X-STANDARD MATHEMATICS
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Transcript of X-STANDARD MATHEMATICS
X-STANDARD MATHEMATICS
MODEL QUESTION
PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21
Section – A
Note: (i) Answer all the 15 questions(ii) Choose the correct answer in each question. Each of these questions contains four options with just one correct option(iii) Each question carries One mark 15 × 1 = 15
1. Let A = { 1, 3, 4, 7, 11 }, B = {–1, 1, 2, 5, 7, 9 } and f : A B be
given by f = { (1, –1), (3, 2), (4, 1), (7, 5), (11, 9) }. Then f is
(A) one-one (B) onto
(C) bijective (D) not a function
Section – A
Section – A
,.......625
54,
125
18,
25
6,
5
2
5
2
5
3
5
4
2. The common ratio of the G.P is
(A) (B) 5
(C) (D)
3. If a1 , a2 , a3 ,……are in A.P. such that then the 13th
term of the A.P. is
(A) (B) 0
(C) 12a1 (D) 14a1
Section – A
2
3
7
4 a
a
2
3
4. The LCM of 6x2 y, 9x2 yz,12x2 y2 z is
(A) 36x2 y2 z (B) 48xy2 z2
(C) 96x2 y2 z2 (D) 72xy2 z
Section – A
5. If b = a + c , then the equation ax2 + bx + c = 0 has
(A) real roots (B) no roots
(C) equal roots (D) no real roots
Section – A
6. If then the order of A is
(A) 2 1 (B) 2 2
(C) 1 2 (D) 3 2
Section – A
2120
11
A
Section – A
Section – A
Section – A
Section – A
Section – A
12.
(A) sin + cos (B) sin cos
(C) sin – cos (D) cosec + cot
Section – A
cottan
1
13. If the total surface area of a solid hemisphere is 12 cm2 then its
curved surface area is equal to
(A) 6 cm2 (B) 24 cm2
(C) 36 cm2 (D) 8 cm2
Section – A
14. Mean and standard deviation of a data are 48 and 12 respectively. The
coefficient of variation is
(A) 42 (B) 25
(C) 28 (D) 48
Section – A
14. If A and B are mutually exclusive events and S is the sample space such
that P(A)=1/3 P(B) and S = A B, then P(A) =
(A) (B) (C) (D)
Section – A
4
1
2
1
4
3
8
3
Section – B Note:
(i) Answer 10 questions(ii) Answer any 9 questions from the first 14 questions. Question No. 30 is Compulsory.(iii) Each question carries Two marks 10 × 2 = 20
16. If A = {4,6,7,8,9}, B = {2,4,6} and C = {1,2,3,4,5,6}, then find
A (B C).
ANSWER:
B C = {2, 4, 6} {1, 2, 3, 4, 5, 6} = {2, 4, 6}
A (B C) = {4, 6, 7, 8, 9} {2, 4, 6} = {2, 4, 6, 7, 8, 9}
X = {1, 2, 3, 4}, g = {(3,1), (4,2), (2, 1)}
g is not a function. since 1 has no image.
17. Let X = { 1, 2, 3, 4 }. Examine whether the relation g = { (3, 1), (4, 2), (2, 1) } is a function from X to X or not. Explain.
1
2
3
4
1
2
3
4
18. Three numbers are in the ratio 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers form an arithmetic sequence. Determine the numbers.
The ratio of three numbers = 2 : 5 : 7
Let the numbers be 2x, 5x, 7x
2x, 5x – 7, 7x are in an AP
5x – 7 – 2x = 7x – (5x – 7)
3x – 7 = 2x – 5x + 7
3x – 7 = 2x + 7
3x – 2x = 7 + 7
x = 14
The numbers are
28, 70, 98
, are the roots of 2x2 – 3x – 1 = 0 a = 2, b = – 3, c = – 1
19. If and are the roots of the equation 2x2 – 3x – 1 = 0, find the value of – if >
2
3 roots theof Sum
a
b
2
1 roots theofProduct
a
c
4)( )( 22
2
14
2
3 )(
22
24
9
4
1
4
89
4
1
2
1
20.
If then find the additive inverse of A.
17
51
59
32A
17
51
59
32A
1579
5312
616
21
Additive Inverse of A = – A
616
21
21. Find the product of the matrices, if exists
12
76
24
014
392
12
76
24
014
392
0780616
36346548
78616
3675414
122
6440
107)1(24)2(0)6()1(44
1)3(7922)2()3()6(942
22. The centre of a circle is at (-6, 4). If one end of a diameter of the circle is at the origin, then find the other end.
2
,2
Midpoint 2121 yyxx
Let the centre of the circle is (– 6, 4)One end = (0, 0), other end = (x, y)Midpoint of diameter = centre of the circle
4,62
0,
2
0
yx
4,62
,2
yx
42
,62
yx
824
1226
y
x
The other end = (– 12, 8)
23. In ABC, DE || BC and . If AE = 3.7 cm, find EC.
Given, AE = 3.7cm
3
2
DB
AD
A
B C
D E
3
2
DB
AD
EC
AE
DB
AD
In ABC, DE || BC
EC
7.3
3
2
2
37.3 EC
85.52
7.11EC
cmEC 85.5
24. A ladder leaning against a vertical wall, makes an angle of 60 with the ground. The foot of the ladder is 3.5 m away from the wall. Find the length of the ladder.
Let AC = Length of the ladder BC = distance of the wall = 3.5m
A
B C60
AC
BC60cosIn ABC, C = 60
AC
5.3
2
1
3.5m
AC = 3.5 2
= 7
Length of the ladder = 7m
25. Prove the identity:1
sec
cos
cos
sin
ec
sec
cos
cos
sin
ecLHS
cos1
cos
sin1
sin
22 cossin
RHS1
26. A right circular cylinder has radius of 14 cm and height of 8 cm. Find its curved surface area.
Radius = 14cm, height = 8cm
Curved surface area = 2rh
8147
222
2
= 2 22 2 8
= 44 16
= 704sq.cm
27. The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.
Height = 12cm
Circumference of the base = 44
2r = 44
447
222 r
2
r = 7cm
222
744
r
Volume = r2h
12777
22
= 22 7 12
= 154 12
= 1848cu.cm
28. Calculate the standard deviation of the first 13 natural numbers.
The standard deviation of the first ‘n’ natural numbers =12
12 n
The standard deviation of the first 13 natural numbers =12
1132
12
1169
12
168
1414. 00
3
39
56 00
.7
7469
31
7.3SD
29. Two coins are tossed together. What is the probability of getting at most one head.
When two coins are tossed the sample space
S = {HH, HT, TH. TT}
n(S) = 4
A = {atmost one head}
A = {HT, TH, TT}
n(A) = 3
)(
)()(
Sn
AnAP
4
3)( AP
30. (a) Simplify: 127
5462
2
xx
x
127
)9(6
127
5462
2
2
2
xx
x
xx
x
)4)(3(
)3)(3(6
xx
xx
)4(
)3(6
x
x
(OR)
30. (b) Show that the lines 2y = 4x + 3 and x + 2y = 10 are perpendicular.
Slope of 2y = 4x + 3 is m1 = 2
Slope of x + 2y = 10 is m2 = - ½
m1 m2 = 2 (– ½ ) = – 1
The given lines are perpendicular.
Note: (i) Answer 9 questions(ii) Answer any 8 questions from the first 14 questions. Question No. 45 is Compulsory.(iii) Each question carries Five marks 9 × 5 = 45
SECTION - C
31.Prove that A \ (B C) = (A \ B) (A \ C) using Venn diagram
A B
C
A A
A A
B B
B B
CC
C C
(1) BC (2) A – ( BC) (3) A – B
(4) A – C (5) (A – B)(A – C)
From the diagram (2)and(5)
A – (BC) = (A – B)(A – C)
32. A function f: [-7, 6)R is defined as follows
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;12
)(
2
x
x
xx
xf
Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
)1(3)6(
)4(2)3(4
ff
ff
32. A function f: [-7, 6)R is defined as follows
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;12
)(
2
x
x
xx
xf
Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
(i) Since -4 lies in the interval -5 x 2 f(x) = x + 5
f(-4) = – 4 + 5
= 1
Since 2 lies in the interval -5 x 2 f(x) = x + 5
f(2) = 2 + 5
= 7
)1(3)6(
)4(2)3(4
ff
ff
2f(-4) + 3f(2)
= 2(1) + 3(7)
= 2 + 21
= 23
A function f: [-7, 6)R is defined as follows
f(x) =
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;122
x
x
xx
Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
(ii) Since -7 lies in the interval -7 x < -5
f(x) = x2 + 2x + 1
f(-7) = (–7)2 + 2(–7) + 1
= 49 – 14 + 1 = 36
Since -3 lies in the interval -5 x 2 f(x) = x + 5
f(2) = -3 + 5
= 2
)1(3)6(
)4(2)3(4
ff
ff
f(-7) – f(–3)
= 36 – 2
= 34
A function f: [-7, 6)R is defined as follows
f(x) =
6 x 2 ; 1
2 x 5- ; 5
5- x 7- ;122
x
x
xx
Find the value of (i) 2f(-4) + 3f(2) (ii) f(-7) – f(– 3), (iii)
(iii) Since -3 lies in the interval -5 x 2 f(x) = x + 5 f(-3) = -3 + 5 = 2
Since -6 lies in the interval -7 x < -5
f(x) = x2 + 2x + 1
f(-6) = (–6)2 + 2(–6) + 1
= 36 – 12 + 1 = 25
)1(3)6(
)4(2)3(4
ff
ff
4 lies in the interval 2 < x < 6
f(x) = x – 1
f(4) = 4 – 1
= 3
Since 1 lies in the interval -5 x 2 f(x) = x + 5
f(1) = 1 + 5
= 6
)6(3)25(
)3(2)2(4
)1(3)6(
)4(2)3(4
ff
ff
1825
68
7
14
2
33. Find the sum of the first 40 terms of the series 12 – 22 + 32 – 42 + ……. .
12 – 22 + 32 – 42 + ……. . = 1 - 4 + 9 - 16 + 25 ….. to 40 terms
= (1 – 4) + (9 – 16) + (25 – 36) + …… to 20 terms.
= (-3) + (-7) + (-11) +………20 terms
It is an AP with a = –3, d = –4, n = 20
dn )1(2a2
nSn
)4)(120(2(-3)2
20S20
)4(19)62
20
7662
20
822
20
10
= 10(-82)
= – 820
40
34. Factorize the following x3 – 5x2 – 2x + 24
1 – 5 – 2 + 24
– 2
0
1
– 2
– 7
14
12
– 24
0 x + 2 is a factor
Other factor = x2 – 7x + 12
= x2 – 4x – 3x + 12
= x(x – 4) – 3(x – 4)
= (x – 4)(x – 3)
x3 – 5x2 – 2x + 24 = (x + 2)(x – 4)(x – 3)
9x4 + 12x3 + 28x2 – nx + m3x2
3x2
9x2(-)
6x2 + 12x3 + 28x2+2x
+ 2x
+ 12x3 + 4x2(–) (-)
24x2– nx + m6x2 + 4x + 4
+ 4
24x2 + 16x + 16
(-) (–) (–)
0 n = -16, m = - 16
35. If m – nx + 28x2 + 12x3 + 9x4 is a perfect square, then find the values of m, n
37. If verify that (AB)T = BTAT .
11
12 ,
37
25BA
LHS = RHS (AB)T = BTAT .
11
12 ,
37
25BA
11
12
37
25AB
13)1(7)1(327
12)1(5)1(225
37314
25210
411
38
LHSAB T ........43
118)(
11
12 ,
32
75 TT BA
32
75
11
12TT AB
317)1(215)1(
3)1(722)1(52
3725
314210
RHSAB TT ........43
118
38. Find the area of the quadrilateral whose vertices are(-4, -2), (-3,-5), (3,-2) and (2, 3)
Area of the quadrilateral ABCD
= ½ {(x1y2 + x2 y3 + x3y4 + x4y1)
– (x2y1 + x3 y2 + x4y3 + x1y4)}
-4
-2
-3
-5
3
-2
2
3
-4
-2
= ½ {(-4)(-5) + (-3)(-2) + (3)(3) + (2)(-2)} –
{(-4)(3) + (2)(-2) + (3)(-5) + (-3)(-2)}
= ½ {(20 + 6 + 9 – 4) – (–12 – 4 – 15 + 6)}
= ½ {31 – (–1)}
= ½ (31 + 1)
= ½ (32)
= 16 sq. units
A(-4,-2)
D(2,3)
B(-3,-5)
C(3,-2)
39. The vertices of ABC are A(2, 1), B(6, –1) and C(4, 11). Find the equation of the straight line along the altitude from the vertex A.
The vertices of the triangle are A(2, 1), B(6, –1), C(4, 11)
Let AD be the altitude.
A
B CD
12
12
xx
yySlope
46
111
BCofSlope 62
12
6
1 ADofSlope
A(2, 1), Slope m = 1/6
Required Equation of the line is
y – y1 = m(x – x1)
)2(6
11 xy
6(y – 1) = x – 2
6y – 6 = x – 2
x – 2 – 6y + 6 = 0
x – 6y + 4 = 0
40. A boy is designing a diamond shaped kite, as shown in the figure where AE = 16cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?
The vertices of the triangle are A(2, 1), B(6, –1), C(4, 11)
Let AD be the altitude.
A
B CD
12
12
xx
yySlope
46
111
BCofSlope 62
12
6
1 ADofSlope
A(2, 1), Slope m = 1/6
Required Equation of the line is
y – y1 = m(x – x1)
)2(6
11 xy
6(y – 1) = x – 2
6y – 6 = x – 2
x – 2 – 6y + 6 = 0
x – 6y + 4 = 0
41. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree.
B C30BC
AC30tan
303
1 AC
A
30m
Let A be the point at which the tree is broken and let the top of the tree touch the ground at B.
In ABC, BC = 30m, mB = 30
41. A vertical tree is broken by the wind. The top of the tree touches the ground and makes an angle 30with it. If the top of the tree touches the ground 30 m away from its foot, then find the actual height of the tree.
B C30
3
30AC
3
330
3
3
3
30
= 10 1.732
= 17.32
A
30m
AB
BC30cos
AB
30
2
3
3
230AB
3
360
3
3
3
60
= 20 1.732
= 34.64
Actual height of the tree
= 17.32 + 34.64
= 51.96m
42. Using clay, a student made a right circular cone of height 48 cm and base radius 12 cm. Another student reshapes it in the form of a sphere. Find the radius of the sphere.
Cone: Radius = 12cm,
Height = 48cm
Volume of the sphere = volume of the cone
hrr 23 3
1
3
4
r3 = 12 12 12
r = 12cm
4812123
1
3
4 3 r12
43.Calculate the standard deviation of the following data.
3 7
8 10
13 15
18 10
23 8
50
x f d = x – 3 d2 fd
fd2 3 – 3 = 0
8 – 3 = 5
13 – 3 = 10
18 – 3 = 15
23 – 3 = 20
0
25
100
225
400
0
50
150
150
160
510
0
250
1500
2250
3200
7200
X 3 8 13 18 23
f 7 10 15 10 8
Standard deviation
22
f
fd
f
fd
2
50
510
50
7200
22.10144
04.104144
96.39
39.966
6
36
396123369
27
.3
3.6
44. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15.
Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards.
Let A be the event that a new car will get an award for its design
B be the event that a new car will get an award for efficient use of fuel.
P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15
P(atleast one of the two awards) = P(AB)
44. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15.
Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards.
Let A be the event that a new car will get an award for its design
B be the event that a new car will get an award for efficient use of fuel.
P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15
P(atleast one of the two awards) = P(AB)
P(AB) = P(A) + P(B) – P(AB)
= 0.25 + 0.35 – 0.15
= 0.60 – 0.15
= 0.45
44. The probability that a new car will get an award for its design is 0.25, the probability that it will get an award for efficient use of fuel is 0.35 and the probability that it will get both the awards is 0.15.
Find the probability that (i) it will get atleast one of the two awards (ii) it will get only one of the awards.
P(A) = 0.25, P(B) = 0.35 P(AB) = 0.15
P(getting only one of the awards)
= P (only A or only B)
B)AP()BP(A
= [P(A) – P(A B)] + [P(B) – P(A B)]
= (0.25 – 0.15) + (0.35 – 0.15)
= 0.10 + 0.20
= 0.30
45. The sum of three consecutive terms in an AP is – 6 and their product is 90. Find the numbers.
Let a – d, a, a + d be three numbers in an AP
Sum = –6
a – d + a + a + d = – 6
3a = –6
a = – 6/3
a = –2
Product = 90
(a – d) a (a + d) = 90
(a2 – d2)a = 90
{(–2)2 – d2 }(–2) = 90
4 – d2 = – 45
– d2 = –45 –4
d2 = 49
d = 7The given numbers
are
– 2, 5, 12
(or) –2, –9, –16
45. A Cylindrical jar of diameter 14cm and depth 20cm is half-full of water . 300 lead shots of same size are dropped into the jar and the level of water raises by 2.8cm. Find the diameter of each lead shots.
Cylindrical jar:
Diameter = 14cm
Radius = 7cm
Height of water level raises = 2.8cm
Volume of 300 lead shots = volume of raised water
45. A Cylindrical jar of diameter 14cm and depth 20cm is half-full of water . 300 lead shots of same size are dropped into the jar and the level of water raises by 2.8cm. Find the diameter of each lead shots.
hrr 23 3
4300
8.277 400 3 r
100
10 400
2877 3
r
7
100
10 1010
7773
r
cmr 7.010
7
Note: 2 ×10 = 20 marks
(i) This section contains Two questions, each with two alternatives.
(ii) Answer both the questions choosing either of the alternatives.
(iii) Each question carries Ten marks
SECTION - D
46. Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents.
O P10
6
A
B
8
8
Length of the tangent segment
22 610
36100
864
46. Draw two tangents from a point which is 10cm away from the centre of a circle of radius 6cm. Also measure the lengths of the tangents.
M
47. (a) Draw the graph of x2 – x – 8 = 0 and hence find the roots of x2 – 2x – 15 = 0.
x2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = –3, 5
In x axis 1cm = 1 unit
y axis 1cm = 2 units.
x -4 -3 -2 -1 0 1 2 3 4 5
x2 16 9 4 1 0 1 4 9 16 25
-x 4 3 2 1 0 -1 -2 - 3 -4 -5
-8 -8 –8 –8 –8 –8 –8 –8 –8 –8 –8
y 12 4 -2 -6 -8 -8 -6 -2 4 12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
Solution set = {– 3, 5}
47. (b) A cyclist travels from a place A to a place B along the same route at a uniform speed on different days. The following table gives the speed of his travel and the
corresponding time he took to cover the distance.
Speed in km/hr
2 4 6 10 12
Time in hrs
60 30 20 12 10
Draw the speed-time graph and use it to find
(i) the number of hours he will take if he travels at a speed of 5 km / hr
(ii) the speed with which he should travel if he has to cover the distance in 40 hrs.
Speed in km/hr 2 4 6 10 12
Time in hrs 60 30 20 12 10
0
10
20
30
40
50
60
70
0 1 2 3 4 5 6 7 8 9 10 11 12 13
If x = 5,
y = 24
If y = 40,
y = 2.4Speed in km/hr
Tim
e in
hr