y PU I Year Trigonometry
Vikasana - CET 2012
Remember:
1. Angle between Minute hand
and Hour hand in X Hr. andY min.
11
is 30X- Y
2
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2. The maximum value of
aCosθ
C θ+bSinθ
S θ+c, is c + a2 +b 2 and
th minimum
the
i i
value
l
iis c − a2 +b 2
Vikasana - CET 2012
3. If Cos
3
C 2A + C
Cos2B = 1 = Si
Sin2A + Si
Sin2B
then A +B = π.
2
4. If aSinx + bCosx = c, then
2
2
2
bSinx -aCosx = ± a +b −c
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1. The angle
g between Hr.hand and
Min. hand of a clock when the time
i 3 : 20
is
1) 10° 2) 20°
3)30°
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1°
4)22
2
11
Solution: Req. = 30(3)- (20)
2
= 90 - 110 = 20°
Ans : (2)
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2. The vertical angle of an isosceles
triangle is 45° then the base angle
in circular
circ lar meas
measure
re is
(1) 67° 30′ (2) 65° 30′ (3)3π (4) 3π
8
16
Vikasana - CET 2012
S o lu tio n: A + B + C = 1 8 0 °
A + B = 1 8 0 -C = 1 8 0 -4 5 = 1 3 5
B = A ∴ 2 A = 1 3 5 °= 3 π
4
∴ A = 3π
8
A n s : (3 )
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3.If the length of a chord of a circle
is equal to that of the radius of the
g subtended in
circle,, then the angle
radians at the centre of the circle by
chord is
1)) 1
π
2))
2
π
3))
3
π
4))
4
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Solution:
∆OAB is an equilateral triangle
∴∠AOB = π
3
A
∴ Ans:(3)
r
o
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r
r
B
1
5
4 If SinA +
4.
= and A is
SinA 2
acute then A is
π
π
1)
2)
6
4
π
3)
4) None of these
3
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1
1
Solution SinA +
= 2+
SinA
2
1
π
∴SinA =
⇒A=
2
6
∴Ans: (1)
:
Vikasana - CET 2012
5.If
5
If S
Secθ + ttanθ = 2
2, th
then th
the
values of Secθ & tanθ
are respectively
1 2
5 3
2 1
1) ,
2) ,
3) ,
4)None
4 3
4 4
3 4
Vikasana - CET 2012
1
Solution : Secθ - tanθ =
2
Q (Secθ + tanθ )(Secθ − tanθ ) = 1
adding and Simplifying
5
3
Secθ =
and tanθ =
4
4
Ans : (2)
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6. The value of
Cos 85 + Cos 5,, is
2
1)0
2
2) - 1
3)1
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1
4)
2
2
2
Solution : Cos A + Cos B = 1
when A + B = 90
∴Ans.(3)1
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7 The maximum value of
7.
4Sinθ + 3Cosθ + 2, is
1)7
)
2)4
)
3)6
)
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4)5
)
Solution : Max.value = c + a + b
2
= 2 + 16 + 9 = 7
∴Ans.(1)7
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2
8. Sinθ + Cosθ = 1,
8
1
then Sin2θ =
1)1
2) - 1
3)0
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4)2
Solution : Sq. both sides,
Sin θ + Cos θ + 2Sinθ Cosθ = 1
⇒ Sin2θ = 0
2
2
Ans.(3)0
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9 If Cosθ + Secθ = 2,
9.
2 then the
value of Cos θ − Sec θ =
100
100
1)0 2)1 3)2 4) - 1
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1
Solution: Cosθ +
= 1+ 1
Cosθ
⇒ Cosθ = 1 = Secθ
∴ G.E. = 1-1 = 0
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10. If Secθ + tanθ = 4 then
Cosθ =
8
1)
15
15
2)
17
8
3)
17
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7
4)
17
1
S l ti : Sec
Solution
S θ − ttanθ =
4
1
adding, we get 2Secθ = 4 +
4
17
⇒ Secθ =
8
8
∴ Cosθ =
17
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11 The value of tan20° + tan40°
11.
+ tan60° + ... + tan180°, is
1)0
2)1
3)2
4)4
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Solution : If A +B = 180° then
tanA = -tanB or tanA + tanB = 0
∴tan20
t 20 + tan160
t 160 + tan40
t 40 +
tan140 +...+ tan180
= 0 + 0 +... = 0
∴ Ans : (1)0
Vikasana - CET 2012
12 The value of
12.The
Sin α
tan α
2
+
+ Cos
C
α
2
2
1 + Cot α (1 + tanα )
2
2
1) - 1 2)0
3)1
4)2
Vikasana - CET 2012
Solution : Put α = 45°
1
1
1
2
G.E. =
+
+
2
1+1 (1+1)
2
=1
Ans : (3) 1
Vikasana - CET 2012
13. A,B,C are the angles of a
∆ABC, then
⎛ 3A + 2B + C ⎞
⎛ A -C⎞
Cos ⎜
⎟ + Cos ⎜
⎟=
2
⎝
⎠
⎝ 2 ⎠
1)1
2)0 3)CosA 4)CosC
Vikasana - CET 2012
S l ti : Put A = B = C
Solution
G.E. = Cos180 + Cos0 = -1+1= 0
Ans : (2) 0
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14. If x =Cos1° and y =Cos1
then
1) x = y
3) x > y
2) x < y
4) 2x = y
Vikasana - CET 2012
S l
Solution
: Cos
C θ iis d
decreasing
i
f 0 <θ <
for
π
2
∴ Cos1° > Cos1
Ans : ((3)x
) >y
Vikasana - CET 2012
15. In a ∆ABC,
15
ABC C = 90°,then
then
the equation whose roots
are tanA & tanB is
2
2
2
2
2
2
1)abx + c x + ab = 0 2)abx + c x - ab = 0
2
2
3)abx - c x - ab = 0 3)abx - c x + ab = 0
Vikasana - CET 2012
Solution : C = 90 ⇒ A +B = 90
(a
2
+b
2
2
=c )
∴ tanAtanB
t At B = 1 ⇒ αβ
β =1
a b
α + β = tanA
t A + ttanB
B= +
b a
a +b
2
ab
2
⎛ -c ⎞
= -⎜
∴ Ans : (4)
⎟
⎝ ab- CET
⎠ 2012
Vikasana
2
16. If 5Sinx + 4Cosx = 3,, then
4Sinx - 5Cosx =
1)4 2)4 2
3)3 2
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4) 2
S l ti : G.E.
Solution
G E = a +b -c
2
2
2
= 16 + 25 - 9 = 32 = 4 2
∴Ans : (2)
( )
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17. If a = Sin1° and b = Sin1
then
th
1)a = b 2)a < b 3)a > b 4)a = 2b
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Solution : Sinθ is increasing
g
in 0 < θ < 90°
∴Ans
A : (2)a
(2) < b
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18. IF CosA = aCosB and
18
SinA = bSinB, then
(b - a )Sin B =
2
1) 1+ a
2
2
2
2) 2 + a 3) 1- a 4) 2 - a
2
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2
2
Solution : squaring we get
Cos A = a Cos B = a (1- Sin B)
2
2
2
2
= a - a Sin B.....(1)
2
2
2
& Sin A = b Sin B.....(2)
2
2
2
Vikasana - CET 2012
2
adding
g ((1)&
) ((2))
1= a - a Sin B + b Sin B
2
2
2
2
⇒ (b - a )Sin B = 1- a
2
2
2
∴Ans : (3)
Vikasana - CET 2012
2
2
19 The
19.
Th maximum
i
value
l
off
4Sin x + 3Cos x is
2
2
1)3 2)4 3)5 4)None
Vikasana - CET 2012
Solution :
G.E. = Sin x + 3(Sin
(
x + Cos x))
2
2
= Sin x + 3
≤ 1+ 3 = 4
2
∴ Ans : (2)4
Vikasana - CET 2012
2
20. The value of
tan100 + tan125 + tan100tan125 =
1)2
2)3
1
3)
3
4) 1
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Sol : tan225 = tan(100 +125) = 1
tan100 + tan125
⇒
=1
1- tan100 tan125
∴ G.E. = 1
Ans : (4) 1
Vikasana - CET 2012
21. If ABCD is a cyclic
quadrilateral then
1)) Sin(A
( + C)) = 1
2)Cos(A
)
( + C)) = -1
3) Sin(B +D) = 1
4) Cos(A + C) = 1
Vikasana - CET 2012
SSolution
l ti :
pp
g
= 180°
Sum of opp.angles
∴ A + C = 180° ⇒ Cos(A + C) = -1
Ans : (2)
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1 a b
22. For a ∆ABC,, 1 c a = 0
1 b c
then the value of
Cos A + Cos B + Cos C =
9
3
4
4
1)
2)
3)
4)
4
4
9
3
2
2
2
Vikasana - CET 2012
Solution : ∆ = 0 if any two rows / columns
are identical ⇒ a = b = c (by inspection)
⇒ A = B = C = 60°
3
G.E. = 3Cos 60 =
4
2
Ans : ((2))
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23
23.
Cos 10° + Cos 110° + Cos 130° =
3
3
1)
4
3
3
2)
8
3 3
3)
8
3
3 3
4)
4
Vikasana - CET 2012
S l i : If α = 60° or 120°or 240° or 300°
Solution
3
then, Cos θ + Cos (α + θ )+Cos (α - θ ) = Cos3θ
4
3
3 3 3 3
3
G.E. = Cos (3x10°) =
=
4
4x2
8
3
3
3
Ans : ((3))
Vikasana - CET 2012
24. Cos 10° + Cos 50° + Cos 70° =
2
1
1)
2
2
2) 1
3
3)
2
2
4)2
Vikasana - CET 2012
Solution :
when α = 60° or 120° or 240°or 300°
3
then Cos θ + Cos (α - θ) + Cos (α + θ) =
2
2
2
2
Ans : (3)
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1- Cosθ
2x
, then,,
=
25. If x =
2
1+ Cosθ
1- x
1) Sinθ 2)Cosθ 3)tanθ
Vikasana - CET 2012
4) Cotθ
θ
Solution : x = tan
2
θ
2tan
2
= tanθ
∴ G.E. =
2 θ
1- tan
2
Ans : (3)
Vikasana - CET 2012
⎛π
⎞
⎞
2⎛ π
C
Cos
Si ⎜ - A ⎟
⎜ - A ⎟ - Sin
4
4
⎝
⎠
⎝
⎠
26.
=
⎞
⎞
2⎛ π
2⎛ π
Cos ⎜ + A ⎟ + Sin ⎜ + A ⎟
⎝4
⎠
⎝4
⎠
1)Cos2A
2)tan2A
3)Si 2A
3)Sin2A
4)C
4)Cot2A
t2A
2
Vikasana - CET 2012
Cos2θ
C
2θ
Solution : G.E. =
1
⎛π
⎞
= Cos ⎜ - 2A ⎟ = Sin2A
⎝2
⎠
∴ Ans : (3)
( )
Vikasana - CET 2012
27. If
27
tanβ
β = 2SinαSinγCosec(α
γ
( + γ),
then Cotα, Cotβ, Cotγ are in
1)A.P. 2)G.P. 3)H.P. 4)A.G.P.
Vikasana - CET 2012
Solution : Taking reciprocals,
Si ( + γ)) Cotγ
Sin(α
C t + Cotα
C t
Cotβ =
=
γ
2
2SinαSinγ
∴ They are in A.P. Ans : (1)
Vikasana - CET 2012
28. If Cos(x
( - y) +
-3
(y - z)) + Cos(z
( - x)) =
Cos(y
2
then, ∑ Cosx =
1)0 2)1
3)2
4)3
Vikasana - CET 2012
-3
Solution : ∑ Cos(x
( - y) =
2
⇒ ∑ 2(CosxCosy
(
y + SinxSiny)
y) = -3
3 + ∑ 2(CosxCosy + SinxSiny) = 0
i.e.,(Cosx + Cosy + Cosz) +
2
(Sinx + Siny + Sinz) = 0
2
∴∑ Cosx = 0 = ∑ Sinx
Vikasana - CET 2012
Minimum off (S
(Sin x + C
Cos x))
29.
=
⎛
2 x
2 x⎞
Ma im m of ⎜ Cos
Maximum
+ Sin
⎟
2
2⎠
⎝
2
1)) - 1
2)1
)
3)2
)
4)) - 2
Vikasana - CET 2012
2
Min.of 1
Solution : G.E.
GE =
Max. of 1
1
= =1
1
∴Ans : (2)
Vikasana - CET 2012
30 3Sin x + 4Cos x ∈
30.
2
2
1) [0,3]
2) [0,4 ]
3)[ 3,4
3 4]
4) [ -4
-4,-3
-3 ]
Vikasana - CET 2012
G E = 3(Sin x + Cos x) + Cos x
G.E.
2
2
= 3 + Cos x ∈ [ 3,4 ]
2
Q Cos x ∈ [0,1]
2
Ans : (3)
Vikasana - CET 2012
2
31.The maximum value of
3
is
5Sinx - 12Cosx +19
1) 1
1
1
2)
3)
2
3
1
4)
4
Vikasana - CET 2012
Sol : G.E. is maximum when
Denominator is Minimum
and
d Min.value
Mi
l
off D
Dr.
n
= 19 - 25 +144 = 6
3 1
∴ G.E.
GE = =
∴ Ans.(2)
Ans (2)
6 2
Vikasana - CET 2012
-4
32. If tanθ =
then, Sinθ =
3
-4
4
-4
4
2))
or
1)) but not
5
5
5
5
4
-4
3) but
b t not
4)None of these
5
5
Vikasana - CET 2012
-4
S l : ttanθ
Sol
θ=
3
⇒ θ ∈ II or IV quadrant
n
4
∴ Sinθ = ±
5
Ans : ((2))
Vikasana - CET 2012
33. The value of
3Cosec20° - Sec20° is
1) 2
2) 4
2Sin20
2Sin20°
3)
Sin40°
4Sin20°
4Sin20
4)
Sin40°
Vikasana - CET 2012
3
1
Sol : G.E. =
Sin20 Cos20
n
3Cos20 - Sin20
=
Sin20 Cos20
Vikasana - CET 2012
⎛ 3
⎞
1
2⎜
Cos20 - Sin20 ⎟
2
2
⎝
⎠
=
1
× 2Sin20 Cos20
2
Sin(60 - 20)
=4
Sin40
= 4 ×1= 4
Ans : ((2))
Vikasana - CET 2012
34. If A = Cos θ + Sin θ, then for
all the values of θ
2
1)1 ≤ A ≤ 2
3
13
3) ≤ A ≤
4
16
4
13
2)
≤ A ≤1
16
3
4)) ≤ A ≤ 1
4
Vikasana - CET 2012
Sol : A = 1- Sin θ + Sin θ
n
2
4
= 1- Sin θ(1- Sin θ) = 1- Sin θCos θ
2
2
2
2
2
⎛ Sin2θ ⎞
= 1- ⎜
⎟ = 1- 0 if Sin2θ is least
⎝ 2 ⎠
1 3
or = 1- = , if Sin2θ is greatest
4 4
3
∴ ≤ A ≤1
Ans : ((4))
4
Vikasana - CET 2012
35.
35
tan20
tan20°+ tan40
tan40°+ 3tan20
3tan20°tan40°
tan40 =
1
1)
3
2) 3
-1
3)
3
Vikasana - CET 2012
4) 3
S l :W
Sol
We h
have, ttan(40
(40 + 20) = 3
n
tan40 + tan20
⇒
= 3
1- tan40 tan20
∴G.E. = 3
Ans : (2)
Vikasana - CET 2012
4
5
36. If Cos(α +β) = , Sin(α - β) =
5
13
and α and β lies between
π
0a
and
d , then
t e tan2α
ta α =
4
16
56
28
1)
2)
3)
4)None
63
33
33
Vikasana - CET 2012
Sol : tan2α = tan(α + β + α - β)
tan(α +β) + tan(α - β)
=
1 tan(α +β) tan(α - β)
13 5
+
56
4
12
=
=
Ans : (2)
3 5
33
1
1×
4 12
n
Vikasana - CET 2012
37. The value of
2 π
2 3π
2 5π
2 7π
Sin
+ Sin
+ Sin
+ Sin
8
8
8
8
1)1
)
2)2
)
1
1
3)1
)
4)2
)
8
8
Vikasana - CET 2012
7π
π
Sol : Sin
= Sin and
8
8
5π
3π
Sin
= Sin
8
8
⎛
2 π
2 3π ⎞
G.E. = 2 ⎜ Sin +Sin
⎟
8
8 ⎠
⎝
n
⎛
2 π
2 π⎞
= 2 ⎜ Sin +Cos ⎟ = 2×1= 2
8 - CET 2012
8⎠
⎝ Vikasana
tan70 - tan20
38.
=
tan50
1) 3
2) 0
3) 1
Vikasana - CET 2012
4) 2
tanA - tanB
Sol :
= 1+ tanA tanB
tan(A - B)
tan70 - tan20
∴ G.E. =
tan(70 - 20)
= 1+ tan70.tan20
= 1+ tan70.Cot70
t 70 C t70 = 1+1=
1+1 2
n
Vikasana - CET 2012
39. If
CosACosB + SinASinBSinC = 1
then a : b : c =
1) 1: 1: 1
2) 2 : 1: 1
3)1: 2 : 1 4)1: 1: 2
Vikasana - CET 2012
Sol :
1= CosACosB + SinASinBSinC
≤ CosA
C A CosB
C B + SinA
Si A Si
SinB
B
(Q SinC ≤ 1)
= Cos(A - B)
1 ≤ Cos(A - B) ⇒ Cos(A - B) = 1
n
Vikasana - CET 2012
⇒ A -B =0 ⇒ A =B
By given expression we get
C = 90 ∴ A = B = 45. Hence
a : b : c = Si
SinA
A : SinB
Si B : Si
SinC
C
1
1
=
:
: 1=
1 1
1: 1
1: 2
A (4)
Ans:(4)
2
2
Vikasana - CET 2012
40. In a ∆ABC, A > B and if the
measures of A and B satisfy
3Sinx - 4Sin x - k = 0,
3
0 < k < 1 then C =
1)30 2)45
3)120 4)None
Vikasana - CET 2012
Sol : 3Sinx - 4Sin x = k
⇒ Sin3x = k. Given that
Sin3A = k and Sin3B = k
∴ Sin3A = Sin3B ⇒
3A = 3B or 3A = 180 - 3B
But A > B means A ≠ B
∴ 3A = 180 - 3B ⇒ A +B = 60
2012 : (3)
∴ C = 120Vikasana - CETAns
n
3
41 ∆ABC is
41.
i right
i ht angled
l d att C
C,
then tanA + tanB =
b
1)
ac
2
2) a + b
a
3)
bc
2
Vikasana - CET 2012
c
4)
ab
2
a
b
Sol : tanA = , tanB =
b
a
2
2
a b a +b
GE = + =
G.E.
B
b a
ab
c
a
2
c
=
Q C = 90
C
b
ab
Ans : (4)
n
Vikasana - CET 2012
A
42. If P1,P2 ,P3 are altitudes
off a ttriangleABC,
i
l ABC ffrom
the vertices A,B,C
, , and
∆, the area of a triangle,
then P +P +P is
-1
1
-1
2
-1
3
s−a
s −b
s−c
s
1)
2)
3)
4)
∆
∆
∆
∆
Vikasana - CET 2012
1
2∆
n
Sol : ∆ = aP ⇒ P =
1 a
2 1
a
-1
∴P =
1
2∆
b
c
-1
1
-1
1
⇒P =
and P =
2
3
2∆
2∆
a + b + c 2s s
H
Hence
G.E.
GE =
=
=
2∆ ∆
2∆
Ans : (4)
Vikasana - CET 2012
43. If in a ∆ABC,,
CosACosB+SinASinBSinC =1,
then the triangle is
1) isosceles
2) right angled
3)isosceles
)
right
g angled
g
4)equilateral
Vikasana - CET 2012
Sol : By Que.No.39
n
A = B = 45° and C = 90°
∴ Ans : (3)
Vikasana - CET 2012
44. If two sides a,b and angle A
b such
be
h th
thatt 2 ttriangles
i
l
are fformed
d
then the sum of two values of third
side is
1) 2b SinA 2) 2b CosA
3)b C
CosA
A 4)(
4)(c+b)CosA
+b)C A
a
Vikasana - CET 2012
a2=b2+c2 -2bcCosA
2bcCosA
∴ c2 -(2bCosA)c+(b2 -a2 )=0
which is Quad. in c,
if c &
&c b
be ttwo roots
t th
then
1 2
c +c =2bCosA
1 2
Vikasana - CET 2012
45. logx y = logy z = logz x,
then
1)x=y=z
1)
3)x=y>z
2) x>y>z
4)x<y<z
Vikasana - CET 2012
k
k
n
k
Sol : y=x , z=y and x=z
∴ xyz=(xyz)k ⇒ k=1
⇒x=y=z
∴ Ans:(1)
Vikasana - CET 2012
46 If log
46.
l
2
2, llog(2x-1)
(2 1) and
d
log(2x+3)
og(
3) a
are
e in A.P.
then x =
-1
1
1)
2)
3)1 4) None
2
2
Vikasana - CET 2012
Soln:
2log(2x-1)
2l
(2 1) = llog2+log(2x+3)
2+l (2 +3)
∴ (2x
(2x-1)
1)2=2(2x+3)
2(2x+3)
⇒x=5 or -1 but,, when x= -1
2
2
2
log(2x-1)
log(2x
1) is not difined.
Ans(4)
Vikasana - CET 2012
47. If x, y, z are in G.P. and
y z
x
a =b =c , then
1)) log
gcb= log
gac 2)log
) gab= log
gcb
3) logac= log a
b
4)log a= logcb
b
Vikasana - CET 2012
Soln: ax = by = cz = k (say)
∴x=logak, y=logbk, z=logck
y z
x,y,z are iin G
G.P.
P ⇒ =
x y
l bk llogck
log
=
⇒logb a=logcb
logak logbk
Ans:(4)
Vikasana - CET 2012
48. If loga + logb=
logb log(a
log(a+b)
b)
then a=
1) b
2) b
b1
b-1
3) b-1
b
Vikasana - CET 2012
4) b
b
b+1
1
Soln: log(ab) = log(a+b)
b
⇒ab=a+b
b +b ⇒ a =
b-1
Ans:(2)
Vikasana - CET 2012
2n
49. 5 - 1 is divisible by
y
1) 10
2) 9
3) 20 4) 24
Vikasana - CET 2012
n
Sol : By inspection, put n=1
2(1)
we get 5 -1 = 24
which is divisible by 24
Ans: (4)
Vikasana - CET 2012
2
50. If x + bx+c =0 and
2
x + cx+ b = 0,
have a common root
and b≠c then b + c =
1) 0
2) -1
3) 2
Vikasana - CET 2012
4) 1
n
Sol : If α be the common root
2
2
then, α +bα+c=0 and α +cα+b=0
solving we get,
get α = 1 & α = -b-1
∴ -b-c = 1 ⇒ b + c = -1
Vikasana - CET 2012
51.
51 The value of
6+ 6+ 6+..... is
1) 3
2) 2
3) 4
Vikasana - CET 2012
4) 5
Sol : x= 6+ 6+ 6+.....
6+
n
= 6+x
2
2
⇒ x = 6+x ⇒ x -x-6=0
∴ x = 3 or -2, but x ≠ -2
∴ x=3
A
Ans : (1)
Vikasana - CET 2012
52 If a,b,c
52.
a b c are the roots of
3
2
x - 6x +2x-7=0
2x 7 0 then,
1 + 1 + 1 =
ab
b bc
b
ca
6
-6
2
-7
1)
2)
3)
4)
7
2
7
7
Vikasana - CET 2012
∑a
a+b+c
S l :G
Sol
G.E.=
E
=
abc
abc
6
=
7
Ans: (3)
n
Vikasana - CET 2012
53. Remainder when
55
24
x + x + 1 is divided by
x + 1 is
1) 0
2) 1
3) 2
Vikasana - CET 2012
4) -1
1
n
Sol : When f(x) is divided
by
y (x-a)
(
) the
remainder = f(a)
= f(-1) = -1 + 1 + 1 = 1
Vikasana - CET 2012
54. The domain of
1) [0,
[0 4]
2
4x-x is
2) (0
(0, 4)
3)) R-(0,4)
( , ) 4)) R-[0,4]
[ , ]
Vikasana - CET 2012
n
2
2
Sol : 4x-x ≥0 i.e., x -4x≤0
2
(x-2) ≤ 4 ⇒ x-2 ≤ 2
⇒ -2 ≤ x-2 ≤ 2, ∴0≤x≤4
A : (1)
Ans
Vikasana - CET 2012
55 Th
55.
The range off th
the ffunction
ti
x-2
f(x)=
, x≠2 is
2-x
1)1 2) -1 3) {1} 4){ -1}
Vikasana - CET 2012
( )
n
x
x-2
2
Sol : f(x)= = -1
1
x-2
∴ Range (f) = { -1}
Ans : (4)
Vikasana - CET 2012
56. The range of the function
Si ([ ] ) ((where
Sin([x]π),
h
[[x]] iis
greatest integer
g
g function)) is
1) 0 2){0}
3)[-1, 1] 4)(0,1)
Vikasana - CET 2012
n
S l : [[x]=integer
Sol
] i t
∴ Sin(kπ)
( ) =0 ∀k∈Z
∴ Range ={0}
Ans: (2)
Vikasana - CET 2012
57. A set A has 6 elements.
Then the number of possible
relations on A is
6
36
1) 6
2) 2
3)2
Vikasana - CET 2012
2
4)6
n
Sol : No.
No of possible relations
mn
from A to B is 2
where m=O(A) and n=O(B)
36
Hence Req.=2
Ans : (3)
Vikasana - CET 2012
58 The number of functions
58.
from a set A containing 7
elements into a set B
containing 3 elements is
7
3
1) 3
2) 7
3) 3
4) 7
Vikasana - CET 2012
n
Sol : No. of functions
from a finite set A
n(A)
(A)
into a finite set B is [n(B)]
7
∴ Req. No. of functions = 3
Ans : ((3))
Vikasana - CET 2012
3x
= 2 + 1
(x 6)(x+a) (x
(x-6)(x+a)
(x-6)
6) x+a
then a =
1) 4
2) 3
3) 2
4) 1
59. If
Vikasana - CET 2012
n
Sol : Put x = 6 in Nr.
Nr of LHS
and Dr. of II term
3(6)
we get,
= 2 ∴ a=3
6+a
Ans: (2)
Vikasana - CET 2012
60. In the expansion of
50
(1 + x) , the sum of the
coefficients
ffi i t odd
dd powers
of x is
1) 0
50
2) 2
49
3) 2
Vikasana - CET 2012
51
4) 2
n
Sol : Req. Sum
= C +C +C +....+C
+ +C
1 3 5
49
50-1
50
49
9
=2
=2
Ans : (3)
( )
Vikasana - CET 2012