Vidyalankar
F.Y. Diploma : Sem. I
Basic Mathematics
Prelim Question Paper Solution
4 [2x  28]  3 [3x  7] + 9 [12  2] = 0
8x  112  9x + 21 + 108  18 = 0
x  1 = 0
x = 1
 1 2
1. (b) x = 
,
 3 4 
34 65 
= 

 9  1 12  3 
 7 11
3x + y = 

 8 9 
al

an
4 5 
y =

 1 3 
 1 2 4 5 
3x + y = 3 


 3 4   1 3 
 3 6  4 5 
= 


 9 12   1 3 
ka




dy
3 9 
2
1. (c)  A = A  A = 

 1 9 
(9)(9)  (9)(9) 
 3.3  (9)(1)
=

(1)(3)  (9)(1) (1)(9)  (9)(9)
0 0 
= 

 6 72
A2

=
0 0
6 72
= 0
Vi
hence A2 is a singular matrix
 1 1
3 2 
=  1 1
B = 3 2 




 4 2 32
 0 4 32
L.H.S. = A + B
3 2 
 1 1


3 2 
1
1


L.H.S. =




 0 4  32  4 2  32
1. (d) We have,


6
r
4 3 9


1. (a) Given:  3 2 7  = 0
 1 4 x 
A
1013/FY/Pre_Pap/Maths_Soln
Prelim Question Paper Solution

 21 
 12 
 42  32

  13 

R.H.S. =   31
  40 

… (1)
ka
 2 1
L.H.S. =  4 1


 4 2 32
R.H.S. = B + A
 1 1
3 2 


R.H.S. = 3 2
 1 1




 4 2  32 0 4 32

r
  31

L.H.S. =  13 
  04 
 12  
 21 
 24 32
an
 2 1

R.H.S. =  4 1


 4 2 32
From eq. (1) and eq. (2) we get,

L.H.S. = R.H.S.

(A + B) = (B + A) is verified.
… (2)
dy
2
al
1
1
A
B
=
=




x
x
1
1
x
x
x x
Multiplying throughout by x (x 1), we get,
1 = A (x  1) + Bx
1. (e) Let
When
x = 1,
1 = 0 + B (1)
When
 A = 1
Vi

x = 0,
1 = A (0  1) + 0
1
1
1
=

2
1
x
x
x x
 B=1
1. (f) We know cos(A + B) = cos A cos B  sin A sin B
put A = B = , cos 2 = cos  cos  sin  sin
= cos2  sin2
2
2
But cos  = 1  sin  & sin2 = 1  cos2
= 1  2sin2 
 cos 2 = (1  sin2 )  sin2 
2
2
 cos 2 = cos   (1  cos ) = 1  2 sin2 
 cos 2 = cos2   (1  cos2 ) = 2 cos2   1
1013/FY/Pre_Pap/Maths_Soln
7
Vidyalankar : F.Y. Diploma  Mathematics
1. (g) Compound Angle
The algebraic sum or difference of two or more angles is called compound angle.
 A B
cos 

 2 
= 80
 (1)
AB
= 20
 (2)
2A = 100
2B = 60
r
A+B
 A = 50
 B = 30
ka
1. (h) Given : 2 sin 40 cos 10 = sin A + sin B
 AB
2 sin 40 cos 10 = 2 sin 

 2 
Equating both sides we get
A B

40 =
2
A B
and
10 =

2
equation (1) + equation (2)

equation (1)  equation (2)

cos 21  sin21
cos 21  sin21
Dividing numerator and denominator by cos 21°, we get,
1  tan21
=
1  tan21
tan45  tan21
=
[ tan 45° = 1]
1  tan45  tan21
 tan A  tanB


= tan (45°  21°)
 1  tan A  tanB  tan A  B 
= tan 24°
= R.H.S.
al
an
1. (i) L.H.S. =
dy
 1
 1
1. (j) L.H.S. = tan1  7  + tan1  13 
In this case,
1
Let x = > 0,
7
1
1 1
1
> 0 and xy = 
=
<1
13
7 13 91
xy
L.H.S. = tan1  1  xy 
Vi

y=
 1 1

 7  13 
= tan1   1   1  
 1     
  7   13  
9
= cot1  2 
= R.H.S.
8
1013/FY/Pre_Pap/Maths_Soln
 13  7 
 91 
= tan1 
1 
 1

91 

 20 / 91
= tan1  90 / 91
2
= tan1  9 


 1
  tan1    cot 1 x 
x


Prelim Question Paper Solution
1. (k) Let the lines be
3 3
=
and
2 2
2 2
2 : 2x  3y  7 = 0
 Its slope, m2 =
=
3 3
 It  is the a cute angle between lines, then
m1  m2
tan  =
1 m1.m2
1 : 3x  2y + 4 = 0
Putting, m1 = 3/2,
m2 = 2/3, we get,
3 2
94

6
2
3
tan  =
=
2
 3  2 
1    
 2  3 
 5 
 = tan1  12 
5
12
ka

r
 Its slope, m1 =

=
an
1. (l) Range = largest Quantity  smallest Quantity
= 31  1
= 30.
al
2. (a) Then writing into D, Dx, Dy and Dz as required in Cramer’s Rule, we have
3 3 1
D = 2 1 2
4 3 2 33
dy
= 3(2  6) 3(4  8) 1(6 + 4)
= 24 + 12  10
11
3
=
=
3(8) 3(4) (10)
22
1
D x = 0 1 2
25 3 2
Vi
= 11(2  6) 3(0  50) 1(0 + 25)
= 88 + 150  25
=
=
11(8)  3(50)  1(25)
37
=
=
150  11(4) 1(50)
200 + 44
3 11 1
Dy = 2 0 2
4 25 2
= 3(0  50)  11(4  8) 1(50)
= 150 + 44  50
= 156
1013/FY/Pre_Pap/Maths_Soln
9
Vidyalankar : F.Y. Diploma  Mathematics
3
3
11
Dz = 2  1 0
4 3 25
al
A
  2  3  3   1
AB = 
3  5

 3 2 4 
AB = 

 8 1 2  23
dy

r
 2 3 
 3 1 2 
= 
and
B = 


 1 5  22
 1 0 0  23
L.H.S. = (AB)T
 2 3 
3 1 2 

AB = 


 1 5  22  1 0 0  23
2. (c) We have
Now
 5 3 7 
7 7 1


an
 (x  4) (5  27) 7 
2. (b) LHS = 
=
(3x  4) (152y) 1 
Comparing a11 and a12
x+4=5 x=1
5 + 2y = 3  y = (35) / 2 = 4
= 75  150 + 110
= 115
ka
= 3(25  0)  3(50  0) + 11(6 + 4)
= 75  40
Then by Cramer’s rule,
D
37
37
x = x =
= 
22
D
22
Dy
156
156
=
y =
=
D
22
22
Dz
115
115
z =
=
=
D
22
22
37
78
115
 x= 
, y=
, and z =
22
11
22


 2  0   4  0  
 1 0   2  0  
Vi
 3 8
= L.H.S.
(AB) =  2 1


 4 2  32
T

… (1)
We have,

10
A
 2 3 
= 

 1 5  22
B
 3 1 2 
= 

 1 0 0  23
R.H.S. = BT  A T
1013/FY/Pre_Pap/Maths_Soln
… (1)
2
 AT = 
 3
3
T
 B =  1

 2
1
5  2
1
0

0  23
Prelim Question Paper Solution

 From eq. (1) and eq. (2) we get
L.H.S. = R.H.S.

(AB)T = BT  A T is proved.
1  4 
0  0 
5 3
8 0 
5
8

0 
… (i)
 2 4
 1 1
BT =  2 3 
AT =  3 5  ,




 1 0 
 4 0 
 2  1 4  1
T
T
A + B =  3  2 5  3


 1  4 0  0 
dy

1
A+B = 
5
1
T
(A + B) =  5

3
4
0 
al
Now,
 1 2
B= 
1 3
2  1 3  2
A+B = 
4  1 5  3
an
 2 3 1
2. (d) A = 

4 5 0 
… (2)
r

 6  3 3  5 


R.H.S. =  2  0   1  0 
  4  0   2  0  
 3 8
R.H.S. =  2 1


 4 2  32
ka

 3 1
 2 1

R.H.S. = 1 0 
 3 5 


 22
 2 0  32 
1 5 
= 5 8 


3 0 
Vi


T
T
A +B
… (ii)
From (i) and (ii)
(A + B)T = AT + BT
2. (e)
x5
x 3  x 2  6x
=
=
x5
x  x2  x  6
x5
x(x  2)(x  3)
1013/FY/Pre_Pap/Maths_Soln
11
Vidyalankar : F.Y. Diploma  Mathematics
A
B
C


x x2 x3
5
x5
5
=
=
=
6
(x  2)(x  3) x 0
(2)(3)
=
=
x5
x(x  3) x  2
=
3
(2)(5)
C
=
x5
x(x  2) x 3
=
8
8
=
3.5
(3)(5)
3x  2
 x  1  x 2  1
=
3x  2
 x  1  x  1  x  1
=
A
B
C


2
x  1  x  1
x 1
=
=
3
25
=
3
10
=
8
15
3x  2
 x  1 2  x  1
ka
2. (f) Let
B
r
A
Multiplying throughout by (x + 1)2 (x  1), we get,
3x + 2 = A (x + 1) (x  1) + B (x  1) + C (x + 1)2
3 + 2 = 0 + 0 + C (2)2
 4C = 5
an
When x = 1,
When x = 1,
3 + 2 = 0 + B (1 1) + 0
 2B = 1
 C = 5/4
 B = 1/2
dy
al
When x = 0, B = 1/2, C = 5/4
1
5
2 = A (1) (1) + (1) + (1)2
2
4
1 5
 2 = A  
2 4
5 1
528
5
 A =  2
=
=
4 2
4
4
3x  2
5
1
5

=


4  x  1 2  x  1 2 4  x  1
 x  1  x 2  1
Vi
 1 3 2


be the given matrix.
3. (a) Let A =  3 2 5 
 2 3 6  33
1

3
2
A = 3 2 5
2 3 6 33
12

A = 1[(6) (2)  (5) (3)] 3[(3) (6)  (2) (5)] + 2[(3) (3)  (2)(2)]

A = [12 + 15]  3[18  10] + 2[9 + 4]

A = 3  3(8) + 2(5)
1013/FY/Pre_Pap/Maths_Soln
Prelim Question Paper Solution

A = 3  24  10

A = 31

A 0
 A1 exists
c1
c2
c3 33
r
To find A1:
1 3 2
a1 b1
= a 2 b2
Let A = 3 2 5
2 3 6 33
a 3 b3
C1 = 
A2 = 
= (18  10)
2 6
3 2
2 3
3
2
3 6
1 2
2 6
= 8
= 9 + 4
= 5
= (18 (6))
= 24
=64
=2
al
B2 = 
3 5
an
B1 = 
ka
Let A1, B1 and C1 ….. are cofactors of elements a1, b1, c1…. respectively.
2 5
A1 = 
= 12 + 15
=3
3 6
C2 = 
3
2
2 5
= (3  6)
=9
= 15 + 4
= 19
= (5  6)
=1
= 2 9
= 11
dy
A3 = +
1 3
2 3
B3 = 
3 5
1 3
3 2
Vi
C3 = +
1 2
8 5 
 A1 B1 C1 
 3



 The matrix of cofactors = A 2 B2 C2 = 24 2
9 




 A 3 B3 C3 
 19
1 11 3
 adj A = Transpose of cofactors matrix
 3 24 19 
 adj A =  8 2
1 


 5 9
11 3
1013/FY/Pre_Pap/Maths_Soln
13
Vidyalankar : F.Y. Diploma  Mathematics
A
We have
 3 24 19 
1
1 
8
=
adjA =
2
1 

31 
A
 5 9
11
… (1)
x + 3y + 2z = 6
3x  2y + 5z = 5
2x  3y + 6z = 7 are given equations.
Writing in matrix form, we get
 1 3 2  x 
 3 2 5  y 

 
 2 3 6   z 
6
= 5
 
 7 
ka
r

1
6
5
 
 7  3
an
 1 3 2
x


X = y
B =
A = 3 2 5


 
 2 3 6  33
 z  3
We have
AX = B

X = A1B
From eq. (1), eq. (2) and eq. (3), we get,
x
 3 24 19 
6
1

5


8
= 
y
2
1


 
 
31 
11 33 7  31
 5 9
 z  31
al
x
 31 
1
y
=   31

 
31 
 62  31
 z  31
dy

 18120133  
x
1

y
=    48107  
 
31
  304577  
 z  31
33

x
 1
y
=  1
 
 
 z  31  2  31
 x = 1, y = 1, z = 2
which is required answer.
Vi

3. (b) We have,
x 2  23x
x  3 x


14
2
x  3 x

=
1
x 2  23x
2

=
1
A
x  3



Bx C
 x 1
2
A x 2 1 Bx C  x  3 
 x  3   x 2 1
x2 + 23x = A(x2) + A + Bx2 + 3Bx + Cx + 3C
1013/FY/Pre_Pap/Maths_Soln
… (2)
… (3)
Prelim Question Paper Solution

x2 + 23x = (A + B)x2 + x(3B + C) + A + 3C
Equating corresponding coefficient

(A + B) = 1
(3B + C) = 23
A + 3C = 0
From eq. (3), we get,
A = 3C
From eq. (1) and eq. (4) we get

 3C + B = 1

B = 1 + 3C
From eq. (2) and eq. (5), we get
3(1 + 3C) + C = 23
3 + 9C + C = 23
10C = 20
C =2
From eq. (4) and eq. (6), we get
A = 6
From eq. (1) and eq. (7) we get
B = 7
2
6
7x  2
x 23x

=
 2
2
 x  3  (x  1)
(x  3)(x  1)
… (1)
… (2)
… (3)
… (4)
ka
r
… (5)
2
=
7x  2
2
(x  1)

A
B
tan 1
=

(tan   2)(tan3)
tan2 tan  3
tan1
1
here
A =
=
tan3 tan2
1
dy
3. (c) Let
B =
… (8)
6
which are required partial fraction.
x  3
al
(x  3) (x  1)
… (7)
an
x 2 23x
… (6)
tan1
tan 2 tan3
=
2
1
= 1
=2
hence,
Vi
1
2
tan 1
=

tan 2 tan 3
(tan 2)(tan  3)
3. (d) We know cos(A + B) = cos A cos B + sin A sin B

put A = ,  = B
2




cos     = cos cos  sin sin 
hence
2

2
2
= 0 + 1 sin 



cos     = sin 
2

1013/FY/Pre_Pap/Maths_Soln
15
Vidyalankar : F.Y. Diploma  Mathematics
3. (e)
L.H.S. =
 L.H.S. =
sinAsin2Asin3Asin4A
cosAcos2Acos 3A cos4A
 sinAsin4A  sin2Asin3A 
 cosAcos4A    cos 2Acos3A 
 4A  A 
 A  4A 
 3A2A 
 2A  3A 
2 sin 
 cos 
2 sin 
  cos 

2
2
 2 
 2 




 L.H.S. =
 A  4A 
 A  4A 
 2A  3A 
 2A  3A 
2 cos 
  cos 
2 cos 
  cos 

2
2
 2 
 2 




ka
r
  5A 
 3A 
 5A 
 A 
2  sin 
 cos 
sin 
 cos   



 2 
 2 
 2 
  2 
 L.H.S. =

 5A 
 3A 
 5A 
 A 
2  cos 
  cos  2 cos  2   cos  2  
2






 

1
cos
2A
=
sin2A
cos 2A
1
al
1  sec 2A
3. (f) L.H.S. =
tan2A
an
 5A  
 3A 
 A 
sin 
cos 
cos   



 2 
 2 
 2 
 L.H.S. =
5A
3A





 A 
cos 
  cos 
cos   
 2 
 2 
 2 
 5A 
 L.H.S. = tan 

 2 
 L.H.S. = R.H.S.
Hence proved.
cos 2A1
sin2A
= cotA
= R.H.S.
=
1  cos 2A
sin2A
dy
=
4. (a) L.H.S. =
=
2 cos2 A
cos A
=
2 sin A.cos A
sin A
1  tan2A tanA
1 tan2A tanA
Vi
sin2A sin A
cos 2A  cos A  sin2A  sin A

cos
2A
cos
A
cos 2A  cos A
=
=
sin2A sin A
cos 2A  cos A  sin2A  sin A
1

cos 2A cos A
cos 2A  cos A
cos 2A  cos A  sin2A  sin A
=
cos 2A  cos A  sin2A  sin A
1
4. (b)
16
L.H.S. = sin 3A
 L.H.S. = sin(2A + A)
 L.H.S. = sin2A  cos A  cos 2A  sin A
 L.H.S. = (2 sin A cos A) cos A + [cos2 A – sin2 A] sin A
1013/FY/Pre_Pap/Maths_Soln
Prelim Question Paper Solution






L.H.S.
L.H.S.
L.H.S.
L.H.S.
L.H.S.
L.H.S.
= 2sin A cos2 A + (cos2 A  sin2 A) sin A
= 2 sin A (1 – sin2 A) + (1 – sin2 A – sin2 A) sin A
= 2 sin A – 2 sin3 A + (1 – 2 sin2 A) sin A
= 2sin A  2sin3A + sin A  2sin3A
= 3sin A  4 sin3A
= R.H.S.
Hence proved.
4. (c) L.H.S. = sin 20. sin 40. sin 60. sin 80
=
=
=
[2 sin 20. sin 40] .sin 80
[Multiplying and dividing by 2]
ka
=
3
4
3
4
3
4
3
4
3
8
[cos (20°  40°)  cos (20° + 40°)] . sin80°
[(cos 20°  cos 60°) . sin80°]
[cos 20° sin 80°  cos 60° sin 80°]
an
=

3
 sin60  

2 
3
. sin 80
2
r
= sin 20. sin 40.
1
sin 80°]
2
[Multiplying and dividing by 2 and cos 60° = 1/2]
[2 cos 20° sin 80°  2.
3
[sin (20° + 80°)  sin (20°  80°)  sin 80°]
8
=
3
[sin 100° + sin 60°  sin 80°]
8
al
=
3
[sin 80° + sin 60°  sin 80°]
8
[ sin 100° = sin (2  90° 80°) = sin 80°]
dy
=
[ sin (60°) =  sin 60°]
=
3
(sin 60°)
8
3
3

8
2
3
=
= R.H.S.
16
Vi
=
4. (d) L.H.S. =
cosec A
cosec A
+
cosec A-1 cosec A+1
 cosecA+1+cosecA-1
= cosecA 

cosec 2 A  1

1
1



= cosecA 
 cosecA-1 cosecA+1
cosecA  2cosecA 
=
cot 2 A
1013/FY/Pre_Pap/Maths_Soln
17
Vidyalankar : F.Y. Diploma  Mathematics
2cosec 2 A
2
sin2 A
.
=
cot 2 A
sin2 A cos2 A
2
=
cos2 A
= 2sec2A
= R.H.S.
 L.H.S. =
 L.H.S. =
al
 L.H.S. =
ka
 L.H.S. =
cotcosec
cotcosec1
cos 
1

1
sin  sin 
cos 
1

1
sin  sin 
cos1  sin 

cos1  sin 



2 cos2  2 sin  cos
2
2
2


2 
2 sin 2 sin  cos
2
2
2



2 cos  cos  sin 
2
2
2



2 sin   sin cos 
2
2
2
an
4. (e)  L.H.S. =
r
=



 cos 2 sin 2 



 cos  sin 
2
2

dy
 L.H.S.

2
=

2 sin
2

cos
2
=

sin
2

= cot
2
= R.H.S.
2 cos
 L.H.S.
Vi
 L.H.S.
 L.H.S.
4
 5 
L.H.S. = cos1  5   sin1  13 
4. (f)
4
Let cos1  5  = 1
18
Hence proved.

sin21 = 1  cos2 1

sin 1 =
3
5
1013/FY/Pre_Pap/Maths_Soln
 cos 1 =
=
1  16
25
4
5
=
9
25
Prelim Question Paper Solution
Also, Let
 cos22
5. (a)
144
169

5
13
12
cos 2 =
13
sin 2 =
cos (1  2) = cos 1 cos 2 + sin 1 sin 2
48 15
 4   12   3   5 
=        =

65 65
 5   13   5   13 
 63 
1  2 = cos1  
 65 
4
5
63 





cos1    sin1   = cos1  
5
 13 
 65 
=
63
65
L.H.S. =
 L.H.S. =
sinA
1  cos A

1  cos A
sin A
sin2 A  1  cos2 A

sin A 1  cos A 

ka

=

an

= 2
r
Next
 5 
sin1  
 13 
25
=1
169
sin2 A  sin2 A
sin A 1  cos A 
 L.H.S. =
2 sin2 A
sin A 1  cos A 
al
 L.H.S. =
 L.H.S. =
sin A 1  cos A 
2 1  cos A 1  cos A 
sin A 1  cos A 
dy
 L.H.S. =
2 1  cos2 A 
Vi
 1  cos A  
 L.H.S. = 2 

 sin A 
 L.H.S. = 2 [cosec A  cot A]
 L.H.S. = R.H.S.
Hence proved.
5. (b) We know sin(A + B) = sin A cos B + cos A sin B
A = C/2,
B = D/2
C
D
C
D
C D
sin 
 = sin cos  cos sin
 2 
2
2
2
2
cos(A  B) = cos A cos B + sin A sin B
Similarly,
C
D
C
D
C D
cos 
 = cos cos sin sin
 2 
2
2
2
2
... (1)
... (2)
1013/FY/Pre_Pap/Maths_Soln
19
Vidyalankar : F.Y. Diploma  Mathematics
Multiplying (1) and (2);
CD
C D
sin 
 cos 

 2 
 2 
C
D
C
D
D
 C
= 2  sin cos cos2  cos2 cos sin
2
2
2
2
2
2

C
D
D
D
C
C

 sin2 cos sin  sin2 cos sin 
2
2
2
2
2
2
r
Multiplying by 2 throughout :
C D
C D
2 sin 
 cos 

 2 
 2 
C
D
C
D

=  sin cos  cos sin 
2
2
2
2

C
D
C
D

 cos cos  sin sin 
2
2
2
2

ka
C
D
D
 C
= 2  sin cos  cos2  sin2 
2
2
2
2

D
D
C
C 
cos  cos2  sin2  
2
2
2
2 
D
C
C
 D
= 2  sin cos  sin cos 
2
2
2
2

 sin
C D
C D
2 sin 
 cos 

 2 
 2 
We know 2 sin  cos  = sin 2.
C  D
C  D
hence 2 sin 
cos 
= [sin D + sin C] = RHS
 2 
 2 
hence proved.
cotAcot2A
cotAcot 2A
cosA cos2A

 L.H.S. = sinA sin2A
cosA cos2A

sinA sin2A
al
L.H.S. =
dy
5. (c)
an

 L.H.S. =
cosA  sin2A cos 2A  sin A 
sin  2AA 
sin  2AA 
Vi
 L.H.S. =
cosA  sin2A cos 2A  sin A 
sinA
sin3A
 L.H.S. = R.H.S.
( sin (A + B) = sin A . cos B + cos A sin B
and sin(A  B) = sin A cos B  cos A sin B)
 L.H.S. =
Hence proved.
5. (d) We have,
L1 : 2x + 3y = 13
L2 : 2x  5y = 7
are given equation of straight lines.
Let m1 and m2 be slopes of line given by eq. (1) and eq. (2)
20
1013/FY/Pre_Pap/Maths_Soln
… (1)
… (2)
Prelim Question Paper Solution
2
2
and m2 =
3
5
If ‘’ be the acute angle between lines, then
m1  m2
 tan  =
1  m1  m2
We get m1 =
=
 tan 
=
 tan 
=
 tan 
=

=
r
 tan 
ka
=
al
 tan 
10  6
15
15  4
15
16
15
11
15
16
11
16
11
16
11
 16 
tan1  
 11 
an
 tan 
2 2

3 5
=
 2 2 
1 
 
 3 5
dy

5. (e) Consider
P (x1, y1) = P (5, 4) and the line is 2x + y + 6 = 0.
Vi
The perpendicular distance from P(x1, y1) on ax + by + c = 0 is
ax  by1  c
= 1
a 2  b2
where a = 2, b = 1, c = 6
x1 = 5, y1 = 4
2  5   1 4   6
10  4  6
20
=
=
=
5
5
4 1
= 4 5 units.
5. (f) Let P(x, y) divide the join of A (+5, 4) and B (2, 3) in the ratio K: 1
 m = K and n = 1
Using external division formula,
mx 2  nx1
my 2  ny1
 Px =
and
Py =
mn
mn
1013/FY/Pre_Pap/Maths_Soln
21
Vidyalankar : F.Y. Diploma  Mathematics

K  2   1 5 
K 1
2K  5
Px =
K 1
K  3   1 4 
K 1
3K  4
=
K 1
 Py =
Px =
Py
ka
an
6. (a) Let for the line, x-intercept = a
y- intercept = b
Given : a = 2b
r
P (x, y) divides the x-axis,
 Py = 0
 0 = 3K + 4
 K = +4/3
8
4
2    5   5
3


3
 Px =
=
4
1
1
3
3
= 23
4 
 The ratio is  : 1 and the point is (23, 0).
3 
 The equation of line in two intercept form is,
x y
 =1
a b
dy
al
Putting a = 2b, we get,
x y
multiplying throughout by 2b.
 = 1,
2b b
 x + 2y = 2b
(i)
This line passes through the point (4, 1)
 Putting x = 4, y = 1, we get,
 4 + 2 (1) = 2b
 2b = 6
 b=3
 a = 2b = 6
Putting b = 3 in equation (1),
 x + 2y = 2 (3)  x + 2y = 6
This is the required line.
Vi
6. (b)
A(3, 4)
(x1, y1)
P(x3, y3)
Q(x4, y4)
B(2, 3)
(x2, y2)
Let P(x3, y3) and Q(x4, y4) be the points of trisection for the join A(3, 4) and B(2, 3)
Then we observe from fig. that the point P(x3, y3) divides the join AB internally in
the ratio 1 : 2.

m = k, n = 2k k  0, k  R

Using internal division formula for x and y we get,
mx 2 nx1
my 2  ny1
Px =
and
Py =
mn
mn
22
1013/FY/Pre_Pap/Maths_Soln
Prelim Question Paper Solution
=
k  3    2k  4
r
k  2k
3k  8k
=
3k
5k
=
3k
5
=
3
ka
Putting values of m, n, x1, y1, x2, y2 we get
k  2  2k  3 
Px =
Py
k  2k
2k6k
Py
Px =
3k
4k
Py =
Py
3k
4
P2 =
Py
3
4 5
 P(x3, y3) =  , 
3 3
Now point Q(x4, y4) is the midpoint of PB.
4
2
3
Qy
Qx =
2
46
Qy
Qx = 3
2
2

3
Qy
Qx =
2
2
Qx = 
Qy
6
1
Qx = 
3
 1 2 

Q = (x4, y4) =  , 
3 3 
an
5
3
3
=
2
4

= 3
2
4
6
2
= 
3
dy
al
= 
hN

  c.f. 
F2

C.I.
F
4
9.5  14.5
6
14.5  19.5
10
19.5  24.5
5
24.5  29.5
7
29.5  34.5
3
34.5  39.5
9
39.5  44.5
6
44.5  49.5
 N = 50
Median class = 24.5  29.5
h=5
F = freq. of median class = 5.
C.f. = 20
Vi
6. (c) M = L 
CF
4
10
20
25
32
35
44
50
1013/FY/Pre_Pap/Maths_Soln
23
Vidyalankar : F.Y. Diploma  Mathematics
Substituting
5
 2520 
5
= 24.5 + 5
= 29.5
Median = 24.5 +
= 11
 xi  x
n
fixi
x =
 fi
60150350810495195
50
2515551525
=
50
 MD
=
1
2
 fi  xi  x 
n

Coefficient of variance =
 100 , where  = 5D =
|x|
Variance
x
=
=
al
6. (f)
3625
3625
2000
50
90
=
50
=
an
6. (e) M  D =
=
r
LS
LS
11
=
61
Coefficient of Range =
ka
6. (d) Range = L  S
= 36  25
= 40
=
9
5
Variance
 fixi
 fi
(9)(15)  (17)(45)  (75)(43)  (105)(82)  (195)(24)  (135)(81)  (165)(44)
300
= 118.7
1
2
Variance =  fi  xi  x 
n
1 
9(15  118.7)2  17(45  118.7)2  43(75  118.7)2
=
300 
Vi
dy
=
 82(105  118.7)2  81(135  118.7)2  24(195  118.7)2 
= 1492.90
Coefficient of variance =

 100
x
=
Variance
 100
118.7
= 32.54

24
1013/FY/Pre_Pap/Maths_Soln