CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-16
Fill in the blanks :
Q.1844/log The solution set of the equation x
x


3
1
4 = x
x


3
2
3 is ______. [Ans.2,4,11]
[Hint: Takinglogonboth the sides
4
1
x 
log | x – 3 | =
3
2
x 
log | x – 3 |
 log | x – 3 | 










 






 
3
2
x
4
1
x
= 0  log | x – 3 | = 0 or 










 






 
3
2
x
4
1
x
= 0
 x = 4 , 2 or x = 11 ]
Q.2846/log If (a2 + b2)3 = (a3 + b3)2 and ab  0 then the numerical value of
a
b
b
a
 is equal to ______
[Hint: a6 + b6 + 3a2b2 (a2 + b2) = a6 + b6 + 2a3b3 [Ans. 2/3 ]

3
2
ab
b
a 2
2



3
2
3
1
3
1
a
b
b
a



 Ans ]
Q.347/log
3
5
log
1
)
1
.
0
(
log
1
10
7
5

 simplifies to ________. [Ans. 2]
[Hint: 3
10
7
5
log
10
/
1
log
1
5

 = 3
1
1
7  = 3
8 = 2 Ans ]
Select the correct alternative : (Only one is correct)
Q.4 The tangents oftwo acute angles are 3 and 2. Thesine of twice theirdifference is :
(A) 7/24 (B) 7/48 (C) 7/50 (D*) 7/25
[Hint: tan  = 3; tan  = 2.
Now sin 2()
= sin2 cos2 cos2 sin2 =
4
1
4
1
.
9
1
3
.
2




4
1
2
.
2
.
9
1
9
1



=
25
7
 D]
Q.5 log3
1
1
3






 + log3
1
1
4






 + log3
1
1
5






 + .............. + log3
1
1
242






 when simplified has the value
equal to
(A) 1 (B) 3 (C*) 4 (D) 5
*Q.6 Whichofthesestatements is false?
(A)Arectangleis sometimes arhombus.
(B)Arhombusis always a parallelogram.
(C*) Thedigonals of a parallelogram always bisect theangles at the vertices.
(D) The diagonals of a rectanglearealways congurent.
Q.7104/QE If f (x) = x2 + 6x + c, where 'c' is an integer, then f (0) + f (–1) is
(A) aneveninteger (B) anodd integer always divisibleby3
(C) an odd integer not divisibleby3 (D*) an odd integer may or not be divisible by3
[Sol. f (0) + f (–1) = c + c – 5 = 2c – 5
Now 2c is always even so 2c – 5 is odd. mayor not be divisible by 3.]

*Q.85/ph-1 Theminimumvalueofthefunction
f (x) = (3sin x – 4 cos x – 10)(3 sin x + 4 cos x – 10), is
(A*) 49 (B)
2
2
60
195
(C) 84 (D) 48
[Sol. f (x) = 9 sin2x – 16 cos2x – 10(3 sin x – 4 cos x) – 10(3 sin x + 4 cos x) + 100
= 25 sin2x – 60 sin x + 84
= (5 sinx – 6)2 + 48
 f (x)min occurs when sin x = 1
minimumvalue=49 ]
Select the correct alternative : (More than one are correct)
Q.9154/log Which of the following are correct ?
(A*) log3 19 . log1/7 3 . log4 1/7 > 2 (B*) log5 (1/23) lies between –2 & – 1
(C*) log10 cosec(160º) is positive (D) log sin . log sin
5
5
5
5







 simplifiestoanirrationalnumber
[Hint: (A) log419 ( since log416 = 2 ) hence log419 > 2
(B) log51/5 = – 1 and log51/25 = – 2 so log51/23 loes between – 1 and – 2
(C) since sin1600 < 1 (cosec1600 > 1)
Hence numberand base on sameside of unity
(D) 2log5sin(/5) . 2logsin(/5)5 = 4 ]
Q.1076/ph-1 It is known that sin  =
5
4
and 0 <  <  then the value of
 









sin
)
cos(
6
cos
2
)
sin(
3
is:
(A*) independent of  for all  in (0, /2) (B*)
3
5
for tan  > 0
(C*)
3 7 24
15
( cot )
 
for tan  < 0 (D) none
[Sol. E =








sin
3
)
cos(
4
)
sin(
3
= 3(sin  · cos  + cos  · sin ) – 4(cos  · cos  + sin  · sin )
= 3 sin  ·
5
3
+ 3 cos  ·
5
4
– 4 cos  ·
5
3
+ 4 sin  ·
5
4
=
5
3
for 0 <  < /2
& to
3 7 24
15
( cot )
 
for /2 <  < ]
Subjective :
Q.1148/06 If cos ( y – z ) + cos ( z – x ) + cos ( x – y ) = –
2
3
,
prove that cosx + cosy + cosz = 0 = sinx + siny + sinz
[Sol. Given 2[cos (x – y) + cos (y – z) + cos (z – x) ] = – 3
or 2[cos (x – y) + cos (y – z) + cos (z – x) + (sin2x + cos2x) + (sin2y + cos2y) + (sin2z + cos2z) ] = 0
or  x
cos2
+ 2 y
cos
x
cos
 +  x
sin2
+ 2 y
sin
x
sin
 = 0
( cosx + cosy + cosz )2 + ( sinx + siny + sinz)2 = 0   x
cos = 0 =  x
sin ]

Q.1251/06 If
1
a
x
sin
=
3
a
x
3
sin
=
5
a
x
5
sin
then show that
3
5
3
1
a
a
a
2
a 

=
1
1
3
a
a
3
a 
.
[Hint: both ratios simplifies to 2(cos 2x  1) ]
[Alternate:
5
3
1 a
x
5
sin
a
x
3
sin
a
x
sin


3
5
3
1
a
a
a
2
a 

=
1
1
3
a
a
3
a 
or 3
a
a
2
a
a
a
a
1
3
3
5
3
1



 or 1
a
a
a
a
a
a
1
3
3
5
3
1


 = 0
0
1
x
sin
x
3
sin
x
3
sin
x
5
sin
x
3
sin
x
sin




x
3
sin
x
5
sin
x
sin 
=
x
sin
x
sin
x
3
sin 
x
3
sin
x
2
cos
x
3
sin
2
=
x
sin
x
sin
.
x
2
cos
2
= 2 cos2x = 2 cos2x ]
Q.135/01 Solve the following equation for x :
6
5
. aA – 3B = 9C
whereA= loga x. log10 a . loga 5, B = log10(x/10) & C = log100 x + log4 2. [Ans. x = 100]
[ Sol. 2
log
x
log
10
x
log
5
log
.
a
log
.
x
log 4
100
10
a
10
a
9
3
a
.
5
6 









 


 2
1
x
log
2
1
x
log
5
log
.
x
log
10
10
a
10
3
3
a
.
5
6
x
log
1
x
log
x
log
x
log 10
10
10
10 3
.
3
3
3
3
1
5
.
5
6




x
log
x
log 10
10
3
3
10
5
.
5
6

2
x
log
3
5
3
5 10













Hence log10 2
x   x = 100 ]

CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-17
Select the correct alternative : (Only one is correct)
*Q.19/ph-1 If the expression, 2 cos10° + sin 100° + sin 1000° + sin 10000°
issimplified,thenitsimplifiesto
(A*) cos 10° (B) 3 cos 10° (C) 4 cos 10° (D) 5 cos 10°
Q.2 If x2 + 2
x
1
= 7 then the value of 3
3
x
1
x  equals (x > 0)
(A*) 18 (B) 21 (C) 24 (D) 27
[Hint:
2
x
1
x 





 – 2 = 7 
2
x
1
x 





 = 9  x +
x
1
= 3,  – 3 (why?) x > 0
3
x
1
x 





 = 27  x3 + 3
x
1
+ 3 






x
1
x = 27  x3 + 3
x
1
= 27 – 9 = 18 ]
*Q.380(i)/aucNumber of positive integers x for which f (x) = x3 – 8x2 + 20x – 13, is a prime number, is
(A) 1 (B) 2 (C*) 3 (D) 4
[Hint: f (x) = (x – 1)(x2 – 7x + 13)
for f (x) to be prime at least one of the factors must be prime.
Hence x – 1 = 1  x = 2 or
x2 – 7x + 13 = 1  x2 – 7x + 12 = 0  x = 3 or 4
 x = 2, 3, 4  (C) ]
*Q.47/ph-1 In the inequality below the value of the angle is expressed in radian measure. Which one of the
inequalitiesbelowistrue?
(A) sin 1 < sin 2 < sin 3 (B) sin 3 < sin 2 < sin 1
(C) sin 2 < sin 1 < sin 3 (D*) sin 3 < sin 1 < sin 2
[Hint: sin 1 – sin 2
= – 2 cos
2
3
· sin
2
1
< 0
 sin 1 < sin 2
sin 1 – sin 3
= – 2 cos 2 sin 1 > 0  sin 1 > sin 3 ]
Q.5 Aparticleismovingalongastraightlineso thatitsvelocityattimet0isv(t)=3t2.Atwhattimetduring
the interval from t = 0 to t = 9 is its velocitythe same as the average velocityoverthe entire interval?
(A) 3 (B) 4.5 (C*) 3(3)1/2 (D) 9
[Sol. 3t2 =
9
1

9
0
2
dt
t
3 =
9
1
 9
0
3
t = 81
t2 = 27  t = 3
3  (C) ]

*Q.68/log If log10sinx + log10cosx = – 1 and log10(sin x + cos x) =
2
1
)
n
(log10 
then the value of 'n' is
(A) 24 (B) 36 (C) 20 (D*) 12
[Sol. Given log10






2
x
2
sin
= – 1 
2
x
2
sin
=
10
1
 sin 2x =
5
1
....(1)
Also log10(sin x + cos x) =
2
10
n
log10 





 log10(sin x + cos x)2 = log10






10
n
 1 + sin 2x =
10
n
 1 +
5
1
=
10
n

5
6
=
10
n
 n = 12  (D) ]
*Q.715/ph-1 The value of x satisfying the equation, x = x
2
2
2 

 is
(A) 2 cos 10° (B) 2 cos 20° (C*) 2 cos 40° (D) 2 cos 80°
[Sol. Note that x  [–2, 2]
Let x = 2 cos  where   [0, ]
x = 


 cos
2
2
2
2
2 cos  =
2
cos
2
2
2


 = 




 


2
cos
1
2
2 =
4
sin
2
2

 = 




 



4
2
cos
2
2
= 












 



4
2
cos
1
2
2 cos  = 2 cos 




 


8
4
;   =
8
4




8
9
=
4

  =
9
2
Hence x = 2 cos
9
2
= 2 cos 40°  (C) Ans. ]
*Q.859/ph-1 If  is eliminated from the equations x = a cos( – ) and y= b cos ( – ) then
)
cos(
ab
xy
2
b
y
a
x
2
2
2
2




 is equal to
(A) sec2 (  – ) (B) cosec2 ( – ) (C) cos2 (  – ) (D*) sin2 ( – )
Sol. ( – ) = ( – ) – ( – )
cos( – ) = cos ( – ) cos ( – ) + sin ( – ) sin( – )
cos( – ) = 2
2
2
2
b
y
1
.
a
x
1
a
x
.
b
y



 



























 2
2
2
2
2
b
y
1
a
x
1
)
cos(
ab
xy
 )
cos(
ab
xy
2
)
(
cos
b
a
y
x 2
2
2
2
2







 = 2
2
2
2
2
2
2
2
b
a
y
x
a
x
b
y
1 



 )
cos(
ab
xy
2
b
y
a
x
2
2
2
2




 = sin2 ( – ) ]
Q.985/QE The quadratic equation X2 – 9X + 3 = 0 has roots r and s. If X2 + bX + c = 0 has roots r2 and s2, then
(b, c) is
(A) (75, 9) (B*) (–75, 9) (C) (–87, 4) (D) (–87, 9)
*Q.10112/ph-1 The sum











 134
sin
133
sin
1
......
50
sin
49
sin
1
48
sin
47
sin
1
46
sin
45
sin
1
is equal to
(A) sec (1)° (B*) cosec (1)° (C) cot (1)° (D) none
[Sol. T1 =

1
sin
1











46
sin
45
sin
)
45
46
sin(
=

1
sin
1
[cot45° – cot46°]
|||ly T2 =

1
sin
1











47
sin
48
sin
)
47
48
sin(
=

1
sin
1
[cot47° – cot48°]

Tl =

1
sin
1











134
sin
133
sin
)
134
133
sin(
=

1
sin
1
[cot133° – cot134°]
Onadding


l
1
r
r
T =

1
sin
1
[{cot45° + cot47° + cot49° + ... + cot133°}– {cot46° + cot48° + cot50° + ... + cot134°}]
= cosec1° [all terms cancelledexcept cot45° remains] ]
SUBJECTIVE
*Q.1116/1 If log10(15) = a and log20(50) = b then find the value of log9(40). [Ans.
2
b
4
a
2
ab
2
b
5




]
[Sol. Since log20(50) = b,it follows that
20
log
2
100
log 





= b 
2
log
1
2
log
2


= b  log102 =
b
1
b
2


....(1)
Since log10(15) = a, we have log 3 + log 5 = a  log 3 + log 





2
10
= a
 log 3 – log 2 = a – 1  log 3 = log 2 + a – 1. Taking (1) into account, we find that
log 3 =
b
1
b
2


+ a – 1. ....(2)
Taking (1)and (2) into consideration,we obtained
log9(40) =
3
log
2
2
log
2
1
=
2
b
4
a
2
ab
2
b
5




Ans ]

*Q.12 Provetheinequality,
sin x +
2
1
sin 2x +
3
1
sin 3x > 0 for 0 < x < 180° [6]
[Sol. LHS =
6
)
x
sin
4
x
sin
3
(
2
x
cos
x
sin
6
x
sin
6 3



=
6
]
x
sin
4
3
x
cos
3
3
[
x
sin
2 2



=
3
]
x
sin
4
3
x
cos
3
3
[
x
sin 2



=
3
]
x
cos
4
x
cos
3
2
[
x
sin 2


=
3
]
x
cos
3
)
x
cos
1
(
)
x
cos
1
[(
x
sin 2
2




which is + ve for 0° < x < 180° ]
Q.13 Let u = 10x3 – 13x2 + 7x and v = 11x3 – 15x2 – 3.
Findthe integralvaluesofxsatisfyingtheinequality,
dx
dv
dx
du
 . [3]
[Sol.
dx
du
= 30x2 – 26x + 7
dx
dv
= 33x2 – 30x

dx
dv
dx
du

30x2 – 26x + 7 > 33x2 – 30x
3x2 – 4x – 7 < 0
(x + 1)(3x – 7) < 0
integral solution set is x {0, 1, 2} ]

CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-18
Select the correct alternative : (Only one is correct)
Q.11/ph-1 a, b,c arethesidesof a triangleABC which is right angled atC, then theminimumvalueof
2
b
c
a
c







is
(A) 0 (B) 4 (C) 6 (D*) 8
[Hint: a = c sin 
b = c cos 
E =
2
b
c
a
c






 =
2
cos
1
sin
1









=



2
sin
)
2
sin
1
(
4
2
= 







 2
sin
1
2
sin
1
4 2
where 0 <  <
2

 Emin = 8 when 2 = 90°   = 45°]
Q.2 Thevalueof m for thezerosof thepolynomial P(x) = 2x2 – mx – 8 differ by(m – 1) is
(A) 4, –
3
10
(B) – 6,
3
10
(C) 6,
3
10
(D*) 6, –
3
10
[Hint: ( – )2 = (m – 1)2
( + )2 – 4 = (m – 1)2 etc.]
Q.322/ph-2 Number ofsolutions of the equation
2 sin3x + 6 sin2x – sin x – 3 = 0 in (0, 2), are
(A) 6 (B*) 4 (C) 3 (D) 2
[Hint: 2 sin2x (sin x + 3) – (sin x + 3) = 0
(2 sin2x – 1)(sin x + 3) = 0
 sin2x =
2
1
Hence 4 solutions  (B) ]
*Q.4 Minimum vertical distance between the graphs of y= 2 + sinx and y=cos x is
(A) 2 (B) 1 (C) 2 (D*) 2 – 2
[Hint: dmin = min(2 + sin x – cos x) = min[2 + 2 sin 




 

4
x ] = 2 – 2 ]
*Q.541/ph-3 Each sideof triangleABC is divided into 3 equal parts.The ratio of
the area of hexagon UVWXYZ to the area of triangleABC is
(A)
9
5
(B*)
3
2
(C)
2
1
(D)
4
3

[Hint: area of hexagon =  –
9
1







 B
sin
ca
2
1
A
sin
bc
2
1
C
sin
ab
2
1
=  –
9
1
[3] =
3
2
 (where  is the area of the triangleABC)

triangle
of
area
heaxgon
of
area
=
3
2
]
Q.688/QE If a, b, c are real numbers such that a2 + 2b = 7, b2 + 4c = – 7 and c2 + 6a = – 14 then the value of
a2 + b2 + c2 is
(A*) 14 (B) 21 (C) 28 (D) 35
[Hint: a2 + 2b = 7
b2 + 4c = – 7
c2 + 6a = – 14
——————
a2 + b2 + c2 + 2b + 4c + 6a = – 14
(a + 3)2 + (b + 1)2 + (c + 2)2 – 14 = – 14
 a = – 3 ; b = – 1 and c = – 2
 a2 + b2 + c2 = 14 ]
*Q.719/ph-1 Find the smallest natural 'n' such that tan(107n)° =






96
sin
96
cos
96
sin
96
cos
.
(A) n = 2 (B*) n = 3 (C) n = 4 (D) n = 5
[Hint:






6
cos
96
cos
6
cos
96
cos
= –




45
sin
51
sin
2
45
cos
51
cos
2
= – cot51°
= tan(90° + 51°) = tan141° = tan(180° + 141°) = tan(321°) = tan(3·107°)  n = 3 ]
SUBJECTIVE
*Q.8 IfA+ B + C =
2

then prove that
 A
sin2
+ 2 A
sin = 1. [3]
[Sol. LHS sin2A + sin2B + sin2C + 2 A
sin
1 – (cos2A – sin2B) + sin2C + 2 A
sin
1 – cos(A + B) cos(A – B) + sin2C + 2 A
sin (A + B = C
2


; cos(A + B) = sinC)
1 – sinC [cos(A – B) – sinC] + 2 A
sin
1 – sinC [cos(A – B) – cos(A + B)] + 2 A
sin
1 – sinC[2sinA sinB] + 2 A
sin
1 – 2 A
sin + 2 A
sin = 1 Hence proved. ]

Q.9 The position vector of a point P in space is given by
k̂
t
cos
4
j
ˆ
t
sin
5
î
t
cos
3
r 



(a) Show that its speed is constant.
(b) Show that its velocityvector v

, is perpendicular to r

. [3]
[Sol.(a)
dt
r
d
v


 = k̂
t
sin
4
j
ˆ
t
cos
5
î
t
sin
3 


speed = |
v
|

= t
sin
16
t
cos
25
t
sin
9 2
2
2

 = 5 unit Ans.
(b) r
·
v


= – 9 (sin t)(cos t) + 25(sin t)(cos t) – 16(sin t)(cos t) = 0
Hence v

, is perpendicular to r

]
Q.10 If x is eliminated from the equation, sin(a + x) = 2b and sin(a – x) = 2c, then find the eliminant.
[Ans.
a
sin
)
c
b
(
2
2

+
a
cos
)
c
b
(
2
2

= 1] [3]
[Sol. adding sin(a + x) + sin(a – x) = 2(b + c)
2 sin a cos x = 2(b + c)  cos x =
a
sin
c
b 
....(1)
sub sin(a + x) – sin(a – x) = 2(b – c)
2 cos a sin x = 2(b – c)  sin x =
a
cos
c
b 
....(2)
squaringand addingboth equation (1) and (2), we get
a
sin
)
c
b
(
2
2

+
a
cos
)
c
b
(
2
2

= 1 Ans. ]
*Q.1133/1 If p, q are the roots of the quadratic equation x2 + 2bx + c = 0, prove that
2 log 
q
y
p
y 

 = log 2 + log 




 


 c
by
2
y
b
y 2
[Sol. x2 + 2bx + c = 0 ; c
pq
b
2
q
p







....(1)
TPT 2 log 
q
y
p
y 

 = log 2 + log 




 


 c
by
2
y
b
y 2
LHS =  
q
y
p
y
2
q
y
p
y
log 





= 




 




 pq
y
)
q
p
(
y
2
)
q
p
(
y
2
log 2
= 




 


 c
by
2
y
2
b
2
y
2
log 2
= log 2 + 




 


 c
by
2
y
b
y
log 2
= RHS ]

Q.1294/6 Solve for '' satisfying cos() · cos () = 1. [Ans.  = 0 ]
[Sol. cos() · cos () = 1
either cos() = 1 & cos () = 1 ....(1)
or cos() = –1 & cos () = –1 ....(2)
Now for (1) to hold, from cos() = 1   = 2m
& cos () = 1   = 2k
which is onlypossible when =0
Equation (2) is not possible for anyas for cos() = – 1,  should be odd multiple of   irrational
and for cos() = – 1   should be odd integer  rational
boththe conditions can not be satisfied
  =0 is theonlysolution. ]
Q.13 Find thevalue of kforwhich thegraphof thequadratic polynomial
P (x) = x2 + (2x + 3)k + 4(x + 2) + 3k – 5 intersects the axis of x at two distinct points. [3]
[Ans. R – {1}]
[Sol. P (x) = x2 + 2(k + 2)x + 3(2k + 1)
for distinct zeros, D > 0
4(k + 2)2 – 4 · 3(2K + 1) > 0
(k2 + 4k + 4) – (6k + 3) > 0
k2 – 2k + 1 > 0
(k – 1)2 > 0 which is always true except at k = 1
whenit touchestheaxis of x.
Hence k = R – {1}]