CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-7
Fill in the blanks :
Q.1866/log If x = 7 5 2
3  
1
7 5 2
3

, then the value of x3 + 3x  14 is equal to ______. [Ans. 0]
[Hint : a =  
7 5 2
1 3

/
, b =
  3
1
2
5
7
1

, then x = a  b , x3 = 14  3x ]
Select the correct alternatives : (More than one are correct)
Q.2509/log If p, q  N satisfy the equation  
x x
x
x
 then p & q are :
(A*) relativelyprime (B) twin prime
(C*) coprime (D*) if logq
p is defined then logp
q is not & vice versa
[Sol.  
x x
x
x
 p, q  N
x log x = x log x
log x 






2
x
x = 0  log x = 0 or 






2
x
x = 0
x = 1 or x = 0 or x = 4 ; x = 0 (rejected) ]
Q.3505/log The expression, logp logp
...... p
p
p
p
p
n radical sign
 

 


where p  2, p  N, when simplified is :
(A*) independent of p, but dependent on n (B) independent of n, but dependent on p
(C) dependent on both p & n (D*) negative .
[Hint: ...... p
p
p
p
= ppn
1
. Therefore logp logp
ppn
1
= logp
1
pn





 . logp (pn) =  n ]
Q.4510/log Which ofthefollowingwhensimplified,reduces tounity?
(A*) log105 . log1020 + 2
log2
10 (B*)
2 2 3
48 4
log log
log log


(C*)  log5 log3 9
5 (D)
1
6
log 3
2
64
27






[Hint: D = – 1 ]
*Q.5508/log The number N =
 
1 2 2
1 2
2
3
3
6
2
2



log
log
log when simplified reduces to :
(A) a prime number (B) anirrational number
(C*) a real which is less than log3 (D*) a real which is greater than log7
6
[Hint:
1 2 2
1 2
2
1 2
1 2
1 2
1
3
3
2
3
2
3
2
3
2
3
2








log
( log )
(log )
( log )
( log )
( log )
]
Subjective :
Q.69/06 If tanA& tan B are the roots of the quadratic equation, ax2 + bx + c = 0 then evaluate
a sin2 (A + B) + b sin (A + B). cos (A + B) + c cos2 (A + B). [Ans:c]
[Sol. tanA + tan B = 
b
a
; tanA. tan B =
c
a
; tan (A + B) =


b
a
c
a
1
=
b
c a


Now E = cos2 (A + B) [a tan2 (A + B) + b tan (A + B) + c ]
=
1
1
2
2


b
c a
( )
a b
c a
b
c a
c
2
2
2
( )









 =
( )
( )
c a
b c a

 
2
2 2
b
c a
a
c a
c
2
1
 






 






=
( )
( )
c a
b c a

 
2
2 2
b c
c a
c
2
2
( )







 E = c ]
Q.718/06 If cos + cos = a and sin + sin= b then prove that, cos2 + cos2 =
  
 
a b a b
a b
2 2 2 2
2 2
2
  

[Sol. squaringand adding, cos(–) =
a b
2 2
2
2
 
....(1)
using C – D relations & dividing, tan
 


2
b
a
 cos( + ) =
1
1
2
2
2



b
a
b
a
=
a b
a b
2 2
2 2


....(2)
now cos2 + cos2 = 2cos( + ) cos ( – ) ....(3)
use (1) & (2) in (3) to get the result ]
Q.8 Establishtricotomyineachofthisfollowingpairsofnumbers
(i) 2
log
3
log 4
27 2
and
3
(ii) )
25
/
1
(
log
and
5
log 16
/
1
4
(iii) 4 and 81
log
10
log 10
3  (iv) )
5
/
1
(
log
and
)
7
/
1
(
log 7
/
1
5
/
1
[(i)(Hint: >); (ii) (Hint:=); (iii) (Hint:<); (iv) (Hint: >) ]
[Hint: log log
3 10
10 4 3 4
4
4
    
y
y
= y
y








 
2
0
2
]
Q.96(sub)/log Given, log712 = a & log1224 = b. Show that, log54168 =
1
8 5


ab
a b
( )
.
[Sol. log712 = a  2 log72 + log73 = a ;
log1224 = b  log724 = ab 3 log72 + log73 = ab
Now log54168 =
log
log
7
7
168
54
=
3 2 3 1
3 3 2
7 7
7 7
log log
log log
 

.
Substituting the value of, log72 & log73 we get the result ]
Q.10 If
b
a
c
log
a
c
b
log
c
b
a
log





, show that aa . bb . cc = 1.
Q.11 Prove that the expression, cos2
8

+ cos2
8
3
+ cos2
8
5
+ cos2
8
7
is not irrational.
[Sol. cos2
8

+ cos2
8
3
+ cos2
8
5
+ cos2
8
7
cos2
8

+cos2 




 


8
8
4
+ cos2 




 


8
8
4
+ cos2 




 


8
 




 


8
sin
8
cos 2
2
+ 




 


8
cos
8
sin 2
2
= 1 + 1 = 2 Ans. ]

CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-8
Fill in the blanks :
Q.1832/log If logx log18  
2 8
 =
1
3
. Then the value of 1000 x is equal to _____. [Ans. 125]
[Hint: logx log18  
2
2
2  = 1/3
logx log18 18 = 1/3  x = 1/8  1000x = 125 ]
Q.2864/log The solution set of the equation
27
log
2
log
x
log 3
9
9 2
x
.
6
4 
 = 0 is ______ .
[Ans {9, 81}and note that x
log9
2 =
2
log9
x ]
[Hint: put 2 9
log x
= y we get y2 – 6y + 8 = 0  y = or 2
 2 9
log x
= 22  log9x = 2  x = 81
or 2 9
log x
= 21  log9x = 1 x = 9 ]
Select the correct alternative : (Only one is correct)
Q.3502/log Which oneof the followingwhensimplified does not reduceto an integer?
(A)
2 6
12 3
log
log log

(B) 243
log
32
log
3
2
(C*)
log log
log
5 5
5
16 4
128

(D) log1/4
1
16
2







[ Hint : A=1, B = 2, C = 2/7, D = – 4]
Q.4 Let u = (log2x)2 – 6 log2x + 12 where x is a real number. Then the equation xu = 256 has
(A) no solution for x (B*) exactlyonesolution for x
(C)exactlytwodistinct solutions for x (D)exactlythreedistinct solutions forx
[Hint:
12
x
log
6
)
x
(log 2
2
2
x


= 256 ; (t2 – 6t + 12) t = 8 where t = log2x ;
 (t – 2)3 = 0  t = 2  x = 4 ]
Q.55/log The equation, log2 (2x2) + log2 x .  
1
x
log
log 2
x
x 
+
1
2
log4
2 (x4) +  
x
log
log
3 2
2
/
1
2
= 1 has:
(A)exactlyonereal solution (B)tworeal solutions
(C)3real solutions (D*)no solution.
[Hint: y3 + 3y2 + 3y + 1 = 1 where y = log2 x  y(y2+3y+3) = 0
 x = 0 (rejected) ; y2 + 3y + 3 (complex roots) ]
Select the correct alternative : (More than one are correct)
Q.6513/log The equation 2
8
2
8
)
x
(log
)
x
8
(
log
= 3 has :
(A)nointegralsolution (B*)onenatural solution
(C*) tworeal solutions (D)oneirrationalsolution
[Hint: 3
y
y
2
1
2


(log8x = y)
3y2 + 2y – 1 = 0  3y2 + 3y – y – 1 = 0  3y (y + 1) – 1(y + 1)
log8x = y = 1/3, – 1  x = 2 , 1/8 ]
Subjective
Q.750/06 Find the exact value of tan2
16

+ tan2
16
3
+ tan2
16
5
+ tan2
16
7
[Ans : 28]

[Sol. Let 16
 = 
tan2 + tan23 + tan25 + tan27
= (tan2  + cot2  ) + (tan23 + cot23) [ Note that tan7 = tan(8 – ) = cot and
tan5 = tan(8 –3) = cot3 ]
= (cot – tan)2 + (cot3 – tan3)2 + 4
= 4 [cot22 + cot26] + 4
= 4 [ cot22 + tan22 ] + 4
= 4 [ (cot2 – tan2)2 + 2] + 4 = 4 (cot2 – tan2)2 + 12
= 4 . 4 cot2 4 + 12 = 16 × 12 + 12 = 28 Ans ]
Q.816/06 In anytriangle, if (sinA+ sin B + sin C) (sinA+ sin B  sin C) = 3 sinAsin B, find the angle C.
[Sol. (sinA+ sin B)2  sin2 C = 3 sinAsin B [Ans. C = 60º]
sin2 A sin2 C + sin2 B = sinAsin B
sin (A+ C) sin (A C) + sin2 B = sinAsin B
sin B [ sin (A C) + sin (A+ C)]= sinAsin B (Using sin(A+C)= sinB)
2 sinAcos C = sinA (sinB  0)
cos C = 1/2  C = 60º ]
Q.9 Whichissmaller?
80
1
log
3
1 or 





 2
15
1
log
2
1
[Hint:
80
1
log
3
1 <
81
1
log
3
1 = 4 or
80
1
log
3
1 < 4 ....(1)
|| ly
2
15
1
log
2
1

>
16
1
log
2
1 = 4; hence
2
15
1
log
2
1

> 4 ....(2)
From (1) and (2)
2
15
1
log
80
1
log
2
1
3
1

 ]
Q.10 Find the possible value(s) of










cot
ec
cos
sec
tan
cos
sin
if tan = –
3
4
. [2.5]
[Ans. in 2nd quadrant
5
23
, in 4th quadrant
35
34
]
[Sol. If  lies in the 2nd quadrant then
tan = –
3
4
 sin =
5
4
; cos = –
5
3
hence E =
4
3
4
5
3
5
3
4
5
3
5
4





= 




 

15
20
9
12








 9
15
20
12
=
4
·
15
12
·
23
=
5
23
Ans.
If  lies in the 4th quadrant
tan = –
3
4
 sin = –
5
4
; cos =
5
3
E =
4
3
4
5
3
5
3
4
5
3
5
4





= 










9
15
20
20
9
12






15
12
=
15
·
14
12
·
17
=
35
34
Ans. ]

Q.11 Solve theequation )
1
x
(
log
)
5
.
0
x
(
log 5
.
0
x
1
x 

 
 . [Ans. x = 1] [5]
[Sol. )
1
x
(
log
)
5
.
0
x
(
log
2
2


= )
5
.
0
x
(
log
)
1
x
(
log
2
2


 [log2(x + 1)]2 = [log2(x – 0.5)2]]
log2(x + 1) = log2(x – 0.5) or – log2(x – 0.5)
if log2(x + 1) = log2(x – 0.5)
x + 1 = x – 0.5  nosolution
if log2(x + 1) = log(x – 0.5)–1
x + 1 = )
2
1
(
x
1
 =
1
x
2
2

(x + 1) (2x – 1) = 2
2x2 + x – 3 = 0
2x2 + 3x – 2x – 3 = 0
(x – 1)(2x + 3) = 0  x = 1 Ans. x = – 3/2 (rejected) ]

Q.9 Compute the value of 81 27 3
1
3 36
4
9
5 9 7
log log log
 
[Hint: T1 = 625; T2 = 216; T3 = 49; Ans : 890 ]
[Sol. 5
log
4
3
log
1
3
5 3
81 
= 54 = 625 ....(1)
36
log
3
36
log 2
3
9 3
27  = 2
3
2
)
36
(
log
6
3
2
/
3
3


= 216 ....(2)
7
log
2
7
log
4
9
log
4
3
9
7 3
3
3 

= 49 ....(3)
625 + 216 + 49 = 890 ]
Q.2819/log The expression log .5
0
2
8 has the value equal to ______. [Ans. 3]
[Hint: Note x2 = x, if x is negative ]

Q.1 Simplifywheneverdefined
))
180
(
cos
)
sin(
)
90
sin(
)
540
(
sin
)
270
sin(
)
720
(
cos
)
270
sin(
2
3
3






















+
)
450
(
ec
cos
)
270
cot(
2






where  is taken suchthat thedenominator appearing inanyfraction in the expression does not vanish.
[Ans. 1 ] [2.5]
[Sol. E =











2
3
3
cos
sin
cos
·sin
cos
)
)(cos
cos
(
+


2
sec
tan
=








2
3
4
cos
cos
sin
sin
cos
cos
+


sec
sin
=
)
cos
(sin
cos
)
cos
(sin
cos 3
3








+ sin cos
= (1 – sin cos) + sin cos = 1 ]
Q.2 Given x2 + 4y2 =12xy, where x>0, y>0 then prove that, log(x + 2y) – 2log2 =
2
1
(log x + log y). [3]
[Sol. x2 + 4y2 + 4xy = 12xy+ 4xy
= 16 xy
(x + 2y)2 = 16xy
2 log(x + 2y) = log216 + log x + log y = 4 log 2 + log x + log y
log (x + 2y) = 2 log 2 +
2
1
(log x + log y) ]
Q.3 Solve the equation, 2
x
log
)
x
log( 
 . [Ans. x = – 1 and x = – 10] [3]
[Sol. Since the equation can be satisfied onlyfor x < 0 hence |
x
|
x2
 = – x
 )
x
log(
)
x
log( 

  log(–x) =[log(–x)]2
log(–x)[1 –log(–x)] = 0
if log(–x) = 0  – x = 1  x = – 1
if log10(–x) = 1
– x = 10  x = – 10 ]
Q.4 Let fn(x) = sinnx + cosnx. Find the number of values of x in [0, ] for which the relation
6f4(x) – 4f6(x) = 2f2(x) holds valid. [Ans. infinitelymany] [3]
[Hint: fn(x) = (sin x)n + (cosx)n (given)
now 6f4(x) – 4f6(x) = 2f2(x)
6[sin4x + cos4x] – 4[sin6x + cos6x] = 2(sin2x + cos2x)
6[1 – 2sin2x cos4x] – 4[1 – 3sin2x cos2x] = 2
2 = 2  it becomesanidentity   solutions ]
Q.5 If 2cos =x +
x
1
, find thevalues of the followingin terms of cosine of the multipleangle of .
(i) x2 + 2
x
1
; (ii) x3 + 3
x
1
and (ii) x4 + 4
x
1
Hence deduce the value of xn + n
x
1
, n  N. [3]
[Ans. (i) 2 cos 2, (ii) 2 cos 3, (iii) 2 cos 4 and 2 cos n ]
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-9

[Sol. (i) LHS = x2 + 2
x
1
=
2
x
1
x 





 – 2 = 4cos2– 2 = 2(1 + cos2) – 2 = 2 cos2
(ii) LHS = x3 + 3
x
1
=
3
x
1
x 





 – 3 






x
1
x = 8cos3 – 6cos = 2(4cos3 – 3 cos) = 2 cos3
(iii) LHS = x4 + 4
x
1
=
2
2
2
x
1
x 





 – 2 = 4cos22 – 2 [usingtheresult of(i)]
= 2(1 + cos4) – 2 = 2 cos4
Since x2 + 2
x
1
= 2 cos2 ; x3 + 3
x
1
= 2 cos3 ; x4 + 4
x
1
= 2 cos4
Hence xn + n
x
1
= 2 cos n ] [11th PQRS, 29-5-2005]
Q.6 If a  b > 1, then find the largest possible value of the expression 





b
a
loga
+ 





a
b
logb . [3]
[Ans. 0 ]
[Sol. Given E = 





b
a
loga
+ 





a
b
logb = logaa – logab + logbb – logba
E = 2 – (logba + logab)
now Ewillbemaximumif logba+logabisminimum
but logba + logab =  2
a
b
b
log
a
log  + 2
Henceitsminimumvalueis2
 Emax = 2 – 2 = 0 Ans. ]
Q.7 Prove that solutionofthe equation,
3
x
27
x
9 4
4
1
log
1
2
1
2
log
2































is an irrational number.
.
[Ans. – log23] [5]
[Sol. 





















 4
4
1
log
3
1
2
1
2
log
2
x
27
x
9  





1
2
1
·
2
log x
3
= 





 4
4
1
log x
3
 1
2
2
x
 = 4
4
1
x
 ; let x
2
1
= y
2y – 1 = y2 – 4
y2 – 2y – 3 = 0  (y – 3)(y + 1) = 0
 y = 3 or y = – 1 (rejected)
x
2
1
= 3  2–x = 3
– x = log23  x = – log23
TPT log23isirrational; let log23 =
q
p
where p  N, q  N, p > q
2p/q = 3; 2p = 3q
LHS is even but RHS is odd. Which is not possible. Hence log23 can not be rational.
Hence x = – log23 is irrational. ] Q.8843/log Solution set of the equation

1 1
6
 log x + 2 = 3 1
6
 log x is _______. [Ans.
1
6
, 





 ]
[Sol. Let log1/6x = a
| 1 – a | + 2 = | 3 – a |
I When a < 1
1 – a + 2 = 3 – a log1/6x < 1
a = 1 x > 1/6
log1/6x = 1
x = 1/6
II When 1  a < 3 1  log1/6x < 3
– (1 – a) + 2 = 3 – a x  1/6, x > 1 / 216
– 1 + a + 2 = 3 – a
2a = 2  a = 1 x = 1/6
III a  3
– (1 – a) + 2 = – (3 – a)
– 1 + a + 2 = – 3 + a
– 1 + 2  – 3 ]
Q.9 Find the value of sin
2

and cos
2

. If sin = –
325
323
and   




 

2
3
, [3]
[Ans. sin
2

=
26
5
19
; cos
2

= –
26
5
17
]
[Sol. Given sin = –
325
323
,   




 

2
3
,

325
36
cos



now sin2
2

=
2
cos
1 

=
650
361
2
325
36
1


 sin
2

=
26
5
19
as
2

 




 

4
3
,
2
|||ly cos2
2

=
2
cos
1 

= 






325
36
1
2
1
=
650
289
 cos
2

= –
26
5
17
Hence sin
2

=
26
5
19
; cos
2

= –
26
5
17
Ans. ]
Q.10 Show that, tan 




 


2
6
tan 




 


2
6
=
1
cos
2
1
cos
2




[3]
[Sol. LHS =
2
tan
3
1
1
2
tan
3
1
·
2
tan
3
1
1
2
tan
3
1








=
2
tan
3
2
tan
3
1
2
2




=
2
sin
2
cos
3
2
sin
3
2
cos
2
2
2
2






=





 








 







 








 


2
cos
2
sin
2
sin
2
cos
2
2
cos
2
sin
2
sin
2
cos
2
2
2
2
2
2
2
2
2
=
1
cos
2
1
cos
2




Hence proved ]

Q.11 Let y =
x
5
cos
x
4
cos
x
2
cos
x
cos
x
5
sin
x
4
sin
x
2
sin
x
sin






. Find the value of y where x =
36

. [3]
[Ans. 2 – 3 ]
[Sol. y =
)
x
4
cos
x
2
(cos
)
x
5
cos
x
(cos
)
x
4
sin
x
2
(sin
)
x
5
sin
x
(sin






=
x
cos
x
3
cos
2
x
2
cos
x
3
cos
2
x
cos
x
3
sin
2
x
2
cos
x
3
sin
2


=
]
x
cos
x
2
[cos
x
3
cos
]
x
cos
x
2
[cos
x
3
sin


= tan 3x
 y = = tan3x =
36
3
tan

=
12
tan

= tan15° = 2 – 3 Ans. ]