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th
10 CBSE{SA – I}
Mathematics
Mock Paper
With
Blue Print of Original Paper
on Latest Pattern
Solution Visits: www.pioneermathematics.com/latest_updates
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10th CBSE First Term {SA- I}
Blue Print
Topic/Unit
MCQs
SA(1)
SA(II)
LA
Total
Number System
2(2)
1(2)
2(6)
–
5(10)
Algebra
2(2)
2(4)
2(6)
2(8)
8(20)
Geometry
1(1)
2(4)
2(6)
1(4)
6(15)
Trigonometry
4(4)
1(2)
2(6)
2(8)
9(20)
Statistics
1(1)
2(4)
2(6)
1(4)
6(15)
10(10)
8(16)
10(30)
6(24)
34(80)
Total
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General instructions:
Time: 3hrs.
M: M: 90
(i) All questions are compulsory.
(ii) The question paper consists of 34 questions divided into sections A, B, C and D.
Section-A comprises of 8 questions of 1 mark each, section-B comprises of 6
questions of two marks each, section-C comprises of 10 questions of three marks
each and section-D comprises of 10 questions of four marks each.
(iii) Question numbers 1 to 8 in section-A are multiple choice questions where you
are required to select one option out of the given four.
(iv) There is no overall choice. However, internal choice has been provided in 1
question of two marks, 3 questions of three marks each and two questions of four
marks each. You have to attempt only one of the alternatives in all such questions.
(v) Use of calculator is not permitted.
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Section-A
Questions number 1 to 8 carry one marks each
1.
The pair of equation x = 2 and x = –4 has :
(a) infinitely many solutions
(b) no solution
(c) two solutions
(d) one solution
Sol : (b)
2.
If sides of two similar triangles are in ratio 4 : 9, then area of these triangles are in the
ratio
(a) 2 : 3
(b) 4 : 9
(c) 81 : 16
(d) 16 : 81
Sol : (d)
Ratio of Areas =
3.
4
9
2
=
16
81
The x-coordinate of the point of intersection of more than and less then ogive is :
(a) mode
(b) mean
(c) median
(d) Variance
Sol : (c)
4.
If LCM (54, 336) = 3024, then HCF (54, 336) is:
(a) 54
(b) 6
(c) 336
(d) 36
Sol : (b)
L.C.M × H.C.F
=Product of two numbers
= 54 × 336 = 3024 × H.C.F
H.C.F = 6
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5.
Value of 5
tan2A–5sec2A
(a) 1
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is :
(b) 0
(c) –5
(d) 5
Sol : (c)
5 tan2A – 5 × (1 + tan2A) = 5 × (tan2A – 1 – tan2A)
= –5
6.
If 2 sin2A = 3 , then A equal to :
(a) 90o
(b) 60o
(c) 45o
(d) 30o
Sol : (d)
7.
sin 2A =
3
2
If sec A =
q
1
, then value of cos A is:
p
p
(a)
Sin2A = Sin600 A = 300
1
p
(b)
1
q
(c)
1
pq
(d)
p
q
Sol : (b)
1
cos A
p
8.
1
p
p
q
1
q
If a positive integer n is divided by 2, then the remainder can be :
(a) 1 or 2
(b) 1, 2, or 3
(c) 0 or 1
(d) 2
Sol : (c)
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Section-B
Questions number 9 to 14 carry two marks each
9.
In the given figures, find measure of X .
Sol :
In
PQR and
PQ
ZY
4.2
8.4
PR
ZX
3 3
6 3
QR
YX
7 1
14 2
PQ
ZY
zyx
1
2
1
2
PR
ZX
QR
YX
PQR  ZYX
PRQ
(By SSS similarity criteria)
ZXY
..(1) (By CPST)
In PQR
PQR +
PRQ +
600 + 700 +
500
PRQ
RPQ = 1800
(Angle sum property of
)
PRQ = 1800
..(2)
From (1) & (2)
ZXY
X
500
10. Is 7 × 11 ×13+13 a composite number? Justify your answer.
Sol :
7 × 11 × 13 + 13
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= 13(7 × 11 × 1 + 1)
= 13(77 + 1)
= 13(78)
= 13 × 2 × 13 × 3
As, the no.7 × 11 × 13 + 13 is having more than 2 prime factors,
The given no. is a composite no.
11. Solve:
3
x
2
2
0 and
y
x
5
19
y
Sol :
3
x
2
y
2
x
5
19
y
Put
1
x
0
..(1)
..(2)
1
b in (1) & (2)
y
a&
3a – 2b = 0
..(3)
2a + 5b = 19
..(4)
Multiply (3) & (4) by 5 & 2 respectively,
(3a – 2b = 0) × 5
15a – 10b = 0
..(5)
& (2a + 5b = 19) × 2
4a + 10b = 38
…(6)
Adding (5) & (6)
15a 10b 0
4a 10b 38
19a
38
a
2
Put = a = 2 in (2)
3(2) – 2b = 0
6 – 2b = 0
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–2b = – 6
b=3
x
1
a
1
2
y
1
b
1
3
&
12. The following distribution give the daily income of 50 workers of a factory:
Daily income in Rs.
100–120 120–140 140–160 160–180
Number of workers
12
14
8
180–200
6
10
Write the above distribution as “more than type” cumulative frequency distribution.
Sol :
Daily income (in
More than type
fi
Cf
Rs)
distribution
100–120
More than 100
12
50
120–140
More than 120
14
38
140–160
More than 140
8
24
160–180
More than 160
6
16
180–200
More than 180
10
10
Total
50
and
13. If
1
are zeroes of polynomial 4x2 –2x + (k –4). Find k.
Sol :
If f(x) = 4x2 – 2x + (k – 4)
where, x
,
1
According to relationship between zeroes & coefficients of a polynomial,
1
c
a
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k 4
4
1
4=k–4
k=8
Value of k = 8.
14. If tan 2A = cot (A –18o), where 2A is an acute angle, find the value of A.
Sol :
cot A 180
tan 2A
cot 900 2A
= cot (A – 18)
90 – 2A = A – 18
90 + 18 = 3A
3A = 108
A = 360
Or
If
is an acute angle and sin
= cos
, find the value of 3 tan2
+ 2sin2
– 1.
Sol :
sin
cos
3 tan2
sin2
=3
cos2
sin
=3
cos
2
= 3 sin
…(1) (Given)
2sin2
1
2
2 sin2
sin2
cos2
[ sin2
cos2
1]
2
2 sin2
sin2
sin2
cos2
 sin
 sin
cos
from(1)
cos
=3
Value of 3tan2
2sin2
1 3
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Section-C
Questions number 15 to 24 carry three marks each
15. A survey regarding the height (in cm) of 50 girls of class X of a school was conducted and
the following data was obtained:
Height(in cm)
120–130
130–140
140–150
150–160
160–170
Total
Number of girls
2
8
12
20
8
50
Find the mode of the data.
Sol :
Model class = 150 – 160
f1 = 20
f0 = 12
f2 = 8
h = 10
l = 150
f1 f0
2f1 f0 f2
Mode = l
= 150
20 12
40 12 8
= 150
8
20
h
10
10
= 150 + 4
= 154 cm
16. Prove that :
cot A cos A
cot A cos A
cosecA 1
.
cosecA 1
cot A cos A
cot A cos A
cosec A 1
cosec A 1
Sol :
To prove :
L. H. S. =
cot A cos A
cot A cos A
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cos A
sin A
=
cos A
sin A
Medical and Non – Medical Classes
cos A
cos A
cos A sin A cos A
sin A
=
cos A sin A cos A
sin A
=
cos A sin A cos A
cos A sin A cos A
=
cos A 1 sin A
cos A 1 sin A
1
cosec A
=
1
1
cosec A
1
cosec A 1
cosec A
=
cosec A 1
cosec A
=
cosec A 1
= R. H. S.
cosec A 1
Hence proved.
17. In the given figure,
ACB 90o and CD AB . Prove that
BC2
AC2
BD
AD
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Sol:
ADC ~ ACB
AC
AB
AD
AC
CD
BC
AC2
AD.AB
……….(1)
CDB ~ ACB
CD
CA
BC
AB
BD
BC
BC2
AB.BD
……….(2)
Equation (2)/(1) we get
BC2
AC2
AB.BD
AD.AB
BC2
AC2
BD
AD
Or
In the given figure, if AD BC, prove that AB2 +CD2 = BD2 +AC2.
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Sol :
Given : In
ABC, AD
BC
To Prove : AB2 + CD2 = BD2 + AC2
Proof : In
ADC, AD
CD
AC2 = CD2 + AD2
AD2 = AC2 – CD2
In
ADB, AD
BD
AB2 = AD2 + BD2
AD2 = AB2 – BD2
(Given)
(By Pythagoras theorem)
..(1)
(Given)
(By Pythagoras theorem)
..(2)
From (1) & (2)
AC2 – CD2 = AB2 – BD2
AC2 + BD2 = AB2 + CD2
AB2 + CD2 = BD2 + AC2
Hence, proved.
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18. In the given figure, the line segment XY is parallel to AC of a ABC and it divides the
triangle into two parts of equal area. Find
AX
.
AB
Sol :
Let Area of
Area of
In
BXY = k
BAC = 2k
XBY and
ABC
XBY =
ABC
(Common angle)
BXY =
BAC
(Corresponding angles)
BXY ~
Area of BXY
Area of BAC
K
2k
BX
AB
BX
AB
1
2
BAC by AA similarity criteria
BX
AB
2
2
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AB AX
AB
2AB
1
2
2AX AB
AX
AB
2 1
2
2 1 AX
=
AB
2
2
2
AX
AB
Medical and Non – Medical Classes
2
2
2
sin 2sin3
19. Prove that:
2cos3
cos
tan
Sol :
sin
2 sin3
To prove :
2 cos3
cos
tan
sin
2 sin3
Proof : =
2 cos3
cos
sin
=
cos
= tan
1 2 sin2
2cos2
cos
1 2 sin2
2 1 sin2
= tan
1 2 sin2
2 2 sin2 1
= tan
1 2 sin2
1 2 sin2
= tan
1
= R. H. S.
Hence, proved.
20. Find the mean of the following data:
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Class- interval
100–120
Frequency
12
Medical and Non – Medical Classes
120–140 140–160 160–180 180–200
14
8
6
10
Or
Find the missing frequency for the given data if mean of distribution is 52.
Wages (in Rs.)
10–20
No. of workers 5
20–30
30–40
40–50
50–60
60–70
70–80
3
4
f
2
6
13
Sol :
Class interval
Frequency (fi)
xi
fixi
100-120
12
110
1320
120-140
14
130
1820
140-160
8
150
1200
160-180
6
170
1020
180-200
10
190
1900
fi
Mean =
=
50
fixi
7260
x ifi
fi
7260
50
= 145.2
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Or
Sol :
Wages in Rs.
No. of workers
xi
di = xi – 45
fidi
10-20
20-30
30-40
40-50
50-60
60-70
70-80
5
3
4
f
2
6
13
15
25
35
45
55
65
75
–30
–20
–10
0
10
20
30
–150
–60
–40
0
20
120
390
fi
Mean = A +
52 = 45 +
7=
33 f
fidi 280
fidi
fi
280
33 f
280
33 f
231 + 7f = 280
7f = 280 – 231
f=7
21. Prove that 2 – 5 is an irrational number.
Sol :
Let us assume the 2
5 is a rational number. Then, according to Euclid’s division
lemma, use can find two co-prime integers, such them,
2
5
5
5
a
b
a
2
b
2b a
b
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As, a & b are integers, this implies, (2b – a)/b is a rational no.
5 is also a rational number. But it contradicts the fact the 5 is irrational. This
contradiction has arisen due to our incorrect assumption that 2
5 is a rational
number.
2
5 is an irrational number.
22. If one zeros of polynomial p(x) = 3x2 – 8x + 2k + 1 is seven times of the other, then find
the zeroes and the value of k.
Sol :
p(x) = 3x2 – 8x + (2k + 1)
x=
,7
b
a
7
8
8
3
8
3
1
3
1
3
x
7
7
3
And,
7
x
1
3
7
3
..(1)
..(2)
c
a
2k 1
3
2k 1
3
(From (1) & (2))
7 = 6k + 3
6k = 4
k=
2
3
Value of k =
2
3
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23. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40
km upstream and 55 km downstream. Find the speed of the stream and that of the boat in
still water.
Sol :
Speed of stream = y km/hr
Speed of boat = x km/hr
5=
D
T
Case I
30
x y
44
x y
10
55
x y
13
Case II
40
x y
1
Let
a,
x y
a
1
,b
5
1
x y
b
1
11
on solving x = 8 km/hr y = 3 km/hr
Or
Solve by cross multiplication method: ax + by =a –b; bx – ay = a + b
Sol :
ax
bx
by
ay
a b
a b
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x
y
ab b2 a2 ab
ab b2 a2 ab
Medical and Non – Medical Classes
1
=
a2 b2
x
y
a2 b2
1
a2 b2
I
II
a2 b2
III
Equating I & III
x
a
2
1
2
b
a
2
b2
x=1
Equating II & III
y
a2 b2
1
a2 b2
y=–1
Values of x = 1 & y = – 1
24. Evaluate: sinA cosA –
sin A cos(90 A)cos A
sec(90 A)
cos A sin(90 A)sin A
.
cosec(90 A)
Sol :
= sin A cos A
sin A sin A cos A
cosec A
cos A cos A sin A
sec A
= sin A cos A - sin3A cos A – cos3A sin A
= sin A cos A – sin A cos A (sin2A + cos2A)
= sin A cos A – sin A cos A = 0
=0
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Section-D
Questions number 25to 34 carry Four marks each
25. Draw ‘more than ogive’ and less than ogive’ for the following distribution and hence
obtain the median.
Class-interval 5–10 10–15
15–20
20–25
25–30
30–35
35–40
Frequency
2
4
3
4
3
2
12
Sol:
Less then type of distribution
Class interval
f.i
c.f
5–10
Less than type
distribution
Less than 10
2
2
10–15
Less than15
12
14
15–20
Less than20
2
16
20–25
Less than 25
4
20
25–30
Less than30
3
23
30–35
Less than35
4
27
35–40
Less than40
3
30
We will plot the points (10, 2); (15, 14); (20, 16); (25, 20); (30, 23) (35, 27); (40, 30)
More than type of distribution
Class interval
f.i
c.f
5–10
More than type
distribution
More than 5
2
30
10–15
More than10
12
28
15–20
More than15
2
16
20–25
More than 20
4
14
25–30
More than25
3
10
30–35
More than30
4
7
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35–40
Medical and Non – Medical Classes
Less than35
3
Total
30
3
We will plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7), (35, 3)
26. Evaluate:
sin2
sin2(90 )
3 sec2 61o cot 2 29o
3cot 2 30o sin2 54o sec2 36o
.
2 cosec265o tan2 25o
Or
Prove that :
1 cos A sin A 1 sin A
sin A cos A 1
cos A
Sol:
sin2
cos2
3 sec2 61 tan2 61
1
=
31
3 3 sin2 54
cos2 36 2 sec2 25 tan2 25
9 sin2 90 36
cos2 36 2 1
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2
=
1
3
9 cos 36
2 cos2 36
=
1
3
=
2 27
6
=
25
6
9
2
Or
To prove :
L.H.S. =
1 cos A sin A 1 sin A
sin A cos A 1
cos A
1 sin A cos A
sin A cos A
1 sin A
2
sin A cos A
1 sin A cos A
sin A cos A 1
cos2 A
2
1
2
1 sin2 2sin A cos2 A
sin2 A cos2 A 2sin Acos A 1
sin2 A sin2 A 2sin A
1 1 2sin Acos A
2sin A sin A 1
2sin Acos A
sin A 1
R.H.S. Hence proved
cos A
27. If tan A + sin A = m and tan A – sin A = n, Show that m2 –n2 = 4 mn .
Sol:
Given : sin A tan A m & tan A sin A n
To prove :
m2 n2 4 mn
Proof:
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2
m
2
n
Medical and Non – Medical Classes
L.H.S.
= sin A tan A
2
tan A sin A
2
= sin2 A tan2 A 2sin A tan A tan2 A sin2 A 2sin A tan A
= 4sinA tanA
R.H.S. = 4 mn
=4
tan A sin A tan A sin A
= 4 tan2 A sin2 A
sin2 A
=4
sin2 A
2
cos A
=4
sin2 A sin2 Acos2 A
cos2 A
= 4 sin2 A
1 cos2 A
cos2 A
= 4 sin2 A
sin2 A
cos2 A
= 4 sin2 A tan2 A
= 4sin A tan A
L.H.S. Hence proved
28. If the median of the distribution given below is 32.5. find x and y.
Class interval
0–10
10–20
20–30
30–40
40–50
50–60
Frequency
x
5
9
12
y
3
60–70 Total
2
40
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Sol:
Class interval
f.i
c.f
0–10
x
x
10–20
5
x+5
20–30
9
x+14
30–40
12
x+26
40–50
y
x+ y+26
50–60
3
x+ y+29
60–70
2
x+ y+31
40
x+ y + 31 = 40
x+y=9
Median = 32.5 = 30 +
20
x 14
12
10
x = 3, x + y = 9
3+y=9
y=6
29. In the given figure, AB||PQ||CD, AB = x units, CD = y units and PQ = z units,
prove that,
1
x
1
y
1
z
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Or
1
In an equilateral triangle ABC, D is a point on BC, such that BD= BC . Prove that 9 AD2 = 7
3
AB2.
Sol :
ABD ~ PQD
x
l m
z
m
xm = z (l + m)
x m = zl + zm
xm – zm = zl
m (x – z) = zl
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m
l
z
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..(1)
x z
BPQ ~ BCP
z
y
l
l m
z(l + m) = yl
zl + mz = yl
mz = yl – zl
mz = l(y – z)
m
l
y z
z
..(2)
From (1) & (2)
z
x 2
y z
z
z2
y z x z
z2 = xy – yz – xz + z2
xy = yz +zx
1
2
1
x
xy
xyz
yz
xyz
zx
xyz
1
y
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Or
Sol:
1
Given : Equilateral ABC , BD= BC
3
To prove : 9AD2 = 7AB2
Construction: Draw AM
Proof: BD =
BM =
BC.
1
BC ……(1)(Given)
3
1
BC
2
…(2) [Altitude of an equilateral
is also its median]
DM = BM –BD
DM =
BC BC
2
3
DM =
3BC 2BC
6
[From (1) and (2)]
BC
6
….(5)
In ADM, AM DM
AD2 = AM2 + DM2
AM2 = AD2 –DM2
(By Pythagoras theorem)
…..(3)
Similarly, in ABM,
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AM2
=
AB2
–BM2
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…(4)
From (3) and (4)
AD2 –DM2 = AB2–BM2
AD2 –AB2 = DM2 –BM2
BC
AD2 –AB2 =
6
AD2
=
–AB2
2
BC2
=
36
BC
2
2
(from (2) and (5)
BC2
4
BC2 9BC2
36
8BC2
=
36
AD2
–AB2
AD2 =
2BC2
=
9
2BC2 9AB2
9
9AD2 = –2AB2 +9AB2
9AD2 = 7AB2 Hence proved
30. Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of the
squares of other two sides.
Sol:
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Given : ABC,
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BAC 900 ; AD BC
To prove : BC2 = AB2 +AC2
Proof: According to a theorem, if a
is drawn from the right
of a
then the two s formed are similar to each other & to the whole
BAD  ACD  BCA
BAD  BCA
AB
BD
BC
AB
AB2 = BC . BD
ACD  BCA
AC
CD
to its hypotenuse,
.
……(1)
From (1)
(by CPST)
……..(2)
(from (1))
BC
AC
AC2 = BC . CD
…….(2)
Adding (1) & (2)
AC2 +AB2 = BC .BD +BC .CD
= BC(BD + CD)
= BC × BC
= BC2
AC2 +AB2 = BC2
Hence proved
31. Use Euclid's division lemma to show that the cube of any positive integer is of the form
9m, 9m +1 or 9m + 8.
Sol:
Let a be any +ve integer and b = 3, the according to Euclid’s division lemma, a can be
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written as,
a = 3q +r, whose, 0 r<3.
a is of the form 3q, 3q+ 1&3q +2.
Case I: a = 3q
cubing both sides
a3 = (3q)3
= 27 q3
= 9(3q3)
=9m, where m = 9q3
a3 is of the of 9m.
Case II: a = 3q + 1
cubing both sides
a3 = (3q +1)3
= 27 q3 + 1 + 27 q2 +9q
= 9(3q3 + 3 q2 +q) +1
= 9m +1, where m = 3q3 +3q2 +q
a3 is of the form 9m +1.
Case III: a = 3q +2
cubing both sides
a3 = (3q+2)3
= 27q3 +8 +54q2 + 36q
= 9(3q3 +6q2 +4q) + 8
= 9m +8, where m = 3q3 + 6q2 +4q
a3 is of the form 9m +8
32. Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by
these lines with x axis. Find the area of the shaded region.
Sol:
2x + y = 6
2x – y =–2
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1
b h
2
Area of
=
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1
4 4 8 cm2
2
33. Find other zeroes of the polynomial p(x) = 2x4– 21x3 + 49x2 – 10x –20, if two of its zeroes
are 5 ± 5 .
Sol:
P(x) = 2x4 –213 + 49x2 – 10x –20
x=5
5
x –5
5= 0
( x –5) – 5 =0
x=5
…….(1)
5
x –5 + 5 = 0
(x –5) + 5 = 0
……….(2)
Multiplying (1) & (2)
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(x – 5) 2 –
5
2
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=0
x2 + 25 –10x – 5 = 0
f(x) = x2 –10x +20 = 0
x2 –10x
2x2 x 1
49x2 10x 20
20 2x 4 21x3
2x 4 20x3 40x2
x3 9x2
10x
x3 10x2 20x
x2
10x 20
x2
10x 20
0
g(x) = 2x2 – x – 1 = 0
= 2x2 – 2x + x + 1 = 0
= 2x ( x –1) + 1(x – 1)
= (2x + 1) ( x–1) = 0
x=
1
,1& 5
2
5, 5
5
Ans.
34. Let days taken by 1 women = x
Let days taken by 1 man = y
2
x
5
y
1
4
……(1)
3
x
6
y
1
3
……(2)
Put
1
1
b in (1) & (2)
=a&
y
x
&
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2a +5b =
3a + 6b =
1
4
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……(3)
1
3
a 2b 1/ 9
2
2a + 4b = 2/9
…..(4)
Subtracting (3) from (4) we get
2a + 5b = ¼
2a + 4b = 2/9
Put b = 1/36 in (3)
2a + 5
72a + 5 =
1
36
1
4
36
9
4
72a = 4
a=
4
1
72 18
x=
1
18 days
a
y
1
36 days
b
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