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Transcript of IIT TRIGNO
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7/30/2019 IIT TRIGNO
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
PAPER B
IITJEE(2013)(Trigonometry and Algebra)
TOWARDS IIT JEE IS NOT A JOURNEY,ITS A BATTLE, ONLY THE TOUGHEST WILL SURVIVE
TIME: 60 MINS MAX. MARKS: 76
MARKING SCHEME
In Section I (Total Marks: 24), for each question you will be awarded 3 marks if you darken ONLYthe bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all
other cases, minus one ( 1) mark will be awarded.
In Section II (Total Marks: 16), for each question you will be awarded 4 marks if you darkenALLthe bubble(S) corresponding to the correct answer(s) ONLYand zeromarks otherwise. There are
no negative marks in this section.
In Section III (Total Marks: 24), for each question you will be awarded 4 marks if you darken ONLthe bubble corresponding to the correct answer and zero marks otherwise. There are no negative
marks in this section.
In Section IV (Total Marks: 12), for each question you will be awarded 2 marks for each row inwhich you have darkenedALL the bubble(s) corresponding to the correct answer(s) ONLYand zer
marks otherwise. Thus, each question in this section carries a maximum of 6 marks. There are no
negative marks in this section.
NAME OF THE CANDIDATE CONTACT NUMBER
L.K. Gupta (Mathematics Classes)FOR SOLUTIONS KINDLY VISIT
www.pioneermathematics.com(In latest Updates)
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
This section contains 8 multiple choice questions. Each question has four choices (a), (b)
(c), (d) out of which ONLY ONE is correct.
1. The equation sin (cos x) = cos (sin x) has(a) only one real solution (b) infinitely many solution
(c) no real solution (d) none of the above
Ans. (c)
Sol :
sin (cos x) = cos (sin x)
cos (sin x) sin (cos x) = 0
cos sin x cos cosx 0
2
cosx sin x cosx sin x2cos . cos 0
4 2 4 2
cosx sin xIf cos 0
4 2
cos x sin x n
4 2 2
2 sin x 4n 1 , n I4 2
4n 1sin x
4 2 2
sin x [ 1,1],
4
which does not certify the original solution,
cos x sin xcos 04 2
cos x sin xThen, cos 0
4 2
cosx sin x
2r 14 2 2
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
2 cos x 4r 14 2
cos x 4r 1 , r I4 2 2
cos x [ 1, 1]4
cos x sin xAlso, cos 0
4 2
LHS 0
and RHS 0
Hence, no real solution.
2. If 2x x 2 is factors of 4 2x x , then 2 2( ) equals(a) 1 (b) 3 (c) 5 (d) 7
Sol: (b)
2x x 2 0
x 2, 1
Then x 2, 1 are the roots of 4 2x x 0
then 4 2(2) (2) 0
4 16
and 4 2( 4) ( 1) 0
1
we get 5 and 4
2 2( ) 25 16 9 3
3. The number of solutions of 5r 1
cos r x = 5 in the interval 0, 2 is
(a) 0 (b) 1 (c) 5 (d) 10
Ans. (b)
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
Sol :
5
r 1
cos r x 5
cos x + cos 2x + cos 3x + cos 4x + cos 5x = 5 which is possible only, when
cos x = cos 2x = cos 3x = cos 4x = cos 5x = 1and is satisfied by x = 0 only.
Hence number of solution = 1.
4. Set a, b , be such that cos (a b) = 1 and cos (a + b) = 1 .e
The number of pair
of a, b satisfying the above system of equation is
(a) 0 (b) 1 (c) 2 (d) 4
Ans. (d)
Sol :
cos a b cos 0
a b 2n, n I
a b 2, 0, 2
a b 2 2a, 2a, 2a 2
1
cos a b cos 2a, cos 2a, cos 2ae
1y cos 2a
e
Hence, number of solutions is 4.
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
5. The number of values of x for which sin 2x + cos 4x = 2 is(a) 0 (b) 1 (c) 2 (d) infinite
Ans. (a)
Sol :
sin 2x + cos 4x = 2
It is possible only, when
sin 2x = 1 and cos 4x = 12x 2n and 2x 2m
2
x n and x m m, n I
4
Then, solution n , n I m, m I4
= 6. If 1, 2, are the three cube roots of unity, then for
2 2
2
, , , R, the expression is
(a) 1 (b) (c) (d) 1
Ans. (b)
Sol :
2 2
2
2 2
3 2 2
3 1 . 7. The number of positive integral solutions of
3 42
5 6
x 3x 4 x 20 is
x 5 2x 7
(a) four (b) three (c) two (d) only one
Ans. (b)
Sol :
3 42
5 6
x 3x 4 x 2Since, 0
x 5 2x 7
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
4 7
x , 7/2, 53 2
x 2, 3, 4.
8. If tan cos cot sin , then the value of cos is4
(a)1
2(b)
1
2 (c)
1
2 2(d) none of these
Ans. (c)
Sol :
tan cos cot sin
tan cos tan sin 2
cos sin
2
1cos sin 2
1 1 1cos sin
2 2 2 2
1cos
4 2 2
Section II (Total Marks: 16)
(Multiple Correct Answer (s) Type)
This section contains 4 multiple choice questions. Each question has four choices (a), (b),
(c), and (d) out of which ONE or MORE may be correct.
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
9. If tan and tan are the roots of the equation 2x px q 0 p 0 , then (a) 2 2sin p sin cos qcos q
(b) tan p/ q 1 (c) cos 1 q
(d) sin p Ans. (a, b)
Sol :
tan tan p, tan tan q q 0
tan tan p
tan 1 tan tan 1 q
p[Alternate.(b)]
q 1
Alternate. (a) :
2 2LHS cos tan p tan q
2
2
tan p tan q
1 tan
2 2
2 22 2
2 2 2
2
p pq
q 1q 1 p q q 1 q q 1
p q 1 p1q 1
22
22
q p q 1
p q 1
= q
Alternate. (c) : p
tan q 1
22
q 1cos
p q 1
22
pAlternate. (d) : sin
p q 1
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
10. The argument and the principal argument of the complex number
22 i
, where i 1 are4i 1 i
(a) 1tan 2 (b) 1tan 2 (c) 11
tan2
(d) 11
tan2
Ans. (a, b)
Sol :
2 2
2 i 2 i 2 i 2 i
4i 1 i 2i 4i 1 1 2i 6i4i 1 i
1 1i
6 3
1 1 1
1
3 tan tan 2 tan 21
6
1 1imaginary part
Argument tan tan 2real part
1and principal arg ument tan 2
11. The roots of the equation, (x2 + 1)2 = x(3x2 + 4x + 3), are given by(a) 2 3 (b) 1 i 3 /2, i 1
(c) 2 3 (d) 1 i 3 /2, i 1
Ans. (a, b, c, d)
Sol :
Given equation is
2
2 2x 1 x 3x 4x 3 4 3 2x 3x 2x 3x 1 0
2 2
2
3 1x x 3x 2 0
x x
x 0
2
2
1 1x 3 x 2 0
x x
21 1
x 3 x 4 0x x
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
1 1x 4 x 1 0
x x
2 2or x 4x 1 x x 1 0
2
1 3or x 2 h 3 x 0
2 4
1 i 3
x 2 3 , .2
12. In a triangletan A + tan B + tan C = 6 and tan A tan B = 2, then the values of tan A, tan B and tan C are
(a) 1, 2, 3 (b) 2, 1, 3 (c) 1, 2, 0 (d) none of these
Ans. (a, b)
Sol :
In a triangle
tan A + tan B + tan C = tan A tan B tan C
or 6 = 2 tan C
tan C = 3 tan A + tan B = 3, tan A tan B = 2 tan A = 1 or 2and tan B = 2 or 1. (i)
Section III (Total Marks: 24)
(Integer Answer Type)
This section contains 6 questions. The answer to each of the questions is a single-digit
integer, ranging from 0 to 9.The bubble corresponding to the correct answer is to be
darkened in the Answer sheet.
13. 4 20 0
1 1If , then the value of 9 81 97 782 must be
cos 290 3 sin250
Ans. 3
Sol :
Here, cos 2900 = cos (2700 + 200) = sin 200 and sin 2500 = sin (2700 200) = cos 200
0 0
1 1The given expression
sin 20 3 cos 20
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
0
0 0 0
1 cos 60
sin 20 sin60 cos 20
0 0 0 0
0 0 0
sin 60 cos20 cos 60 sin20
sin 20 cos 20 sin60
0 0
0sin 60 20 sin40 3
2 2
4
3
2 163
4 2 256 16Then, 9 81 97 9 81 979 3
= 256 + 432 + 97-782
= 3
14. The least degree of a polynomial with integer coefficient whose one of the roots may bcos 120 is
Ans. 4
Sol :
0 012 5 60
0let 12
05 60
0
3 2 60
0cos 3 2 cos 60
1cos 3 cos 2 sin 3 sin 2
2
3 2 31
4 cos 3 cos 2cos 1 3 sin 4sin 2 sin cos 2
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
Let cot x
3 2 2 2 14x 3x 2x 1 2x 3 4 1 x 1 x2
5 3 3 21
8x 10x 3x 2x 2x 4x 12
5 3 3 216x 20x 6x 4x 4x 4x 1 1 0
5 332x 40x 10x 1 0
4 3 21
x 32x 16x 32x 16x 2 02
4 3 21but x , 16x 8x 16x 8x 1 02
Degree is 4.
15. The three angles of a quadrilateral are 600, 60g and 5 ,6
if fourth angle is then 0,
the value of -90 must be
Ans. 6
Sol :
First angle =600,second angle=60g=
0
0
0
0
9060 54
100
5 180Third angle 150
6
Fourth angle=3600-(600+540+1500)
=960= 0-90=6
16. The number of solutions of 3sec 5 4 tan in [0,4 ] must beSol. 8
The given equation can be written as
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
3 4sin5
cos cos
4 sin 5 sin 3,cos 0
4 5 3sin cos
41 41 41 (i)
Let5 4
cos , then sin41 41
Now, from Eq. (i),
3
cos cos41
32n , where cos
41
or 2n
or 2n
Now, in [0,2 ], 2n gives two solutions.
So, in total in [0,4 ]
we have , 2 + 2 + 2 + 2 = 8 solutions
17. The sum of the roots of equation cos 4x + 6 = 7 cos 2x over the interval [0, 314] is , then the numerical quantity 4949 must be
Ans. 1
Sol :
On putting cos 2x = 1 we get
2t2 1 + 6 = 7t
5t 1,
2
5t (impossible)
2
t 1
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
cos 2x 1
2x 2n
x n , n I
The roots over [0, 314] are , 2, 3, ...., 99
100 314
Sum of roots 2 3 .... 99 4950
4950
18. The sides of a cyclic quadrilateral are in AP, the shortest is 6 and the difference of thelongest and the shortest is also 6. The square of the area of the quadrilateral is n-5756,find
the value of n.
Ans.4
Sol :
Let sides are a, a + d, a + 2d, a + 3d
Given, a = 6 and a + 3d a = 6 d 2
Sides 6, 8, 10, 12
2s = 6 + 8 + 10 + 12
= 36
s 18
Then, (Area)2 = (18 6) (18 8) (18 10) (18 12)
= 12 10 8 6
= 5760
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
Section IV (Total Marks: 12)
(Matrix-Match Type)
This section contains 2 questions. Each question has three statements (a, b and c) given in
Column I and five statements (p, q, r, s and t) in Column II. Any given statements in
Column I can have correct matching with ONE or MORE statements(s) given in Column II
For example, if for a given question, statement B matches with the statements given in qand r, then for the particular question, against statement B, darken the bubbles
corresponding to q and r in the ANSWER SHEET.
19. Observe the following columns :Column I Column II
(A)If maximum and minimum values of
2
2
7 6tan tan for
1 tan
all real values of
are and respectively , then2
(P) 2
(B) If maximum and minimum values of
5cos 3cos 3
3
for all real values of are and
respectively, then
(Q) 6
(C) If maximum and minimum values of
1 sin 2cos
4 4
for all real values of
are and respectively, then
(R) 6
(S) 10
(T) 14
Ans. A R, S ; B R, T ; C P, Q Sol :
2
2
7 6 tan tan A Let y
1 tan
2 27 cos 6 sin cos sin
1 cos 2 1 cos 27 3 sin 2
2 2
3 sin 2 4 cos 2 3
2 2 2 23 4 3 sin 2 4 cos 2 3 3 4 3
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
2 y 8 8, 2
6, 10 R, S
B Let y 5 cot 3 cos /3 3
1 35 cos 3 cos sin 3
2 2
13 3 3cos sin 3
2 2
2 22 213 3 3 13 3 3 13 3 3
3 cos sin 3 32 2 2 2 2 2
3 7 y 3 7 4 y 10
10, 4 6, 14 R, T
C Let y 1 sin 2 cos 4 4
1 cos 2 cos
2 4 4
1 cos 2 cos
4 4
1 3 cos
4
1 cos 1
4
3 3 cos 3
4
1 3 1 3 cos 1 3
4
2 y 4 4, 2
2, 6 P,Q
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L.K. Gupta (Mathematic Classes) wwwwww..ppiioonneeeerrmmaatthheemmaattiiccss..ccoomm MOBILE: 9815527721, 4617721
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40D, CHANDIGARH
SOLUTIONS ON www.pioneermathematics.com
20. Observe the following columns :Column I Column II
(A) 1If , are the solutions of sin x in
2 0, 2 and , are
the solutions of cos x =
3in 0,2 , then
2
(P)
(B) If , are the solutions of cot x 3 in 0,2 and
, are the solutions of cosec x 2 in 0, 2 , then
(Q)
(R)
(C)
1If , are the solutions of sin x in 0, 2 and ,
2
are the solutions of tan x = 1
in 0, 2 , then3
(S) 3
(T) 2
Ans. A Q, S ; B P,T ; C R, S, T Sol :
1
A sin x2
sin
6
sin , sin 2
6 6
7 11x ,6 6
..(i)
3 and cos x cos
2 6
cos , cos
6 6
5 7x ,
6 6 ..(i)
From Eqs. (i) and (ii). It is clear that7 11 5
, , y6 6 6
3 S , Q
B cot x 3
cot6
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PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S C O 320 SECTOR 40 D CHANDIGARH
cot , cot 2
6 6
5 11x ,
6 6 .(i)
and cosec x 2 cosec 6
cosec , cosec 26 6
7 11x ,
6 6 (ii)
From Eqs. (i) and (ii), It is clear that
11 5 7 , , 6 6 6
2 T , P
1
C sin x2
sin
6
sin , sin 26 6
7 11x ,
6 6 .(i)
1and tan x
3
tan
6
tan , tan 6 6
7
x ,6 6
(ii)
From Eqs. (i) and (ii), it is clear that
7 11 , ,
6 6 6
3, S , 2 T , R