) If A and B are two sets such that n(A∪B) = 17 and n(A∩B) = 5,
find n(A) + n(B)
iii) In a group of 100 peoples, 40 people like cinema, 10 like both drama and cinema. How many like drama? How many like drama only not cinema?
Answer:
i) A∪A’ = ∪
ii) We have; A(A∪B)=n(A) + n(B) – n(A∩B)
=> 17 = n(A) + n(B) – 5
=> n(A) + n(B) = 17 + 5 = 22
iii) Let D – is the set of Drama, C is the set of Cinema.
Given;
n(D∪C) = 100, n{C) = 40, n(D∩C) = 10,
n(D ∪ C) = n(D) + n(C) – n(D ∩ C)
=>100 = n(D) +40 – 10
=> n(D) = 100-40 +10 = 70
n(Drama not Cinema) = n(D ∩ C’)
= n(D) – n(D∩C) = 70-10 = 60

Question 2.
i) If A and B are two sets such that A ⊂ B, (IMP-2013)
a) A ∪B is
b) Draw the Venn diagram of B – A
ii) In a committee, 60 people speak English,30 speak Hindi and 15 speak both English and Hindi. How many speak atleast one of these two languages?
Answer:
i) a) B
b)

ii) Let the events be E – English and H – Hindi
n(E) = 60; n(H) = 30; n(E∩H) = 15
People speak atleast one of these two languages = n(E∪H)
= n(E) + n(H)-n(E∩H)
= 60 + 30 – 15 = 75

Question 3.
Consider the set A = {2,3,5,7} and B = {1,2,3,4,6,12} (MARCH-2014)
i) Find A∩B.
ii) Find A – B,B – A and hence show that (A∩B)∪(A – B)∪(B – A)=A∪B
iii) Write the power set of A ∩ B
Answer:
i) A ∩ B = {2,3}
ii) A – B = {5,7} ,B-A = {1,4,6,12}
(A∪ B) = {1,2,3,4,5,6,712}
(A ∩B) ∪(A – B) ∪ (B – A) = {1,2,3,4,5,6,7,12}
iii) P(A∩B) = {𝟇,{2},{3},{2,3}}

Question 4.
i) If A = {1,2,3,4} and B = {3,4,5,6} then (MARCH-2014)
A – B= ………
(ii) In a group of 70 people, 37 like coffee, 52 like tea and each person likes atleast one of the two drinks. How many people like both coffee and tea?
iii) Let U = {1,2,3,4,5,6}, A= {2,3} and B = {3,4,5}. Find A’,B’and hence show that (A ∪ B)’ = A’ ∩ B’
Answer:
i) A – B = {1,2}
ii) n(C∪T) = 70,n(C) = 37,n(t) = 52 ,n(C∪T) = n(C) + n(T) – n(C∩T)
=> 70 = 37 + 52 – n(C ∩ T)
=> n(C ∩ T) = 89 – 70 = 19
iii) A= {1,4,5,6},B’ = {1,2,6}
A’∩B’ = {1,6}
{A∪B)’ = ({2,3,4,5})’ = {1,6}
Hence; (A∪ B)’= A’ ∩ B’

Question 5.
Let A= {1,2,3,4,5}, B = {1,2,6} and C = {1,6,7,8} (IMP-2014)
i) Find A∪B and B∩C
ii) Show that
A∪(B∩C) = (A ∪ B)∩(A ∪ C)
iii) Find A – (B∩C) and (A ∪ B) – (B∩C)
Answer:
i) A∪B = {1,2,3,4,5,6}
(B∩C) = {1,6}
ii) A∪ (B∩C) = {1,2,3,4,5} ∪ {1,6}
= {1,2,3,4,5,6}
A∪C = {1,2,3,4,5,6,7,8}
(A ∪ B) ∩ (A ∪ C) = {1,2,3,4,5,6}
Hence A∪(B∩C) = (A∪B)∩(A∪C)
iii) A- (B∩C) = {1,2,3,4,5} – {1,6} = {2,3,4,5}
(A∪B) – (B∩C) = {2,3,4,5}

Question 6.
i) A: {x : x is a prime number x ≤ 6} (IMP-2015)
a) Represent A in Roster form.
b) Write the Power set of A.
ii) Out of 25 members in an office 17 like to take tea, 16 like to take coffee. Assume that each takes at least one of the two drinks. How many like (a) Both coffee and tea? (b) Only tea and not coffee?
Answer:
i) A ={2,3,5}
ii) P(A) = {𝟇,{2},{3},{5},{2,3},{2,5},{3,5},{2,3,5}}
iii) Let the sets be defined as follows;
T- Tea; C – Coffee.
n(T) = 17; n(C) = 16; n(T ∪ C) = 25
n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
Coffee and Tea = n(T∩C) = 17 + 16 – 25 = 8
Only tea and not coffee =
n(T ∩C’) = n(T) – n(T∩C) = 17 – 8 = 9

Question 7.
i) If A is a subset of the set B, then (MARCH-2016)
A∩B = _____
ii) Represent the above set A∩B by Venn diagram.
iii) In a school, there are 20 teachers who teach Mathematics or Physics. Of These, 12 teach Mathematics, 12 teach Physics. How many teach both the subject?
Answer:
i) A
ii)

iii) Let the sets be defined as follows;
M- Mathematics; P – Physics.
n(M) = 12
n(P) = 12;
n(M∩P) = 20
n(M ∪ P) = n(M) + n(P) – n(M ∩ P)
n(M ∪ P) = 12 +12- 20 = 4

Question 8.
i) If U is the universal set and A is any set U∩A (MARCH-2017)
(a) U
(b) A
(c)𝟇
(d) A’
ii) Consider the sets
U = {a,b,c,d,e,f,g}, A = {b,c,d,e}, B = {a,c,g). Find A’and B’ the verify (A∪B)’ = A’∩B’
iii) In a group of 400 people, 250 speaks Hindi, 200 can speak Malayalam. How many people can speak both Malayalam and Hindi?
Answer:
i) b) A
ii) A∪B = {a,b,c,d,e,g}
A’ = {a,f,g} ;
B’ = {b,d,e,f};
(A∪B)’ = {f}; A’∩B’ = {f}
iii) n(A∩B) = 400,n(A) = 250
n(B) = 200
n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 250 + 200 – 400
= 50

HSE Kerala Board Syllabus HSSLive Plus One Maths Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala Plus One Chapter Wise Previous Questions and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus One Maths Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus.

Board	SCERT, Kerala
Text Book	NCERT Based
Class	Plus One
Subject	Maths
Chapter	All Chapters
Category	Kerala Plus One

Kerala Plus One Maths Chapter Wise Previous Year Questions and Answers

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Chapter 10 Straight Lines
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability

We hope the given HSE Kerala Board Syllabus HSSLive Plus One Maths Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus One Maths Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSE Kerala Board Syllabus HSSLive Plus One Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of Plus One Kerala SCERT. Here HSSLive.Guru has given Higher Secondary Kerala Plus One Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus.

Board	SCERT, Kerala
Text Book	NCERT Based
Class	Plus One
Subject	All Subjects
Chapter	Plus One Chapter Wise Previous Questions
Category	Kerala Plus One

Kerala Plus One Chapter Wise Previous Year Questions and Answers

• Plus One Maths Chapter Wise Previous Questions and Answers
• Plus One Physics Chapter Wise Previous Questions and Answers
• Plus One Chemistry Chapter Wise Previous Questions and Answers
• Plus One Botany Chapter Wise Previous Questions and Answers
• Plus One Zoology Chapter Wise Previous Questions and Answers
• Plus One Economics Chapter Wise Previous Questions and Answers
• Plus One Business Studies Chapter Wise Previous Questions and Answers
• Plus One Accountancy Chapter Wise Previous Questions and Answers
• Plus One Computer Science Chapter Wise Previous Questions and Answers
• Plus One Computer Application Chapter Wise Previous Questions and Answers

We hope the given HSE Kerala Board Syllabus HSSLive Plus One Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus One Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus One