VIKAS PRE – UNIVERSITY COLLEGE, MANGALURU
SECOND PUC – BASIC MATHEMATICS
ANSWER KEY
PART A
1
3
5
2. nP4 = 36
!
4 !
n=6
3. p^q
4. 5:20 = 3:x
x=12
6
2
4
1. A′ =
Therefore
5
30
15 10
25 20
.c
5%
ol
le
117 / 125
8/3
27
–cosec 2x
tan x – sec x + c
100
as
c
6.
7.
8.
9.
10.
!"#
ge
100
5. Rate of interest =
om
360
PART B
!
w
- *.!
.v
3
,
4
1 1
+
2 2
w
w
|A|=4-6=-2
1
adj A = +
2
ik
11. A-1 = |&| '() *
3
,
4
1/2
+
1
3/2
,
2
12. a. 2 people are selected. From remaining 10, we choose 4 more which can be done in 10C4 ways
= 210 ways
b. 2 people are rejected, from remaining 10 we need to choose 6, which can be done in 10 C 6 =
210 ways
13. P(A0 B) = P(A) +P(B) – P(A1 B) = 1/2 + 1/3 – 7/12 = ¼
!3
P(B\A) = !32
4
!
2
14. Converse q 5p
“If x is a prime number then x<1”
Contrapositive ~q5 ~p
“If x is not a prime number then x71”
15.
89:;
;9:;
2
8
3(3x + 5) = 2 (5x + 5) or X = 5
Numbers are 3x = 15 and 5x = 25
16. TD = BD – BG = 1250 – 50 = Rs. 1200
Rs.30000
2 @AB &
2 @AB &
17. LHA = sin 2A=2 sin A cos A = CD@ & . FGH 2 * = CD@ & . I
RS
Q
4
cos P T RS
Q/4.Z:Q/4:Z
S . HY
2
Q
2Z
= 2 sin . sin (4
2
!
= -2 2 sin R =√2sin α =
√
P
(U cos *
Q/4.Z.Q/4.Z
S
2
RHS
2 KLB &
J &=RHS
= !:
cos V
2 sin
ol
= 2 HY P
J&
le
Q
4
18. LHS = cos P
!
om
!2;# !2##
=
;#
ik
as
c
19. (x1, y1) = (3, 4)
(x2, y2) = (2, -5)
Equation of circle is (x-x1) (x-x2)+ (y-y1) (y-y2) = 0
i.e. (x – 3) (x – 2) +(y -4 )(y + 5) = 0
i. e. x2 + y2 – 5x + y – 14 = 0
w
w
w
.v
20. For x = 0, f(0) = k
^ 29 1
^ 29 1
^ 29 1
lim
lim
. 2 2 lim
95#
95#
95#
_
2_
2_
Uf(x) is continuous at x = 0, lim95# ` _
` 0
-k=2
21. y = x sin x
Taking log on both sides,
log y = sin x (log x)
1 b
1
.a
sin _ . T cGd_ cos _
a
_
sin
_
a b x @AB f
T cGd_ FGH_
_
22. A. C = Tx + 2 + 3/x
M C = 10x + 2
&:<
2
.c
<= >=
<?
ge
F=
2 1
2
sin
&.<
2
23. g
!
√9:9
(_
Put 1 T √_
-k
2 (h
h
!
g
(_
√9:√9√9
j9
h i 9
√
2k
(h
h
2
24. g! _ 8 T ^ 9 (_
g
2 (h
2 log h
o
9
n
4
!
√9 √9:√9
T ^9p
2
!
(_
2 mYd 1 T √_ T F
!;
4
T ^2
^TF
3T 1
1 2 2T 1 4
5 0
,=+
,
3T4
1
1 2T4 4
7 18
5
7
(AB)’ = +
,
0 18
3
1 2 1
5
7
B’A’ = +
,+
,=+
,
2
4
1 4
0 18
Therefore (AB)’ = B’ A’
ge
.c
2
25. AB = +
1
om
PART – C
'! T _ '2 T a '8 T u
v2
v8 t
t v!
F!
F2
F8
= (a1 + x)(b2c3 – b3c2) – (a2 + y) (b1 c3 – b3 c1) + (a3 + z)(b1 c2 –b2 c1)
= a1(b2c3 – b3c2) – a2(b1c3 – b3 c1) + a3(b1 c2 –b2 c1) + x(b2c3 – b3c2) – y(b1 c3 – b3 c1) + z)(b1 c2 –b2 c1)
_ a u
'! '2 '8
v
v
v
v
Therefore Δ t !
2
8 t T t ! v2 v8 t
F! F2 F8
F! F2 F8
Hence the result
.v
ik
as
c
ol
le
26. Let ∆
w
w
w
27. i. There are six letters which can be permuted in 6!/2 ways = 720/2 = 360 ways
ii. Remaining 3 letters can be permuted in 3! Ways = 6 ways
iii. there are two consonants which can be selected in 2P1 ways and other 5 letters can be
arranged in 5!/2! Ways. Hence total number of ways is, 2P1 x5!/2!=120 ways
28. A: getting a queen in first draw
B: getting a queen in second draw
i.
P(A0B) = P(A).P(B) = 4/52 x 4/52 = 1/169
ii.
P(A0B) = P(A).P(B\A) = 4/52 x 3/51 = 1/221
iii.
29.
Earnings
Carpenter
Days
360
3
6
x
8
12
hours
9
6
360 : x = 3 : 8
6 : 12
9:6
x=
8x# " !2 x
8 x
30. BG =
20 =
yH. 1280
J
z
!:
z
{
|J
!:
#.#4 J
{
|J
#.#4
F=51000
;!###
z
{
!:
|J
TD=!:
{
|J
#.#4
#.#4
=1000
om
BD = BG+TD = 1020
.c
31.
Second investment
MV
Income
100
5
x
I2
Third investment
MV
Income
100
6
x
I3
I2 = 0.05 x
I3 = 0.06 x
le
ge
First investment
MV
Income
100
4
x
I1
as
c
ol
I1 = 0.04 x
.v
ik
I1 + I3 + I3 = 3600
0.04 x+0.05x + 0.06x = 3600
x = 24000
Total investment 3(24000) = Rs.72000
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w
32. SP = (100-10)% of 12000 = Rs.10800
Payment = (100+8)% of 10800 = Rs11000
w
33. Since the point (1/2, 2) lies in first quadrant, equation of parabola is x2 = 4 ay ------(1)
Substitute (x, y) = (1/2, 2) in (1),
! 2
P2 S
4' 2 i'
!
Therefore, x2 = "y or
!
82
8 x2 – y = 0
34. x=a (θ – sin θ)
j9
j}
j•
j9
' 1
j•/j}
j9/j}
y = a(1 – cos θ)
FGH~
I }
!. I}
=
I }
!. I}
j•
j}
=
€
€
2 @AB CD@
J
J
€
2I J
J
' HY ~
cot
}
2
35. Surface Area A = 4 π r3
j&
j
(*
(h
0.6 F‚2 /H^F
(„
4 ƒ 3 „ 2 … † , „ 3 F‚
(h
(„
(„ 0.0055
0.6 4 ƒ 3 9
i
(h
(h
ƒ
4
Volume = 8 ƒ „ 8
4
(„
ƒ 3 „ 2 … † , „ 3 F‚
3
(h
4
0.0055
ƒ3 9
0.198 FF/H^F
3
ƒ
!
g_
2
(_
le
9J
2
ol
log x
log x
9J
2
! 9J
2 2
T F
(2x + 5) dx = dt
ncGd hŠ!# n cGd _ 2 T 5_ T 3 Š!# = log 9 -1
.v
38. Put x2 + 5x + 3 = t and
! 29:;
!j
g# 9 J :;9:8 (_ g#
9J !
(_
2 9
J
!9
T F
2 4
g
as
c
9J
2
9J
log x 2
log x
ik
37. g _ log _ (_
ge
.c
36. f(x) = 9 x2 + 12 x + 2
f' (x) = 18x + 12
f’(x) = 0 i 18x + 12 = 0 or x = -2/3
f”(x) = 18 >0. Therefore, f(x) = f(-2/3) = -2
Hence f(x) is minimum at x = -2/3 and minimum value is -2.
om
(‰
(h
(‰
(h
:!
:!
&
<
T 9:8 J
9.8
T 9.2
PART D
w
w
w
39. Middle terms are 2 and 2 T 1. Hence the 8th and 9th terms are middle terms.
r+1 = 8 or r = 7
T8 = 15C7 (3x) 15 -7 (–2/x2)7 =–15C738 27 x8-14 = –15C738 27 / x6
T7 = 15C8 (3x) 15 -8 (–2/x2)8 =15C837 28 x7-16 = 15C837 28 / x9
40.
89:2
9:8 J 9.2
3x + 2 = A (x+3)(x-2) + B(x-2) + c(x+3)3
Put x=-3, -7 =A(0) + B(-5) + C(0) or B = 7/5
Put x=2, 8 = A(0) + B(0) + C(25) or C = –8/25
Put x=0, 2=–6A – 2B + 9C
or A = 8/25
89:2
"
"
‹
Hence 9:8 J 9.2
T ; 9.2
2; 9.8
2; 9:8 J
41. To check for tautology construct the following table.
p
q
p
p q
T
T
F
T
T
F
F
T
F
T
T
T
F
F
T
F
Therefore it is a tautology
p ^ (p q)
F
F
T
F
42. Total amount A + B+ C = 5625 ---------------(1)
A=
or A = Rs.1875
w
.v
w
w
44. 3x + 5y = 15
x=0, y = 3
y = 0, x = 5
ik
as
c
ol
le
1
1
1
2
2
4
4
8
Total hour for 8 units = 2048
Time required for 1 unit = 500
Hence total hour for 7 units is (2048-500) = 1580
Total cost = 40 x 1548 = Rs.61920
Cumulative average
time per lot (hours)
500
500x80% = 400
400 x 80% = 320
320 x 80% = 256
.c
Total output in units
ge
43. No of units 7
Units produced
om
or B = Rs.1125
B=
Substituting A and B in (1), C = Rs.2625
5x + 2y = 10
x = 0, y = 5
y = 0, x = 2
Scale X axis: 1 cm = 1 unit
Y axis: 1 cm = 1 unit
Total hours
500
800
1280
2048
′→q
T
T
T
T
From the graph, corner points are,
Corner point
Z = 5x + 3y
O(0, 0)
0
A(0, 3)
9
B(20/19, 45/19)
235/19 ----------Maximum
C(2, 0)
10
Solution is x = 20/19, y = 45/19 and maximum value is 235/19
w
w
w
.v
ik
le
as
c
ol
46. Let the required equation of the circle be
x2+y2+2gx+2fy+c=0
.............(1)
Put (x,y) = (1,-4) in (1);
2g-8f + c = -17
.............(2)
Put (x,y) = (5,2) in (1);
10g+4f + c = -29
.............(3)
(3) – (2) ;
2g + 3f= -4
.............(4)
Centre (-g,-f) lies on x-2y+9=0
Put (x,y)=(-g,-f) on above eqn;
-g + 2f= -9
.............(5)
On solving (4) and (5),
f = -3 , g = 3, c = -47
Hence , the required equation is :
x2+y2+6x-6y-47=0
ge
.c
om
45. LHS= sin2A+sin2B=sin2C
=2sin(A+B)cos(A – B) – sin2C
=2sinC cos(A- B ) – 2sinCcosC
=2sinC[ cos(A-B) – cosC]
=2sinC[cos(A-B)+cos(A+B)]
=2sinC.2cosA.cosB
=4cosAcosBcosC
= RHS
47. y = log(x – √_ 2 T 1)
y1=
y1=
√_ 2
√_ 2
!
9:√9 J :!
!
[1 +
√9 J :!
T 1.y = 1
T 1.y2 +y1[
√_ 2 T 1.y2 +
!
2√9 J :!
!
2√9 J :!
9Œ|
√9 J :!
.2x]
.2x] = 0
=0
Multiply both sides by √_ 2 T 1 ;
(1+x2)y2 + xy1 = 0
48. 4y = 3x2
y = x2 .......... (1)
Given 3x – 2y +12 =0
y=
.......... (2)
On solving (1) and (2),
x2 =
(x+2) (x -4 )=0
x =-2 , 4.
Point of intersections are A(-2,3) and B(4,12)
dx
om
Area =
ge
Part E
.c
= 27 sq.units
as
c
ol
le
49. a) 9x + 10y + 2z = 800
15x + 5y + 4z = 900
6x + 10y + 3z = 850
X =A-1B
|A| = -195
ik
Adj A=
.v
X=
w
w
x= 17.95 , y = 43.05 , z = 103.85
w
b) (1.01)5 = 5C0.15 + 5C1.14 (0.01)1 + 5C2.13 (0.01)2+ 5C3.12 (0.01)3+ 5C4.11(0.01)4+ 5C5.10 (0.01)5
= 1+ 0.05 + 0.001+ 0.00001+....
=1.05101
50. a) Let O be centre of unit circle and θ be an angle at O
From A , draw a perpendicular meeting at OB produced to D.
Let OB = OA = radius
From the figure ,
Area of ∆OAB ≤ Area of sector AOB ≤Area of ∆OAD
i.e., ½.OA.BC ≤ ½.OA2. θ ≤ 1/2.OA.DA ..........(1)
From ∆OAD, tanθ = DA/OA
⇒ DA = r.tanθ
From ∆BOC, sinθ = BC/OB
⇒ BC = r.sinθ
(1) becomes,
BC ≤ θ ≤ DA
sinθ ≤ θ ≤ tanθ
Dividing by sinθ ;
1≤
≤
Applying limit as θ→0,
≤
≤
om
=1
⇒
tan 600 =
ol
DC = h
From right angled triange ABD,
le
ge
.c
b) Let AB represents the height of the hill
CD represents height of the tower.
By data,
as
c
--------(1)
From right angled triange aec
w
w
.v
ik
tan 300 =
w
Hence height of the hill is 3h/2