CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-10
Q.1 Number of real x satisfying the equation | x – 1 | = | x – 2 | + | x – 3 | is
(A) 1 (B*) 2 (C) 3 (D) more than 3
[Hint: x = 2 and x = 4 only solution ]
Q.2 A diameter and a chord of a circle intersect at a point inside the circle. The two parts of the chord are
length 3 and5 and one part of the diameter is lengthunity.The radius ofthe circle is
(A*) 8 (B) 9
(C) 12 (D) 16
[Hint: x × 1 = 3 · 5  x = 15
 diameter = 16
radius = 8 cm]
Q.3 Smallest positive solution of the equation, 4 




 x
sin2
16 =  
x
sin
6
2 , is
(A)
2

(B)
3
2
(C)
6
5
(D*) none
[Hint: x =
6

,
6
5
or
2


6

 (D) ]
Q.4log If log3(x) =p and log7(x) =q, whichof the followingyields log21(x)?
(A) pq (B)
q
p
1

(C*) 1
1
q
p
1



(D) 1
1
q
p
pq



[Hint: log3x = p and log7x = q
now log21x = 21
log
1
x
= 7
log
3
log
1
x
x
 =
q
1
p
1
1

= 1
1
q
p
1



Ans. ]
*Q.5ph-1 The value of the expression
1
)
44
cos
.......
..........
2
cos
1
(cos
2
)
89
sin
.......
3
sin
2
sin
1
(sin
2















equals
(A*) 2 (B)
2
1
(C)
2
1
(D) 1
[Sol. Nr = 2[(sin 1° + sin 89°) + (sin 2° + sin 88°) + ......... + (sin 44° + sin 46°) + sin 45°]
= 2[sin 45°(2(cos 44° + cos 43° + ...... + cos1°) + 1] (using sin C + sin D)
 r
r
D
N
= 2 sin 45° = 2 Ans ]
*Q.6 If x satisfies log2x + logx2 = 4, then log2x can be
(A*) tan(/12) (B) tan(/8) (C*) tan (5/12) (D) tan(3/8)

Q.7 Find the number of degree in the acute angle  satisfying cos  =
2
1
2
2  ? [Ans.  = 22.5°]
[Sol. 4cos2 – 2 = 2  2(cos2 – 1) = 2  cos 2 =
2
1
 2 =
4

  =
8

= 22.5° Ans.]
Ans.]
Q.8 Find all values of asuch thatthe three equations
ax + y = 1
x + y = 2
x – y = a
aresimultaneouslysatisfied bysame orderedpair (x,y). [Ans. a = 0 or a = – 1]
[Sol. from (1) and (2)
adding, x =
2
2
a 
subtracting y=
2
a
2 
substituting in(1)
a(a + 2) + (2 – a) = 2
a2 + 2a – a = 0  a2 + a = 0
 a = 0 or a = 1 Ans. ]
Q.9 Let Dbeanypoint on thebaseof anisosceles triangleABC.AC is
extended to E so that CE = CD. ED is extended to meetAB at F.
If angle CED = 10°, find the cosine of the angle BFD.
[Ans. – 2
3 ]
[Sol. Asshown
cos(180 – ) = – cos 
= – cos 30° = –
2
3
]
Q.10 Inthefigure,EisthemidpointofABandFisthemidpointofAD. Ifthe
area of FAEC is 13 sq. units, find the area of the quadrilateralABCD.
[Ans. 26]
[Sol. Area = 2(A1 +A2) (Thing !)
= 2 × 13 = 26 ]
Q.11 In the figure, 'O' is the centre of the circle and A, B and C are three
points on the circle. Suppose that OA =AB = 2 units and angle
OAC = 10°. Find the length of the arc BC.
[Ans.
9
10
units]
[Sol. arc BC = l = r
l = 2 ·
180

· 100
=
9
10
Ans. ]

Q.1 If logab+ logbc + logca vanishes where a, b andc are positivereals different than unitythen thevalue
of (logab)3 + (logbc)3 + (logca)3 is
(A*) an odd prime (B) aneven prime
(C) an odd composite (D)anirrationalnumber
[Hint: x + y + z = 0  x3 + y3 + z3 = 3xyz  3  (A) ]
*Q.2 Each ofthe four statements givenbelow areeitherTrueor False.
I. sin765° = –
2
1
II. cosec(–1410°) = 2
III. tan
3
13
=
3
1
IV. cot 




 

4
15
= – 1
Indicate the correct order of sequence, where 'T' stands for true and 'F' stands for false.
(A) F T F T (B) F F T T (C) T F F F (D*) F T F F
Q.3 The value of p which satisfies the equation 122p–1 = 5(3p ·7p)is
(A) 12
n
21
n
12
n
5
n
l
l
l
l


(B) 21
n
12
n
5
n
12
n
l
l
l
l


(C*) 21
n
144
n
12
n
5
n
l
l
l
l


(D) 21
n
5
12
n
12
n
l
l
l

*Q.482/ph-1 The expression





20
sin
·
20
tan
20
sin
20
tan
2
2
2
2
simplifies to
(A)arationalwhichis notintegral (B) a surd
(C)anaturalwhichis prime (D*) anaturalwhich is not composite
[Hint: tan220° – sin220° = tan220° (1 – cos220°) = tan220° sin220°
Hence Nr = Dr  (D) ]
Q.5 If tan
2

= m, then the value of




sin
1
)
2
/
(
sin
2
1 2
is
(A)
m
1
m
2

(B*)
m
1
m
1


(C)
m
1
m
1


(D)
m
2
m
1
[Sol. y =



sin
1
cos
=
)
2
/
sin(
)
2
/
cos(
)
2
/
sin(
)
2
/
cos(






=
)
2
/
tan(
1
)
2
/
tan(
1




=
m
1
m
1


Ans. ]
*Q.668/ph-1 The value of
3 76 16
76 16
  
  
cot cot
cot cot
is :
(A*) tan46º (B) tan44º (C) cot 46º (D) cot2º
[Sol. Using 0
0
0
0
0
0
0
0
16
cos
76
sin
16
sin
76
cos
16
cos
76
cos
16
sin
.
76
sin
3


= 0
0
0
0
0
0
0
92
sin
]
16
cos
76
cos
16
sin
76
[sin
16
sin
76
sin
2 

=






92
sin
60
cos
92
cos
60
cos
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-11

= 0
0
92
sin
92
cos
1
= 0
0
0
2
46
cos
46
sin
2
46
sin
2
= tan460 = cot440 Ans ]
*Q.7 An unknown polynomial yields a remainder of 2 upon division by x – 1, and a remainder of 1 upon
division byx – 2. Ifthis polynomial is divided by(x – 1)(x – 2), then theremainderis
(A) 2 (B) 3 (C*) – x + 3 (D) x + 1
[Sol. We have
P (x) = Q1(x – 1) + 2 ....(1)
P (x) = Q2(x – 2) + 1 ....(2)
P (x) = Q3(x – 1)(x – 2) + ax + b ....(3)
From (1), P(1) = 2 = a + b ....(4)
from (2), P (2) = 1 = 2a + b .....(5)
hence a = – 1 and b = 3
Hence the remainder is 3 – x  (D) ]
Q.8 The difference (sin8 75° – cos8 75°) is equal to
(A) 1 (B)
8
3
3
(C)
16
3
3
(D*)
16
3
7
[Hint: (sin4  – cos4 )(sin4  + cos8 ) where  = 75°
(– cos150°)(1 – 2 sin2 cos2)
+
2
3







 150
sin
2
1
1 2
=
2
3







8
1
1 =
2
3
·
8
7
=
16
3
7
Ans. ]
Ans. ]
*Q.9circle There is an equilateral triangle with side 4 and a circle with the centre on one of the vertex of that
triangle.Thearcofthat circledivides thetriangleinto twoparts of equal area. Howlongis the radius of
thecircle?
(A*)

3
12
(B)

3
24
(C)

3
30
(D)

3
6
[Sol.
4
3
· 16 = 2 ·
2
1
r2 ·
3

[11th
J-Batch (15-01-2006)]
r2 = 12 3  r =

3
12
Ans. ]
Q.10 In a triangleABC, BC = 8, CA= 6 andAB = 10.Aline dividing the triangleABC into two regions of
equal area is perpendicular toAB at the point X. Find the length BX.
[Sol. 2 ·
2
y
·
x
=
2
6
·
8
= 24
x · x tan B = 24
x2 ·
4
3
= 24
x2 = 32  x = 2
4 Ans. ]

Q.11 If sec x + tan x =
7
22
, find the value of tan
2
x
. Use it to deduce the value of cosec x + cot x. [3]
[Sol.
x
cos
x
sin
1
=
7
22
[Ans.
29
15
,
15
29
]
7
22
t
1
t
1
t
1
t
2
1
2
2
2





where t = tan
2
x
2
2
t
1
)
t
1
(


=
7
22

t
1
t
1


=
7
22
)
t
1
(
)
t
1
(
)
t
1
(
)
t
1
(






=
7
22
7
22


=
15
29

t
2
2
=
15
29
 t =
29
15
Ans.
now cosec x + cot x
=
x
sin
x
cos
1
=
 
   
2
x
cos
2
x
sin
2
2
x
cos
2 2
= 





2
x
cot =  
2
x
tan
1
=
15
29
 cosec x + cot x =
15
29
Ans. ]

Q.1 





 x
sin
3
4
x
sin
dx
d 3
when x = 12° is
(A) 0 (B)
4
2
6 
(C*)
4
1
5 
(D)
4
1
5 
[Hint: y =
3
x
sin
4
x
sin
3 3

=
3
x
3
sin
; y' = cos 3x = cos 36° =
4
1
5 
]
Q.2 A rectanglehas its sides oflength sin x and cos x for some x. Largest possible area which it can have,
is
(A)
4
1
(B) 1 (C*)
2
1
(D) can not be determined
[Hint: A= sin x cos x =
2
1
sin 2x, henceA
Amax =
2
1
]
Q.3 If logAB + logBA2 = 4 and B <A then the value of logAB equals
(A) 1
2  (B) 2
2
2  (C) 3
2  (D*) 2
2 
[Hint: Let logA(B) =x
x +
x
2
= 4  x2 – 4x + 2 = 0  x =
2
8
16
4 

= 2 ± 2
 x = 2 – 2 ;  B <A hence logB < logA ; logAB < 1 ]
Q.4 The sum of3 real numbers is zero. If the sum of theircubes is C then their product is
(A) a rational greater than 1 (B) a rational less than 1
(C*) an irrational greater than 1 (D) anirrational less than 1
[Hint: a + b + c = 0  a3 + b3 + c3 = 3abc
 abc =
3

which is irrational > 1  (B) ]
Q.5 The sides of a triangleABC are as shown in the given figure.
Let D be any internal point of this triangle and let e, f, and g
denote the distance between the point D and the sides of the
triangle. The sum (5e + 12f + 13g) is equal to
(A) 120 (B) 90
(C*) 60 (D) 30
[Hint:
2
f
12
2
g
13
2
e
5

 = 30 ( =
2
12
·
5
= 30)
hence 5e + 12f + 13g = 60 Ans. ]
Q.6 The value of tan27° + tan18° + tan27° tan18°, is
(A)anirrationalnumber (B)rationalwhichisnotinteger
(C)integerwhichisprime (D*) integer whichis not aprime.
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-12

[Sol. Consider tan(27° + 18°) =






18
tan
27
tan
1
18
tan
27
tan
1 =






18
tan
27
tan
1
18
tan
27
tan
 1 – tan27° tan18° = tan27° + tan18°
 tan27° + tan18° + tan27° tan18° = 1 ]
Q.7 In a triangleABC, the value of
C
sin
B
sin
A
cos
+
A
sin
C
sin
B
cos
+
B
sin
A
sin
C
cos
(A*)is prime (B)is composite
(C)isrational whichis notaninteger (D*)an integer
[Hint: – [(cos B cos C – 1) + (cos C cosA – 1) + (cos A cos B – 1)] = 2
Q.869/6 If cos( + ) + sin( – ) = 0 and tan  =
2006
1
. Find tan . [3]
[Ans. – 1]
[Sol. cos  · cos  – sin  · sin  + sin  · cos  – cos  · sin  = 0
cos (cos  – sin ) + sin (cos  – sin ) = 0
(cos  – sin ) × (cos  + sin ) = 0
if cos  – sin  = 0  tan  = 1 which is not possible
 sin  + cos  = 0
 tan  = – 1 Ans. ]
Q.9 ABC is aright angledtriangle.Show that
sinA·sinB·sin(A–B)+sinB·sinC·sin(B–C)+sinC·sinA·sin(C–A)+sin(A–B)·sin(B–C)·sin(C–A)=0. [3]
[Sol. Let B = 90°
T1 = sinA· sin B · sin(A– B)
= (sinA) · (1) · sin(A– 90°)
T1 = – sinAcosA
T2 = sin B · sin C · sin(B – C)
= (1) (cos A) · sin(90° – C) = cos Asin A
T3 = sin C · sinA· sin(C –A)















A
sin
C
cos
A
cos
C
sin
0
B
cos
/
90
B
sin
= sin(90° –A) · sinA· sin(90° – 2A)
T3 = sinA· cosA· cos2A
T4 = sin (A– B) · sin (B – C) · sin(C –A)
= – cos A · sin A · cos 2A ]
Q.10(a) Solve (x + 2)(x – 2)(x – 13) = (x + 2)(x – 7)(x – 11) for x.
(b) Solve (x – 3)(x – 2)(x – 13) = (x – 3)(x – 4)(x – 11) for x.
[Sol. (a) (x + 2) [(x2 – 15x + 26) – (x2 – 18x + 77)] = 0
(x + 2)(3x – 51) = 0  x = – 2 or x = 17
[Ans. (a) x = – 2 or 17, (b) x = 3 ]
Q.11 If m, n > 1 and for all x > 0 and x  1
lognx = 3 logmx.
Writeanequationexpressingmexplicitlyin termsof n.
[Sol. lognx = 3
m
log
x
log
n
n
if x  1  lognm = 3
m = n3 ]