Kerala Plus One Maths Question Paper March 2019 with Answers

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf March 2019 helps in understanding answer patterns.

Kerala Plus One Maths Previous Year Question Paper March 2019

Time: 2 Hours
Total Scores: 60

Answer any 6 from questions 1 to 7. Each question carries 3 points.

Question 1.
Let A = {x: x is a prime number less than 11} and B = {x: x is an integer such that 2 ≤ x ≤ 8 }.
(i) Write C = A ∩ B.
(ii) Find the number of subsets of C which have 3 elements.
(iiii) What is the probability of getting a subset of 3 elements from the power set of C.
Answer:
(i) A = {2, 3, 5, 7}; B = {2, 3, 4, 5, 7, 8}
C = {2, 3, 5, 7} ∩ {2, 3, 4, 5, 7, 8} = {2, 3, 5, 7}
(ii) Number of subsets of C which have 3 elements = ⁴C₃ = 4
(iii) Probability of getting a subset of 3 elements from the power set of C = \(\frac{\text { favourable cases }}{\text { total number of cases }}=\frac{{ }^4 C_3}{2^4}=\frac{4}{16}=\frac{1}{4}\)

Question 2.
(i) Find (a + b)⁴ – (a – b)⁴.
(ii) Hence evaluate (√3 + √2)⁴ – (√3 – √2)⁴.
Answer:
(i) (a + b)⁴ – (a – b)⁴ = 2 × 4a³b + 2 × 4ab³ = 8ab(a² + b²)
(ii) (√3 + √2)⁴ – (√3 – √2)⁴ = 8√3 √2 ((√3)² + (√2)²)
= 8√3 √2(3 + 2)
= 40√6

Question 3.
Find the square root of the complex number 3 + 4i.
Answer:
Let x + iy = \(\sqrt{3+4 i}\)
Then (x + iy)² = 3 + 4i
⇒ x² – y² + 2xyi = 3 + 4i
Equating real and imaginary parts, we have
x² – y² = 3 ……….(1)
2xy = 4
We know the identity
(x² + y²)² = (x² – y²)² + (2xy)²
= 9 + 16
= 25
Thus, x² + y² = 5 ………….(2)
From (1) and (2), x² = 4 and y² = 1
or x = ±2 and y = ±1
Since the product xy is positive, we have
x = 2, y = 1 or, x = -2, y = -1
Thus, the square roots of 3 + 4i are 2 + i and -2 – i.

Question 4.
The sum of the first three terms of a Geometric Progression is \(\frac{13}{12}\) and their product is -1. Find the common ratio and the terms.
Answer:

Question 5.
Find the solution of the equation sin x + sin 3x + sin 5x = 0
Answer:
sin x + sin 3x + sin 5x = 0
⇒ 2 sin 3x cos 2x + sin 3x = 0
⇒ sin 3x (2 cos 2x + 1) = 0
⇒ sin 3x = 0; 2 cos 2x + 1 = 0

Question 6.
Consider the graph of the function f(x)

(i) Identify the function f(x)
(a) f(x) = sin x
(b) f(x) = cos x
(c) f(x) = tan x
(d) f(x) = cosec x
(ii) Using the function f(x) find \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{f(x)-f\left(\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}\)
Answer:

Question 7.
(i) Find the general term in the expansion of \(\left(x^2+\frac{1}{x}\right)^5\)
(ii) If the expansion of \(\left(x^2+\frac{1}{x}\right)^n\) has a term independent of x, then which of the following can be the value of n?
(a) 18
(b) 16
(c) 22
(d) 13
Answer:

Which means n should be a multiple of 3 and divisible by 2.
Hence (a) 18.

Answer any 8 from questions 8 to 17. Each question carries 4 scores.

Question 8.
In a school, a survey among 400 students, 100 were listed as taking apple juice, 150 as taking orange juice, and 75 were listed as taking both apple juice as well as orange juice.
(i) How many students take apple juice or orange juice?
(ii) How many take apple juice alone but not orange juice?
(iii) How many students were taking neither apple juice nor orange juice?
Answer:
Let A: Apple juice, O: Orange juice.
(i) n(A ∪ O) = n(A) + n(O) – n(A ∩ O)
= 100 + 150 – 75
= 175
(ii) n(A ∩ O’) = n(A) – n(A ∩ O)
= 100 – 75
= 25
(iii) n(A’ ∩ O’) = 400 – n(A ∪ O)
= 400 – 175
= 225

Question 9.
The figure shows the graph of the function f(x).

(i) Write the domain and range of f(x).
(ii) Find f(0) and f(-0.01).
(iii) Check the existence of \(\lim _{x \rightarrow 0} f(x)\).
Answer:
(i) Domain = R
Range = (-∞, 0) ∪ {1}
(ii) f(0) = 1, f(-0.01) = -0.01
(iii) \(\lim _{x \rightarrow 0^{-}} f(x)=0, \lim _{x \rightarrow 0^{+}} f(x)=1\)
\(\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)\)
Therefore, the limit does not exist.
OR
There is a break in the graph of f(x) at x = 0.
So the limit does not exist at x = 0.

Question 10.
Consider the set A = {-1, 1}
(i) Write all elements in A × A.
(ii) How many relations are there from A to A?
(iii) Write all functions from A to A which have Range = {-1, 1}.
Answer:
(i) A × A = {(-1, -1),(-1, 1),(1, -1),(1, 1)}
(ii) Number of relations = \(2^{n(A \times A)}\) = 24 = 16
(iii) There are two functions
f1 = {(-1, -1), (1, 1)}
f2 = {(-1, 1), (1, -1)}

Question 11.
Using the principle of mathematical induction, prove that n(n + 1)(n + 5) is a multiple of 3 for all n ∈ N.
Answer:
p(1): 1(1 + 1)(1 + 5) = 12 divisible by 3, hence true.
Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k² + 8k + 12)
= (k + 1){k² + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore, by using the principle of mathematical induction, true for all n ∈ N.

Question 12.
If z is a complex number with |z| = 2 and arg(z) = \(\frac{4 \pi}{3}\), then
(i) express z in a + ib form.
(ii) Find \(\bar{z}\)
(iii) Verify that (\(\bar{x}\))² = 2z
Answer:

Question 13.
Seven cards are drawn from a well-shuffled pack of 52 playing cards.
(i) How many ways can this be done?
(ii) What is the probability that the selection contains all kings?
(iii) What is the probability that selection does not contain a king card?
Answer:

Question 14.
(i) Write the contrapositive of the given statement.
“If a number is divisible by 9, then it is divisible by 3.”
(ii) Verify by the method of contradiction:
“p: √7 is irrational.”
Answer:
(i) If a number is not divisible by 3, it is not divisible by 9.
(ii) Assume that √7 is rational. Then √7 can be written in the form √7 = \(\frac{p}{q}\), where p and q are integers without common factors.
Squaring; 7 = \(\frac{p^2}{q^2}\)
⇒ 7q² = p²
⇒ 7 divides p²
⇒ 7 divides p
Therefore, p = 7k for some integer k.
⇒ p² = 49k²
⇒ 7q² = 49k²
⇒ q² = 7k²
⇒ 7 divides q²
⇒ 7 divides q
Hence p and q have a common factor 7, which contradicts our assumption.
Therefore, √7 is irrational.

Question 15.
Calculate the mean deviation about median for the following data:

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequencies	6	7	15	16	4	2

Answer:

Question 16.
Consider the word ASSASSINATION
(i) How many different ways can the letters of the word be arranged?
(ii) How many of these words have all vowels together?
Answer:
(i) In the word ASSASSINATION there are 13 letters, of which A appears 3 times, S appears 4 times, N appears 2 times, I appears 2 times and the rest all are different.
Therefore the total number of ways is \(\frac{13!}{3!\times 4!\times 2 \times 2!}\) = 10810800.
(ii) Vowels are A, A, A, I, I, O
Vowels can be arranged = \(\frac{6!}{3!\times 2!}\)
Vowels are taken as one unit, it can be arranged = \(\frac{8!}{4!\times 2!}\)
Total arrangement in which vowels are together = \(\frac{8!}{4 \times 2!} \times \frac{6!}{3!\times 2!}\)

Question 17.
Let A(0, 7, 10), B(-1, 6, 6,) and C(-4, 9, 6) are the vertices of a triangle.
(i) Show that is a right triangle.
(ii) Find the coordinate of the centre of the circle passing through the points A, B, C.
Answer:
(i) AB² = (-1 – 0)² + (6 – 7)² + (6 – 10)²
= 1 + 1 + 16
= 18
BC² = (-4 – (-1))² + (9 – 6)² + (6 – 6)²
= 9 + 9 + 0
= 18
AC² = (-4 – 0)² + (9 – 7)² + (6 – 10)²
= 16 + 4 + 16
= 36
∴ AB² + BC² = AC²
Therefore triangle ABC is a right triangle.
Since the triangle is a right triangle, the circumcentre will lie on the midpoint of the side AC.
Centre = \(\left(\frac{0-4}{2}, \frac{7+9}{2}, \frac{10+6}{2}\right)\) = (-2, 8, 8)

Answer any 5 from questions 18 to 24. Each question carries 6 scores.

Question 18.
The figure shows a unit circle and a line L which makes 30° with the positive direction of x-axis.

(i) Write the equation of the line L.
(ii) Write the coordinates of the points A and B.
(iii) Find the equation of the tangent line to the circle at A.
Answer:
(i) The slope of the line L is
m = tan 30° = \(\frac{1}{\sqrt{3}}\)
Equation of the line is y = mx
⇒ y = \(\frac{1}{\sqrt{3}}\)x
(ii) Point A = (r cos θ, r sin θ)
= (cos 30°, sin 30°)
= \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)
Point B lies in the third quadrant = \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\)
(iii) Equation of tangent is y – y₁ = m(x – x₁)
⇒ \(y-\frac{1}{2}=-\sqrt{3}\left(x-\frac{\sqrt{3}}{2}\right)\)
⇒ 2y – 1 = -2√3x + 3
⇒ √3x + y = 2

Question 19.
Consider two lines L₁: 2x + y = 4 and L₂: (2x – y = 2)
(i) Find the angle between L₁ and L₂.
(ii) Find the equation of the line passing through the intersection of L₁ and L₂ which makes an angle 45° with the positive direction of x-axis.
(iii) Find the x and y intercepts of the third line obtained in part (ii).
Answer:
(i) The slope of the line L₁ is -2.
The slope of the line L₂ is 2.
tan θ = \(\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{2-(-2)}{1+2 \times-2}\right|=\frac{4}{3}\)
(ii) The family of lines is of the form L₁ + λL₂ = 0
⇒ (2x + y – 4) + λ(2x – y – 2) = 0
⇒ (2 + 2λ)x + (1 – λ)y – 4 – 2λ = 0
⇒ Slope = \(-\frac{2+2 \lambda}{1-\lambda}\) = tan 45 = 1
⇒ -2 – 2λ = 1 – λ
⇒ λ = -3
Equation of the line is 2x – 2y = 1
(iii) 2x – 2y = 1
⇒ \(\frac{x}{1 / 2}+\frac{y}{-1 / 2}\) = 1
∴ x intercept is \(\frac{1}{2}\)
∴ y intercept is \(-\frac{1}{2}\)

Question 20.
If an ellipse passes through (3, 1) having foci (±4, 0), then
(i) Find the length of the major axis.
(ii) Find the standard equation of the ellipse.
(iii) Find the eccentricity and length of the latus rectum.
Answer:
(i) Length of the major axis = Sum of the distances of the point (3, 1) from the foci (±4, 0)

Question 21.
(i) Find sin 75°.
(ii) The figure shows ∆ABC with side AC = 4√2 units inscribed in a circle of radius 4 units. The length of the arc BDC is \(\frac{10 \pi}{3}\) units.

(a) Write ∠A in degree measure. (2)
(b) Find the length of the sides AB and BC. (2)
Answer:

Question 22.
(i) Solve \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\).
(ii) Solve thq inequalities 2x + 3y ≤ 12; x ≥ 1; y ≥ 2 graphically.
Answer:
(i) \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
⇒ 9(x – 2) ≤ 25(2 – x)
⇒ 9x – 18 ≤ 50 – 25x
⇒ 34x ≤ 68
⇒ x ≤ 2

Question 23.
(i) Find the derivative of y = x² using the first principle. (3)
(ii) Find \(\frac{d y}{d x}\) if y = \(\frac{x}{1+\tan x}\). (3)
Answer:

Question 24.
Consider the sequence 3, 6, 9, 12,………, 99
(i) How many terms are there in the given sequence? (1)
(ii) Find the mean of the sequence. (2)
(iii) Find the sum of squares of each term of the given sequence. (2)
(iv) Find the variance of the sequence. (1)
Answer:
(i) 3 × 1, 3 × 2, 3 × 3,………, 3 × 33
Hence there are 33 terms in the given sequence.
(ii) S₃₃ = 3 + 6 + 9 +…… 99 = 3(1 + 2 + 3 +……+ 33)