# Plus One Maths Model Question Paper 1

## Kerala Plus One Maths Model Question Paper 1

Time Allowed: 2 1/2 hours
Cool off time: 15 Minutes
Maximum Marks: 80

General Instructions to Candidates :

• There is a ‘cool off time’ of 15 minutes in addition to the writing time .
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• .Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Questions 1 to 7 carry 3 score each. Answer any 6.

Question 1.
Write the set $$\left\{ \frac { 1 }{ 2 } ,\frac { 2 }{ 3 } ,\frac { 3 }{ 4 } ,\frac { 4 }{ 5 } ,\frac { 5 }{ 6 } ,\frac { 6 }{ 7 } \right\}$$ r in the set- builder form.
a. If A and B are two sets such that A ⊂ B , then what is A ∪ B
b. There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C1 50 to the chemical C2, and 30 to both the chemicals C1 and C2. Find the number of individuals exposed to Chemical C1 but not chemical C2.

Question 2.
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 poiny B. Determine the height of the lamp post.

Question 3.
Find square root of the complex number 8- i6.

Question 4.
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 first examination?

Question 5.
a. A die is thrown until a six comes up. Write down the sample space for this experiment.
b. What is the probability of getting a doublet in throws of a pair of dice.

Question 6.
(a) A hyperbola with a = b is known as…………..

A. rectangular hyperbola
B. isosceles hyperbola
C. equilateral hyperbola
D. None of these

(b) Find the equation of an ellipse whose length of major axis is 26 and foci (± 5, 6)

Question 7.

Questions from 8 to 17 carry 4 score each. Answer any 8.

Question 8.

Question 9.
Let P(n) be the statement : 13 + 23 + 33 +………+n3
$$={ \left[ \frac { n(n+1) }{ 2 } \right] }^{ 2 }$$
a.Verify whether the statement is true for n = 1
b. Prove the result by using mathematical induction

Question 10.
a. Express (5 – 3i)3 in the form a + ib
b. Solve the equation -x+ x – 2 = 0

Question 11.
Solve the folowing linear inequalities graphically:

Question 12.
a. How many 3 digit even number can be formed form the digits 1,2,3,4,5 and 6, if the digits can be repeated?
b. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Question 13.
a. The base of an equilateral triangle with side 2 a lies along they-axis such that the mid-point of the base is at the origin. Find vertices of the triangle….
b. Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6)

Question 14.
a. find the centere and radius of the circle x2+ y– 8x + 10y – 12 = 0
b. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Question 15.
a. If p is the length of perpendicular from the origin to the line whose intercepts on the exes are a and b, then show that

b. If the lines 2x + y-3=0, 5x + ky -3=0 and 3x – y-2=0 are concurrent, find the value of k.

Question 16.
a. Find the octant in which the points (-3,1,2) and (-3,1,-2) lie.
b. Find the coordinates of a point on y-axis which are a distance of 5√2 from the point P (3,-2,5)

Question 17.
a. Write the negation of the following statement: If you do all the exercises in this book, you get an A grade in the class,
b. Verify by the method of contradiction, p: √2 is irrational.

Questions from 18 to 24 carry 6 score each. Answer any 5.

Question 18.
a. Draw the graph of the function f: R→ R defined by f(x) = xx ∈ R
b. If set A has 2 elements and set B has 3 elements, then the number of function from A to B is
(i) 64
(ii) 6
(iii) 9
(iv) 8

Question 19.
a. If sec x = 13/5, x lies in fourth quadrant, find other t-functions.
b. Prove that show that tan 3x tan 2x tan x = tan 3x – tan 2x – tan x
c. Find the value of tan π/8

Question 20.
a. Find the sum of the sequence 7, 77, 777,  7777, ………….. to n terms.
b. Find the sum of first n terms of the series: 3 12 + 5 x 22 + 7 x 32+…….

Question 21.
a. Find the middle term in the expansion of

b. Find the term independent in the expansion of

Question 22.
a. Find the derivative of x sin x form first principles.

Question 23.
a. Find mean deviation about median for the following data.

b. Calculate mean, variance and standard deviation of the following distribution:

Question 24.
Four cards are drawn from a well shuffled pack of 52 cards. Find the probability that it contains
a. all aces
b. atmost 2 aces
c. atleast two aces

– 15 – 8i
Let $$\sqrt { 8-6i }$$ = x + iy……….. (1)
Squaring we have,
8 – i6=(x + iy)2 = x2 + i2xy + i2y2 = (x– y2)  +i2xy
Equating the real and imaginary parts, we have
x– y2= 8 ………….. (2)
2xy = – 6 ……………….. (3)
We know that (x+ y2)2
= (x– y2)2 + 4xy2
= (8)2+(-6)2= 64+36 = 100
∴ x+ y= 4m = √100 = 10 ………….. (4)
(2) + (4) we have,
x2 -y = 8 x2 + y= 10 2x2 = 18
x2 = 9 ⇒ x = ± 3
In (4), we have, 32+y2=10 ⇒ y2
=10 – 9 = 1 ⇒ y = ± 1
Since 2xy = -6,
When x = 3, y = -1 and when x = -3, y = 1
3-i and -3 + i are the square roots.

Let the mark in the first exam = x
Mark in the second exam = x+5
Mark in the third exam = x + 10 x + x + 5 + x + 10.

x >75
Minimum mark Arathi should get in the first exam = 75.

a. s = {6,(1,6),(1, 1, 6),….,(2, 6),(2, 1, 6),…. }
b. n (s) = 36
doublets are (1,1), (2,2), (3,3),
(4,4),(5,5) and (6,6)
∴ P (getting a doublet) =  $$\frac { 6 }{ 36 }$$ = $$\frac { 1 }{ 6 }$$

a. C.
b. 2a = 26 => a = 13
foci = (±5,0) => c = 5
But c2 = a2– b2 ⇒ b=a– c= 132-5= 144

a. (5 – 3i)= 5– 3 x 52(3i)+3 x 5(3i)2– (3i)3
= 125 – 225i+ 135i2 – 21 i3
= 125 – 225i+ 135(-1)-27(-i)
= 125 – 225i – 135 + 27i
= -10 – 198i
b. -x+ x + 2 = 0
a = 1 ; b = 1 ; c = 2
D = b2 – 4ac = (1)2 – 4 (-1) (2)
= 1 – 8 = -7 < 0

The shaded region is the solution region.

a. Here unit place can filled in three ways (i.e. by 2, 4, 6), whereas tens and hundred place can be filled in 6 ways.
b. Out of available nine courses, two are compulsory, Hence the student is free select 3 courses out of 7 remaining courses. Then the number if ways of selecting courses out of 7 courses

(a) Let ABC be the given equilateral triangle with side 2a. Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. i.e., BO = OC = a, where O is the origin Now, it is clear that the coordinates of point B are (-a, 0) and C are (a,0), while the coordinates of point. Using Pythagoras theorem to ΔAOC, we obtain
AC2 = OA2 + OC(2a)2
= OA2+ a2 4a2– a2 = OA2
⇒ OA=3a2 ⇒ OA = √3a

∴ co-ordinates of A are (±√3a ,0). Thus, the vertices of the given equilat­eral triangle are (0, a),(0,-a), and (√3a,0) or (0, a), (0, -a), and (-√3a ,0)
(b) Any line through (-3,5) is y-5 = m(x-(-3)) = m(x + 3) Slope of line joining (2, 5) and (-3, 6) =

∴  slope of a line⊥ to it = 5
∴reqd. line is y-5 = 5(x+3)
⇒ 5x – y + 20=0

a. centre = (-g, -f) = (4, -5)

b. Equation is
(x1 – x) (x – x2)+(y1 – y) (y – y1)
(x – a)(x – 0)+(y – 0)(y – b) = 0
x2-ax + y2 – by = 0
x2 + y2 – ax -by = 0

a. The equaton of line making intercepts a and b on the axes is

(intercept from) ….(i)
Given p = perpendicular distance from

b. Three lines are said to be concurrent, if they pass through a common point, i.e., point of intersection of any two lines lies on the third line. Here given lines are
2x + y – 3 = 0 ……….. (1)
5x + ky – 3 = 0……….. (2)
3x – y – 2 = 0 …………. (3)
Solving (1) and (3) by cross – multipli­cation method, we get,

Therefore, the point of intersection of two lines is (1, 1). Since above three lines are concurrent, the point (1, 1) will satisfy equation (2) so that
5.1+k.1- 3 = 0 ⇒ k = -2.

a. (-3,1,2) – 2nd octant
(-3,1,-2) – 6th octant
b. Let A (0,y,0) be a point on y-axis having distance 5 √2 from (3,-2,5)

⇒ y+ 4y – 12 = 0
⇒ (y + 6) (y-  2) =0
⇒ y = -6,2
∴ A(0,-6,0) and ,4(0,2,0) are the required points

a. If you do not do all the exercises in this book, you will not get an a grade in the class.
b. Let us assume that be rational.
∴ √2 = a/b, where a and b are co-prime,
i.e., a an b have no other common factors except 1.
Then 2b2 = a2 ⇒ 2 divides a.
∴ there exists an integer tk> such that a – 2k
∴ 2b2 = 4k2 ⇒ 4k= 2b= 2k2 = ⇒ b2=⇒ 2 divides b.
i.e., 2 devides both a and b, which is contradiction to our assumption that a and b have no common factor.
∴ our supposition is wrong.
∴  √2 is an irrational number.

a.

b. (iii) n(b)n(A) =3= 9

a. This is not a G.P., however, we can relate it to a G.P by writing the terms as Sn =7 + 77 + 777 + 7777 + …….. to n terms

b. Let Tn denote the nth term of the given series

a. Here n = 10 is even
Middle term $$\frac { 10 }{ 2 }$$ +1 = 6th term

Since we have to find a term indepen­dent of x, i.e., term not having x, so