Annual Test XI
Time Allowed : 3 Hours
[Maximum Marks : 100]
General Instructions:
(i) All question are compulsory.
(ii) There is no overall choice. However internal choice has been provided in some questions.
(iii) Q. Nos. 1 to 10 are of 3 marks each, Q. Nos. 11 to 20 are of 4 marks each and Q. Nos. 21 to 25 are of 6 marks each.
(iv) Write the serial number of the question before attempting.
(v) If you wish to answer any questions already answered for any reason, cancel the previous answer.
(vi) Use of calculator is not permitted. However, you may ask for logarithmic tables.
1.
2.
3.
4.
5.
In a school, 50 people speak English , 20 speak Tamil and 10 speak both English and Tamil How
many speak at least one of these two languages?
sin A  sin 3 A
Prove that :
 tan 2 A .
cos A  cos3 A
3
Prove that : sin10 sin 50 sin 60 sin 70 
.
16
34 n  2  52 n 1 is divisible by 14 for all n  N.
A straight line passes through (1, 1) and portion of the line intercepted between the axes is
divided at this point in the ratio 3 : 4. Find the equation of the line.
Or
Find the acute as well as obtuse angle between the lines :




y  2  3 x  6 and y  2  3 x  9 .
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
1 x  1 x
.
sin x
4 x  5sin x
Find the derivative of :
with respect to x
3x  7cos x
(a) Write the contrapositive of the following statement “If number is divisible by 9, then it is
divisible by 3”.
(b)
Write the converse of the statement “If a number n is even, then n2 is even.”
A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with
number ‘3’. If die is rolled once, determine.
(a) P(2)
(b) P(1 or 3)
(c) P(not 3).
A card is drawn from a pack of cards. Find the probability of getting a king or a heart or a red card.
Find the general solution of the equation : sin 2x + cos x = 0.
Find the equation of the circle through (2, 4) and having centre at the intersection of the lines
x – y = 4 and 2x + 3y + 7 = 0.
3
12
If sin x  ,cos y   , where x and y both lie in second quadrant, find the value of sin(x + y).
5
13
1  7i
Convert the complex number z 
in the polar form.
 2  i 2
Evaluate : lim
x 0
Find the solution of quadratic equation in the complex number system : 5x2 – 4ix + 9 = 0.
Solve the given system of inequations graphically: 2x + y  4; x + y ≤ 3, 2x – 3y ≤ 6.
Or
FTM-01
Page 1
17.
18.
19.
20.
21.
22.
A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution
must be added to it so that acid content in the resulting mixture will be more than 15% but less
than 18%?
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 ?
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come
together ?
If S1, S2, S3 be the sums of the first n natural numbers, their squares, their cubes respectively,
show that 9 S 22  S3 1  8S1  .
3 3
3069
, ,......... are needed to give the sum
?
2 4
512
Or
Prove that the product of 2nd and 3rd terms of an A.P. exceeds the product of 1st and 4th by twice
the square of the difference between the 1st and 2nd.
Prove that : tan x + tan(60 + x) + tan(120 + x) = 3 tan 3x.
(a) Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, B = {3, 4, 5, 6}. Find (A – B)′.
(b) Draw the graph of the function : f: R  R such that f(x) = x - 2.
How many terms of G.P. 3,
23. Mean of 50 observations is 32. Mean of 60 observations is 52 .Show that combined mean is
24.
25.
472
.
11
(a) Find the co-ordinates of the vertices, e (eccentricity) and the length of latus rectum of the
ellipse: 4x2 + 9y2 = 36.
(b) Find the ratio in which the line joining the points (2, 4, 5) and (3, 5, -4) is divided by the xy-plane.
Find n, if the ratio of the fifth term from the beginning to the fifth term from end in the
 1

1
expansion of  2 4  1  is 6 :1 .


34 

Or
If three consecutive coefficients in the expansion of (1 + x)n are in the ratio 6 : 33 : 110, find n and r.
ANSWERS
4 x  3 y  7  0; 120 or 60
6.
1,7 Quotient rule
1.
60
5.
8.
(a)
(b)
if a number is not divisible by 3 then it is not divisible by 9
If n 2 is even then n is even
9.
1 1 5
, ,
2 2 6
7
13
11.
12.
x 2  y 2  2 x  6 y  40  0
13.
16.
between 120 liters and 300
17.
10.
 π
nπ  (1) n   
 6
56
3π
3π 

14.
2  cos  i sin  15.
65
4
4 

441
18. 6090 24. 5:4
25.
9i
, i
5
n=10 or 12:2
SOLUTIONS
FTM-01
Page 2
1.
F : People speaking English,
n(F) = 50, n(S) = 20; n(F  S) = 10,
2.
Consider L.H.S
sin 3 A  sin A 2sin 2 A cos A
=
= tan 2 A R.H.S.

cos3 A  cos A 2cos 2 A cos A
Consider L.H.S.
3
= sin 10 sin 50 sin 60 sin 70 =
sin10 sin  60  10  sin  60  10 
2
3 3
3


sin10  sin 3 10

sin10 sin 2 60  sin 2 10 

2
2 4

3
3
3
3sin10  4sin 3 10 


sin  3  10  
 sin 30


8
8
8
3 1
3
= R.H.S

 
8 2 16
Or
Consider L.H.S.
= a sin A – b sin B = k sin2 A – k sin2 B
[Using sine rule]
2
2
=k[sin A – sin B] = k sin(A + B) sin(A – B) = k sin( - C) sin(A – B) [ A + B + C = ]
= k sin C sin(A – B) = c sin(A – B) = R.H.S [Using sine rule]
x y
Let line be   1
a b
The coordinates of A and B are (a, 0), (0, b).
P(1, 1) divides AB in the ratio 3 : 4
3.
5.
S : People speaking Tamil
n(F  S) = n(F) + n(S) – n(F  S)
= 5 0 + 20 – 10 = 60
7
7
 4a  0 0  3b 
 a  and b  .

,
  1,1
7 
4
3
 7
Substituting in (1), we get
x
y
4x 3y

1,


1
7/4 7/3
7
7
 4x + 3y – 7 = 0 is the required equation.
Or




Given lines are y = 2  3 x  6 and y  2  3 x  9
m1  2  3 and m2  2  3
Let  is angle between the lines, then
FTM-01
Page 3
tan  
6.
m1  m2
2 3 2 3

1  m1m2
1 2  3 2  3

dy d  4 x  5sin x 
4 x  5sin x

 

dx dx  3x  7cos x 
3x  7cos x
d
d
 3x  7cos x   4 x  5sin x    4x  5sin x   3x  7cos x 
dx
dx
=
2
 3x  7cos x 
=
=
=
9.
10.
11.

Let y 
=
8.

2 3
  3  3
1 4  3
  = 120 or 60
Hence obtuse angle is 120 and acute angle is 60
1  x   1  x 
1 x  1 x
Consider Lt
= Lt
x 0
x 0 sin x  1  x  1  x 
sin x


2x
x
1
= Lt
= 2  Lt

x 0 sin x  1  x  1  x 
x 0 sin x  1  x  1  x 




1
2
= 2 1
 1
2
1 0  1 0

7.

=
 3x  7 cos x  4  5cos x    4 x  5sin x  3  7sin x 
 3x  7 cos x 2
12 x  15 x cos x  28cos x  35cos 2 x  12 x  15sin x  28 x sin x  35sin 2 x


 3x  7 cos x 2
35 cos2 x  sin 2 x  15 x cos x  28 x sin x  28cos x  15sin x
 3x  7cos x 2
35  15  x cos x  sin x   28  x sin x  cos x 
 3x  7 cos x 2
(a) If number is not divisible by 3, then it is not divisible by 9.
(b) If n2 is even then the number n is even.
Two faces with ‘1’; three faces with ‘2’ and one face with ‘3’.
2 1 3 1
23 5
(a)
P(2) = 3/6 = ½ (b)
P(1 or 3) =
P(not 3) =
  (c)

6
6
6
6 2
n(S) = 52,
Event A : king;
Event B : heart; Event C : red card.
P(A or B or C)
= P(A) + P(B) + P(C) – P(A  B) – P(B  C) – P(C  A) + P(A  B  C)
4 13 26 1 13 2
1 28 7
=
 
   


52 52 52 52 52 52 52 52 13
Consider the equation sin 2x + cos x = 0
 2 sin x cos x + cos x = 0
 cos x(2 sin x + 1) = 0  cos x = 0 or 2 sin x + 1 = 0
 cos x = 0 or sin x = - ½
FTM-01


 x   2n  1 , n  I or sin x   sin
2
6
Page 4
12.
 
 
 sin x = sin   
 x = n + (-1)n    , n  I is the general solution
 6
 6
Centre is at the intersection of lines x – y = 4 and 2x + 3y + 7 = 0
Solving the two equations, we get x = 1, y = -3
 Centre is (1, -3)
Also circle passes through the points (2, 4)
 Radius =
 2  12   4  32
= 1  49  50
 Equation of circle is (x – 1)2 + (y + 3)2 =
13.

50

2
 x2 – 2x + 1 + y2 + 6y + 9 = 50
 x2 + y2 – 2x + 6y – 40 = 0 is the required equation.
3
9
16
4
sin x   cos x   1 

 ,
5
25
25
5
as x  2nd quadrant.
12
cos y = 
13
2
144
25
5
 12 
 sin y  1      1 

 as y  2nd quadrant.
169
169 13
 13 
Consider sin (x + y) = sin x cos y + cos x sin y.
3  12   4  5  36  20 56
 sin  x  y            

5  13   5  13 
65
65
14.
15.
16.
Consider
1  7i  3  4i  3  4i  21i  28 25  25i
1  7i
1  7i
1  7i
=


z


2
9  16
25
25
 2  i  4  1  4i 3  4i
 z = -1 + i = r(cos  + i sin )
……………..(1)
r cos  = -1, r sin  = 1
 r  11  2
1
1
 cos   
,sin  
  =  - /4 = 3/4, as   2nd quadrant.
2
2
Hence from (1),
3
3 

z  2  cos
 i sin 
4
4 

Consider the equation 5x2 – 4ix + 9 = 0,
a = 5, b = -4i, c = 9, D = b2 – 4ac = (-4i)2 – 4.5.9 = -16 – 180 = -196
  4i   196 4i  14i
 Solution is x 

10
10
4i  14i 4i  14i 9i
 x
,
 , i is the required solution.
10
10
5
Consider the inequations: 2x + y  4, x + y ≤ 3, 2x – 3y ≤ 6
Corresponding equations are
2x + y = 4;
x + y = 3;
2x – 3y = 6
FTM-01
Page 5
x
0
2
1
17.
18.
y
x
y
x
4
0
3
0
0
3
0
3
2
1
2
6
Substituting (0, 0) in inequations, we get
0+04
 0  4, false
0+0≤3
 0 ≤ 3, true
0–0≤6
 0 ≤ 6, true
y
-2
0
2
Plotting the above information, we get shaded portion as the solution.
Or
Let x litres of 30% of acid solution is added.
Total solution = (600 + x) litres.
We have,
15
12
30
18
 600  x    600   x   600  x 
100
100
100
100
 5(600 + x) ≤ 2400 + 10x ≤ 6(600 + x)
 3000 + 5x ≤ 2400 + 10x ≤ 3600 + 6x
 3000 + 5x ≤ 2400 + 10x and 2400 + 10x ≤ 3600 + 6x
 600 ≤ 5x; 4x ≤ 1200
 120 ≤ x; x ≤ 300
 120 ≤ x ≤ 300
30% acid solution should be between 120 litres to 300 litres.
Total group: 4 girls and 7 boys
For a team of 5 members having at least 1 boy and 1 girls we have (1) 1 boy, 4 girls (2) 2 boys,
3 girls (3) 3 boys, 2 girls (4) 4 boys, 1 girl.
 Total possibilities are
4
C4  7C1 + 4C3  7C2 + 4C2  7C3 + 4C1  7C4
= 1  7 + 4  21 + 6  35 + 4  35
= 7 + 84 + 210 + 140 = 441
For MISSISSIPPI we have one ‘M’, four ‘I’s, four ‘S’s, two ‘P’s.
11!
1110  9  8  7  6  5  4!
 Total distinct permutation are

 6930
4!4!2!
4! 24  2
8! 4!
Total distinct permutation with four ‘I’s together
  840
4!2! 4!
FTM-01
Page 6
19.
Therefore, distinct permutation, so that four I’s are not together = 6930 – 840 = 6090
n  n  1
Given S1   n 
2
 n  n  1 
S3   n  

 2 
n  n  1 2n  1
S2   n 
2
2
3
6
Consider R.H.S. = S3(1 + 8S1)
 n  n  1 
=

 2 
2
 8n  n  1 
1 

2


 n  n  1 
 n  n  1 
2
2
=
  4n  4n  1 = 
  2n  1
2
2




2
2
 n  n  1 2n  1 
 9
 9S2 = L.H.S
23


3069
3 3
Let Sn 
Given G.P. is 3, , ,............
12
2 4
a = 3, r = ½
n
  1  
3 1    
n
  1 n 
3069
3069
  2  
1

1  


 6 1     
1
512  6
512
2
  2  
1
2
n
10
1023
1
1
1

 
   1
 n = 10.
1024 1024  2 
2
Hence 10 terms are needed.
Or
Let A.P. be a, a + d, a + 2d, …………………
We have to show that
(a + d)(a + 2d) – a(a + 3d) = 2 {(a + d) – a}2
Consider
L.H.S = (a + d)(a + 2d) –a(a + 3d) = a2 + 3ad + 2d2 – a2 – 3ad = 2d2 = 2{(a + d) – a}2 = R.H.S
L.H.S = tan x + tan(60 + x) + tan(120 + x)
= tan x + tan(60 + x) + tan{180 – (60 – x)}
tan 60  tan x
tan 60  tan x
= tan x + tan(60 + x) – tan(60 – x)
= tan x 

1  tan 60 tan x 1  tan 60 tan x
3  tan x
3  tan x

= tan x 
1  3 tan x 1  3 tan x
2
20.
21.
= tan x 


 
3 tan x 1 
3  tan x 1  3 tan x  1  3 tan x
1 
3 tan x


3  tan x

3  3tan x  tan x  3 tan 2 x  3  tan x  3tan x  3 tan 2 x
1  3tan 2 x
 3tan x  tan 3 x 
8tan x
tan x  3tan 3 x  8tan x
9 tan x  3tan3 x
3
= tan x 
=
=



2
1  3tan 2 x
1  3tan 2 x
1  3tan 2 x
 1  3tan x 
= 3 tan 3x = R.H.S.
(a) Given U  {1,2,3,4,5,6,7,8,9}
A  {2,4,6,8}, B  {3,4,5,6}
= tan x 
22.
FTM-01
Page 7
( A  B) U  ( A  B)
 U  ({2,4,6,8}  {3,4,5,6})
 {1,2,3,4,5,6,7,8,9}  {2,8}
 {1,3,4,5,6,7,8,9}
(b)
 x  2, x  2
f ( x)  x  2  
 x  2, x  2
Given
x
y
0
2
1
1
24.
(a)
2
0
3
1
4
2
x2 y2

1
9
4
Here a 2  9 , b2  4 vertices are ( a,0)  ( 3,0)

Given ellipse is 4 x 2  9 y 2  36
a 2  b2
94
5
2b 2 2  4 8
,
Length
of
latus
rectum


 .


a
3
3
9
3
a2
Let xy  plane divides the join of A(2,4,5) and B(3,5, 4) in the ratio k :1.
Eccentricity (e) 
(b)
 3k  2 5k  4 4k  5 
The point of division is 
,
,
 as this point lies on xy  plane.
 k 1 k 1 k 1 
4k  5
5
5
Ratio is :1  5: 4 internally.
 0  4k  5  k 
k 1
4
4
n
25.
4
n4 
 1


 1
1  th
1
n

4
4
Given 2  1 5 term from beginning is
T5  T41  C4  2    1 
 


 
 
34 

 34 
Also 5th term from end = (n – 5 + 2)th term from beginning
n4
n4
1
n

4 
1 nn 4 
C
2


4


1 
n
3  6

4
 Tn 3  T n 41  Cn 4   2 
Given
 1
n4
 
 
1
 
4
4
1


n
3 
Cn  4  2    
3


1
n
2
24
1
 
3
n
24
n
34
4

9
n4
4
 6

 6
24  1 

 
4  3
[ C4  Cn  4 ]
n
n
n
  6  4  36 6
n
5
  6 4  6 2
Or
Cr - 1 : nCr : nCr + 1 = 6 : 33 : 110
n
r ! n  r ! 2
n!
C
6
 n r 1 



33
n!
11
Cr
 r  1! n  r  1!

2

4
 6
n 5
  n  10
4 2
n
FTM-01

r
2

n  r  1 11
 2n  2r  2  11r
Page 8
 2n – 13r + 2 = 0
n
C
33
Also n r 
Cr 1 110

n!

r ! n  r !
………(1)
 r  1! n  r  1!
n!
r 1 3
 3n  13r  10

 3n  3r  10r  10
n  r 10
ss
Solving (1) and (2), we get
n = 12, r = 2

FTM-01

3
10
……….(2)
Page 9