mathematical tools 1st test.pdf

Upload
aanchalmittal 
Category
Documents

view
232 
download
0
Transcript of mathematical tools 1st test.pdf

7/29/2019 mathematical tools 1st test.pdf
1/13
a
\[Page No.1]
Master SolutionPremier Institute for the preparation of IITJEE /
Date 01 05 2011 Class : 10 + 1 M.M:171 Time: 1:30 hrs
Test Code 4001 TEST : Mathematical Tools (Master solution)
SECTIONA (ONE OPTION CORRECT TYPE QUESTIONS)
The value ofo2o4
o42
20cos20sin
20cos20sin
++
isQ.1
(a.) 1 (b.) 2 (c.)
2
1
(d.) None of these
Ans. (a)
o2o4
o4o2
20cos20sin
20cos20sin
++
=o2o22
o2o2o2
20cos]20cos1[20sin
)20sin1(20cos20sin
++
120cos20sin1
20cos20sin1o2o2
2o2
==
sinP2P 20 + sinP2 P70Q.2
(a.) 0 (b.) 1 (c.) 2 (d.) sinP2Px
Ans. (b)
Let p = sinP2P 20 + sinP2P 70= sinP2P20 + sinP2P(90  20)= 1
The value of8
7sin
8
5sin
8
3sin
8sin 2222
+
+
+
isQ.3
(a.) 1 (b.) 2 (c.)
8
11
(d.)
8
12
Ans. (b)= sinP2P
8
+ sinP2
8sin
8
5sin
8
3 22 7++
=
+
+
+
8
3sin
8
3sin
8sin
8sin 2222
=
+
8
3sin
8sin2 22
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327

7/29/2019 mathematical tools 1st test.pdf
2/13
a
\[Page No.2]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
=
+
82sin
8sin2 22
=
+
8cos
8sin2 22 = 2
The value of cos 1 cos 2 cos 3.. cos 179 is equal toQ.4
(a.)
2
1
(b.) 0 (c.)
2
1
(d.) 2
Ans. (b)
cos1 cos2 cos3 cos 179= 0 {as cos 90 = 0
If tan A =2
1and tan B =
3
1, then the value of A + B is
Q.5
(a.)
4
(b.) 0 (c.) (d.)
4
Ans. (a/d)
Tan (A + B) =Btan.Atan1
BtanAtan
+
Put tan A =2
1; tan
3
1B =
1)BAtan( =+A + B = /4The maximum and minimum values of 6 sin x cos x + 4 cos 2x are:Q.6
(a.) 5, 5 (b.) 4, 4 (c.) 7,7 (d.) 132,132
Sol: (a)
6 sin x cos x + 4 cos 2x = 3sin2x + 4 cos 2x
Max = 22 43 + 5
The greatest value of sinx. cosx isQ.7
(a.) 1 (b.) 2 (c.)
2
1
(d.) 1
Sol: (c)
y = sin x cos x = x2sin2
1 greatest value =
2
1
The third term of a G. P. is 4. The product of the first five terms isQ.8
(a.) 4P3P (b.) 4P5P (c.) 4P4P (d.) 4P2P
Sol: (b)
TB3B = arP2
P = 4
Product of first five terms
= a ar arP2P arP3P arP4
P aP5PrP10P = (arP2P)P5P = 4P5P

7/29/2019 mathematical tools 1st test.pdf
3/13
a
\[Page No.3]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
Which term of the sequence 72, 70, 68, 66,.. is 40?Q.9
(a.) 17 (b.) 15 (c.) 19 (d.) 16
Sol: (a)
72, 70, 68, 66 ., 40 A. P
a = 72, d = 2, TB
nB
= 40 40 = 72 + (n 1) (2)32 = 2 (n 1) n = 17
The 10PthP and 18PthP terms of an A. P. are 41 and 73 respectively. Find 26PthP term.Q.10
(a.) 100 (b.) 105 (c.) 110 (d.) 108
Sol: (b)
aB10B = 41 and aB18B = 73
a + 9d = 41 .(1) and a + 17 d = 73 ..(2)
Solving eq (1) & (2), we get
8d = 32 d = 4 a + 9 4 = 41 a = 41 36 = 5, a = 5
aB26B = a + 25 d = 5 + 25 4 = 5 + 100
aB2BB6B = 105
One root of the equation 5xP2P + 13x + K = 0 is the reciprocal of the other, ifQ.11
(a.) K = 0 (b.) K = 5 (c.)K =
6
1
(d.) 6
Ans. (b)5xP2P + 13x + K = 0
Let one root be Then other root be
1
Product of roots ==
1
5
K
K = 5For what value of p the difference of the roots of the equation xP2P px + 8 = 0 is 2?Q.12
(a.) 2 (b.) 4 (c.) 6 (d.) 8
Ans. (c)xP2P px + 8 = 0
Let its root be &
+ = p = 8(  ) = 2
222 p2 =++
16p222 =+
4222 =+
48216p2 =p = 6

7/29/2019 mathematical tools 1st test.pdf
4/13
a
\[Page No.4]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
If , are the roots of the equation xP2P p(x + 1) c = 0, then ( + 1) ( + 1) =Q.13
(a.) c (b.) c 1 (c.) 1 c (d.) None of these
Ans. (c)xP2P p(x + 1) c = 0xP2P px (p + c) = 0
)cp( +=
.1)1()1( +++=++= p c + p + 1= 1  c
The roots of the equation 2P2xP 10. 2PxP + 16 = 0 areQ.14
(a.) 2, 8 (b.) 1, 3 (c.) 1, 8 (d.) 2, 3
Ans. (b)
2P2xP 10 2PxP + 16 = 0Let 2PxP = y
yP2P 10y + 16 = 0yP2P 8y 2y + 16 = 0
y(y 8)  2(y 8) = 0(y 2) (y 8) = 0y = 2, 8
2PxP = 2P1 P x = 12PxP = 2P3P x = 3
The angle between the lines 2x y +3 = 0 and x + 2y + 3 = 0 isQ.15
(a.) 90 (b.) 60 (c.) 45 (d.) 30
Sol: (a)
2x y + 0
x + 2y + 3 = 0
mB1B = + 2, mB2B =2
1
0
2/5
m.m1
mmtan
12
12 =+
=
= /2
The equation of the line which has Yintercept 4 units and is parallel to the line 2x 3y 7 = 0. Will cuts
the Xaxis.
Q.16
(a.) (6, 0) (b.) (5, 0) (c.) (5, 0) (d.) (4, 0)Ans. (a)
2x 2y 7 = 0Line parallel to above line2x 3y + k = 0 (1)y intercept = 4yintercept is obtained by putting x = 0
y =3
k

7/29/2019 mathematical tools 1st test.pdf
5/13
a
\[Page No.5]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
12k;43
k==
Eqn (1) become ; 2x 2y + 12 = 0It will cuts the x ax is it y = 02x = 12x = 6 pt, (6, 0)The inclination of the straight line passing thro the point (3, 6) and the midpoint of the line joining the
point (4, 5) and (2, 9) is
Q.17
(a.)
4
(b.)
6
(c.)
3
(d.)
4
3
Ans. (d)
xB1B=2
24 yB1B=
2
59
xB1B = 1 yB1B = 2xB2B = 3 yB2B = 6
113
26m ==
tan =4
3
The ratio in which the line segment joining (2, 3) and (4, 1) is divided by the line joining (1, 2) and ( 4, 3)
is
Q.18
(a.) 1 : 2 (b.) 1 :3 (c.) 2 : 3 (d.) None of these
Sol: (b)
eq. of CD is( ) ( )112
121 xx
xx
yyyy
=
(4,1
y 2 = ( 1x1423
) y 2 = ( )1x
31
(x, y)
(4,3)
(2,3)
m n
P
D
C(xB1B, yB1B)
(1,2)
A B
(xB2B, yB2B)
3(y 2) = (x 1) .(1)
nm
n2m4
nm
2n4m
++
=++
P(x,y) is x =
nm
n3m
nm
3n1my
++=
++
=
Put in eq. (1), we get
++= ++1
nmn2m4
12
nmn3m3 + += + + nm
nmn2m4nm
n2m2n3m3
nm31
nm3 +=
+
3m + 3n = 3m + n 3m 3m = n 3n
6
2
n
m= m : n = 1 : 36m =  2n
Q.19 The vertices of a triangle ABC are (1, 2), (0, 1) and (2, 0) respectively. The equation of the median

7/29/2019 mathematical tools 1st test.pdf
6/13
a
\[Page No.6]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
through (1, 2) is
(b.) (c.) (d.) None of these(a.) 5x 4y 3 = 0 x + 2y 2 = 0 4x + y 1 = 0
Sol: (a)
++=
2
01,
2
20D
=
2
1,1D
B
A CD(0,1) (1,1/2)
(1,2)
(2,0)
)xx(xx
yy1
12
12
eq of BD is y yB1B =
( )1x11
2
12
2
1y
= ( )1x
22
14
2
1y
=
( 1x4
5
2
1y2=
( ) 5x51y22 =)
5x 5 4y +2 = 05x52y4 =
5x 4y 3 = 0
If 7 times the 7 PthP term of an A.P is equal to 11 times its 11PthP term. Then 18PthP term of the A.P. isQ.20
(a.) Zero (b.) 7 (c.) 13 (d.) 9
Ans. (a)
7(a + 6d) = 11(a + 10d)
7a + 42d = 11a + 10d
4a + 68d = 0
4(a + 17d) = 0
A + 17d = 0
18PthP term = a + 17d
= 0
SECTIONB (ONE OR MORE THAN ONE CORRECT TYPE QUESTIONS)5
12Q.1If cot = , is in quadrant III then
13
5
13
12cos=
(b.)(a.)sin =
1312cos =
13cossin 7=+(c.) (d.)
Ans. (a,b)
cot =5
12
sin [ ]vesinquadrantIIIrdtheinAs13
5=
13
5
12
cos = ]quadrantIIIrdinveis[cos13
12

7/29/2019 mathematical tools 1st test.pdf
7/13
a
\[Page No.7]
Master SolutionPremier Institute for the preparation of IITJEE /
The fourth, seventh and last term of a G. P. are 10, 80 and 2560 respectively. Which of the following is/are
correct.
Q.2
Ratio of 5PthP term of series to the 4 PthP term is 2.(a.)
(b.) The common ratio of the series is 2.
(c.) Total number of terms of the series is 12
(d.) The fifth term of the series is 20
Sol: (a, b, c, d)
aB4B = 10, aB7B = 80, aBnB = 2560
arP3P = 10 ..(1)
arP6P = 80 ..(2)
Divide (2) by (1), we get
10
80
ar
ar3
6
= rP3P = 8 r = 2
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
a(2)P3P =10 a 8 = 10 a =4
5
8
10 =4
5a =
24
5 aB2B = ar = aB2B =
2
5
2560)2(4
5 1n=
Also arPn1P = 2560
5(2)Pn1P = 2560 4
(2)Pn1P = 512 4
(2)Pn1P = (2)P11P n 1 = 11 n =12
aB5B = arP4P
= ( )424
5 16
4
5= = aB5B =20
Q.3
4
1If sin = , then which of the following is/are correct?
8
7
15
1
15
4(c.)(b.) (d.)(a.)
4
3 Tan = Sec =Cos 2 =Sin 2 =
Sol: (b, c, d)
sin =4
1
sin 2 = 2 sin. cos = 2 4
15
4
1
sin2 =8
15 .(1) A B
C
1
4
15
4P2P = 1P2P + ABP2Pcos 2 = 1 2 sinP2
P
= 1 2 2
4
1
16
1
8
18 AB116 = = 1 2 =
AB15 =

7/29/2019 mathematical tools 1st test.pdf
8/13
a
\[Page No.8]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
cos 2 =8
7
tan =15
1
Q.4 If 3 sin + 4 cos = A sin ( + ), then which of the following is/are correct?
(d.)(a.) A = 5 (b.) A = 7 (c.)
3
4tan =
4
3tan =
Sol: (a, c)
3sin + 4 cos = A sin ( + )22 baA += 22 43 + 169 + 25= = = = 5
As [ ] + cos4sin3
=
+ cos
5
4sin
5
35
5
4sin,
5
3=cos =
3
5
5
4 tan =
tan =3
4
Q.5 The equation of straight line is given as 3x 4y + 10 = 0 . Then
4
3
4
3
(a.) (b.)Its slope is Its slope is
3
10
(c.) (d.) It will pass through (6, 7)Its xintercept is
Ans. (b, c, d)3x 4y + 10 = 0 (1)4y = 3x + 10
y =4
10x
4
3+
slope =4
3
to find xintercept put y = 03x = 10
x =3
10
Since pt (6,7) satisfy eq(1) so the eqn pass through (6,7)
Q.6 The right bisector of segment joining (3, 4) and (1, 2) has
(a.) Slope 2 (b.) Slope 2
(c.) Equation 2x + y = 5 (d.) Equation 2x + y + 5 = 0
Ans. (a, c)A (3, 4) B = (1, 2)
2
24,
2
13 +Pt. from which bisector pass =

7/29/2019 mathematical tools 1st test.pdf
9/13
a
\[Page No.9]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
= (1, 3)& slope of right bisector = 2
Eqn = y 3 = 2(x 1)y 3 =2x + 2
2x + y 3 2 = 0
2x + y 5 = 0Q.7 It and are the roots of 4xP2P + 3x + 7 = 0, then
7
3
7
3
16
47
16
47
(a.) (b.) (c.) (d.) + = + = P2P + P2P = P2P + P2P =Sol: (d)
4xP2P + 3x + 7 = 0
4
3=+
4
7=
+=+ 2)( 222
=4
72
16
9
2
7
16
9=
16
47=22 +
Q.8 A person standing on the bank of a river, observes that the angle of elevation of the top of a tree on the
opposite bank of the river is 60 and when he retires 40 m away from the tree the angle of elevation
becomes 30. Then
(a.) width of the river is 20 m (b.) width of the river is 60 m
(c.) (d.)320 3/20Height of the tree is Height of the tree is
Ans. (a, c)
6040 x
30y
o60tanx
y= x3y =
o30tan40x
y=+
3
40xy+= 40x)x3(3 +=
3x = x + 40 x = 20
320y =
Width of the river = 20 m
3 Height of tree = 20

7/29/2019 mathematical tools 1st test.pdf
10/13
a
\[Page No.10]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
Q.9 The sequence 9, 12, 15, 18, is an A.P. Then
(a.) Its general term is 3n + 6 (b.) Its general term is 3n 6
(c.) Its 16PthP term is 54 (d.) Its 10PthP term is 36
Ans. (a,c, d)
9, 12, 15, 18 .
a = 9d = 3general term = a +(n 1) d
= 9 + (n 1)3= 3n + 6
10PthP term 9 +(10 1)9 + 27 = 36
16PthP term = 9 + 15 3= 54
Q.10 A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3.Then
31
34(a.) (b.)Its yintercept is Its yintercept is
Its equation is x 3y + 4 = 0 Its equation is x 3y = 4(c.) (d.)
Ans. (b,c)
3
1 Slope of required line =
3
1Eqn of line = y 2 = (x 2)
3y 6 = x 2 x 3y + 4 = 0
3
4
Its yintercept =SECTIONC (SUBJECTIVE TYPE QUESTIONS)Q.1
=
+
2tan
2tan
3 sin = 5 sin , then
Sol: [4]
= sin5sin3
3
5
sin
sin=
35
35
sinsin
sinsin
+=+
2sin.
2cos2
2cos.
2sin2
+
+
= 4

7/29/2019 mathematical tools 1st test.pdf
11/13
a
\[Page No.11]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
2.tan
2tan
+
= 4
Q.2 Find the sum of the given series up to 2 decimal places
..........321
161
81
41
211 +++
Sol: [0.66 or 0.67]
As it is infinite G.P. series; usingr1
aS= =
2
11
1=
3
2= 0.67
Q.3 Find the slope of the equation of the perpendicular bisector of the line segment joining the points A(2, 3)
and B( 6, 5)
Sol: [0.5 ]
Sol: 0.5
12
12
xx
yy
4
826
35
mB1B = 2= =Slope of AB =
Using mB1B mB2B = 1
2 mB2B = 1
mB2B =2
1 mB2B = 0.5
Q.4 If one root of the equation (x 1) (7 x) = m is three times the other, then m is equals to
Ans. [5](x 1) (7 x) = m
Let one root = Other root = 37x xP2P 7 + x = mxP2P 8x + 7 + m = 0
4 = 8 = 27 + m = 3P2Pm + 7 = 12m = 5
Q.5 The xintercept of equation of a line perpendicular to x 2y + 3 = 0 and passing through the pt(3, 2) is
Ans. [2]
Slope of required line = 2
Eqn of line y + 2 = 2(x 3)y + 2 = 2x + 62x + y 4 = 0
For its xintercept put y = 0x = 2

7/29/2019 mathematical tools 1st test.pdf
12/13
a
\[Page No.12]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
SECTIOND (MATCH THE COLUMN TYPE QUESTIONS)(Items of Column I can match one or more than one Items of column II with individual marking scheme)
Q.1 In the following options symbols have their usual meaning. Match the column I with columnII
ColumnI Column II
r1a
(i.) Sum of n terms in arithmetic progression (a.)
( )r1
r1a n
(ii.) nPthP term of arithmetic progression (b.)
(iii.) nPthP term of geometric progression (c.) a + (n 1)d
arPn1P(iv.) Sum of infinite series of geometric progression (d.)
( )[ ]d1na22
n+
(e.)
Sol: (i) (e), (ii) (c), (iii) (d), (iv) (a)
ColumnI Column IIQ.2
(a.) 0(i.) If sinB1B + sinB2B + sinB3B = 3 then1
cos B1B + cos B2B + cos B3B =
(b.) 1(ii.) tan(180 + ) . tan(90 + ) =
(c.) 2(iii.) cotA + tan (180 + A) + tan(90 + A) +
tan(360  A) =
(iv.) (d.) 3
o
o
o
o
70cot
20tan
36tan
54cot+ =
(e.) 1
Ans. (i a), (ii e), (iii a), (iv c)(i) sin B1B + sinB2B + sinB3B = 3
2
So, B1B = B2B = B3B =
So, cosB1B + cosB2B + cosB3B
= 0
(ii) tan(180 + ) . tan(90 + )
=tan.[cot ] = 1
(iii) cotA + tan(180 + A) + tan(90 + A) +tan(360  A)
= cotA + tanA + [cotA) tanA
= 0
o
o
o
o
70cot
20tan
36tan
54cot+(iv)

7/29/2019 mathematical tools 1st test.pdf
13/13
a
\[Page No.13]
PiSCHOLASTICS:S.C.O 1617, 1st & 2nd Floor, Sector 20D, Chandigarh.Ph: 6544444, 3910727, 5025027, 9216144327
Master SolutionPremier Institute for the preparation of IITJEE /
=oo
o
o
oo
2090cot(
20tan
36tan
)3690cot(
+
= 1 +1 = 2