1. Select the correct alternative : (Only one is correct)
Q.190/-1 The value of cot 7
1
2
0
+ tan 67
1
2
0
– cot 67
1
2
0
– tan7
1
2
0
is :
(A)arationalnumber (B*)irrationalnumber (C) 2(3 + 2 3 ) (D) 2 (3 – 3 )
Q.2817/ph-1  x  R the greatest and the least values of y =
1
2
cos 2x + sin x are respectively
(A)
3
4
1
2
, (B*)
3
4
3
2
,  (C)
1
2
3
2
,  (D) 1 , -
3
2
[Sol. E =
2
x
sin
2
x
2
cos 
=
2
x
sin
2
x
sin
2
1 2


=
2
1
x
sin
2
x
sin
2 2



=

















4
3
2
1
x
sin
2
=
















2
2
1
x
sin
4
3
Emax = 3/4
Emin = 3/4 – (–3/2)2
= 3/4 –9/4 = – 3/2 ]
Q.370/-1 If tan  =
1
x
x
x
x
2
2



and tan  =
1
x
2
x
2
1
2


(x  0, 1), where 0 < ,  <
2

, then tan ( + )
has the value equal to :
(A*) 1 (B) – 1 (C) 2 (D)
4
3
[Sol. x2 – x = t ; tan  =
1
t
t

; tan  =
1
t
2
1

 tan ( + ) = 





tan
tan
1
tan
tan
=
1
t
2
1
.
1
t
t
1
1
t
2
1
1
t
t






= t
)
1
t
(
)
1
t
2
(
1
t
)
1
t
2
(
t






=
1
t
2
t
2
1
t
2
t
2
2
2




= 1
 tan ( + ) = 1 ]
Q.4100/-1 In a triangle ABC, angle Ais greater than angle B . If the measures of angles A& B satisfy the
equation, 3sinx  4sin3 x  K = 0, 0 < K < 1, then the measure of angle C is
(A) /3 (B) /2 (C*) 2/3 (D) 5/6 [ JEE ’90 , 2 ]
[Sol. k = sin3A = sin3B
3A = – 3B
A+ B = /3
C = 2/3 Ans ]
CLASS : XI (J-Batch) DATE : 06-07/10/2006 TIME : 60 Min. DPP. NO.-31

2. Q.5 Thevaluesofxsmallerthan3inabsolutevaluewhichsatisfytheinequality )
ax
2
x
(
log
)
x
a
2
( 2 

>1 for
all a > 5 is
(A) –2 < x < 3 (B) –3 < x < 3 (C) –3 < x < 0 (D*) –3 < x < –1
[Hint: log ( )
( )
2 2 2
a x
x ax

 > 1, a > 5 & –3 < x < 3
note that 2a – x2 > 1 for all a > 5 & x  ( –3, 3)
Hencetheaboveinequalitywill betrueif
(x – 2ax) > 2a – x2
or x – 2ax + x2 – 2a > 0
x(1+x) –2a (1 + x) > 0
(x + 1) (x – 2a) > 0  x  (–3, –1)  D ]
Q.68/-1 The exact value of cos cos cos cos cos cos
2
28
3
28
6
28
9
28
18
28
27
28
     
ec ec ec
  is equal to
(A) – 1/2 (B) 1/2 (C) 1 (D*) 0
[Hint: put

28
= x , T1 =
cos
sin
2
3
x
x =
cos sin
sin sin
2
3
x x
x x =
1
2
3
3
sin sin
sin sin
x x
x x

L
N
M O
Q
P
=
1
2
3
cos cos
ecx ec x
 etc. ]
Q.722 If 60a = 3 and 60b = 5 then the value of )
b
1
(
2
b
a
1
12 


equals
(A*) 2 (B) 3 (C) 3 (D) 12
[Sol. 60a = 3  a = log603
60b = 5  b = log605
let x = )
b
1
(
2
b
a
1
12 


log12x =
)
b
1
(
2
b
a
1



=
 
 
5
log
1
2
15
log
1
60
60


 log12x = log1444 = log122  x = 2 Ans. ]
Q.820 Smallest positive x satisfyingthe equation cos33x + cos35x = 8 cos34x · cos3x is
(A) 15° (B*) 18° (C) 22.5° (D) 30°
[Sol. cos33x + cos35x = (cos 5x + cos 3x)3
cos33x + cos35x = cos35x + cos33x + 3 cos 5x cos 3x (cos 5x + cos 3x)
 (3 cos 3x · cos 5x) (2 cos 4x · cos x) = 0
 cos x · cos 3x · cos 4x · cos 5x = 0
 x = (2n + 1)
2

, (2n + 1)
6

, (2n + 1)
8

, (2n + 1)
10

 smallest + ve values of x is
10

i.e. 18°Ans. ]
Q.920 LetABCDEFGHIJKL be a regular dodecagon, then the value of
AF
AB
+
AB
AF
is
(A*) 4 (B) 3
2 (C) 2
2 (D) 2

3. [Sol. In  OAB
 
12
sin
a
 = 2R ....(1) (AB = a)
AB = a = 2R sin
12

again In  OAF
 
12
5
sin
b
 = 2R (AF = b)
 AF = b = 2R sin
12
5
Hence
AF
AB
+
AB
AF


75
sin
15
sin
+


15
sin
75
sin
=


15
cos
15
sin
+


75
cos
75
sin
= tan 15° + tan 75° =    
3
2
3
2 

 = 4 Ans.]
Ans.]
Q.105 If














2
a
sin
2
a
tan
a
cos
1
2
2
=
a
cos
p
w
a
cos
k

where k,w and phaveno common factorotherthan 1, then the
value of k2 + w2 + p2 is equal to
(A) 3 (B) 4 (C) 5 (D*) 6
[Sol.








2
a
sin
)
2
a
(
cos
)
2
a
(
sin
a
cos
1
2
2
2
=













2
a
sin
)
2
a
(
cos
)
2
a
(
sin
2
a
sin
2
2
2
2
2
=































2
a
cos
2
a
sin
1
2
a
cos
2
2
a
sin
2
2
2
2
=
a
cos
1
a
cos
2

k = 2; w = 1; p = 1  k2 + w2 + p2 = 4 + 1 + 1 = 6 Ans.]
SUBJECTIVE
Q.1147/6 If 15 sin4 + 10cos4 = 6, evaluate 8cosec6 + 27sec6
[Sol. Dividingbycos4
15 tan4 + 10 = 6 sec4
15 tan4 + 10 = 6(1 + tan2)2
9 tan4 – 12 tan2 + 4 = 0
(3 tan2 – 2)2 = 0  tan2 =
3
2
Now 8 cosec6 + 27 sec6
8 (1 + cot2)3 + 27 (1 + tan2)3
8
3
2
3
1 





 + 27
3
3
2
1 







4. 125 + 125 = 250 Ans ]
Q.1279/1 John has 'x' children byhis first wife. Maryhas x + 1 children byher first husband. Theymarryand
have children of their own. The whole family has 24 children.Assuming that the children of the same
parents donotfight,findthemaximumpossiblenumberoffights thatcan take place.
[ Ans. : 191 ]
[Sol. f(x)representingthenumberoffightfunctions
f (x) = x ( x + 1) + (x + 1) (23 – 2x) + (23 – 2x) x
= (x2 + x) + ( – 2x2 + 21x + 23) – 2x2 + 23x
f (x) = – 3x2 + 45x + 23
f (x) = – 6x + 45 (X > 0 , N
x )
hence x = 7.5 gives maximabut N
x
hence practical maxima occurs at x = 7 or 8
 f (7) = 191 ]
Q.1357/6 If x, y, z be all positive acute angle then find the least value of
tanx (cot y + cot z) + tany (cot z + cot x) + tanz (cot x + cot y) [Ans: 6]
[Sol. x, y, z )
2
/
,
0
( 
E =  x
tan (cot y + cot z)
Let tanx = a ; tany = b ; tanz = c  a , b, c > 0
 E = 






















b
1
a
1
c
a
1
c
1
b
c
1
b
1
a

b
c
a
c
a
b
c
b
c
a
b
a




 

 



 


 

2
min
2
min
2
min
c
a
a
c
b
c
c
b
a
b
b
a


























 Emin = 6 ]

5. Select the correct alternative : (Only one is correct)
Q.122/ph-3 In an isosceles ABC A= ,AB =AC = 
sin , and BC = sin , then the area of ABC, is
(A) 1/2 (B*) 8/25 (C) 9/16 (D) none
[Sol. Using sine law
C
sin
AB

=
A
sin
BC







 


2
180
sin
sin
=


sin
sin
= 1
sin2 




 

2
90 = sin  
2
cos
1 

= sin ; (2 sin – 1)2 = 1 – sin2
 5 sin2 – 4 sin = 0  sin = 4/5
  =
2
1
bc sinA =
2
1

sin 
sin ·sin2 =
2
1
sin2 = 8/25 ]
Q.27 If x and yare real numbers such that x2 + y2 = 8, the maximum possible value of x – y, is
(A) 2 (B) 2 (C)
2
2
(D*) 4
[Sol. Let, x = 
cos
2
2 , y = 
sin
2
2
x – y = )
sin
(cos
2
2 


 (x – y)max = 4 ]
Q.33(D) Let a, b, c are distinct reals satisfying a3 + b3 + c3 = 3abc. If the quadratic equation
(a + b – c)x2 + (b + c – a)x + (c + a – b) = 0 has equal roots then a root of the quadratic equation is
(A)
3
c
b
a 

(B*) 1 (C)
2
c
b
a 

(D) 3
abc
[Sol. a3 + b3 + c3 – 3abc = (a + b + c) 






 




2
)
a
c
(
)
c
b
(
)
b
a
( 2
2
2
since a, b, c are distinct
 a3 + b3 + c3 – 3abc = 0  a + b + c = 0
now put x = 1 in the quadratic equation, we get a + b + c = 0
 x = 1 is the root of the quadratic equation ]
Q.41 If x,yRandsatisfythe equation xy(x2 –y2)=x2 +y2 where x  0 then theminimum possiblevalue
of x2 + y2 is
(A) 1 (B) 2 (C*) 4 (D) 8
[Hint: If x = r cos  and y = r sin  then E = x2 + y2 = r2. Hence we have to minimise r2. Now in the given
equation substituting x = r cos  and y= r sin ,
we get r2 = 4 cosec 4 
min
2
r = 4 Ans. ]
Q.515 If S = 12 + 32 + 52 + ....... + (99)2 then the value of the sum 22 + 42 + 62 + ....... + (100)2 is
(A) S + 2550 (B) 2S (C) 4S (D*) S + 5050
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-32

6. [Sol. Let S = 12 + 32 + 52 + ....... + (99)2
and x = 22 + 42 + 62 + ....... + (100)2
 (x – S) = (22 – 12) + (42 – 32) + ....... + (1002 – 992)
= (1 + 2) + (3 + 4) + ........ + (99 + 100) = 1 + 2 + 3 + 4 + ...... + 99 + 100
= 5050
 x = S + 5050 Ans. ]
Q.621 Let A = { x | x2 + (m – 1)x – 2(m + 1) = 0, x  R}
B = { x | (m – 1)x2 + mx + 1 = 0, x  R}
Numberofvalues of m such thatA Bhas exactly3 distinct elements, is
(A) 4 (B) 5 (C) 6 (D*) 7
[Sol. Case-I: when B is a quadratic equation
D1 = (m + 3)2; D2 = (m – 2)2
roots of 1st equation are 2, – (m + 1) setA
roots of 2nd equation are – 1,
m
1
1

set B
For exactlythere elements inA B two of the roots must be same note that 2  – 1
possibilitiesare
2 = – (m + 1)  m = – 3
2 =
m
1
1

 2 – 2m = 1  m = 1/2
– m – 1 = – 1  m = 0
– (m + 1) =
m
1
1

 1 – m2 = – 1  m = ± 2
m
1
1

= – 1  m – 1 = 1  m = 2.
Case-II: Now if m = 1, then B becomes linear
roots of B as x = – 1
roots of Aare 2 and – 2
 3elementsincommon
 all permissible m are {– 3,
2
1
, 2 , – 2 , 2, 0, 1} ]
Q.732/QE PQRS is a common diameter of three circles. The area of the middle
circle is the average of the area of the other two. If PQ = 2 and RS = 1
thenthelengthQR is
(A) 1
6  (B*) 1
6 
(C) 5 (D) 4
[Sol. Let QR = x
then the diameters are 2, x + 2, x + 3 
2
)
1
x
(
2 2
2


= (x + 2)2
 2(x + 2)2 = 22 + (x + 3)2
2(x2 + 4 + 4x) = 4 + (x2 + 6x + 9)
x2 + 2x – 5 = 0  x = 1
6  Ans.]

7. Q.8 Numberofprincipalsolutionoftheequation
tan 3x – tan 2x – tan x = 0, is
(A) 3 (B) 5 (C*) 7 (D) more than 7
[Hint: 0, , 2,
3

,
3
2
,
3
4
,
3
5
as given expression = tan 3x tan 2x tan x]
More than one are correct:
Q.930 Let 2 sin x + 3 cos y = 3 and 3 sin y + 2 cos x = 4 then
(A*) x + y = (4n + 1)/2, n  I
(B) x + y = (2n + 1)/2, n  I
(C) x and ycan be the two non right angles of a 3-4-5 triangle with x > y.
(D*) x and ycan be the two non right angles of a 3-4-5 triangle with y> x.
[Sol. Squaringandadding
4 + 9 + 12 sin(x + y) = 25  sin(x + y) = 1 = sin
2

 x + y = 2n +
2

= (4n + 1)
2

n  I  (A)
if x + y =
2

 y =
2

– x
5 sin x = 3  sin x =
5
3
or cos x =
5
4
also cos y =
5
3
and sin y =
5
4
; hence y> x  (D)]
Q.1032 For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are
(A*) there is onlyone integral k for which f (x) is non negative  x R
(B*) for k< 0 the numberzero lies between the zeros of the polynomial.
(C*) f (x) = 0 has two distinct solutions in (0, 1) for k  (1/4, 4/7)
(D) Minimum value of y  k  R is k(1 + 12k)
[Sol. (A) f (x) is non negative  x R
 f (x)  0  x  R  D  0
 64k2 – 16k  0
4k2 – k  0
k (4k – 1)  0
 integral value ofk = 0
(B) f (0) < 0
k < 0  (B)
(C) for distinct roots in (0, 1)
D > 0  k (4k – 1) > 0 ....(1)
0 < –
a
2
b
< 1  0 < k < 1 ....(2)
f (0) > 0  k > 0 ....(3)
f (1) > 0  4 – 7k > 0
 k < 4/7 ....(4)
(1)  (2)  (3)  (4)  k  (1/4, 4/7) ]

8. SUBJECTIVE:
Q.11 In a triangleABC, if sinAsinB sinC + cosAcosB = 1, then find the value of sin C. [Ans. 1]
Q.12 In a ABC, if a, b, c are inA.P, then prove that
cos(A – C) + 4cosB = 3 [4]
[Sol. Given 2b = a + c
2 sin B = sinA+ sin C
2 2 sin
2
B
cos
2
B
= 2 sin
2
C
A 
cos
2
C
A 
2 sin
2
B
= cos
2
C
A 
Onsquaring
4 sin2
2
B
=
2
)
C
A
(
cos
1 

8 sin2
2
B
= 1 + cos(A – C)
4 (1 – cos B) = 1 + cos(A – C)
 cos (A – C) + 4 cos B = 3 Hence Proved ]
Q.1387/6 Find the general solution of the trigonometric equation cosec x – cosec 2x = cosec 4x
[Ans: x = (2n+1)
7

, where 4
m
7
n 
 , m, n  I ]
[Sol. cosec 2x + cosec 4x = cosec x
x
sin
1
x
4
sin
x
2
sin
x
2
sin
x
4
sin


( note
4
n
and
2
n
,
4
n



 )
2sin3x cosxsinx =sin2x sin4x
or sin3x=sin4x )
0
x
2
(sin 
4x = n + (–1)n (3x)
If n is odd 4x = (2n+1) – 3x
7
)
1
n
2
(
x



 ( ....
,
17
,
10
,
3
n  ) ]

9. CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-33
Select the correct alternative : (Only one is correct)
Q.15/ph-3 LetABC be a triangle with circumradius R, perimeter Pand area k. The maximum value of 3
R
kP
,is
(A*)
4
27
(B)
2
3
9
(C)
4
3
9
(D)
8
27
[Sol. a = 2R sinA, b = 2R sinB and c = 2R sinC
P = 2R  A
sin ; k = 2R2
 A
sin (using  =
R
4
abc
)
 3
R
kp
= 4 sinA sinB sinC(sinA + sinB + sinC) = 4 ·
2
3
3
·
8
3
3
=
4
27
Thiswillbemaximum iftheisequilateral.]
Q.2ph-1 The value of, (tan 18°)(sin 36°)(cos 54°)(tan 72°)(tan 108°)(cos 126°)(sin 144°)(tan162°)(cos180°)
is k sin2 18° then k has the value equal to
(A)
16
5
(B*)
4
5
(C)
4
1
(D)
16
1
[Sol. (tan 18°)(tan 72°)(tan 108°)(tan 162°) = 1
hence weareleft with
( – ) (sin 36°)(cos 54°)(cos 126°)(sin 144°)
( – ) (sin 36°)(sin 36°)(– sin 36°)(sin 36°) = sin436°
=
2
2
72
cos
1





 

=
4
)
18
sin
1
( 2


=
2
4
1
5
1
4
1







 
 =
2
4
5
5
4
1







 
=
2
4
1
5
4
5







 
=
4
5
sin218°  k =
4
5
Ans. ]
Q.327/ph-3 Let a  b  c be the lengths of the sides of a triangleT. If a2 + b2 < c2 then which one of the following
must be true?
(A)All 3 angles ofTare acute. (B*) Some angleof T is obtuse.
(C) Oneangle of T is a right angle. (D)Nosuchtriangle canexist.
[Hint: a2 + b2 < c2
a2 + b2 < a2 + b2 – 2ab cosC
 cosC < 0
 C is obtuse  (B) ]
Q.463/matrices Identifywhether the statement is True or False.
Therecanexisttwotrianglessuchthatthesidesofonetrianglearealllessthan1cmwhilethesidesofthe
other triangle are all bigger than 10 metres, but the area of the first triangle is larger than the area of
secondtriangle.
[Hint: Consider 21 = b1h1
22 = b2h2
b1h1 > b2h2
first triangle is each with 1 cm, 21 =
2
3

10. 2nd triangle
22 =
2
10
10
20 10
3 


10–6  1 > 2 ]
Q.540/ph-3 ABC is an acute angled triangle with circumcentre 'O'orthocentre H. If AO =AH then the measure
of theangleAis
(A)
6

(B)
4

(C*)
3

(D)
12
5
[Hint: OA = HA
R = 2R cosA (distance of orthocentrefrom the vertexAis 2R cosA)
 cosA =
2
1
 A =
3

 (C)] [12th & 13th 25-09-2005]
*Q.66/ph-3 Triangle ABC is right angled at A. The points P and Q are on the hypotenuse BC such that
BP = PQ = QC. IfAP = 3 andAQ = 4 then the length BC is equal to
(A) 27 (B) 36 (C*) 45 (D) 54
[Sol. In  ABP
9 = c2 + x2 – 2cx cosB; but cosB =
x
3
c
[Quiz]
9 = c2 + x2 – 2cx
x
3
c
9 = x2 +
3
c2
....(1) ; |||ly 16 = x2 +
3
b2
....(2)
(1) + (2)
25 = 2x2 +
3
1
(b2 + c2) = 2x2 + 3x2
 5x2 = 25  x = 5 ;  BC = 5
3 = 45 Ans. ]
Q.7 Sides a, b,cofatriangleABC are in A.P. and
c
a
b
cos
,
c
b
a
cos 2
1





 ,
b
a
c
cos 3


 ,then the
value of
2
tan
2
tan 3
2
1
2 


, is
(A) 0 (B) 1 (C*) 2/3 (D)
3
5
[Sol. 2b = a + c
tan2
2
1

+ tan2
2
2

=
3
3
1
1
cos
1
cos
1
cos
1
cos
1









=
b
a
c
1
c
a
c
1
c
b
a
1
c
b
a
1









=
c
b
a
c
b
a
c
b
a
a
c
b









=
)
c
b
a
(
b
2


=
b
3
b
2
=
3
2
Ans. ]

11. Q.84/ph-3Atrianglehassides6, 7,8.Thelinethroughitsincentreparallel to theshortest sideis drawnto meetthe
other two sides at P and Q. The length of the segment PQ is
(A)
5
12
(B)
4
15
(C*)
7
30
(D)
9
33
[Sol.  =
2
h
·
6
= 3h ; (where h is the altitude fromA); Also  =
2
r
·
21
(using  = r·s)

2
r
·
21
=
2
h
·
6
= 3h 
h
r
=
7
2
now APQandABC are similiar
h
r
h 
=
6
PQ

h
r
1 =
6
PQ

7
2
1 =
6
PQ

7
5
=
6
PQ
 PQ =
7
30
Ans. ]
Ans. ]
Q.9106/circle A square and an equilateral triangle have the same perimeter. Let A be the area of the circle
circumscribedabout thesquare and Bbetheareaof thecirclecircumscribed about thetrianglethen the
ratioA/Bis
(A)
16
9
(B)
4
3
(C*)
32
27
(D)
8
6
3
[Hint: Radius ofthe circlecircumscribingthesquare =
2
a
2
3
A =
4
a
18
·
2
 =
2
a
9 2

radius of the circle circumscribing triangle = 2a sec 30° = 2a ·
3
2
=
3
a
4
B =
3
a
16
·
2
 ; Hence
B
A
=
16
3
·
2
9
=
32
27
]
Q.10ph-2 The sum of angles that satisfythe equation 2sin2x – sin x + cos x – sin 2x = 0 where x  (0, 2), is
(A)
6
7
(B*)
2
5
(C)
2

(D)
12
7
[Hint: 2 sin x (sin x – cos x) – (sin x – cos x) = 0
(sin x – cos x)(2 sin x – 1) = 0  tan x = 1 or sin x = 1/2
hence x =
4

,
4
5
,
6

,
6
5
=
2
5
Ans. ]
SUBJECTIVE:
*Q.1141/1 If cos A, cos B and cos C are the roots of the cubic x3 + ax2 + bx + c = 0 where A, B, C are the
angles of a triangle then find the value of a2 – 2b – 2c. [Ans. 1]
[Sol.  A
cos = – a;  B
cos
A
cos = b and  A
cos = – c
now (cos A + cos B cos C)2 =  
 A
cos2
+ 2 
 B
cos
A
cos [Quiz]
 cos2A + cos2B + cos2C = a2 – 2b
1 – 2 cos A cos B cos C = a2 – 2b2
1 – 2c = a2 – 2b2  a2 – 2b – 2c = 1 Ans. ]

12. Q.12119/6 If d1, d2, d3 arediameters of the excircles of ABC, touching thesides a, b, crespectivelythenprove
that
a
d
d
a
1
1





 


 .
[Sol. TPT
a
d
b
d
c
d
d d d
a b c
1 2 3
1 2 3
  
 
 
LHS
a
r
b
r
c
r
2 2 2
1 2 3
 
1
2
. [a(s – a) + b(s – b) + c(s – c)]
=

2
1
[s(a + b + c) – (a2 + b2 + c2)] =

4
1
[(a + b + c)2 – 2(a2 + b2 + c2]
2
4
2 2 2
( ) ( )
ab bc ca a b c
    

....(1)














c
s
b
s
a
s
s
1
= 

















)
c
s
)(
b
s
)(
a
s
(
)
b
s
)(
a
s
(
)
c
s
)(
a
s
(
)
c
s
)(
b
s
(
s
similarly RHS =
d d d
a b c
r r r
s
1 2 3 1 2 3
 
 

 
can be simplified equal to 1 ]
*Q.13115/6 In a  ABC , if cos A + cos B = 4 sin2
C
2
, prove that tan
A
2
. tan
B
2
=
1
3
. Hence deduce that the
sides of thetriangle are inA.P.
[Sol. 2cos
A B

2
cos
A B

2
= 4 sin2
C
2
[Quiz]
or cos
A B

2
= 2 sin
C
2
sin cos
C A B
2 2








= 2 cos
A B

2
or cos
A B

2
 cos
A B

2
= cos
A B

2
2 sin
A
2
. sin
B
2
= cos
A
2
. cos
B
2
 sin
A
2
. sin
B
2
3 sin
A
2
. sin
B
2
= cos
A
2
. cos
B
2
or tan
A
2
. tan
B
2
=
1
3
Now

s s a
( )

.

s s b
( )

=
1
3
;
s c
s

=
1
3
 2 s = 3 c a + b + c = 3 c
 a + b = 2 c  a , c , b are in A.P. ]

13. Select the correct alternative : (Only one is correct)
Q.16/ph-2 The equation , sin2 
4
1
3
sin  
= 1 
4
1
3
sin  
has :
(A) no root (B) one root (C) two roots (D*) infiniteroots
[Sol. sin2 = 1 [sin 1

 ]  sin = 1   = 2n + /2  infinite roots ]
Q.21/ph-3 In a triangleABC, R(b + c) = bc
a where R is the circumradius of the triangle.Then the triangle is
(A) Isoscelesbut not right (B)rightbut not isosceles
(C*)rightisosceles (D)equilateral
[Hint: R(b + c) = bc
a
R(b + c) = 2RsinA bc
 sinA=
bc
2
c
b 
now applyingAM  GM for b and c
bc
2
c
b 
 bc ; 
bc
2
c
b 
 1
hence sinA 1 whichis not possible.
hence sinA= 1  A = 90°
 A = 90° and b = c  (C)]
Q.313/ph-2 The general value of x satisfying the equation, 2cot2x + 2 3 cotx + 4 cosecx + 8 = 0 is
(A) n –

6
(B) n +

6
(C*) 2n –

6
(D) 2n +

6
[Hint: E = (cot ) cot cos
x x ec x
    
3 4 5 0
2 2
= (cot ) cos cos
x ec x ec x
    
3 4 4 0
2 2
= (cot ) (cos )
x ecx
   
3 2 0
2 2
cotx =  3 or cosecx = – 2
 x =
5
6

or
11
6

;  x =
7
6
11
6
 
or  2n –

6
 C ]
Q.4ph-3 In the equilateralABC,AB= 12. One vertex of a square is at the midpoint of the side BC, and the two
adjacent vertices are on the other two sides of the triangle. The length of a side of the square maybe
expressed as 6
q
2
p  where p and q are integers. The ordered pair (p, q) is
(A) (6, –2) (B*) (9, – 3) (C) (8, – 4) (D) (5, – 1)
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-34

14. [Sol. Usingsinelaw inBME,

60
sin
x
=

75
sin
6
 x =


75
sin
60
sin
6
=
 
1
3
2
2
·
3
3

=
 
2
1
3
6
6 
=  
6
18
3 
= 6
3
2
9 
 p = 9, q = – 3
 (9, – 3) Ans. ]
Q.5st.lineAtriangle with integral sides has perimeter 8 cm.Then the area of the triangle, is
(A*) 2
cm
2
2 (B)
2
cm
3
9
16
(C) 2
cm
3
2 (D) 2
cm
2
4
[Hint: Onlypossibilityfor the sides can be 3, 3, 2 (think !)
A = )
c
s
)(
b
s
)(
a
s
(
s 

 = 2
1
1
4 

 = 2
cm
2
2 Ans. ]
Q.6ph-3 Let ABC be theright triangle withvertices ofA(0, 2), B(1,0)and C(0, 0). IfDis the point onABsuch
that the segment CD bisects angleC, then the length of CD is
(A)
2
1
(B)
2
5
(C)
2
3
(D*)
3
2
2
[Sol.
2
1
× 1 × x sin 45° +
2
1
× x × 2 sin 45° =
2
1
× 1 × 2
3x sin 45° = 2  x =
3
2
2
]
Q.7QE All values of k such that the quadratic equation – 2x2 +kx + k2 +5 = 0 has two distinct roots and only
one of the roots satisfies 0 < x < 2, is
(A*) – 3 < k < 1 (B) – 3 < k < 0 (C) – 2 < k < 0 (D) – 1 < k  3
[Sol. f (0) · f (2) < 0 [12th 17-9-2006]
(k2 + 5)(k2 + 2k – 3)
k2 + 2k – 3 < 0
(k + 3)(k – 1) < 0
– 3 < k < 1
if one root is x = 2
then f (2) = 0
k2 + 2k – 3 = 0
 k = 1 or k = – 3
if k = 1
+ 2x2 – x – 6 = 0  2x2 – 4x + 3x – 6 = 0  2x(x – 2) + 3(x – 2) = 0
 x = 2 or x = – 3/2
other roots does not line in (0, 1)
|||ly when k = – 3
roots are 2, – 7  k = – 3 is also not possible]

15. Q.8113/ph-1 The base angles of a triangle are 22.5° and 112.5°.The ratio of the base to the height of the triangle
is
(A) 2 (B) 2 2 –1 (C) 2 2 (D*) 2
[Sol. fromtriangleABD cot /8 = b+x/h ....(1)
fromtriangleACD x = h cot3/8 ....(2)
(1) – (2) gives result Ans ]
SUBJECTIVE:
Q.9110/6 Let the incircle of the ABC touches its sides BC , CA&AB atA1 , B1 & C1 respectively. If 1 , 2
& 3 are the circum radii of the triangles , B1 IC1 , C1 IA1 andA1 IB1 respectively, then prove that,
2 1 2 3 = Rr2 where R is the circumradius and r is the inradius of the ABC.
[Sol. AC1 IB1 is a cyclic quadrilateral.
Hence in  B1 C1 I , AI = 21
in  B1C1I
21 = r cosec
A
2
=
r
A
sin
2
 1 =
r
A
2
2
sin
 2 1 2 3 =
r
A
3
4
2
 sin
=
r R
r
3
.  






4
2
 sin
A r
R
= r2 R ]
Q.10 Find the general solution of the equation, sin42x + cos42x = sin 2x cos 2x. [3]
[Ans. x =
8
2
n 


, n  I ]
[Hint: (sin22x + cos22x)2 – 2 sin22x cos22x = sin 2x cos 2x
1 – 2t2 = t
2t2 + t – 1 = 0
2t2 + 2t – t – 1 = 0
2t(t + 1) – (t + 1) = 0
t = – 1 (rejected), t = 1/2
4 sin 2x cos 2x =
2
1
sin 4x = 1 = sin
2

x =
8
)
1
(
4
n n 



Ans. ]
Q.11112/6 Theratiosofthelengths ofthesides BC &AC ofa triangleABCto theradius ofacircumscribed circle
areequalto2&3/2respectively. Showthat theratio ofthelengthsofthe bisectors oftheinteriorangles
B & C is ,
 
7 7 1
9 2

.

16. [Sol.
a
R
= 2 ;
b
R
=
3
2

2 R A
R
sin
= 2 ; sin B =
3
4
 sin A = 1 ; c2 = 4 R2 
9
4
2
R
A = 90º ; c =
7
2
R
Now l1 =
2a c
a c

cos
B
2
=
2a c
a c

1
2
 cos B
and l2 =
2a b
a b

cos
C
2
=
2a b
a b

1
2
 cos C

l
l
1
2
=
a b
a c


.
c
b
1
1


cos
cos
B
C
=
c a b
b a c
( )
( )


1
1


c
a
b
a
=
c
b
a b
a c


Substituting a = 2 R ; b =
3
2
R & c =
7
2
R, we get the desired result ]
Q.12s&p If the sum 2
2
2
1
1
1
1 
 + 2
2
3
1
2
1
1 
 + 2
2
4
1
3
1
1 
 + ....... + 2
2
)
2000
(
1
)
1999
(
1
1 
 equal
to n –
n
1
where n  N. Find n. [Ans. 2000]
[Sol. S = 
 


1999
1
a
2
2
)
1
a
(
1
a
1
1 = 
 




1999
1
a
2
2
2
2
2
2
)
1
a
(
a
a
)
1
a
(
)
1
a
(
a
= 
 






1999
1
a
2
2
2
2
2
2
)
1
a
(
a
a
)
1
a
2
a
(
)
1
a
2
a
(
a
= 
 




1999
1
a
2
2
2
3
4
)
1
a
(
a
1
a
2
a
3
a
2
a
= 
 


1999
1
a
2
)
1
a
(
a
1
a
a
= 









1999
1
a
2
a
a
1
1
= 1999 + 









1999
1
a 1
a
1
a
1
, a telescopic sum
= 1999 + 1 –
2000
1
= 2000 –
2000
1
 n = 2000 Ans. ]

17. Q.17/1 Solve the inequality, x
log
4
x
log 2
2
2
/
1  < 2 (4  log16x4).
( Ans. : [1, 4)  (0, 1/4] )
[Sol. log log
2
2
2
2
x x
 < 2 (4  log2 x)
Let log2 x = y  y y
2
2
 < 2 (4  y)
This will be true if ; y2 + 2 y  0 and 4  y  0
and y2 + 2y < 2(4  y)2  y2 – 18y + 32 < 0
i.e. (y  16) (y  2) > 0
Hence solution set is y  (–, –2]  [, 2)
Hence log2 x < 2  x < 4
log2 x  0  x  1. Solution is [1, 4) ;
again log2 x  2  x  1/4 and
log2 x >  x > 0
 x  (0, 1/4] ]
Q.2105/6 If p, q, r be the roots of x3 – ax2 + bx – c = 0, show that the area of the triangle whose sides are p, q
& r is
1
4
[a(4ab – a3 – 8c)]1/2.
[Sol. given p + q + r = a, pq = b, pqr = c
 = s p q r
(s )(s )(s )
   where 2s = p + q + r
 =
1
4
( )( )(q )( )
p q r p q r r p r q p
        =
1
4
2 2 2
a r p q
(a )(a )(a )
  
=  
1
4
2 4 8
3 2
a a a p q r pq qr rp pqr
       
( ) ( )a
=
1
4
2 4 8
3 3
a a ab c
   
(a ) =  
1
4
4 8
3 1/2
a ab a c
  
( ) ]
Q.3118/1 ThesumofaninfiniteGPis2&thesumoftheGPmadefromthecubesofthetermsofthisinfiniteseries
is 24.Find the series. [ REE ’89, 6 ]
[ Ans. : 3  (3/2) + (3/22)  (3/23) +..... ]
[Sol. Let the G.P. be
a , ar , ar2 , ar3 , ..................
now
a
r
1
2

 ....(1)
also
a
r
3
3
1
24

 ....(2)
from (1)
 
8
r
1
a
3
3


....(3)
(2)  (3) give
( )
1
1
3
3
3



r
r
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-35

18. ( ) ( )
1 3 1
2 2
   
r r r ( )
 
r 1
1 + r2 – 2r = 3 + 3r + 3r2
2r2 + 5r + 2 = 0
2r2 + 4r + r + 2 = 0
2r ( r+ 2) + (r + 2) = 0
r = – 2 or r = –1/ 2
r  2 ( rejected)
Hence r = – 1/ 2
   
a
a
.2
3
2 3
Hence theseries is, 3
3
2
3
2
3
2
2 3
    
.................. ]
Q.465/1 Let therebe a quotient oftwo natural numbers in which the denominator is one less than the square of
the numerator. If we add 2 to both numerator & denomenator, the quotient will exceed 1/3 & if we
subtract 3 from numerator & denomenator, the quotient will lie between 0 & 1/10. Determine the
quotient. [ REE ’90 , 6 ] [Ans:4/15]
[Sol. Let Q =
x
y
, x, y  N
given x2 – y = 1
Hence Q =
x
x2
1

(x  1, –1)
given
x
x



2
1
1
3
2 .............(1)
and 0 <
10
1
4
x
3
x
2



.............(2)
common solution to 1 and 2 is x = 4 Hence Q =
4
15
Ans. ]
Q.5102/1 The number of terms of anA.P. is even; the sum of the odd terms is 310 ; the sum of the even terms is
340 ; the last term exceeds the first by57. Find the number of terms and the series. [5]
[ Ans: 20 ; a = 4 ; d = 3 , S = 4 + 7 + 10 + .... ]
[Sol. Let the number of terms be 2m N =2m
a , a + d , a + 2d , ..... , a + (2m – 2)d , a + (2m – 1)d
a + (a + 2d) + ..... + (a + (2m – 2)d) = 310 ....(1)
a + d + a + 3d + ....... + (a + (2m – 1) d) = 340 ....(2)
a + (2m – 1)d – a = 57 (2m – 1)d = 57....(3)
2
m
(a + a + (2m – 2) d ) = 310  m (a + (m – 1) d) = 310 ....(4) from (1)
Also
2
m
[2a + 2md] = 340  m [ a + md ] = 340 ....(5)from (2)
(5) – (4)  md = 30
from (3) 2. 30 – d = 57  d = 3
 d = 3 ; m = 10
 Number of terms 2m = 20 and a = 4
 Series S = 4 + 7 + 10 + ...... ]

19. Q.6120/6 If x, y, z are perpendicular distances of the vertices of a ABC from the opposite sides and  is the
area of thetriangle, then prove that
)
C
cot
B
cot
A
(cot
1
z
1
y
1
x
1
2
2
2





 [ REE ’90 , 6 ]
[Hint: a = x(cotB + cotC) ; =
1
2
ax =
1
2
x2(cotB + cotC)
b = y(cotC + cotA) ; =
1
2
by =
1
2
y2(cotC + cotA)
c = z(cotA + cotB) ;=
1
2
cy =
1
2
z2(cotA + cotB)
& adding
1 1 1 1
2
2 2 2
x y z
  

[cotB + cotC + cotC + cotA + cotA + cotB]
=
2
2
(cot cot cot )
A B C
 

Hence proved]
Q.7117/6 Ifp,q, rbethelengths ofthebisectorsoftheangles ofatriangle ABC fromthe angular pointsA,Band
C respectively, prove that
(i)
1
2
1
2
1
2
1 1 1
p
A
q
B
r
C
a b c
cos cos cos
     and (ii)
pqr abc a b c
a b b c c a
4

 
  
( )
( )( )( )
[Sol. p =
2
2
bc
A
b c
cos

etc.
Hence
LHS of (i)
b c
bc
A
A c a
ca
a b
ab





2
2
2 2 2
cos
.cos =
1
2
2
1 1 1 1 1 1
a b c a b c
 











   
(ii) pqr bc
b c
ca
c a
ab
a b
A B C
R
abc
4
2 2 2 2 2 2








cos cos cos
=
abc
a b b c c a
R
A
( )( )( )
cos
  
 
8
2
But 4 cos
A
2
cos
B
2
cos
C
2
= sinA + sinB + sinC
Hence
pqr abc
a b b c c a
4

  
( )( )( )
2R (sinA + sinB + sinC) =
abc a b c
a b b c c a
( )
( )( )( )
 
  
]
Q.893/6 Find all the solutions of the equation )
x
1
sin(  = x
cos
which satisfythe condition x [0, 2] [x =
2
1
4
7


]
[Sol. sq. both sides, cos x = sin (1 – x) = cos 








)
x
1
(
2
x = 2n ± 








)
x
1
2

20. (+) ve sign x = 2n +
2

– 1 + x nosolution
(–) ve sign x = 2n –
2

+ 1 – x
2x = 2n –
2

+ 1
x =
2
1
4
3


; x =
2
1
4


(Both rejected)
x =
2
1
4
7


Ans. ]