TrigonometricFunctions
Question1
Letthesetofalla∈Rsuchthattheequationcos2x+asinx=2a−
7hasasolutionbe[p,q]and
[27-Jan-2024Shift1]
Answer:48
Solution:
-------------------------------------------------------------------------------------------------
Question2
If2tan2θ−5secθ=1hasexactly7solutionsintheinterval[0,
nπ/2],fortheleastvalueofn∈Nthen isequalto:
[27-Jan-2024Shift2]
Options:
A.
B.
C.
D.
Answer:D
Solution:
-------------------------------------------------------------------------------------------------
Question3
Ifα, isthesolutionof4cosθ+5sinθ=1,thenthevalue
oftanαis
[29-Jan-2024Shift1]
Options:
A.
B.
C.
D.
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question4
Thesumofthesolutionsx∈ℝoftheequation
[29-Jan-2024Shift2]
Options:
A.
0
B.
1
C.
-1
D.
3
Answer:C
Solution:
-------------------------------------------------------------------------------------------------
Question5
If2sin3x+sin2xcosx+4sinx−4=0hasexactly3solutionsinthe
interval[0,nπ/2],n∈N,thentherootsoftheequation
x2+nx+(n−3)=0belongto:
[30-Jan-2024Shift1]
Options:
A.
(0,∞)
B.
(−∞,0)
C.
D.
Z
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question6
Thenumberofsolutionsoftheequation4sin2x−4cos3x+9−4cosx=
0;x∈[−2π,2π]is:
[1-Feb-2024Shift2]
Options:
A.
1
B.
3
C.
2
D.
0
Answer:D
Solution:
Question7
Forα,β∈(0,π/2),let3sin(α+β)=2sin(α−β)andareal
numberkbesuchthattanα=ktanβ.Thenthevalueofkisequalto:
[30-Jan-2024Shift2]
Options:
A.
B.
-5
C.
2/3
D.
5
Answer:B
Solution:
Question8
[1-Feb-2024Shift1]
Options:
A.
C
B.
π−C
C.
2π−C
D.
π/2−C
Answer:A
Solution:
-------------------------------------------------------------------------------------------------
Question9
Ifmandnrespectivelyarethenumbersofpositiveandnegativevalue
ofθintheinterval[−π,π]thatsatisfytheequation
cos 2 θ cos θ
2=cos 3 θ cos 9θ
2,thenmnisequalto________.
[25-Jan-2023Shift2]
Answer:25
Solution:
cos 2 θ ⋅cos θ
2=cos 3 θ ⋅cos 9θ
2
⇒2 cos 2 θ ⋅cos θ
2=2 cos 9θ
2⋅cos 3 θ
Question10
LetS = {θ∈ [0,2π) : tan(π cos θ) + tan(πsin θ) = 0}.
Then ∑
θ∈Ssin 2θ+π
4isequalto
[24-Jan-2023Shift2]
Answer:2
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question11
Letf (θ) = 3 sin 43π
2−θ+sin 4(3π +θ) − 2(1−sin 22θ)and
S=θ∈ [0,π] : f′(θ) = − √3
2.If4β =∑
θ∈Sθ,thenf (β)isequalto
[29-Jan-2023Shift1]
Options:
A. 11
8
B. 5
4
( )
( ( ) )
{ }
⇒cos 5θ
2+cos 3θ
2=cos 15θ
2+cos 3θ
2
⇒cos 15θ
2=cos 5θ
2
⇒15θ
2=2kπ ±5θ
2
5θ =2kπ or 10θ =2kπ
θ=2kπ
5
∴θ= −π,−4π
5,−3π
5,−2π
5,−π
5,0,π
5,2π
5,3π
5,4π
5,π
∴m⋅n=25
{ }
tan(π cos θ) + tan(πsin θ) = 0
tan(π cos θ) = −tan(πsin θ)
tan(π cos θ) = tan(−πsin θ)
π cos θ =nπ −πsin θ
sin θ +cos θ =n where n ∈I
possiblevaluesaren=0,1and−1because
−√2≤sin θ +cos θ ≤ √2
Nowitgivesθ∈0,π
2,3π
4,7π
4,3π
2,π
So ∑
θ∈S
sin 2θ+π
4=2(0) + 41
2=2
{ }
( ) ( )
C. 9
8
D. 3
2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question12
Thesetofallvaluesofλforwhichtheequation
cos22x −2sin 4x−2cos2x=λ
[29-Jan-2023Shift2]
Options:
A.[−2, −1]
B. −2, − 3
2
C. −1, − 1
2
D. −3
2, −1
Answer:D
Solution:
[ ]
[ ]
[ ]
f(θ) = 3 sin 43π
2−θ+sin 4(3x +θ) − 2(1−sin 22θ)
S=θ∈ [0,π] : f′(θ) = − √3
2
⇒f(θ) = 3(cos4θ+sin 4θ) − 2cos22θ
⇒f(θ) = 3 1 −1
2sin 22θ −2cos22θ
⇒f(θ) = 3−3
2sin 22θ −2cos2θ
=3
2−1
2cos22θ =3
2−1
2
1+cos 4 θ
2
f(θ) = 5
4−cos 4 θ
4
f′(θ) = sin 4θ
⇒f′(θ) = sin 4θ = − √3
2
⇒4θ =nπ + (−1)nπ
3
⇒θ=nπ
4+ (−1)nπ
12
⇒θ=π
12,π
4−π
12 ,π
2+π
12 ,3π
4−π
12
⇒4β =π
4+π
2+3π
4=3π
2
⇒β=3π
8⇒f(β) = 5
4−
cos 3π
2
4=5
4
( ( ) )
{ }
( )
( )
( ) ( ) ( )
Solution:
-------------------------------------------------------------------------------------------------
Question13
Iftan 15∘+1
tan 75∘+1
tan 105∘+tan 195∘=2athenthevalueof a +1
ais:
[30-Jan-2023Shift1]
Options:
A.4
B.4 −2√3
C.2
D.5 −3
2√3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question14
Ifthesolutionoftheequation
( )
λ=cos22x −2sin 4x−2cos2x
convertallinto cos x .
λ= (2cos2x−1)2−2(1−cos2x)2−2cos2x
=4cos4x−4cos2x+1−2(1−2cos2x+cos4x)−
2cos2x
=2cos4x−2cos2x+1−2
=2cos4x−2cos2x−1
=2 cos4x−cos2x−1
2
=2 cos2x−1
2
2−3
4
λmax =21
4−3
4=2× − 2
4= −1 (maxValue)
λmin =2 0 −3
4= − 3
2(MinimumValue)
So,Range = − 3
2, −1
[ ]
[ ( ) ]
[ ] ( )
[ ]
[ ]
Option(1)
tan 15∘=2− √3
1
tan 75∘=cot 75∘=2− √3
1
tan 105∘=cot(105∘) = −cot 75∘= √3−2
tan 195∘=tan 15∘=2− √3
∴2(2− √3) = 2a ⇒a=2− √3
⇒a+1
a=4
logcos x cot x +4logsin x tan x =1,x∈0,π
2,issin −1α+ √β
2,whereα,β
areintegers,thenα +βisequalto:
[30-Jan-2023Shift1]
Options:
A.3
B.5
C.6
D.4
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question15
Thevalueoftan 9∘−tan 27∘−tan 63∘+tan 81∘is_______:
[6-Apr-2023shift2]
Answer:4
Solution:
Solution:
-------------------------------------------------------------------------------------------------
( ) ( )
logcos x cot x +4logsin x tan x =1
⇒ln cos x −ln s i n x
ln cos x +4ln s i n x −ln cos x
ln s i n x =1
⇒ (ln s i n x)2−4(ln s i n x)(ln cos x) + 4(ln cos x)2=1
⇒ln s i n x =2 ln cos x
⇒sin 2x+sin x −1=0⇒sin x =−1+ √5
2
∴α+β=4
Correctoption(4)
(tan 9∘+cot 9∘) − (tan 27∘+cot 27∘)
1
sin 9∘cos 9∘−1
sin 27∘cos 27∘
2
sin 18∘−2
sin 54∘
2(4)
√5−1−2(4)
(√5+1)
8(√5+1)
4−8(√5−1)
4
2[(√5+1) − (√5−1)]
=4
Question16
Thevalueof36(4cos29∘−1)(4cos227∘−1)(4cos281∘−1)(4cos2243∘−1)
is
[8-Apr-2023shift2]
Options:
A.27
B.54
C.18
D.36
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question17
96 cos π
33 cos 2π
33 cos 4π
33 cos 8π
33 cos 16π
33 isequalto:
[10-Apr-2023shift1]
Options:
A.4
B.2
C.3
D.1
Answer:C
Solution:
Solution:
4cos2θ−1=4(1−sin 2θ) − 1=3−4sin 2θ=sin 3θ
sin θ
sogivenexpressioncanbewrittenas
36×sin 27∘
sin 9∘×sin 81∘
sin 27∘×sin 243∘
sin 81∘×sin 729∘
sin 243∘
36×sin 729∘
sin 9∘=36
96 cos π
33 cos 2π
33 cos 22π
33 cos 23π
33 cos 24π
33
∵cos A cos 2 A cos 22A...cos 2n−1A=sin (2nA)
2nsin A
HereA=π
33,n=5
-------------------------------------------------------------------------------------------------
Question18
LetS =x∈ − π
2,π
2:91−tan2x+9tan2x=10 andb =2
∑
x∈S tan
x
3,then
1
6(β−14)2isequalto
[10-Apr-2023shift2]
Options:
A.16
B.32
C.8
D.64
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question19
Thenumberofsolutionsof|cos x| = sin x,suchthat−4π ≤x≤4πis:
{ ( ) } ( )
=
96sin 25π
33
25sin π
33
=
96sin 32π
33
32sin π
33
=
3sin π −π
33
sin π
33
=3
( )
( )
( )
( )
( )
( )
Let9tan2x=P
9
P+P=10
P2−10P +9=0
(P−9)(P−1) = 0
P=1,9
9tan2x=1,9tan2x=9
tan2x=0,tan2x=1
x=0, ± π
4∴x∈ − π
2,p
2
β=tan2(0) + tan2+π
12 +tan2−π
12
=0+2(tan 15∘)2
2(2− √3)2
2(7−4√3)
Than 1
6(14 −8√3−14)2=32
( )
( ) ( )
[25-Jul-2022-Shift-1]
Options:
A.4
B.6
C.8
D.12
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question20
LetS =θ∈ [0,2π] : 82sin2θ+82cos2θ=16 .Then
n(s) + ∑
θ∈Ssec π
4+2θ cosec π
4+2θ isequalto:
[26-Jul-2022-Shift-1]
Options:
A.0
B.−2
C.−4
D.12
Answer:C
Solution:
Solution:
{ }
( ( ) ( ) )
Periodof|cos x| = π
Andperiodofsin x =2π
Graphofsin xand|cos x|cutseachotherattwopointsAandBin[0,2π]
So,in[−4π,4π],total4similargraphwillbepresentandgraphofsin xand|cos x|willcut4×2=8times.
∴Totalpossiblesolutions = 8
S=θ∈ [0,2π] : 82sin2θ+82cos2θ=16
NowapplyAM≥GMfor82sin2θ,82cos2θ
{ }
-------------------------------------------------------------------------------------------------
Question21
Ifthesumofsolutionsofthesystemofequations2sin2θ−cos 2 θ =0
and2cos2θ+3 sin θ =0intheinterval[0,2π]iskπ,thenkisequalto
________.
[26-Jul-2022-Shift-2]
Answer:3
Solution:
82sin2θ+82cos2θ
2≥82sin2θ+2cos2θ
1
2
8≥8
⇒82sin2θ=82cos2θ
or sin2θ=cos2θ
∴θ=π
4,3π
4,5π
4,7π
4
n(S) + ∑
θ∈S
sec π
4+2θ cos e c π
4+2θ
4+∑
θ∈S
2
2 sin π
4+2θ cos π
4+2θ
=4+∑
θ∈S
2
sin π
2+4θ
=4+2∑
θ∈S
cosec π
2+4θ
=4+2 cosec π
2+π cosec π
2+3π +cosec π
2+5π +cosec π
2+7π
=4+2−cosec π
2−cos e c π
2−cos e c π
2−cosec π
2
=4−2(4)
=4−8
= −4
( )
( ) ( )
( ) ( )
( ) ( )
[ ( ) ( ) ( ) ( ) ]
[ ]
Equation(1)2sin2θ=1−2sin2θ
⇒sin2θ=1
4
⇒sin θ = ± 1
2
⇒θ=π
6,5π
6,7π
6,11π
6
Equation(2)2cos2θ+3 sin θ =0
⇒2sin2θ−3 sin θ −2=0
⇒2sin2θ−4 sin θ +sin θ −2=0
⇒(sin θ −2)(2 sin θ +1) = 0
⇒sin θ =−1
2
⇒θ=7π
6,11π
6
∴Commonsolutions = 7π
6;11π
6
Sumofsolutions = 7π +11π
6=18π
6=3π
∴k=3
Question22
LetS = {θ∈ (0,2π) : 7cos2θ−3sin2θ−2cos22θ =2}.Then,thesumof
rootsofalltheequationsx2−2(tan2θ+cot2θ)x+6sin2θ=0,θ∈S,is
_______.
[29-Jul-2022-Shift-1]
Answer:16
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question23
2 sin π
22 sin 3π
22 sin 5π
22 sin 7π
22 sin 9π
22 isequalto:
[25-Jul-2022-Shift-2]
Options:
A. 3
16
B. 1
16
C. 1
32
D. 9
32
Answer:B
Solution:
Solution:
( ) ( ) ( ) ( ) ( )
7cos2θ−3sin2θ−2cos22θ =2
⇒41+cos 2 θ
2+3 cos 2 θ −2cos22θ =2
⇒2+5cos2θ−2cos22θ =2
⇒cos 2 θ =0 or 5
2(rejected )
⇒cos 2 θ =0=1−tan2θ
1+tan2θ⇒tan2θ=1
∴Sumofroots =2(tan2θ+cot2θ) = 2×2=4
Butastan θ = ±1for π
4,3π
4,5π
4,7π
4intheinterval(0,2π)
∴Fourequationswillbeformed
Hencesumofrootsofalltheequations = 4×4=16.
( )
2 sin π
22 sin 3π
22 sin 5π
22 sin 7π
22 sin 9π
22
-------------------------------------------------------------------------------------------------
Question24
LetS =θ∈0,π
2:∑
m=1
9sec θ + (m−1)π
6sec θ +mπ
6= − 8
√3.
Then
[27-Jul-2022-Shift-2]
Options:
A.S =π
12
B.S =2π
3
C. ∑
θ∈S
θ=π
2
D. ∑
θ∈S
θ=3π
4
Answer:C
Solution:
Solution:
{( ) ( ) ( ) }
{ }
{ }
=2 sin 11π −10π
22 sin 11π −8π
22 sin 11π −6π
22 sin 11π −4π
22 sin 11π −2π
22
=2 cos π
11 cos 2π
11 cos 3π
11 cos 4π
11 cos 5π
11
=
2 sin 32π
11
25sin π
11
=1
16
( ) ( ) ( ) ( ) ( )
s=0∈0,π
2:
9
∑
m=1
sec θ + (m−1)π
6sec θ +mπ
6= − 8
√3.
9
∑
m=1
1
cos θ + (m−1)π
6
cos θ +mπ
6
1
sin π
6
9
∑
m=1
sin θ +mπ
6−θ+ (m−1)π
6
cos θ + (m−1)π
6cos θ +π
6
=2
9
∑
m=1
tan θ +mπ
6−tan θ + (m−1)π
6
Now,
m=1 2 tan θ +π
6−tan(θ)
m=2 2 tan θ +2π
6−tan θ +π
6
.
.
.
m=9 2 tan θ +9π
6−tan θ +8π
6
∴ = 2 tan θ +3π
2−tan θ =−8
√3
{( ) ( ) ( ) }
( ) ( )
( )
[ ( ) ( ) ]
( ) ( )
[ ( ) ( ) ]
[ ( ) ]
[ ( ) ( ) ]
[ ( ) ( ) ]
[ ( ) ]
-------------------------------------------------------------------------------------------------
Question25
LetS = −π,π
2− − π
2, − π
4, − 3π
4,π
4.Thenthenumberofelementsin
theset∣A= {θ∈S:tan θ(1+ √5tan(2θ)) = √5−tan(2θ)}is_______.
[28-Jul-2022-Shift-2]
Answer:5
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question26
LetS =θ∈ [−π,π] − ± π
2:sin θ tan θ +tan θ =sin 2 θ .
IfT =∑
θ∈Scos 2 θ,thenT +n(S)isequalto:
[24-Jun-2022-Shift-1]
Options:
A.7 + √3
[ ) { }
{ { } }
= −2[cot θ +tan θ] = −8
√3
= − 2×2
2sinθ cos θ =−8
√3
=1
sin2θ =2
√3
⇒sin2θ =√3
2
2θ =π
3
2θ =2π
3
θ=π
6
θ=π
3
∑θi=π
6+π
3=π
2
Lettan α = √5
∴tan θ =tan α −tan 2 θ
1+tan α tan 2 θ
∴tan θ =tan(α−2θ)
α−2θ =nπ +θ
⇒3θ =α−nπ
⇒θ=α
3−nπ
3;n∈Z
Ifθ∈ [−π,π∕2]thenn=0,1,2,3,4areacceptable
∴5solutions.
B.9
C.8 + √3
D.10
Answer:B
-------------------------------------------------------------------------------------------------
Question27
Thenumberofvaluesofxintheinterval π
4,7π
4forwhich
14cosec2x−2sin2x=21 −4cos2xholds,is____
[25-Jun-2022-Shift-1]
Answer:4
Solution:
( )
Question28
Thenumberofelementsintheset
S= {θ∈ [−4π,4π] : 3cos22θ +6 cos 2 θ −10cos2θ+5=0}is
[29-Jun-2022-Shift-1]
Answer:32
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question29
Thenumberofsolutionsoftheequation
2θ −cos2θ+ √2=0inRisequalto
[29-Jun-2022-Shift-1]
Answer:1
Solution:
Solution:
3cos22θ +6 cos 2 θ −10cos2θ+5=0
3cos22θ +6 cos 2 θ −5(1+cos 2 θ) + 5=0
3cos22θ +cos 2 θ =0
Cos 2 θ =0 OR cos 2 θ = −1∕3
θ∈ [−4π,4π]
2θ = (2n +1)⋅ π
2
∴θ= ±π∕4. ± 3π ∕4....... ± 15π ∕4
Similarlycos 2 θ = −1∕3gives16solution
Given,
2θ −cos2θ+ √2=0
⇒2θ + √2=cos2θ
⇒2θ + √2=1+cos 2 θ
2
⇒4θ +2√2=1+cos 2 θ =y (Assume)
∴y=4θ +2√2 and
y=1+cos 2 θ
Fory=1+cos 2 θ
whenθ=0,y=1+1=2
whenθ=π
4,y=1+cos π
2=1
-------------------------------------------------------------------------------------------------
Question30
Thenumberofsolutionsoftheequationsin x =cos2xintheinterval
(0,10)is____
[29-Jun-2022-Shift-2]
Answer:4
Solution:
θ=π
2,y=1+cos π =1−1=0
Fory=4θ +2√2
whenθ=0,y=2√2
whenθ=π
2,y=2π +2√2
=2(π+ √2)
=2(3.14 +1.41)
=2(4.55)
=9.1
whenθ= − π
2,y= −2π +2√2
=2(−π+ √2)
=2(−3.14 +1.41)
= −3.46
∴Twographcut'satonlyonepointsoonesolutionpossible.
Question31
Thenumberofsolutionsoftheequation
cos x +π
3cos π
3−x=1
4cos22x,x∈ [−3π,3π]is:
[24-Jun-2022-Shift-2]
Options:
A.8
B.5
C.6
D.7
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question32
Thevalueof2 sin(12∘) − sin(72∘)is:
[25-Jun-2022-Shift-2]
Options:
A. √5(1− √3)
4
B. 1− √5
8
C. √3(1− √5)
2
D. √3(1− √5)
4
Answer:D
( ) ( )
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question33
Ifsin2(10∘)sin(20∘)sin(40∘)sin(50∘)sin(70∘) = α−1
16 sin(10∘),then
16 +α−1isequalto____
[26-Jun-2022-Shift-1]
Answer:80
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question34
16 sin(20∘)sin(40∘)sin(80∘)isequalto:
[26-Jun-2022-Shift-2]
2 sin 12∘−sin 72∘
=sin 12∘+ (−2 cos 42∘⋅sin 30∘)
=sin 12∘−cos 42∘
=sin 12∘−sin 48∘
=2 sin 18∘⋅cos 30∘
= −2√5−1
4⋅√3
2
=√3(1− √5)
4
( )
Options:
A.√3
B.2√3
C.3
D.4√3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question35
Thevalueofcos 2π
7+cos 4π
7+cos 6π
7isequalto:
[27-Jun-2022-Shift-1]
Options:
A.−1
B.−1
2
C.−1
3
D.−1
4
Answer:B
Solution:
Solution:
( ) ( ) ( )
cos 2π
7+cos 4π
7+cos 6π
7 =
sin 3 π
7
sin π
7
cos
2π
7+6π
7
2
=
sin 3π
7⋅cos 4π
7
sin π
7
=
2 sin 4π
7cos 4π
7
2 sin π
7
( ) ( )
( ) ( )
( )
-------------------------------------------------------------------------------------------------
Question36
α=sin 36∘isarootofwhichofthefollowingequation?
[27-Jun-2022-Shift-2]
Options:
A.16x4−10x2−5=0
B.16x4+20x2−5=0
C.16x4−20x2+5=0
D.16x4−10x2+5=0
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question37
Ifcot α =1andsec β = − 5
3,whereπ <α<3π
2and π
2<β<π,thenthe
valueoftan(α+β)andthequadrantinwhichα +βlies,respectivelyare
:
[28-Jun-2022-Shift-2]
Options:
A.−1
7andIV th quadrant
B.7andI st quadrant
C.−7andI V th quadrant
D. 1
7andI st quadrant
Answer:A
Solution:
=
sin 3π
7
2 sin π
7
=
−sin π
7
2 sin π
7
=−1
2
( )
α=sin 36∘=x(say)
∴x=10 −2√5
4
⇒16x2=10 −2√5
⇒(8x2−5)2=5
⇒16x4−80x2+20 =0
∴4x4−20x2+5=0
√
Solution:
-------------------------------------------------------------------------------------------------
Question38
cosec 18°isarootoftheequation
[31Aug2021Shift1]
Options:
A.x2+2x −4=0
B.4x2+2x −1=0
C.x2−2x +4=0
D.x2−2x −4=0
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question39
Thevalueof
2 sin π
8sin 2π
8sin 3π
8sin 5π
8sin 6π
8sin 7π
8is
[26Aug2021Shift2]
Options:
A. 1
4√2
( ) ( ) ( ) ( ) ( ) ( )
∵cot α =1,α∈π,3π
2
thentan α =1
andsecβ = − 5
3,β∈π
2,π
thentan β = − 4
3
∴tan(α+β) = tan α +tan β
1−tan α ⋅tan β
=
1−4
3
1+4
3
= − 1
7
α+β∈3π
2,2π i.e.fourthquadrant
( )
( )
( )
cosec 18° = 1
sin 18°=4
√5−1= √5+1
Ifx= √5+1,then
(x−1)2=5
⇒x2−2x −4=0
B.1
4
C.1
8
D. 1
8√2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question40
Ifsin θ +cos θ =1
2,then16(sin(2θ) + cos(4θ) + sin(6θ))isequalto:
[27Jul2021Shift1]
Options:
A.23
B.−27
C.−23
D.27
Answer:C
Solution:
Solution:
2 sin π
8sin 2π
8sin 3π
8sin 5π
8sin 6π
8sin 7π
8
=2 sin π
8sin 2π
8sin 3π
8sin π −3π
8sin π −2π
8sin π −π
8
=2 sin π
8sin 2π
8sin 3π
8sin 3π
8sin 2π
8sin π
8
=2sin2π
8sin22π
8sin23π
82sin2π
8
1
√2
2sin2π
2−π
8
=2sin2π
8×1
2×cos2π
8=sin2π
8cos2π
8
=1
42 sin π
8cos π
8
21
4sin2π
4=1
4
1
√2
2=1
8
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( )
sin θ +cos θ =1
2
sin2θ+cos2θ+2 sin θ cos θ =1
4
sin 2 θ = − 3
4
Now:
cos 4 θ =1−2sin22θ
=1−2−3
4
2
=1−2×9
16 = −1
8
sin 6 θ =3 sin 2 θ −4sin32θ
( )
-------------------------------------------------------------------------------------------------
Question41
Thevalueofcot π
24is:
[25Jul2021Shift2]
Options:
A.√2+ √3+2− √6
B.√2+ √3+2+ √6
C.√2− √3−2+ √6
D.3√2− √3− √6
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question42
If15sin4α+10cos4α=6,forsomeα ∈R,thenthevalueof
27sec6α+8cosec6αisequalto
[18Mar2021Shift2]
Options:
A.350
= (3−4sin22θ) . sin 2 θ
=3−49
16 . −3
4
⇒3
4× −3
4 = − 9
16
16[sin 2 θ +cos 4 θ +sin 6 θ]
16 −3
4−1
8−9
16 = −23
[ ( ) ] ( )
[ ] ( )
( )
cot θ =1+cos 2 θ
sin 2 θ =
1+√3+1
2√2
√3−1
2√2
θ=π
24
⇒cot π
24 =
1+√3+1
2√2
√3−1
2√2
=(2√2+ √3+1)
(√3−1)×(√3+1)
(√3+1)
=2√6+2√2+3+ √3+ √3+1
2
= √6+ √2+ √3+2
( )
( )
( ) ( )
( )
B.500
C.400
D.250
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question43
Ifforx ∈0,π
2,log10 sin x +log10 cos x = −1and
log10(sin x +cos x) = 1
2(log10n−1), n>0,thenthevalueofnisequalto
[16Mar2021Shift1]
Options:
A.20
B.12
C.9
D.16
Answer:B
Solution:
Solution:
( )
Given,15sin4α+10cos4α=6
⇒15sin4α+10cos4α=6(sin2α+cos2α)2
⇒15sin4α+10cos4α=6(sin4α+cos4α+2sin2αcos2α)
⇒9sin4α+4cos4α−12sin2αcos2α=0
⇒ (3sin2α−2cos2α)2=0
⇒3sin2α−2cos2α=0
⇒3sin2α=2cos2α
⇒tan2α=2
3
∴cot2α=3∕2
Now,
27sec6α+8cosec6α=27(sec2α)3+8(cosec2α)3
=27(1+tan2α)3+8(+cot2α)3
=27 1 +2
3
3+8 1 +3
2
3=250
( ) ( )
log10 sin x +log10 cos x = −1,x∈ (0,π/2)
log10(sin x cos x) = −1
⇒sin x cos x =10−1=1/10
log10(sin x +cos x) = 1/2(log10n−1), n>0
2log10(sin x +cos x) = (log10n−log1010)
⇒log10(sin x +cos x)2=log10(n/10)
⇒ (sin x +cos x)2=n/10
-------------------------------------------------------------------------------------------------
Question44
If0 <a,b<1andtan−1a+tan−1b=π
4,thenthevalueof
(a+b) − a2+b2
2+a3+b3
3−a4+b4
4+ ...is
[26Feb2021Shift2]
Options:
A.loge2
B.e2−1
C.e
D.loge
e
2
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question45
Allpossiblevaluesofθ ∈ [0,2π]forwhichsin 2 θ +tan 2 θ >0liein
[25Feb2021Shift1]
Options:
A. 0,π
2∪π,3π
2
( ) ( ) ( )
( )
( ) ( )
⇒sin2x+cos2x+2 sin x cos x =n
10
⇒1+2(1/10) = n/10 ⇒12/10 =n/10
∴n=12
Given,tan−1a+tan−1b=π
4
⇒tan−1a+b
1−ab =π
4
⇒a+b
1−ab =tan π
4=1
⇒a+b=1−ab
⇒ (1+a)(1+b) = 2...(i)
Now,(a+b) − a2+b2
2+a3+b3
3−a4+b4
4+ ...
=a−a2
2+a3
3−a4
4+ ... + b−b2
2+b3
3−b4
4+ ...
=loge(1+a) + loge(1+b) [ Usingexpansionofloge(1+x) ]
=loge(1+a)(1+b) [∴log a +log b =log a b]
=loge2[useEq.(i)]
[ ]
( ) ( ) ( )
[ ] [ ]
B. 0,π
2∪π
2,3π
4∪π,7π
6
C. 0,π
4∪π
2,3π
4∪3π
2,11π
6
D. 0,π
4∪π
2,3π
4∪π,5π
4∪
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question46
If0 <x,y<πandcos x +cos y −cos(x+y) = 3
2,thensin x +cos yisequal
to
[25Feb2021Shift2]
Options:
A. 1
2
B. √3
2
C. 1− √3
2
D. 1+ √3
2
Answer:D
Solution:
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
sin 2 θ +tan 2 θ >0
⇒sin 2 θ +sin 2 θ
cos 2 θ >0
⇒sin 2 θ 1 +1
cos 2 θ >0
⇒sin 2 θ cos 2 θ +1
cos 2 θ >0
⇒tan 2 θ(2cos2θ) > 0
cos 2 θ ≠0
⇒1−2sin2θ≠0
sin θ ≠ ± 1
√2
Now,tan 2 θ(1+cos 2 θ) > 0
⇒tan 2 θ >0as1+cos 2 θ >0
⇒2θ ∈0,π
2∪π,3π
2∪2π,5π
2∪3π,7π
2
∴θ∈0,π
4∪π
2,3π
4∪π,5π
4∪3π
2,7π
4
Since,sin θ ≠ ± 1
√2
( )
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
]
Solution:
-------------------------------------------------------------------------------------------------
Question47
Ifnisthenumberofsolutionsoftheequation
2 cos x 4 sin π
4+x sin π
4−x−1 =1,x∈ [0,π]andSisthesumof
allthesesolutions,thentheorderedpair(n,S)is
[1Sep2021Shift2]
Options:
A. 3,13π
9
B. 2,2π
3
C. 2,8π
9
D. 3,5π
3
Answer:A
Solution:
Solution:
( ( ) ( ) )
( )
( )
( )
( )
Given,cos x +cos y −cos(x+y) = 3
2
⇒2 cos x+y
2cos x−y
2−2cos2x+y
2−1=3
2[Useformula,
cos a +cos b =2 cos a+b
2cos a−b
2
cos 2 x =2cos2x−1]
⇒2 cos x+y
2cos x−y
2−2cos2x+y
2
=3
2−1=1
2
⇒4 cos x+y
2cos x−y
2−4cos2x+y
2
=1
2×2=1=cos2(x−y)
2+sin2(x−y)
2
⇒cos x−y
2−2 cos x+y
2
2+sin2x−y
2=0
⇒sin x−y
2=0andcos x−y
2−2 cos x+y
2=0
⇒x=yandcos 0 −2 cos x =0
Gives,cos x =1
2=cos y
∴sin x =1−cos2x=1−1
4=√3
2
∴sin x +cos y =√3
2+1
2=1+ √3
2
( ) ( ) [ ( ) ]
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
[ ( ) ( ) ] ( )
( ) ( ) ( )
√√
2 cos x 4 sin π
4+x sin π
4−x−1
⇒2 cos x 2 cos(2x) − 2 cos π
2−1 = 1
⇒2 cos x(4cos2x−3) = 1
( ( ) ( ) )
( ( ) )
-------------------------------------------------------------------------------------------------
Question48
Thenumberofsolutionsoftheequation32tan2x+32sec2x=81,0≤x≤π
4
is
[31Aug2021Shift2]
Options:
A.3
B.1
C.0
D.2
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question49
SectionB:NumericalTypeQuestions
LetSbethesumofallsolutions(inradians)oftheequation
sin4θ+cos4θ−sin θ cos θ =0in[0,4π].Then,85
πisequalto
[27Aug2021Shift2]
Answer:56
⇒cos 3 x =1
2
⇒3x =π
3,5π
3,7π
3
⇒x=π
9,5π
9,7π
9
Numberofsolutions=n=3
Sumofsolutions = S=13π
9
32tan2x+32sec2x=81
⇒32tan2x+321+tan2x=81
⇒33 ×32tan2x=81
⇒32tan2x=27
11
⇒tan2x=ln32
27
11
tan x =ln32
27
11 ∈ (0,1)
⇒Onesolutionin 0,π
4
( )
√( )
[ ]
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question50
Thesumofsolutionsoftheequation cos x
1+sin x
= | tan 2 x ,x∈ −π
2,π
2−π
4, −π
4is
[26Aug2021Shift1]
Options:
A.−11π
30
B. π
10
C.−7π
30
D.−π
15
Answer:A
Solution:
Solution:
| ( ) { }
sin4θ+cos4θ−sin θ cos θ =0in[0,4π]
⇒1−2sin2θcos2θ−sin θ cos θ =0
⇒2−sin22θ −sin 2 θ =0
⇒sin22θ +sin 2 θ −2=0
⇒(sin 2 θ +2)(sin 2 θ −1) = 0
⇒sin 2 θ =1,2θ ∈ [0,8π]
θ=π
4,5π
4,9π
4,13π
4
SumofsolutionsS=28π
4
Then,8S
π=8
π×28π
4=56
Wehave,cos x
1=tan 2 x
⇒
cos2x
2−sin2x
2
cos2x
2+sin2x
2+2 cos x
2sin x
2
=tan 2 x
⇒
cos x
2−sin x
2cos x
2+sin x
2
cos x
2+sin x
2
2=tan 2 x
⇒
1−tan x
2
1+tan x
2
=tan 2 x
⇒tan π
4−x
2=tan 2 x
⇒tan2π
4−x
2=tan22x
| |
( ) ( ) | |
[ ( ) ( ) ] [ ]
( ) | |
| |
( ) | |
( )
-------------------------------------------------------------------------------------------------
Question51
Thenumberofsolutionsofsin7x+cos7x=1,x ∈ [0,4π]isequalto
[22Jul2021Shift2]
Options:
A.11
B.7
C.5
D.9
Answer:C
Solution:
Solution:
⇒2x =nπ ±π
4−x
2
⇒2x =nπ +π
4−x
2
or2x =nπ −π
4−x
2
⇒5x
2=n+1
4π
or3x
2=n−1
4π
⇒−π
2<2
5n+1
4π<π
2
or−π
2<2
3n−1
4π<π
2
⇒− 5
4<n+1
4<5
4
or−3
4<n−1
4<3
4
⇒−6
4<n<1
or−1
2<n<1
⇒n= −1,0
orn=0
Whenn= −1,x=−3
10 π
orwhenn=0,x= −π
6
n=0,x=1
10 π
∴Requiredsum = −3
10 π+1
10 π+−1
6π=−11
30 π
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( ) ( ) ( )
sin7x≤sin2x≤λ1.......(1)
andcos7x≤cos2x≤1........(2)
alsosin2x+cos2x=1
⇒equalitymustholdfor(1)&(2)
⇒sin7x=sin2x&cos7=cos2x
⇒sin x =0&cos x =1
or
cos x =0& sin x =1
⇒x=0,2π,4π,π
2,5π
2
-------------------------------------------------------------------------------------------------
Question52
Letα =max
x∈R
{82 sin 3 x .44 cos 3 x}andbeta =min
x∈R
{82 sin 3 x .44 cos 3 x}.If
8x2+bx +c=0isaquadraticequationwhoserootsareα1∕5and
β1∕5,thenthevalueofc −bisequalto:
[27Jul2021Shift2]
Options:
A.42
B.47
C.43
D.50
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question53
Thenumberofsolutionsoftheequation|cot x| = cot x +1
sin xinthe
interval[0,2π]is
[18Mar2021Shift1]
Answer:1
Solution:
⇒5solutions
α=max{82 sin 3 x .44 cos 3 x}
=max{26 sin 3 x .28 cos 3 x}
=max{26 sin 3 x +8 cos 3 x}
andβ=min{82 sin 3 x .44 cos 3 x} = min{26 sin 3 x +8 cos 3 x}
Nowrangeof6 sin 3 x +8 cos 3 x
= − 62+82, + 62+82= [−10,10]
α=210&β =2−10
So,α1∕5=22=4
⇒β1∕5=2−2=1∕4
quadratic8x2+bx +c=0,c−b=
8× [(productofroots] + (sumofroots)
=8×4×1
4+4+1
4 = 8×21
4=42
[√ √ ]
[ ] [ ]
-------------------------------------------------------------------------------------------------
Question54
Thenumberofsolutionsoftheequationx +2 tan x =π
2intheinterval
[0,2π]is
[17Mar2021Shift2]
Options:
A.3
B.4
C.2
D.5
Answer:A
Solution:
Solution:
Given,|cot x| = cot x +1
sin x....(i)
Ifcot x >0,then|cot x| = cot x
FromEq.(i), cot x =cot x +1
sin x
⇒1
sin x =0 (notpossible)
Ifcot x <0,then|cot x| = −cot x
FromEq.(ii), − cot x =cot x +1
sin x
⇒2 cot x +1
sin x =0⇒2 cos x = −1
⇒x=2π
3or 4π
3
Here,x=4π
3rejectedbecause 4π
3∈thirdquadrantandinthirdquadrantcot xispositive.Since,weconsidered
cot x <0.∴x=2π/3istheonlyonesolution.
Given,x+2 tan x =π
2
⇒2 tan x =π
2−x⇒tan x =π
4−x
2
⇒tan x = − 1
2x+π
4...(i)
ApproachInthistypeofproblemsolving,graphicalapproachisbestbecausewehavetofindonlynumberofsolutions,
notthesolution(i.e.notthevalue(s)ofx).
ConceptTofindthenumberofsolution(s)forEq.(i),firstofall,lety=tan x...(ii)andy=−1
2x+π
4...(iii)andthen
drawthegraphofEqs.(ii)and(iii).
Now,totalnumberofsolution(s)=Totalnumberofpoint(s)ofintersectionofthegraph(ii)and(iii).
( )
( )
-------------------------------------------------------------------------------------------------
Question55
Thenumberofrootsoftheequation,(81)sin2x+ (81)cos2x=30inthe
interval[0,π]isequalto
[16Mar2021Shift1]
Options:
A.3
B.4
C.8
D.2
Answer:B
Solution:
Solution:
y= − 1
2x+π
4intersectsy=tan xatthreedistinctpointsin[0,2π].
∴Totalnumberofsolutions = 3
Given,81sin2x+81cos2x=30
⇒81sin2x+81(1−sin2x)=30
⇒81sin2x+81
81sin2x=30
Let81sin2x=y.
∴y+81
y=30
⇒y2−30y +81 =0
⇒ (y−27)(y−3) = 0
⇒y=3ory=27
81sin2x=3or81sin2x=27
34sin2x=3or34sin2x=33
⇒4sin2x=1or4sin2x=3
⇒sin2x=1/4orsin2x=3∕4
⇒sin2x=sin2(π/6)orsin2x=sin2(π/3)
⇒x=nπ ±π/6orx=nπ ±π/3
From[0,π],
x=π/6,5π/6orx=π/3,2π/3
Hence,thetotalnumberofsolutions = 4
-------------------------------------------------------------------------------------------------
Question56
If√3(cos2x) = (√3−1)cos x +1,thenumberofsolutionsofthegiven
equationwhenx ∈0,π
2is..........
[26Feb2021Shift1]
Answer:1
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question57
Thenumberofintegralvaluesofkforwhichtheequation
3 sin x +4 cos x =k+1hasasolution,k ∈Ris..........
[26Feb2021Shift1]
Answer:11
Solution:
Solution:
[ ]
Given,√3cos2x= (√3−1)cos x +1,x∈ [0,π∕2]
Letcos x =t,then
√3t2= (√3−1)t+1
⇒ √3t2− √3t +t−1=0
⇒ (√3t2− √3t) + (t−1) = 0
⇒ √3t(t−1) + 1(t−1) = 0
⇒ (t−1)(√3t +1) = 0
Thisgivest=1andt=−1
√3
Put,t=cos x,then
cos x =1andcos x =−1
√3
cos x = −1∕ √3isrejectedasx∈ [0,π∕2]
cos x = −1∕ √3 isrejectedas x ∈ [ 0,π
∴cos x =1
Since, x ∈0,π
2,then cos x =cos 0
Thisgivesx=0isonlysolution.
Therefore,numberofsolutionwhenx∈ [0,π∕2]is1.
[ ]
Given,3 sin x +4 cos x =k+1...(i)
MultiplyanddivideLHSofEq.(i)by 32+42=5
√
-------------------------------------------------------------------------------------------------
Question58
Thevalueofcos3π
8⋅cos 3π
8+sin3π
8⋅sin 3π
8is
[Jan.9,2020(I)]
Options:
A. 1
√2
B. 1
2√2
C. 1
2
D. 1
4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
( ) ( ) ( ) ( )
i.e.53
5sin x +4
5cos x =k+1
⇒5(cos α sin x +sin α cos x) = k+1
[Letcos α =3∕5thensin α =1− (3∕5)2=4
5
⇒5 sin(x+α) = k+1[Use sin(a+b) = sin a cos b +cos a sin b]
⇒sin(x+α) = k+1
5
Letx+α=θ
Then,sin θ =k+1
5
∵ − 1≤sin θ ≤1
⇒ − 1≤k+1
5≤1
⇒ − 5≤k+1≤5
⇒ − 6≤k≤4
∴Possibleintegralvaluesofkare−6, −5, −4, −3, −2, −1,0,1,2,3and4.i.e.
Total11integralvaluesofkarepossibleforwhichEq.(i)hassolution.
( )
√]
cos3π
84cos3π
8−3 cos π
8+sin3π
83 sin π
8−4sin3π
8
=4cos6π
8−4sin6π
8−3cos4π
8+3sin4π
8
=4 cos2π
8−sin2π
8
sin4π
8+cos4π
8+sin2π
8cos2π
8;
-3 cos2π
8−sin2π
8cos2π
8+sin2π
8;
=cos π
44 1 −sin2π
8cos2π
8−3
=1
√21−1
2=1
2√2.
[ ] [ ]
[ ( ) ]
[( ) ]
[ ( ) ( ) ]
[ ( ) ]
[ ] ]
Question59
IfL =sin2π
16 −sin2π
8andM =cos2π
16 −sin2π
8,then:
[Sep.05,2020(II)]
Options:
A.L = − 1
2√2+1
2cos π
8
B.L =1
4√2−1
4cos π
8
C.M =1
4√2+1
4cos π
8
D.M =1
2√2+1
2cos π
8
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question60
If √2sin α
√1+cos 2 α =1
7and 1−cos 2 β
2=1
√10 α,β∈0,π
2,thentan(α+2β)is
equalto_______
[Jan.8,2020(II)]
Answer:1
Solution:
Solution:
( ) ( ) ( ) ( )
√( )
L+M=1−2sin2π
8=cos π
4=1
√2........(i)
andL−M= −cos π
8.........(ii)
Fromequations(i)and(ii),L=1
2
1
√2−cos π
8=1
2√2−1
2cos π
8andM=1
2
1
√2+cos π
8=1
2√2+1
2cos π
8
( ) ( )
√2sin α
√2cos α =1
7and 1−cos2β
2=1
10
⇒√2sin β
√2=1
√10
∴tan α =1
7andsin β =1
√10
tan β =1
3
√
-------------------------------------------------------------------------------------------------
Question61
Thesetofallpossiblevaluesofθintheinterval(0,π)forwhichthe
points(1,2)and(sin θ,cos θ)lieonthesamesideofthelinex +y=1is:
[Sep.02,2020(II)]
Options:
A. 0,π
2
B. π
4,3π
4
C. 0,3π
4
D. 0,π
4
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question62
Iftheequationcos4θ+sin4θ+λ=0hasrealsolutionsforθ,thenλlies
intheinterval:
[Sep.02,2020(II)]
Options:
A. −5
4, −1
( )
( )
( )
( )
( )
∴tan 2 β =2 tan β
1−tan2β=
2⋅1
3
1−1
9
=
2
3
8
9
=3
4
tan(α+2β) = tan α +tan 2 β
1−tan α tan 2 β
=
1
7+3
4
1−1
7⋅3
4
=
4+21
28
25
28
=1
Letf(x,y) = x+y−1
Given(1,2)and(sin θ,cos θ)areliesonsameside.
∴f(1,2) ⋅ f(sin θ,cos θ) > 0
⇒2[sin θ +cos θ −1] > 0
⇒sin θ +cos θ∣ > 1⇒sin θ +π
4>1
√2
⇒θ+π
4∈π
4,3π
4⇒θ∈0,π
2
( )
( ) ( )
B. −1, − 1
2
C. −1
2, − 1
4
D. −3
2, − 5
4
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question63
Thenumberofdistinctsolutionsoftheequation,
log1∕2sin x | = 2−log1∕2|cos x∣intheinterval[0,2π],is_______
[Jan.9,2020(I)]
Answer:8
Solution:
Solution:
[ ]
( ]
[ ]
Givenequationiscos4(θ) + sin 4(θ) + λ=0
⇒λ= −[cos4(θ) + sin 4(θ)]
⇒λ= −[(cos2(θ) + sin 2(θ))2−2cos2(θ)sin 2(θ)]
[∵a2+b2= (a+b)2−2ab]
⇒λ= −(1)2+1
2(2 cos(θ)s i n (θ))2
[∵cos2(θ) + sin 2(θ) = 1]
⇒λ=1
2sin 2(2θ) − 1
Weknowthat −1≤sin (x) ≤ 1
Therefore 0 ≤sin 2(x) ≤ 1
So 1
2sin 2(2θ) ∈ 0,1
2
for 1
2sin 2(2θ) = 0⇒λ=0−1= −1
for 1
2sin 2(2θ) = 1
2⇒λ=1
2−1=−1
2
∴λ∈ −1, − 1
2
Hence,Option(B)i.e. −1,−1
2iscorrect
[ ]
[ ]
[ ]
log1∕2|sin x | = 2−log1∕2∣cos x
⇒log1∕2|sin x cos x | = 2
⇒ | sin x cos x =1
4
⇒sin 2 x = ± 1
2
|
-------------------------------------------------------------------------------------------------
Question64
Foranyθ ∈π
4,π
2theexpression
3(sin θ −cos θ)4+6(sin θ +cos θ)2+4sin6θequals:
[Jan.9,2019(I)]
Options:
A.13 −4cos2θ+6sin2θcos2θ
B.13 −4cos6θ
C.13 −4cos2θ+6cos4θ
D.13 −4cos4θ+2sin2θcos2θ
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question65
Theangleofelevationofthetopofaverticaltowerstandingona
horizontalplaneisobservedtobe45∘fromapointAontheplane.LetB
bethepoint30mverticallyabovethepointA.Iftheangleofelevationof
thetopofthetowerfromBbe30∘,thenthedistance(inm)ofthefootof
thetowerfromthepointAis:
[April12,2019(II)]
Options:
A.15(3+ √3)
( )
Hence,totalnumberofsolutions=8.
3(sin θ −cos θ)4+6(sin θ +cos θ)2+4sin6θ
=3(1−2 sin θ cos θ)2+6(1+2 sin θ cos θ) + 4sin6θ
=3(1+4sin2θcos2θ−4 sin θ cos θ) + 6
−12 sin θ cos θ +4sin6θ
=9+12sin2θcos2θ+4sin6θ
=9+12cos2θ(1−cos2θ) + 4(1−cos2θ)3
=9+12cos2θ−12cos4θ+4(1−cos6θ−3cos2θ+3cos4θ)
=9+4−4cos6θ
=13 −4cos6θ
B.15(5− √3)
C.15(3− √3)
D.15(1+ √3)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question66
Thevalueofcos210∘−cos 10∘cos 50∘+cos250∘is
[April9,2019(II)]
Options:
A. 3
4+cos 20∘
B..3/4
C. 3
2(1+cos 20∘)
D.3
2
Answer:B
Solution:
LettheheightofthetowerbehanddistanceofthefootofthetowerfromthepointAisd.Bythediagram,
tan 45∘=h
d=1
h=d......(i)
tan 30∘=h−30
d
√3(h−30) = d........(ii)
Putthevalueofh
from(i)to(ii),
√3d=d+30√3
d=30√3
√3−1=15√3(√3+1) = 15(3+ √3)
-------------------------------------------------------------------------------------------------
Question67
Twopolesstandingonahorizontalgroundareofheights5mand10m
respectively.Thelinejoiningtheirtopsmakesanangleof15∘withthe
ground.Thenthedistance(inm)betweenthepoles,is:
[April.09,2019(II)]
Options:
A.5(2+ √3)
B.5(√3+1)
C. 5
2(2+ √3)
D.10(√3−1)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question68
Thevalueofsin 10∘sin 30∘sin 50∘sin 70∘is:
[April.09,2019(II)]
cos210∘−cos 10∘cos 50∘+cos250∘
=1+cos 20∘
2+1+cos 100∘
2−1
2(2 cos 10∘cos 50∘)
=1+1
2(cos 20∘+cos 100∘) − 1
2[cos 60∘+cos 40∘]
=1−1
4+1
2[cos 20∘+cos 100∘−cos 40∘]
=3
4+1
2[2 cos 60∘×cos 40∘−cos 40∘]
=3
4
( ) ( )
( )
Bythediagram.
tan 15∘=5
d⇒d=5
tan 15∘=5(√3+1)
√3−1
=5(4+2√3)
2=5(2+ √3)
Options:
A. 1
16
B. 1
32
C. 1
18
D. 1
36
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question69
Ifcos(α+β) = 3
5,sin(α−β) = 5
13and0 <α,β<π
4,thentan(2α)isequal
to:
[April8,2019(I)]
Options:
A. 63
52
B. 63
16
C. 21
16
D. 33
52
Answer:B
Solution:
Solution:
∵sin(60∘+A) ⋅ sin(60∘−A)sin A =1
4sin 3 A
∴sin 10∘sin 50∘sin 70∘=sin 10∘sin(60∘−10∘)
sin(60∘+10∘) = 1
4sin 30∘
⇒sin 10∘sin 30∘sin 50∘sin 70∘=1
4sin230 =1
16
∵α+βandα−βbothareacuteangles.
cos(α+β) = 3
5,thensin(α+β) = 1−3
5
2=4
5
tan(α+β) = 4
3
Andsin(α−β) = 5
13,then
√( )
-------------------------------------------------------------------------------------------------
Question70
Ifsin4α+4cos4β+2=4√2sin α cos β;α,β∈ [0,π],then
cos(α+β) − cos(α−β)isequalto:
[Jan.12,2019(II)]
Options:
A.0
B.-1
C.√2
D.−√2
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question71
Letf k(x) = 1
k(sinkx+coskx)fork =1,2,3, ...Thenforallx ∈R,the
valueoff 4(x) − f6(x)isequalto:
[Jan.11,2019(I)]
cos(α−β) = 1−5
13
2=12
13
⇒tan(α−β) = 5
12
Now,tan 2 α =tan((α+β) + (α−β))
=tan(α+β) + tan(α−β)
1−tan(α+β) ⋅ tan(α−β)=
4
3+5
12
1−4
3⋅5
12
=63
16
√( )
∵Thegivenequationis
sin4α+4cos4β+2=4√2 sin α ⋅cos β,α,β∈ [0,π]
Then,byA.M.,G.M.ineqality;
A.M.≥G.M.
sin4α+4cos4β+1+1
4≥ (sin4α⋅4cos4β⋅1⋅1)
1
4
sin4α+4cos4β+1+1≥4√2 sin α⋅ | cos β|
Inequalitystillholdswhencos β <0butL.H.S.ispositivethancos β >0,then
L.H.S. = R.H.S
∴sin4α=1 and cos4β=1
4
⇒α=π
2and β =π
4
∴cos(α+β) − cos(α−β)
=cos π
2+β−cos π
2−β
= −sin β −sin β = −2 sin π
4= −√2
( ) ( )
Options:
A. 1
12
B. 1
4
C. −1
12
D. 5
12
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question72
Thevalueof
cos π
22⋅cos π
23⋅... ⋅ cos π
210 ⋅sin π
210
[Jan.10,2019(II)]
Options:
A. 1
512
B. 1
1024
C. 1
256
D. 1
2
Answer:A
Solution:
fk(x) = 1
k(sinkx+coskx)
f4(x) = 1
4[sin4x+cos4x]
=1
4(sin2x+cos2x)2−(sin 2 x)2
2
=1
41−(sin 2 x)2
2
f6(x) = 1
6[sin6x+cos6x]
=1
6sin2x+ (cos2x) − 3
4(sin2x)2
=1
61−3
4(sin 2 x)2
Nowf4(x) − f(6)(x) = 1
4−1
6−(sin 2 x)2
8+1
8(sin 2 x)2
=1
12
[ ]
[ ]
[ ( ]
[ ]
Solution:
-------------------------------------------------------------------------------------------------
Question73
Thenumberofsolutionsoftheequation
1+sin4x=cos23x,x∈ − 5π
2,5π
2is:
[April12,2019(I)]
Options:
A.3
B.5
C.7
D.4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question74
LetSbethesetofallα ∈Rsuchthattheequation,cos 2 x
+α sin x =2α −7hasasolution.ThenSisequalto
[April12,2019(II)]
Options:
A.R
B.[1,4]
C.[3,7]
[ ]
A=cos π
22⋅cos π
23...cos π
210 ⋅sin π
210
=1
2cos π
22⋅cos π
23...cos π
29sin π
29
=1
28cos π
22⋅sin π
22=1
29sin π
2
=1
512
( )
( )
Considerequation,1+sin4x=cos23x
L.H.S. = 1+sin4xandR⋅H.S. = cos23x
∵L.H.S. ≥ 1andR.H.S. ≤ 1⇒L.H.S. = R.H.S. = 1
sin4x=0,andcos23x =1
⇒sin x =0and(4cos2x−3)2cos2x=1
⇒sin x =0andcos2x=1⇒x=0, ±π, ±2π
Hence,totalnumberofsolutionsis5.
D.[2,6]
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question75
If[x]denotesthegreatestinteger≤x,thenthesystemoflinear
equations
[sin θ]x+ [−cos θ]y=0
[cot θ]x+y=0
[April12,2019(II)]
Options:
A.haveinfinitelymanysolutionsifθ ∈π
2,2π
3andhasauniquesolutionifθ ∈π,7π
6.
B.hasauniquesolutionifθ ∈π
2,2π
3∪π,7π
6.
C.hasauniquesolutionifθ ∈π
2,2π
3andhaveinfinitelymanysolutionsifθ ∈π,7π
6
D.haveinfinitelymanysolutionsifθ ∈π
2,2π
3∪π,7π
6
Answer:A
Solution:
Solution:
( ) ( )
( ) ( )
( ) ( )
( ) ( )
Givenequationis,cos 2 x +α sin x =2α −71−2sin2x+α sin x =2α −7
2sin2x−α sin x + (2α −8) = 0
⇒sin x =α±α2−8(2α +8)
4
⇒sin x =α± (α−8)
4⇒sin x =α−4
4
[sin x =2(rejected )]
∵equationhassolution,then α−4
4∈ [−1,1]
⇒α∈ [2,6]
√
Accordingtothequestion,therearetwocases.
Case1:θ∈π
2,2π
3
Inthisinterval,[sin θ] = 0, [−cos θ] = 0and[cot θ] = −1Thenthesystemofequationswillbe;
0⋅x+0⋅y=0and−x+y=0
Whichhaveinfinitelymanysolutions.
Case2:θ∈π,7π
6
Inthisinterval,[sin θ] = −1and[−cos θ] = 0
Thenthesystemofequationswillbe;
−x+0⋅y=0and[cot θ]x+y=0
Clearly,x=0andy=0whichhasuniquesolution.
( )
( )
-------------------------------------------------------------------------------------------------
Question76
LetS = {θ∈ [−2π,2π] : 2cos2θ+3 sin θ =0}.
ThenthesumoftheelementsofSis:
[April9,2019(I)]
Options:
A. 13π
6
B. 5π
3
C.2π
D.π
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question77
If0 ≤x<π
2,thenthenumberofvaluesofxforwhich
sin x −sin 2 x +sin 3 x =0,is:
[Jan.09,2019(II)]
Options:
A.3
B.1
C.4
D.2
2cos2θ+3 sin θ =0
(2 sin θ +1)(sin θ −2) = 0
⇒sin θ = − 1
2orsin θ =2→Notpossibe
Therequiredsumofallsolutionsin[−2π,2π]is
=π+π
6+2π −π
6+ − π
6+ −π+π
6=2π
( ) ( ) ( ) ( )
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question78
Thenumberofsolutionsofsin 3 x =cos 2 x,intheinterval π
2,π is
[OnlineApril15,2018]
Options:
A.3
B.4
C.2
D.1
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question79
Ifsumofallthesolutionsoftheequation
8 cos x ⋅cos π
6+x⋅cos π
6−x−1
2−1in[0,π]iskπthenkisequal
to:
[2018]
( )
( ( ) ( ) )
sin x −sin 2 x +sin 3 x =0
⇒sin x −2 sin x ⋅cos x +3 sin x −4sin3x=0
⇒4 sin x −4sin3x−2 sin x ⋅cos x =0
⇒2 sin x(1−sin2x) − sin x ⋅cos x =0
⇒2 sin x ⋅cos2x−sin x ⋅cos x =0
⇒sin x ⋅cos x(2 cos x −1) = 0
∴sin x =0,cos x =0,cos x =1
2
∴x=0,π
3∵x∈0,π
2
[ )
sin 3 x =cos 2 x
⇒3 sin x −4sin3x=1−2sin2x
⇒4sin3x−2sin2x−3 sin x +1=0
⇒sin x =1,−2±2√5
8
Intheinterval π
2,π,sin x =−2+2√5
8
So,thereisonlyonesolution.
( )
Options:
A. 13
9
B. 8
9
C. 20
9
D. 2
3
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question80
If5(tan2x−cos2x) = 2 cos 2 x +9,thenthevalueofcos 4 xis
[2017]
Options:
A.−7
9
B.−3
5
C. 1
3
D. 2
9
∵8 cos x cos2π
6−sin2x−1
2=1
⇒8 cos x 3
4−1
2−sin2x=1
⇒8 cos x 1
4− (1−cos2x) = 1
⇒8 cos x 1
4−1+cos2x=1
⇒8 cos x cos2x−3
4=1
⇒84cos3x−3 cos x
4=1
⇒2(4cos3x−3 cos x) = 1
⇒2 cos 3 x =1⇒cos 3 x =1
2
∴3x =2nπ ±π
3,n∈1
⇒x=2nπ
3±π
9
Inx∈ [0,π] : x=π
9,2π
3+π
9,2π
3−π
9,only
Sumofallthesolutionsoftheequation
=1
9+2
3+1
9+2
3−1
9π=13
9π
( )
( )
( )
( )
( )
( )
( )
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question81
IfmandM aretheminimumandthemaximumvaluesof
4+1
2sin22x −2cos4x,x∈R,thenM −misequalto:
[OnlineApril9,2016]
Options:
A. 9
4
B. 15
4
C. 7
4
D. 1
4
Answer:B
Solution:
Solution:
Wehave
5tan2x−5cos2x=2(2cos2x−1) + 9
⇒5tan2x−5cos2x=4cos2x−2+9
⇒5tan2x=9cos2x+7
⇒5(sec2x−1) = 9cos2x+7
Letcos2x=t
⇒5
t−9t −12 =0
⇒9t2+12t −5=0
⇒9t2+15t −3t −5=0
⇒(3t −1)(3t +5) = 0
⇒t=1
3ast≠ − 5
3
cos 2 x =2cos2x−1=21
3−1= − 1
3
cos 4 x =2cos22x −1=2−1
3
2−1= − 7
9
( )
( )
4+1
2sin22x −2cos4x
4+2(1−cos2x)cos2x−2cos4x
−4 cos4x−cos2x
2−1+1
16 −1
16
−4 cos2x−1
4
2−17
16
0≤cos2x≤1
−1
4≤cos2x−1
4≤3
4
0≤cos2x−1
4
2≤9
16
{ }
{ ( ) }
( )
-------------------------------------------------------------------------------------------------
Question82
If0 ≤x<2π,thenthenumberofrealvaluesofx,whichsatisfythe
equationcos x +cos 2 x +cos 3 x +cos 4 x =0is:
[2016]
Options:
A.7
B.9
C.3
D.5
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question83
Thenumberofx ∈ [0,2π]forwhich
2sin4x+18cos2x−2cos4x+18sin2x=1is
[OnlineApril9,2016]
Options:
A.2
B.6
C.4
D.8
|√ √ |
−17
16 ≤cos2x−1
4
2−17
16 ≤9
16 −17
16
17
4≥ −4 cos2x−1
4
2−17
16 ≥1
2
M=17
4
m=1
2
M−m=17
4−2
4=15
4
( )
{ ( ) }
cos x +cos 2 x +cos 3 x +cos 4 x =0
⇒2 cos 2 x cos x +2 cos 3 x cos x =0
⇒2 cos x 2 cos 5x
2cos x
2=0
cos x =0,cos 5x
2=0,cos x
2=0
x=π,π
2,3π
2,π
5,3π
5,7π
5,9π
5
( )
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question84
Ifcos α +cos β =3
2andsin α +sin β 1
2andθisthethearithmeticmeanof
αandβ,thensin 2 θ +cos 2 θisequalto:
[OnlineApril11,2015]
Options:
A. 3
5
B. 7
5
C. 4
5
D. 8
5
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question85
2sin4x+18cos2x−2cos4x+18sin2x=1
2sin4x+18cos2x−2cos4x+18sin2x= ±1
2sin4x+18cos2x= ±1+2cos4x+18sin2x
bysquaringboththesideswewillget8solutions
|√ √ |
√ √
√ √
Letcos α +cos β =3
2
⇒2 cos α+β
2cos α−β
2=3
2........(i)
andsin α +sin β =1
2
⇒2 sin α+β
2cos α−β
2=1
2.........(ii)
Ondividing(ii)by(i),
wegettan α+β
2=1
3
Given:θ=α+β
2⇒2θ =α+β
Considersin 2 θ +cos 2 θ =sin(α+β) + cos(α+β) =
2
3
1+1
9
+
1−1
9
1+1
9
=6
10 +8
10 =7
5
( )
Letf k(x) = 1
k(sinkx+coskx)wherex ∈Randk ≥1Thenf 4(x) − f6(x)
equals
[2014]
Options:
A. 1
4
B. 1
12
C. 1
6
D. 1
3
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question86
If2 cos θ +sin θ =1 θ ≠π
2then7 cos θ +6 sin θisequalto:
[OnlineApril11,2014]
Options:
A. 1
2
B.2
C. 11
2
D. 46
5
Answer:D
Solution:
Solution:
( )
(b)Letfk(x) = 1
k(sinkx+coskx)Considerf4(x) − f6(x) = 1
4(sin4x+cos4x)
−1
6(sin6x+cos6x)
=1
4[1−2sin2xcos2x] − 1
6[1−3sin2xcos2x]
=1
4−1
6=1
12
Given2 cos θ +sin θ =1
Squaringbothsides,
weget
-------------------------------------------------------------------------------------------------
Question87
Ifcosecθ =p+q
p−q(p≠q≠0),then cot π
4+θ
2isequalto:
[OnlineApril9,2014]
Options:
A. p
q
B. q
p
C.√pq
D.pq
Answer:B
Solution:
Solution:
| ( ) |
√
√
(2 cos θ +sin θ)2=12
⇒4cos2θ+sin2θ+4 sin θ cos θ =1
⇒3cos2θ+ (cos2θ+sin2θ) + 4 sin θ cos θ =1
⇒3cos2θ+not x +4 sin θ cos θ =not
⇒3cos2θ+4 sin θ cos θ =0
⇒cos θ(3 cos θ +4 sin θ) = 0
⇒3 cos θ +4 sin θ =0⇒3 cos θ = −4 sin θ
⇒−3
4=tan θ =sec2θ−1=−3
4
∵tan θ =sec2θ−1
⇒sec2θ−1=−3
4
2=9
16
⇒sec2θ=9
16 +1=25
16 ⇒sec θ =5
4
orcos θ =4
5......(1)
Now,sin2θ+cos2θ=1⇒sin2θ+4
5
2=1
sin2θ+4
5=1⇒sin2θ=1−16
25 =9
25
sin θ = ± 3
5.. . (2)
Taking sin θ = + 3
5because sin θ = − 3
5cannot
satisfythegivenequation.
Therefore;7 cos θ +6 sin θ
=7×4
5+6×3
5=28
5+18
5=46
5
√
(√)
( )
( )
( ) ( )
cosecθ =p+q
p−q,sin θ =p−q
p+q
cos θ = ± 1−sin2θ=1−p−q
p+q
2=2√pq
(p+q)
cot π
4+θ
2=
cot π
4cot θ
2−1
cot π
4+cot θ
2
=
cot θ
2−1
cot θ
2+1
√√( )
| ( ) | | | | |
-------------------------------------------------------------------------------------------------
Question88
Thenumberofvaluesofαin[0,2π]forwhich
2sin3α−7sin2α+7 sin α =2,is:
[OnlineApril9,2014]
Options:
A.6
B.4
C.3
D.1
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question89
Theexpression tan A
1−cot A +cot A
1−tan Acanbewrittenas:
[2013]
cos θ
2−sin θ
2
cos θ
2+sin θ
2
Onrationalizingdenominator,weget
cos θ
2−sin θ
2
cos θ
2+sin θ
2
cos θ
2+sin θ
2
cos θ
2+sin θ
2
=cos θ
sin2θ
2+cos2θ
2+2 sin θ
2cos θ
2
=cos θ
1+sin θ =2√pq ∕ (p+q)
1+(p−q)
p+q
=√pq
p=q
p
| |
| ( ) ( ) |
| |
| | ||√
2sin3α−7sin2α+7 sin α −2=0
⇒2sin2α(sin α −1) − 5 sin α(sin α −1)
+2(sin α −1) = 0
⇒(sin α −1)(2sin2α−5 sin α +2) = 0
⇒sin α −1=0or2sin2α−5 sin α +2=0
sin α =1orsin α =5± √25 −16
4=5±3
4
α=π
2orsin α =1
2,2
Now,sin α ≠2
for,sin α =1
2
α=π
3,2π
3
Therearethreevaluesofαbetween[0,2π]
Options:
A.sin A cos A +1
B.sec A c o s e c A +1
C.tan A +cot A
D.sec A +cosecA
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question90
LetA = {θ:sin(θ) = tan(θ)}andB = {θ:cos(θ) = 1}betwosets.Then:
[OnlineApril25,2013]
Options:
A.A =B
B.A ⊆B
C.B ⊆A
D.A ⊂BandB −A≠φ
Answer:B
Solution:
Solution:
Givenexpressioncanbewrittenas
sin A
cos A×sin A
sin A −cos A +cos A
sin A ×cos A
cos A −sin A
∵tan A =sin A
cos A and
cot A =cos A
sin A
=1
sin A −cos A
sin3A−cos3A
cos A sin A
∵a3−b3= (a−b)(a2+ab +b2)
=sin2A+sin A cos A +cos2A
sin A cos A
=1+secAcosecA
( )
{ }
LetA= {θ:sin θ =tan θ}
andB= {θ:cos θ =1}
Now,A=θ:sin θ =sin θ
cos θ
= {θ:sin θ(cos θ −1) = 0}
= {θ=0,π,2π,3π, .....}
ForB:cos θ =1⇒θ=π,2π,4π, ......
ThisshowsthatAisnotcontainedinB.i.e.A not ⊂B.butB⊂A
{ }
-------------------------------------------------------------------------------------------------
Question91
Thenumberofsolutionsoftheequationsin 2 x −2 cos x +4 sin x =4in
theinterval[0,5π]is:
[OnlineApril23,2013]
Options:
A.3
B.5
C.4
D.6
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question92
Statement-1:Thenumberofcommonsolutionsofthetrigonometric
equations2sin2θ−cos 2 θ =0and2cos2θ−3 sin θ =0intheinterval
[0,2π]istwo.
Statement-2:Thenumberofsolutionsoftheequation,2
cos2θ−3 sin θ =0intheinterval[0,π]istwo.
[OnlineApril22,2013]
Options:
A.Statement-1istrue;Statement-2istrue;Statement-2isacorrectexplanationforstatement-
1.
B.Statement-1istrue;Statement-2istrue;Statement-2isnotacorrectexplanationfor
statement-1.
C.Statement-1isfalse;Statement-2istrue.
D.Statement-1istrue;Statement-2isfalse.
Answer:B
Solution:
sin 2 x −2 cos x +4 sin x =4
⇒2 sin x ⋅cos x −2 cos x +4 sin x −4=0
⇒(sin x −1)(cos x −2) = 0
∵cos x −2≠0, ∴ sin x =1
∴x=π
2,5π
2,9π
2
Solution:
-------------------------------------------------------------------------------------------------
Question93
Thevalueofcos 255∘+sin 195∘is
[OnlineMay26,2012]
Options:
A. √3−1
2√2
B. √3−1
√2
C.−√3−1
√2
D. √3+1
√2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
( )
2sin2θ−cos 2 θ =0
⇒2sin2θ− (1−2sin2θ) = 0
⇒2sin2θ−1+2sin2θ=0
⇒4sin2θ=1⇒sin θ = ± 1
2
∴θ=π
4,3π
4,5π
4,7π
4,θ∈ [0,2π]
∴θ=π
6,5π
6,7π
6,11π
6
Now2cos2θ−3 sin θ =0
⇒2(1−sin2θ) − 3 sin θ =0
⇒−2sin2θ−3 sin θ +2=0
⇒ − 2sin2θ−4 sin θ +sin θ +2=0
⇒2sin2θ−sin θ +4 sin θ −2=0
⇒sin θ(2 sin θ −1) + 2(2 sin θ −1) = 0
Now2cos2θ−3 sin θ =0
⇒2(1−sin2θ) − 3 sin θ =0
⇒−2sin2θ−3 sin θ +2=0
⇒−2sin2θ−4 sin θ +sin θ +2=0
⇒2sin2θ−sin θ +4 sin θ −2=0
⇒sin θ(2 sin θ −1) + 2(2 sin θ −1) = 0
⇒sin θ =1
2, −2
Butsin θ = −2,isnotpossible∴ sin θ =1
2, −2⇒θ=π
6,5π
6
Hence,therearetwocommonsolution,thereeachofthestatement-1and2aretruebutstatement-2isnotacorrect
explanationforstatement-1.
Considercos 255∘+sin 195∘
=cos(270∘−15∘) + sin(180∘+15∘)
= −sin 15∘−sin 15∘
= −2 sin 15∘= −2√3−1
2√2= − √3−1
√2
( ) ( )
Question94
Letf (x) = sin x,g(x) = x
Statement1:f (x) ≤ g(x)forxin(0,∞)
Statement2 :f(x) ≤ 1forxin(0,∞)butg(x) → ∞asx →∞.
[OnlineMay7,2012]
Options:
A.Statement1istrue,Statement2isfalse.
B.Statement1istrue,Statement2istrue,Statement2isacorrectexplanationforStatement1
.
C.Statement1istrue,Statement2istrue,Statement2isnotacorrectexplanationfor
Statement1.
D.Statement1isfalse,Statement2istrue.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question95
Theequationesin x −e−sin x −4=0has:
[2012]
Options:
A.infinitenumberofrealroots
B.norealroots
C.exactlyonerealroot
D.exactlyfourrealroots
Answer:B
Solution:
Solution:
Letf (x) = sin xandg(x) = x
Statement-1:f(x) ≤ g(x) ∀x∈ (0,∞)
i.e.,sin x ≤x∀x∈ (0,∞)
whichistrue
Statement-2:f(x) ≤ 1∀x∈ (0,∞)
i.e.,sin x ≤1∀x∈ (0,∞)
Itistrueand
g(x) = x→∞asx→∞alsotrue.
Givenequationisesin x −e−sin x −4=0
Putesin x =tinthegivenequation,weget
t2−4t −1=0
-------------------------------------------------------------------------------------------------
Question96
IfA =sin2x+cos4x,thenforallrealx:
[2011]
Options:
A. 13
16 ≤A≤1
B.1 ≤A≤2
C. 3
4≤A≤13
16
D. 3
4≤A≤1
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question97
Thepossiblevaluesofθ ∈ (0,π)suchthatsin(θ) + sin(4θ) + sin(7θ) = 0
are
[2011RS]
Options:
⇒t=4± √16 +4
2=4± √20
2
=4±2√5
2=2± √5
⇒esin x =2± √5(∵t=esin x)
⇒esin x =2− √5and esin x =2+ √5
⇒esin x =2− √5<0
and sin x =ln(2+ √5) > 1
So,rejected.
Hence,givenequationhasnosolution.
∴ Theequationhasnorealroots.
A=sin2x+cos4x
=sin2x+cos2x(1−sin2x)
=sin2x+cos2x−1
4(2 sin x ⋅cos x)2
=1−1
4sin2(2x)
∵−1≤sin 2 x ≤1
⇒0≤sin2(2x) ≤ 1
⇒0≥ − 1
4sin2(2x) ≥ − 1
4
⇒1≥1−1
4sin2(2x) ≥ 1−1
4
⇒1≥A≥3
4
A. π
4,5π
12,π
2,2π
3,3π
4,8π
9
B. 2π
9,π
4,π
2,2π
3,3π
4,35π
36
C. 2π
9,π
4,π
2,2π
3,3π
4,8π
9
D. 2π
9,π
4,4π
9,π
2,3π
4,8π
9
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question98
Letcos(α+β) = 4
5andsin(α−β) = 5
13where0 ≤α,β≤π
4.Thentan 2 α =
[2010]
Options:
A. 56
33
B. 19
12
C. 20
7
D. 25
16
Answer:A
Solution:
Solution:
sin 4 θ +2 sin 4 θ cos 3 θ =0
sin 4 θ(1+2 cos 3 θ) = 0
sin 4 θ =0
or cos 3 θ = − 1
2
4θ =nπ;n∈I
or3θ =2nπ ±2π
3,n∈I
θ=π
4,π
2,3π
4or θ=2π
9,8π
9,4π
9[∵θ, ∈(0,π)]
cos(α+β) = 4
5⇒tan(α+β) = 3
4
sin(α−β) = 5
13 ⇒tan(α−β) = 5
12
tan 2 α =tan[ (α+β) + (α−β)
=tan(α+β) + tan(α−β)
1−tan(α+β)tan(α−β)=
3
4+5
12
1−3
4⋅5
12
=56
33
-------------------------------------------------------------------------------------------------
Question99
LetAandBdenotethestatements
A:cos α +cos β +cos γ =0
B:sin α +sin β +sin γ =0
I f cos(β−γ) + cos(γ−α) + cos(α−β) = − 3
2,then:
[2009]
Options:
A.AisfalseandBistrue
B.bothAandBaretrue
C.bothAandBarefalse
D.AistrueandBisfalse
Answer:B
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question100
Ifpandqarepositiverealnumberssuchthatp2+q2=1thenthe
maximumvalueof(p+q)is
[2007]
Options:
A. 1
2
B. 1
√2
C.√2
D.2.
Giventhat
cos(β−γ) + cos(γ−α) + cos(α−β) = − 3
2
⇒2[cos(β−γ) + cos(γ−α) + cos(α−β)] + 3=0
⇒2[cos(β−γ) + cos(γ−α) + cos(α−β)]
+sin2α+cos2α+sin2β+cos2β
+sin2γ+cos2α=0
⇒ [ sin2α+sin2β+sin2γ+2 sin α sin β
+2 sin β sin γ +2 sin γ sin α ] + [ cos2α+cos2β
+cos2γ+2 cos α cos β +2 cos β cos γ
+2 cos γ cos α ] = 0
[∵cos(A−B) = cos A ⋅cos B +sin A ⋅sin B]
⇒ [sin α +sin β +sin γ]2+ (cos α +cos β +cos γ)2=0
⇒sin α +sin β +sin γ =0andcos α +cos β +cos γ =0
∴AandBbotharetrue.
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question101
Atriangularparkisenclosedontwosidesbyafenceandonthethird
sidebyastraightriverbank.Thetwosideshavingfenceareofsame
lengthx.Themaximumareaenclosedbytheparkis
[2006]
Options:
A. 3
2x2
B. x3
8
C. 1
2x2
D.πx2
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question102
If0 <x<πandcos x +sin x =1
2,thentan xis
[2006]
√
Giventhatp2+q2=1
∴p=cos θandq=sin θsatisfythegivenequation
Thenp+q=cos θ +sin θ
Weknowthat
−a2+b2≤a cos θ +b sin θ ≤a2+b2
∴−√2≤cos θ +sin θ ≤ √2
Hencemax.valueofp+qis√2
√ √
Area = 1
2x2sin θ
Maximumvalueofsin θis1atθ=π
2
Amax =1
2x2
Options:
A. (1− √7)
4
B. (4− √7)
3
C.−(4+ √7)
3
D. (1+ √7)
4
Answer:C
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question103
Thenumberofvaluesofxintheinterval[0,3π]satisfyingtheequation
2sin2x+5 sin x −3=0is
[2006]
Options:
A.4
B.6
C.1
D.2
Answer:A
Solution:
Solution:
cos x +sin x =1
2⇒1+sin 2 x =1
4
⇒sin 2 x = − 3
4
∴π<2x <2π
⇒π
2<x≤π........(i)
2 tan x
1+tan2x= − 3
4
⇒3tan2x+8 tan x +3=0
∴tan x =−8± √64 −36
6= − −4± √7
3
for π
2<x<π,tanx<0
∴tan x =−4− √7
3
-------------------------------------------------------------------------------------------------
Question104
Ifu =a2cos2θ+b2sin2θ+a2sin2θ+b2cos2θ
thenthedifferencebetweenthemaximumandminimumvaluesofu2is
givenby
[2004]
Options:
A.(a−b)2
B.2 a2+b2
C.(a+b)2
D.2(a2+b2)
Answer:A
Solution:
Solution:
-------------------------------------------------------------------------------------------------
√ √
√
2sin2x+5 sin x −3=0
⇒(sin x +3)(2 sin x −1) = 0
⇒sin x =1
2andsin x ≠ −3
∴In[0,3π], xhas4values.
u2=a2+b2+2a4+b4cos2θsin2θ; +a2b2(cos4θ+sin4θ) ........(1)
Now,(a4+b4)cos2θsin2θ+a2b2(cos4θ+sin4θ)
= (a4+b4)cos2θsin2θ+a2b2(1−2cos2θsin2θ)
= (a4+b4−2a2b2)cos2θsin2θ+a2b2
= (a2−b2)2⋅sin22θ
4+a2b2.......(2)
∵0≤sin22θ ≤1
⇒0≤ (a2−b2)2sin22θ
4≤(a2−b2)2
4
⇒a2b2≤ (a2−b2)2sin22θ
4+a2b2
≤(a2−b2)2⋅1
4+a2b2...........(3)
From(1)
a2+b2+2a2b2≤u2≤a2+b2+2
2(a2+b2)2
(a+b)2≤u2≤2(a2+b2)
∴Max.value−Min.value
=2(a2+b2) − (a+b2) = (a−b)2
√( ) )
√ √
Question105
Letα,βbesuchthatπ <α−β<3π.
Ifsin α +sin β = − 21
65andcos α +cos β = − 27
65 ,thenthevalueofcos α−β
2
[2004]
Options:
A. −6
65
B. 3
√130
C. 6
65
D.−3
√130
Answer:D
Solution:
Solution:
-------------------------------------------------------------------------------------------------
Question106
Thefunctionf (x) = log x +x2+1,is
[2003]
Options:
A.neitheranevennoranoddfunction
B.anevenfunction
C.anoddfunction
D.aperiodicfunction.
Answer:C
(√)
π<α−β<3π
⇒π
2<α−β
2<3π
2⇒cos α−β
2<0......(1)
sin α +sin β = − 21
65
⇒2 sin α+β
2cos α−β
2= − 21
65.......(2)
cos α +cos β = − 27
65
⇒2 cos α+β
2cos α−β
2= − 27
65......(3)
Squaringandadding(2)and(3),weget
4cos2α−β
2=(21)2+ (27)2
(65)2=1170
65 ×65
∴cos2α−β
2=9
130 ⇒cos α−β
2= − 3
√130 [from(1)]
Solution:
-------------------------------------------------------------------------------------------------
Question107
Theperiodofsin2θis
[2002]
Options:
A.π2
B.π
C.2π
D.π ∕2
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
Question108
Whichoneisnotperiodic?
[2002]
Options:
A.|sin 3 x| + sin2x
B.cos √x+cos2x
C.cos 4 x +tan2x
D.cos 2 x +sin x
Answer:B
Solution:
Givenf(x) = log x +x2+1
f(−x) = log −x+x2+1=log x2−x2+1
x+x2+1
= −log x +x2+1= −f(x)
⇒f(x)isanoddfunction.
(√)
{√}{√}
(√)
Weknowthatsin2θ=1−cos 2 θ
2;
Sinceperiodofcos 2 θ =2π
2=π
Henceperiodofsin2θisalsoπ.
-------------------------------------------------------------------------------------------------
Question109
Thenumberofsolutionoftan x +sec x =2 cos xin[0,2π)is
[2002]
Options:
A.2
B.3
C.0
D.1
Answer:B
Solution:
-------------------------------------------------------------------------------------------------
weknowthatcos √xisnonperiodic
∴cos √x+cos2xcannotbeperiodic.
∵tan x +sec x =2 cos x
⇒sin x +1=2cos2x
⇒sin x +1=2(1−sin2x)
⇒2sin2x+sin x −1=0
⇒ (2 sin x −1)(sin x +1) = 0
⇒sin x =1
2, −1.
⇒x=30∘,150∘,270∘
Numberofsolution = 3