Document 6528454

Transcription

Document 6528454

Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Latest Pattern Sample Paper {Mathematics} Class-X
Term-I Examination (SA-I)
Time: 3hours
Max. Marks: 90
General Instructions:
(i) All questions are compulsory.
(ii)The question paper consists of 31 questions divided into 4 sections A, B, C and D.
Section A comprises of 4 questions of 1 mark each, Section B comprises of 6 questions of
2 marks each, Section C comprises of 10 questions of 3 marks each and Section D
comprises of 11 questions of 4 marks each.
(iii) There is no overall choice. However, internal choice has been provided in 1 question
of 2 marks, 3 questions of 3 marks each and 2 questions of 4 marks each. You have to
attempt only one of the alternatives in all such questions.
(iv) Use of calculator is not permitted.
Section-A
1.
Find The number of zeroes lying between –2 to 2 of the polynomial f(x), whose graph is
given in the figure,
Solution:
Number of zeroes lying between –2 to 2 of the polynomial f(x) is 2. Because in this
interval, the curve cut the X-axis at two points.
2.
In the given figure , ACB APQ , if BA = 6m and BC = 8 cm and PQ = 4 cm then find the
length of AQ.
www.pioneerpsa.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
1
Solution:
As ACB
So,
APQ
AB BC
6
8



AQ PQ
AQ 4
(AB = 6cm, BC = 8cm and PQ = 4cm given)
 AQ 
64
8
 AQ = 3 cm
3.
Check whether the rational number
129
is a terminating decimal expansion or non225775
terminating repeating decimal expansion.
Solution:
Let
p
129
 5 7 5
q 2 .5 .7
i.e., q = 25.57.75, which is not of the form 2n. 5m.
Hence,
4.
129
will have a non-terminating repeating decimal expansion.
25.57.75
The abscissa of the point of intersection of the less than type and of the ‘more than type’
cumulative frequency curve of grouped data gives, which types of central tendency?
Solution:
Median.
2
Section-B
5.
In the given figure, find the value of x for which DE
AB.
In figure, PA, QB, RC and SD are all perpendiculars to a line l, if AB = 6 cm, BC = 9 cm, CD
= 12 cm and SP = 36 cm, then find PQ, QR and RS.
Solution:
Given, DE

AB
AD BE

DC EC
(by converse of Thale's theorem)

3x  19 3x  4

x 3
x
 x (3x +19) = (3x + 4)(x + 3)
 3x2 +19x = 3x2 + 9x+4x+12
 3x2 +19x – 3x2 –13x = 12
 6x = 12
3
 x
12
2
6
OR
Given , AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. It is clear from the figure that
AP, BQ, CR and DS is perpendicular to l.
 AP BQ CR DS By converse of basic proportionality theorem,
PQ: QR: RS = 6 : 9 : 12
Let PQ = 6y, QR = 9y and RS = 12y
PQ + QR + RS = PS
 6y + 9y +12y = 36
 27y = 36  y =
 PQ = 6y = 6 ×
QR = 9y = 9 ×
36 4
=
27 3
4
= 8cm
3
4
= 12cm
3
and RS = 12y = 12 ×
6.
4
= 16cm
3
To find out the concentration of SO 2 in the air (in parts per million, i.e., ppm), the data
was collected for 30 localities in a certain and is presented below.
Concentration of SO2
(in ppm)
0.00-0.04
0.04-0.08
0.08-0.12
0.12-0.16
0.16-0.20
0.20-0.24
Frequency
4
9
9
2
4
2
Find the mean concentration of SO2 in the air.
4
Solution:
Let, assumed mean A = 0.10
(Class width) h = 0.04, ui =
xi  A
h
Table for step deviation and their product with corresponding frequency is given below
Concentration
of SO2
fi
(in ppm)
ui 
Class
xi  A
h
mark  x i 
x  0.10
 i
0.04
fi  ui
0.00-0.04
4
0.02
-2
–8
0.04-0.08
9
0.06
-1
–9
0.08-0.12
9
0.10 = a
0
0
0.12-0.16
2
0.14
1
2
0.16-0.20
4
0.18
2
8
0.20.0.24
2
0.22
3
6
Total
N = 30
(Total observations) N = 30 and
–1
6
f u
i1
i
i
 1
By step-deviation method,
Mean = A + h ×
1 6
  fiui
N i1
= 0.10 + 0.04 ×
1
0.04
  1  0.10 
30
30
= 0.10-0.001
= 0.099 ppm
5
7.
Find the HCF of 18 and 24 by using factor tree method.
Solution:
Using factor tree method, we have
 18 = 2× 3× 3 = 2×32
and
24 =2×2×2×3 = 23 ×3
HCF = Product of common prime factors with lowest powers
 HCF(18, 24) = 3 × 2 = 6
8.
Express 0. 16 as a rational number in the simplest form.
Solution:
Let x =1. 16 =1.161616....
...(i)
On multiplying Eq (i) both sides by 100, we get
100x = (1.161616 ....) × 100
 100x = 116.161616....
...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
100x – x = (116.161616…….–1.161616 ....)
 99x  115  x 
Thus,
9.
1.16 
115
99
115
99
Evaluate:
tan2 600  4sin2 450  3sec2 300  5 cos2 900
.
cosec 300  sec600  cot 2 300
6
Solution:
Now,
tan2 600  4sin2 450  3sec2 300  5cos2 900
cosec300  sec600  cot 2 300
2
2
 2 
 1 
0
( 3)  4 
 3  
 5  0
 2
 3
=
2  2  ( 3)2
2
1
4
3 4  3  0
32 4
2
3
=

9
4 3
1
10. Show that



cos2 450    cos2 450  

 
tan 600   tan 300  

  1.
Solution:
LHS =



cos2 450    cos2 450  


 

sin [90   45   ]  cos 45   
=
cot[90  60   ]tan 30   
tan 60   tan 30  
0
2
0
0
0
0
0
2
0


 cos   sin 900  

and cot   tan 900  


=
=
(


0






sin2 450    cos2 450  

 
cot 300   tan 300  


1
 1  RHS
1
sin 2  cos2  = 1 and tan  cot   1 )
7
Section-C
11. If the polynomial
6x4 +8x3 +17x2 +21x + 7 is divided by another polynomial 3x2 + 4x +1, the remainder comes
out to be (ax + b). Find the values of a and b.
Or
Without drawing the graph, state whether the following pair of linear equations will
represent intersecting lines, coincident lines or parallel lines 6x–3y + 10 = 0and
2x – y+9 = 0.
Justify your answer.
Solution:
Let,
p(x) = 6x4 +8x3 +17x2 +21x + 7
and
g(x) = 3x2 + 4x +1

3x2  4x  1 6x 4  8x3  17x 2  21x  7 2x 2  5
6x 4  8x3  2x3
 

15x2  21x  7
15x2  20x  5



x 2
 Remainder = x + 2
 ax + b = x + 2
On comparing the coefficients of x and constant terms, we get
a = 1 and b = 2
Or
The given system of linear equation is
6x – 3y + 10 = 0
...(i)
8
and 2x – y + 9 = 0
...(ii)
On comparing Eqs. (i) and (ii) with
a1x + b1y + c1 =0 and a2x + b2y + c2 =0, we get
a1 =6,b1 =– 3 and c1=10
a2 – 2, b2 = – 1and c2 =9
We have,
and
a1 6
b
3
  3, 1 
3
a2 2
b2 1
c1 10

c2 9
Clearly,
a1 b1 c1


a2 b2 c2
Thus, given pair of linear equations will not intersect, so the given system of equations
will represent parallel lines.
12. The following table shows the age distribution of cases of a certain disease admitted
during a year in a particular hospital.
Age (in years)
5-14
15-24
25 – 34
35 – 44
45-54
55-64
Total
Number of cases
6
11
21
23
14
5
80
Find the modal age.
Solution:
As the distribution is discontinuous, converting it to a continuous distribution by using
9
adjustment factor of
15  14
= 0.5, we get the following
2
Age (in years)
Number of cases
4.5-14.5
6
14.5-24.5
11
24.5-34.5
21
34.5-44.5
23
44.5-54.5
14
54.5-64.5
5
As the class 34.5-44.5 has maximum frequency, it is the modal class.
Here, l = 34.5, f1, = 23, f2 = 14, f0 = 21 and h = 10
 Mode = l +
f1  f0
h
2f1  f0  f2
= 345 
23  21
 10
2  23  21  14
= 345 
2  10
20
 345 
46  35
11
= 345+1.82 [approx]
= 36.32
Hence, the modal age is 36.32 yr.
13. In the given figure,
AB  BC, FG  BC and DE  AC, prove that,
ADE
GCF .
10
Solution:
Given,  ABC in which  B = 90°
GF  BC and DE  AC
To prove  ADE
 GCF
Proof, In  ABC and  GFC,
B = F
[Each 90°, given]
C = C
[common angle]
 ABC
GCF ,
[by AA similarity criterion]
Also,
 A =  FGC
[by CPCT]
Now, in  ADE and  GCF,
E = F
(Each 90°)
 DAE =  FGC
[
 A =  DAF]
[proved above]
  ADE  GCF
Hence proved.
14.  ABC and  DBC are on the same base BC and on opposite sides of BC and O is the point
of intersection of AD and BC. Prove that
ar  ABC  AO

ar  DBC  DO
11
Solution:
Given  ABC and  DBC are on the same base BC and O is the point of intersection of AD
and BC.
To prove

Construction Draw AM  BC and DN  BC
Proof In  AMO and  DNO,
 AMO =  DNO
[each 900]
 AQM =  DON
[vertically opposite angles]
  MOA -  NOD

AM AO

DN DO
...(i)
[ratio of corresponding sides of a similar triangles]
1
Now, ar (  ABC) =  BC  AM
2
ar (  DBC) =
1
BC × DN
2
...(ii)
...(iii)
On dividing Eq. (ii) by Eq. (iii), we get
12
1
ar  ABC  2  BC  AM

ar  DBC  1  BC  DN
2

ar  ABC  AM

ar  DBC  DN


[from Eq. (i)]
Hence proved.
15. A school conducted a sport meet, the number of players in football, hockey
and athletics are 48, 60 and 132, respectively.
Find the minimum number of rooms required. If in each room the same number of
players are to be seated and all of them being in the same sports.
Solution:
The number of room will be minimum, if each room accommodate maximum number of
players.
Since in each room the players of same sports are to be accommodated.
Therefore, the number of players in each room must be HCF of 48, 60 and 132.
By prime factorisation of 48, 60 and 132, we have
48 = 24 × 3
60 = 22 × 3 × 5
and 132 = 22 × 3 × 11
 HCF of 48, 60 and 132 = 22 ×3 = 12
So, in each room 12 players of same sports can be accommodated.
 Number of rooms required
=
Total number of players
Number of players in a room
13
=
48  60  132
12
=
240
 20
12
16. A person can row a boat at the rate of 5km/h in still water. He takes thrice as much time
in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
OR
If  and  are zeroes of the polynomial x2 – 2x –15, then form a quadratic polynomial
whose zeroes are 2  and 2  .
Solution:
Let the speed of the stream = x km/h
and speed of the boat in still water = 5 km/h
 Speed of boat in downstream = (5 + x) km/h
and speed of boat in upstream = (5 –x) km/h
Distance covered = 40 km
Time taken by boat in downstream =
40
5 x
and time taken by boat hi upstream =
40
5 x
According to the question,
 40 
 40 

 3 

 5 x 
 5 x 

40
120

5 x 5 x
 40(5 + x) = 120(5 – x)
 200 + 40x = 600 – 120x
 40x + 120x = 600 – 200
 160x = 400
14
 x
400 5
 = 2.5 km/h
160 2
Hence, the speed of stream is 2.5 km/h
Or
Let, p(x) = x2 – 2x –15
On comparing with ax2 + bx + c, we get
a =1, b = – 2 and c = – 15
Given,  and  are the zeroes of p (x).
Sum of zeroes, (    ) = –
   
b
a
 2
1
   2
..(i)
and product of zeroes, ( . ) =
 
c
a
15
1
   15
...(ii)
Given, zeroes are 2 and 2 .
 Sum of zeroes = 2  + 2  = 2(    ) = 2 × 2 = 4
[using Eq. (i)]
and product of zeroes = 2.2
= 4  4   15  60
[using Eq. (ii)]
 Required polynomial
= x2 – (Sum of zeroes) x + (Product of zeroes)
= x2 – 4x + (– 60)
= x2 –4x –60
15
17. The mean of the following frequency distribution is 25. Find the value of P.
Class interval
0-10
10-20
20-30
30-40
40-50
Frequency
2
3
5
3
P
OR
Find the median of the following data.
Class interval
0–10
10–20
20–30
30–40
40–50
50–60
60–70
70–80
80–90
90–100
Solution:
Frequency
5
3
4
3
3
4
7
9
7
8
Table for mid value and their product with corresponding frequency is given below
Class Interval
Frequency
 fi 
Mid value
 xi 
fi  x i
0-10
10-20
20-30
30-40
40-50
2
3
5
3
p
5
15
25
35
45
10
45
125
105
45P
 f  13  p
Here, mean,  x   25,  f  13  p
Total
i
 f x = 285 + 45p
i i
i
and
f x
i i
 285  45p
16
 x 
fi x i
f
i

25 285  45p

1
13  p
 25(13 + p) = 285 + 45p
 325 + 25p = 285 + 45p
 25p – 45p = 285 – 325
 –20p = – 40
 p
40
2
20
OR
Table for cumulative frequency is given below
Class interval
Frequency
Cumulative frequency
0-10
5
5
10-20
3
8
20-30
4
12
30-40
3
15
40-50
3
18
50-60
4
22 cf
60-70
7f
29
70-80
9
38
80-90
7
45
90-100
8
53
Total
N = 53
17
Here, N = 53
Now,
N 53
=
= 26.5 which lies in cumulative frequency 29.
2
2
So, the median class is (60-70),
f =7cf = 22
and l = 60
N

 2  cf 
Median= l  
h
f




= 60 
26.5  22
 10
7
= 60 
4.5
45
 10  60 
7
7
= 60 + 6.43 =66.43
18. In  ABC, right angled at B, if tan A =
1
3
m then find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
Solution:
Given, tan A =
Let P = 1k, B =
perpendicular  p  1
1


Base  B
3
3
3k
In right angled  ABC,
B2 + P2 =H2
18
[by Pythagoras theorem]
 ( 3k)2  1k   H2
2
 3k2  k2  H2
 H2  4k 2
 H = 2k [Since, side cannot be negative]
(i) sin A cos C + cos A sin C
 BC   BC   AB   AB 
=

 


 AC   AC   AC   AC 
 1k  1k   k 3  k 3 
=     


 2k  2k   2k 
 2k 
=
1k 2 3k 2 4k 2 4


 1
4k 2 4k 2 4k 2 4
(ii) cos – A cos C – sin A sin C
 BC  AB   AB  BC 
=

 


AC
AC


  AC  AC 
 1k   k 3   k 3   1k 
=   
 
 
 2k   2k   2k   2k 
19
=
=
1
3
3 1



2 2
2 2
3
3

0
4
4
19. Prove that, if a, b, c and d are positive rationals such that, a  b  c  d , then either
a = c and b = d or b and d are squares of rationals.
Solution:
Given, a  b  c  d
If a = c, then b  d
 b=d
[squaring both sides]
If a  c, then, there exists a positive rational number .v such that, a = c + x.
Now, a  b  c  d
 cx b c d
[
a cx]
 x b  d

 x b
  
2

d
2
[squaring both sides]
 x2 + b + 2x b = d
 x2  2x b  b  d  0
 2x b  d  b  x2

b
d  b  x2
2x
Hence, d, x and b are rationals.
d  b  x2
So,
is rational.
2x
Then,
b is a rational number.
Hence, b is the square of a rational number.
20
From Eq. (i),
d  x b
Also,
d is a rational.
So, d is the square of a rational number.
Hence, either a = c and b = d or b and d are the squares of rationals.
20. Determine the value of x, such that 2 cosec2 30° + x sin2 60° –
3
tan2 300  10
4
Solution:
Given,
3
2cosec2 30° + x sin2 60° –  tan2 30o = 10
4
2
2
 3  3 1 
 22  x 
  
  10
 2  4 3 
2
 0
3
1 
0
and tan 300 
30  2, sin 60 

2
3


 8
3x 3 1
   10
4 4 3
 8
3x 1
  10
4 4

3x
1
 10   8
10
4

3x
1
 2
4
4

3x 8  1

4
4

3x 9

4 4
 3x = 9
x=3
21
Section-D
21. The median of the following data is 50. Find the value of p and q, if the sum of all the
frequencies is 90.
Marks
20-30
30-40
40-50
50-60
60-70
70-80
80-90
Total
Solution:
Frequency
p
15
25
20
q
8
10
90
Table for cumulative frequency is given below
Marks
Frequency  fi 
Cumulative frequency (cf)
20-30
30-40
40-50
50-60
60-70
70-80
80-90
p
15
25
20
q
8
10
P
15 + p
40 + p
60 + p
60 + p + q
68 + p + q
78 + p + q
Total
 f  78  p  q
Given,
Now,
i
N = 90
N 90

= 45, which lies in the interval (50-60).
2 2
Here, l = 50, f = 20, cf = 40 + p, h = 10
N

  cf 
2
h
 Median = l + 
f
22
 50  50 
{45   40  p }
 10
20
 50  50 
45  40  p
 10
20
 50  50 
5 p
2
 5–p=0
p=5
Also,
78 + p + q = 90 (given)
p + q = 90 – 78
 p + q = 12
 5 + q = 12(put p = 5)
 q = 12 – 5
q = 7
Hence, p = 5 and q = 7
22. Size of agricultural holdings in a survey of 200 families is given in the following table.
Size of agricultural
Number of families
holdings (in hec)
0-5
10
5-10
15
10-15
30
15-20
80
20-25
40
25-30
20
30-35
5
Total
200
Compute median and mode size of the holdings.
23
Solution:
Table for cumulative frequency, is given below
Size of agricultural
holdings (in hec)
Number of families (fi)
Cumulative
frequency
(cf)
0-5
5-10
10-15
15-20
20-25
25-30
30-35
10
15
30
80
40
20
5
10
25
55
135
175
195
200
Total
(i)Here, N = 200
Now,
200
N 200

= 100, which lies in the interval 15 – 20.
2
2
l =15, h = 5, = 80, cf = 55
N

 2  cf 
 Median = l  
h
f




45
 100  55 
= 15  
  5  15 
16
 80 
= 15 + 2.81 =17.81 hec
(ii) In a given table, 80 is the highest frequency. So, the modal class is 15–20.
Here, l = 15, f1 = 80, f0 = 30, f2 = 40, h = 5
 f f

 Mode = l   1 0   h
 2f1  f0  f2 


80  30
= 15  
5
 2  80  30  40 
24
50
 50 
= 15  
5
  5  15 
90
 160  70 
= 15 
250
15+ 2.77 =17.77 hec
90
23. The students of a class are made to stand in rows. If 3 students are extra in
a row, there would be 1 row less. If 3 students are less in a row, there would
be 2 rows more, then find the number of students in the class.
OR
For which values of a and b, are the zeroes of g(x) = x3+2x2 + a, also the zeroes of the
polynomial p(x) = x5 –x4 – 4x3+3x2 + 3x + b?
Which zeroes of p(x) are not the zeroes of q(x)?
Solution:
Let the number of students in the class be x and the number of rows be y.
 Number of students in each row =
x
y
According to the question,
x

x    3   y  1
y

 xx

x
 3y  3
y
x
 3y  3  0
y
..(i)
x

and x    3   y  2
y

 xx
2x
 3y  6
y
25

2x
 3y  6  0
y
On putting
..(ii)
x
= u in Eqs. (i) and (ii), we get
y
u –3y+3 = 0
...(iii)
and 2u–3y–6 = 0
..(iv)
On subtracting Eq. (iii) from Eq. (iv), we get
u – 9 = 0 u = 9
On substituting u = 9 in Eq. (iii), we get
9 – 3y + 3 = 0
 3y=12
 y=4
...(v)
Now, u = 9

x
9
y

x
9
4

x
 put u  y 


[from Eq. (v), y = 4]
 x = 36
 The number of students in the class x = 36.
OR
Using division algorithm,
26
x2  3x  2
x3  2x 2  a x5  x 4  4x3  3x 2  3x  b 
x5 2x 4  ax 2
 

3x 4  4x3  3  a  x 2  3x  b
3x 4  6x3  3ax



2x3  3  a  x 2  3  3a  x  b
2x3  4x 2  2a
 

 1  a  x2  3  3a  x   b  2a 
If (x3 + 2x2 + a) is a factor of
(x5 – x4 – 4x3 + 3x2 + 3x + b), then remainder should be zero.
i.e., –(1 + a)x3 + (3 + 3a) x + (b – 2a) = 0
= 0.x2 + 0.x + 0
On comparing the like coefficient of x, we get
a+1=0
 a=–1
and b – 2a = 0
 b = 2a
 b = 2(–1) = – 2
[
a = – 1]
For a = –1 and b = –2, the zeroes of q(x) also the zeroes of the polynomial p(x).
 q(x) = x3 + 2x2 – 1
and p(x) = x5 – x4 – 4x3 + 3x2 + 3x – 2
Now using,
Dividend = Divisor × Quotient + Remainder
27
p(x) = (x3 + 2x2 – 1) (x2 – 3x + 2)
= (x3 + 2x2 – 1) (x2 – 2x – x + 2)
= (x3 + 2x2 – 1) x (x – 2)–1(x – 2)
= (x3 + 2x2 – 1) (x – 2) (x – 1)
Hence, the zeroes 1, 2 of p(x) are not the zeroes of q(x).
24. For going to a city B from city A. There is a route via. city C such that AC  CB, AC =
2xkm and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly
connects the two cities A and B. Find how much distance will be saved in reaching city
B from city A after the construction of the highway.
Solution:
Here, A and B be the position of two cities.
As AC  BC, therefore joining all these points formed a right angled  ABC.
In  ABC,
AB2 = AC2 + BC2
[by Pythagoras theorem]
 (26)2 = (2x)2 + [2(x + 7)]2
 676 = 4x2 + 4(x2 + 49 + 14x)
 169 = x2 + (x2 + 49 + 14x)
[divide both sides by 4]
 2x2 + 14x – 120 = 0
 x2 + 7x – 60 = 0
 x2 + 12x – 5x – 60 = 0
 x (x + 12) – 5 (x + 12) = 0
 (x – 5) (x + 12) = 0
 x = 5 km
28
x  12 , as x cannot be negative]
[
 AC = 2x = 2 × 5 = 10 km
CB = 2(x + 7) = 2(5 + 7) = 24 km
 Distance saved directly from city a to city B
= AC + CB – (AB)
= 10 + 24 – 26 = 8 km
Hence, the distance 8 km saved directly to reach from city A to city B.
25. If two scalene triangles are equiangular, then prove that the ratio of the corresponding
side is the same as the ratio of the corresponding angle bisector segments.
Given  ABC and  PQR are equiangular and AD and PS are angle bisectors of  A and
 P, respectively.
AD BC AB AC



PS QR PQ PR
To prove
Proof In  ABC and  PQR,
A = P
B =Q
and
C = R
 ABC
 PQR
[by AAA similarity criterion]

AB BC AC


PQ QR PR
..(i)
Also, A  P
29

1
1
A  P
2
2
[
AD and PS are angle bisectors]
 1 = 2
In  ABD and  PQS,
1  2
B  Q
 ABD
PQS
AB AD

PQ PS
..(ii)
From Eqs. (i) and (ii) ,
AD AB BC AC



PS PQ QR PR
Hence proved.
26. For which values of a and b does the following pair of linear equations have an infinite
number of solutions?
2x + 3y = 7 and
(a – b)x + (a + b)y = 3a + b – 2
Solution:
(i) We have, 2x + 3y – 7 = 0
and (a – b)x + (a + b) y– (3a + b – 2) = 0
Here,
a1 = 2, b1 = 3,c1 = – 7
and a2 = a–b, b2 =a + b, c2 = 3a + b – 2
For infinite number of solutions,
2
3
7


a  b a  b  3a  b  2




a1 b1 c1 
  
a2 b2 c2 
a  b a  b 3a  b  2


2
3
7
30
(reciprocal the above equation)
From first and second terms,
ab ab

2
3
 3a – 3b = 2a + 2b
 a = 5b
From second and third terms
a  b 3a  b  2

3
7
 7a + 7b = 9a + 3b – 6
 4b = 2a – 6
 2b = a – 3
[divide both sides by 2] ...(ii)
From Eqs. (i) and (ii), eliminating a,
2b = 5b – 3  b = 1
On substituting b = 1 in Eq. (i), we get
 a=5
a=5×1
27. Prove that,
sin   cos   1 1  sin 
.

sin   cos   1
cos 
Without using trigonometric tables, evaluate the following.
cosec2 650  tan2 250
1
+
(tan 100. tan 300. tan 800).
0
2
0
sin17  sin 73
3
Solution:
To prove,
LHS =
sin   cos   1
 sec  tan 
sin   cos   1
sin   cos   1
sin   cos   1
31
On dividing numerator and denominator by cos  , we get
sin  cos 
1


cos  cos  cos 
=
sin  cos 
1


cos  cos  cos 
=
tan   1  sec 
tan   1  sec 
=
 tan   sec    1
 tan   sec    1




sin 
 tan 
cos 

 tan   sec    sec2   tan2  
=
 tan   sec   1
[
=
 tan   sec    sec   tan  sec   tan  
 tan   sec   1
=
 sec   tan  1  sec   tan  
 tan   sec   1
=
 sec   tan   tan   sin   1 
 tan   sec   1
sec2   tan2   1 ]
[ a2  b2   a  b a  b]
= sec   tan 
=
1
sin  1  sin 


cos  cos 
cos 
LHS = RHS
Hence proved.
OR
We have,
cosec2 650  tan2 250 1

sin2 170  sin2 730
3
(tan 100 tan 300 tan 800)
32
=
=


cosec2 900  250  tan2 250

sin2 170  sin2 900  170
1
[tan 100. tan 300. tan(900 – 100)]
3
sec2 250  tan2 250 1

[tan100 tan300 cot100 ]
2
0
2
0
sin 17  cos 17
3






 cosec 900    sec 


sin 900    cos 

and tan 900    cot 

=

+







1 
cot



tan  

1 1 
1 
0
0

 tan 10 .tan30 .

1
tan100 
3
= 1
1 1
1 4
.
1 
3 3
3 3
28. In an acute angled  ABC, if sin2(A + B– C) = 1 and tan(B+C – A) = 3 , then find the
value of A, B and C.
Solution:
We have,
2(A + B – C) = 1
 sin 2(A + B – C) = sin 900
[
 2(A + B – C) = 450
..(i)
sin 900 = 1]
and tan (B + C – A) = 3
 tan (B + C – A) = 600
..(ii)
On adding Eqs. (i) and (ii), we get
A + B – C + B + C – A = 45°+60°
 2B = 105°
 B
1050
10
 52
2
2
33
1050
On putting B =
in Eq. (ii), we get
2
1050
 C  A  600
2
1050
 C  A  60 
2
0
1200  1050
 CA 
2
150
 CA 
2
..(iii)
In  ABC,
A + B + C = 180°
[Since, sum of three angles of a triangle is 180°]
1050
 C  1800
 A
2
1050
 A  C  180 
2
0
3600  1050 2550
=

2
2
 C A 
2550
2
..(vi)
On adding Eqs. (iii) and (iv), we get
150 2550
2C 

2
2
2700
 1350
 2C 
2
 C
1350
10
 67
2
2
On putting the value of C in Eq. (iv), we get
34
1350
2550
A
2
2
A
2550 1350 1200


 600
2
2
2
10
10
 A  60 , B  52 and C  67
2
2
0
29. In the following figure,  ABC is right angled at B,  BSC is right angled at S,  BRS is
right angled at R, AB = 18 cm, BC =7.5 cm, RS = 5 cm, RB = 6 cm,  BSR = x° and  SAB
= y°, then find
(i) tan x° (ii) sin y°
(iii) cos y0
Solution:
Given,
 CBA = 90°,  BRS = 90°,  BSC = 90°, BC = 7.5 cm.
RS = 5 cm, BR = 6 cm and AB = 18 cm
Here, AR = AB – RB = 18 – 6 = 12cm
In  ARS, using Pythagoras theorem,
AS2 =AR2 +RS2 =(12)2 +(5)2
 AS2 =144 + 25=169
 AS2 =169
 AS  169
[Taking positive square root]
 AS = 13 cm
[since, side cannot be negative]
35
(i) In AB7?5, tanx° BRS, tan x0 
[
BR 6

RS 5
perpendicular = BR and base =SR]
(ii) In ARS , sin y 0 
[
SR 5

AS 13
perpendicular = SR and hypotenuse = AS]
(iii) In ARS , cos y 0 
AR 12

AS 13
30. The progressive club help needy students by giving free text books and stationery item.
The club purchased 5 books and 7 pens together cost Rs 79, whereas 7 books and 5 pens
together cost Rs 77.
(i) Find the total cost of 100 books and 100 pens.
(ii) Which mathematical concept is used in the above problem?
(iii) By donating books and stationery items to needy students which value is depicted
by progressive club?
Solution:
(i) Let the cost of one book be Rs and that of one pen be Rs y.
Cost of 5 books + Cost of 7 pen = Rs 79
Then,
5x + 7y-79
...(i)
Also, cost of 7 books + Cost of 5 pens = Rs77
 7x + 5y =77
...(ii)
On multiplying Eq. (i) by 7 and Eq. (ii) by 5 and then subtracting, we get
35x  49y  553
35x  25y  385



24y  168
 y
168
7
24
36
On putting y = 7 in Eq. (i), we get
5x + (7×7) = 79
 5x + 49 = 79
 5x = 79 – 49 = 30
 x
30
6
5
Cost of 100 books = 7 × 100 = Rs 700
and Cost of 100 pens = 6 × 100 = Rs 600
 Cost of 100 books + Cost of 100 pens
= 700+600 = Rs 1300
(ii) Pair of Linear Equation in two variables.
(iii) Charity
31. If sides AB, AC and median AD of a  ABC are respectively, proportional to sides PQ and
PR and median PM of another  PQR, then show that  ABC
 PQR.
Solution:
Given In  ABC and  PQR, AD and PM are the medians, respectively.
Then,
AB AC PD


PQ PR PM
To prove  ABC
...(i)
 PQP
Construction Produce AD to E such that AD = DE and produce PM to N such that
PM = MN.
Join BE, CE, QN and RN.
37
Proof Quadrilaterals  BEC and  QNR are parallelograms because their diagonals
bisect each other at D and M, respectively.
 BE = AC and QN = PR

BE AC
BE AB



QN PR
QN PQ
AB BE

PQ QN
i.e.,
[from Eq. (i)]
..(ii)
From Eq. (i),
AB AD 2AD AE



PQ PM 2PM PN
[
D is a mid-point of BC and M is a mid-point of QR also, diagonals bisect each other]
AB AE

PQ PN
i.e.,
..(iii)
From Eqs. (i) and (iii), we get
AB BE AE


PQ QN PN
 ABE
PQR
 1 = 2
..(iv)
[since, corresponding angles of two similar triangles are equal]
Similarly, we can prove that
 ACE
 PRN
38
 3 = 4
...(v)
On adding Eqs. (iv) and (v), we get
1 + 3 = 2 + 4
 A = P
 ABC
 PQR
[by SAS similarity criterion]
Hence proved.
39

Document 6528454

Transcription

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