Daily Practice Problems
CLASS : XI (J-Batch) TIME : 50 Min. DPP. NO.-19
Select the correct alternative : (Only one is correct)
Q.148/QE The equation, x =  2x2 + 6x  9 has:
(A*)nosolution (B)onesolution (C)twosolutions (D)infinitesolutions
[Hint: Left always +ve& right always –ve]
Q.2129/QE If a > b > 0 are two real numbers, the value of ,
a b a b a b a b a b a b a b
      
( ) ( ) ( ) ....... is :
(A*) independent of b (B) independent of a
(C) independent of both a & b (D) dependent on both a & b .
[Hint: We have, y = ab a b y
 
( )
y2 = ab + (a – b)y
 y2 – (a – b)y – ab = 0  (y – a) (y + b) = 0  y = a or y = –b (not possible)  A ]
Q.3 Numberofintegerswhichsimultaneouslysatisfiestheinequalities
| x | + 5 < 7 and | x – 3 | > 2, is
(A) exactly1 (B*) exactly2
(C)more than2but finite (D)infinitelymany
[Hint: The onlyone integer that satisfy| x | + 5 < 7 are the ones that satisfy| x | < 2, namely, – 1, 0, 1.
The integer that satisfy | x – 3 | > 2 are 6, 7, 8, ..... and 0, – 1, –2, .....
So, the integer that satisfyboth are 0, –1, and there are 2 of them. ]
Q.473/ph-1 If tan A & tan B are the roots of the quadratic equation x2  ax + b = 0, then the value of
sin2 (A +B) is :
(A*)
a
a b
2
2 2
1
 
( )
(B)
a
a b
2
2 2

(C)
a
b a
2
2
( )

(D)
a
b a
2
2 2
1
( )

[Sol. tanA + tanB = a ; tanA.tanB = b
tan(A+B) =
b
1
a

, hence sin2(A+B) =
a
a b
2
2 2
1
 
( )
Ans: ]
*Q.566/QE If themaximum and minimum valuesof y=
c
x
3
x
c
x
3
x
2
2




are 7 and
7
1
respectivelythen the value
of c is equal to
(A) 3 (B*) 4 (C) 5 (D) 6
[Hint: y =
c
x
3
x
c
x
3
x
2
2




= 7 or
7
1
must give two coincident values of x  c = 4 ]
Subjective
*Q.6 Find the value of x satisfying the equation 3
2
4
x
x2



 = x2 + x – 12. [Ans. 11/2]
[Hint: 3
2
4
x
x2



 = x2 + x – 12

3
2
x
x2


 = x2 + x – 12
1
x
x2

 = x2 + x – 12
2x = 11  x = 11/2 Ans. ]
*Q.721/1 Find all values of 'a' for which
1
x
x
2
ax
x
2
2




lies between –3 and 2 for all real values of x.
[Ans: )
2
,
1
(
a 
 ]
[Sol. – 3 <
1
x
x
2
ax
x
2
2




< 2
 – 3x2 – 3x – 3 < x2 – ax – 2 < 2x2 + 2x + 2 R
x

x2 + x(a + 2) + 4 > 0  (a + 6) (a – 2) < 0 ....(1)
and 4x2 + x(3 – a) + 1 > 0
(3 – a)2 – 16 < 0  (a + 1) (a – 7) < 0 ....(2)
from (1) and (2) )
2
,
1
(
a 
 Ans. ]
Q.8 If un
= sinn
 + cosn
, prove that
u u
u
u u
u
3 5
1
5 7
3



[Sol. consider
   
   
u u
u u
5 7
3 5
5 5 7 7
3 3 5 5



  
  
sin cos sin cos
sin cos sin cos
   
   
=
   
   
sin sin cos cos
sin sin cos cos
5 2 5 2
3 2 3 2
1 1
1 1
   
   
  
  
=
 
 
sin cos sin cos
sin cos sin cos
2 2 3 3
2 2
   
   


=
u
u
3
1
]
*Q.929/6 Show that the triangleABC isright angles if andonlyif sinA+ sinB + sinC= cosA+cosB+ cosC+ 1.
[Sol. 1st part : LetABC is aright angle.
Also let A = 90°
C = 90 – B
now LHS, 1 + sinB + cosB
RHS = cos90° + cosB + sinB + 1
= 1 + sinB + cosB = LHS
henceifanyoneangle is 90° thenthegivenrelation is true.
2nd Part : Let sinA + sinB + sinC = cosA + cosB + cosC + 1
2
B
A
cos
2
B
A
sin
2


+
2
C
cos
2
C
sin
2 =
2
B
A
cos
2
B
A
cos
2


+
2
C
cos
2 2





 



2
B
A
cos
2
B
A
sin
2
B
A
cos – 






2
C
sin
2
C
cos
2
C
cos = 0








2
C
sin
2
C
cos
2
B
A
cos – 






2
C
sin
2
C
cos
2
C
cos = 0
















2
C
cos
2
B
A
cos
2
C
sin
2
C
cos = 0
if
2
C
sin
2
C
cos  = 0 
2
C
tan = 1 
2
C
=
4

 C = 90° ....(1)
if
2
B
A
cos

=
2
C
cos = cos 






2
C
then A – B = C or A – B = – C
A = B + C A + C = B
here A + B + C = 
A =
2

....(2)  B =
2

....(3)
from (1), (2)and (3) one of the angle must be 90° henceABC is a right angled triangle.]
Q.106/6From the relation, 3 (cos 2 – cos 2) = 1 – cos 2 cos 2, we get
tan = k tan where ,   




 
2
,
0 . Find the value of k.
[Ans. k = 2 ]
[Sol. Fromgivenrelation,
3cos 2 – 3cos 2 = 1 – cos 2 cos 2
or (3 + cos2)cos 2 = 1 + 3cos 2 or cos 2 =




2
cos
3
2
cos
3
1
applycomponendoanddividendo,




2
cos
1
2
cos
1
=




2
cos
1
2
cos
1
·
2
1
or tan2 =
2
1
tan2.
But given tan = k tan wherebyk2 = 2, and since  and  are positive acute angles k = 2 ]

Daily Practice Problems
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-20
Select the correct alternative : (Only one is correct)
*Q.179/ph-1 The value of the expression (sinx + cosecx)2 + (cosx + secx)2 – ( tanx + cotx)2 wherever defined
is equal to
(A) 0 (B*) 5 (C) 7 (D) 9
Q.287/QE The numberofintegers 'n' such that the equation nx2 +(n + 1)x +(n + 2) =0 has rational roots only, is
(A) 1 (B) 2 (C*) 3 (D) 4
[Hint: roots rational D must be perfect square and D > 0
D = (n+1)2 –4n(n+2)
D = –3n2– 6n + 1 = 0  3n2 + 6n – 1 = 0 or D = 4 – 3(n + 1)2 (now interpret)
 =
  
6 36 12
6
=
 
3 2 3
6
=    
1
2 3
3
1
2 3
3
or
Hence possible integral values are –2 , –1 , 0 and all of them make D a perfect square. ]
Q.3 If cos + cos= a and sin + sin= b, then the value of cos·coshas the value equal to
(A)
 
 
a b a
a b
2 2
2
2
2 2
4
4
 

(B)
 
 
a b b
a b
2 2
2
2
2 2
4
2
 

(C)
 
 
a b a
a b
2 2 2 2
2 2 2
4
2
 

(D*)
 
 
a b b
a b
2 2
2
2
2 2
4
4
 

[Hint: square and add, cos( – ) =
a b
2 2
2
2
 
using C – D relation tan
 


2
b
a
Now E =
1
2
[cos( + ) + cos( – )]
=
1
2
1
1
2
2
2
2
2
2 2
2



 












b
a
b
a
a b
simplifyto gettheresult]
Select the correct alternative : (More than one are correct)
Q.4511/QE The value(s) of 'p' for which the equation ax2  px + ab = 0 and x2  ax  bx + ab = 0 mayhave
a common root, given a, b are non zero real numbers, is
(A) a + b2 (B*) a2 + b (C*) a(1 + b) (D) b(1 + a)
[Sol. x2  (a + b) x + a b = 0 or (x  a) (x  b) = 0  x = a or b
if x = a is the root of other equation , a3  a p + a b = 0  p = a2 + b
if x = b is the root of the other equation , then a b2  p b + a b = 0
p = a (1 + b) ]
Subjective:

*Q.584/1 Findthe solutionset ofthe inequality 2
1
|
x
|
2
|
x
|
3



.
[ Ans:(–  , – 1)  







5
4
,
1  {0}  





1
,
5
4
 (1, )]
[Sol. Let |x| = y ( note |x|  1 )
2
1
y
2
y
3



Hence 2
1
y
2
y
3



....(1) or 2
1
y
2
y
3




....(2)
From (1) 0
2
1
y
2
y
3




0
1
y
2
y
2
2
y
3





0
1
y
y


)
,
1
(
]
0
,
(
y 



But |x|  0
Hence |x|  (1, ) x (– , –1)  (1, ) ....(3)
From (2) 0
2
1
y
2
y
3




0
1
y
4
y
5




 













 1
,
5
4
|
x
|
1
,
5
4
y
 














 1
,
5
4
5
4
,
1
x ....(4)
For (3) & (4) together with the fact that x = 0 is the obv. solution as equalityholds hence
x  )
,
1
(
1
,
5
4
}
0
{
5
4
,
1
)
1
,
( 



















 ]
Q.656/QE Let a, b, c be the three roots of the equation x3 + x2 – 333x – 1002 = 0 then find the value of
a3 + b3 + c3. [Ans. 2006]
[Sol. Let t be the root of the given cubic where t can take values a, b, c
hence t3 + t2 – 333t – 1002 = 0 or t3 = 1002 + 333t – t2
  3
t = 1002 + 333t –  2
t = 3006 + 333 t –   ]
[ 2
1
2
t
t
2
t 
 
but t = – 1 ;  2
1t
t = – 333
 a3 + b3 + c3 = 3006 – 333 – [1 + 666] = 3006 – 333 – 667 = 3006 – 1000 = 2006 Ans. ]
*Q.734/1 A polynomial in x of degree greater than 3 leaves the remainder 2, 1 and –1 when divided by
(x–1);(x+2)&(x+1)respectively.Findtheremainder,ifthepolynomialisdivided by,(x2 1)(x +2).
[ Ans. :
7
6
x2 +
3
2
x

2
3
]
[Sol. f (x) = Q1 (x  1) + 2 = Q2 (x + 2) + 1 = Q3 (x + 1)  1

 f (1) = 2 ; f ( 2) = 1 ; f ( 1) =  1
again f (x) = Qr (x2  1) (x + 2) + ax2 + bx + c
Hence a + b + c = 2 ; 4 a  2 b + c = 1 and a  b + c =  1 ]
Q.852/6 Prove the inequality
sin
sin
x
x


1
2
+
1
2

2
3


sin
sin
x
x
 x  R.
Q.933/6 Show that the sum to n terms of the series :
sin cos 3 + sin 3 cos 5 + ..... + sin(2n – 1). cos(2n + 1)=
sin ( ) .sin
sin
2 1 2
2 2
n n
  

–
n
2
sin2
[Sol: We have, 2S = 2sin cos 3 + 2 sin3 cos 5 + ..... + 2 sin(2n – 1) × cos (2n + 1)
Hence, sin4 – sin2
+
sin8 – sin2
2S = +
sin12 – sin2
+

sin4n – sin2
________________________________________________
2S =
sin sin sin ........ sin sin
4 8 12 4 2
1
    
    
n n
say S
 
       
      
now, 2 sin2.S1 = 2 sin2 sin4 + 2 sin2 sin8 + ....... + 2 sin2 sin4n
= cos2 – cos6
+
= cos6 – cos10

+
= cos(4n – 2) – cos(4n + 2)
_________________________________________________________________
2 sin2.S1 = cos 2 – cos (4n + 2) = 2 sin(2n + 2). sin2n
S1 =
 

2 2 2 2
2 2
sin( ).sin
sin
n n
  

Hence, S =
sin ( ) .sin
sin
2 1 2
2 2
n n
  

–
nsin2
2

]
Q.1045/6 Let
)
sin(
)
sin(






=
b
a
&
)
cos(
)
cos(






=
d
c
then prove that cos () =
bc
ad
bd
ac


.
[Hint:
a
sin ( )
 

=
b
sin ( )
 

= k1 ;
c
cos ( )
 

=
d
cos ( )
 

= k2
=
a c bd
a d bc


=
k k
k k
1 2
1 2
































)
(
cos
)
(
sin
)
(
cos
)
(
sin
)
(
cos
)
(
sin
)
(
cos
)
(
sin
=     )
(
sin
)
(
2
sin
)
(
sin
)
(
2
sin
)
(
2
sin
)
(
2
sin


























=
 
 
)
(
2
sin
2
)
(
cos
)
(
2
sin
2













= cos () ]

Daily Practice Problems
CLASS : XI (J-Batch) TIME : 60 Min. DPP. NO.-21
Select the correct alternative : (Only one is correct)
*Q.1 The set of values of 'p' for which theexpression x2
 2px +3p + 4 is negative for atleast one real x is:
(A)  (B) (1, 4)
(C*) (,  1)  (4, ) (D) {1, 4}
[Put D > 0 ]
Q.2 The value of f (x) = x2 + (p – q)x + p2 + pq + q2 for real values of p, q and x (p + q  0)
(A)isalways negative (B*)isalways positive
(C) is sometimes zero for non zero value of x (D) none of these
[Hint: f (x) = x2 + (p – q)x + p2 + pq + q2
D = (p – q)2 – 4(p2 + pq + q2)
= p2 + q2 – 2pq – 4p2 – 4pq – 4q2
= – 3p2 – 3q2 – 6pq
= – 3(p + q)2
D < 0 ]
Q.360/QE For R
x
 , the difference between the greatest and the least value of y =
1
x
x
2

is
(A*) 1 (B) 2 (C) 3 (D)
2
1
Q.4 If the roots of the cubic, x3 + ax2 +bx + c = 0 arethreeconsecutive positive integers. Then the value
of
1
b
a2

is equal to
(A*) 3 (B) 2 (C) 1 (D) none of these
[Sol. n, n + 1, n + 2
sum = 3(n + 1) = – a
 a2 = 9(n + 1)2
sum of the roots taken 2 at atime = + b
 n(n + 1) + (n + 1)(n + 2) + (n + 2)n + 1 = b + 1
n2 + n + n2 + 3n + 2 + n2 + 2n + 1 = b + 1
 b + 1 = 3n2 + 6n + 3
b + 1 = 3(n + 1)2 =
3
a2
; 
1
b
a2

= 3  (A) ]
Q.577/ph-1 If x sin  = y cos  then
x
sec2
+
y
cosec2
is equal to
(A*) x (B) y (C) x2 (D) y2
[Hint: sec 2 = 2
2
2
2
y
x
y
x


; cosec 2 =
xy
2
y
x 2
2

]

Q.6112/QE The equation (x  R) x
x
2
2
1
1
5
3
 

= x
(A) has no root (B*) exactlyone root (C) two roots (D) four roots
[Hint: only x =  4/3 satisfies ; x = 4/3 is rejected ]
[Sol.
x x
x
2
2
1
1
5
3
  
 ....(1)
or x x x
2 2
1
5
3
    ....(2)
(2) – (1) given 2x = x
x
2
2
5
3
1
5
3
 

squaring 4x2 = x
x
2
2
5
3
1
5
3
2
 


3
1
5
3
11
3
0
2
2
x
x


  ; 3
1
5
3
11
3
y
y


 = 0
3
5
3
11
3
5
3
1 0
y y y

F
H
G I
K
J 
F
H
G I
K
J 
3y2 – 5y + (11/3)y – (55/9) – 1 = 0
27y2 – 12y – 6y = 0
27y2 – 48y + 36y – 6y = 0
3y(9y–16) + 4(9y–16) = 0
y = 
4
3
or y =
16
9
 x2 = –
3
4
(rejected) or x2 =
9
16
 x =
3
4
or –
3
4
. Only x = –
3
4
satisfies. ]
*Q.7280/flcd The function f (x) is defined by f (x) = cos4x + K cos22x + sin4x, where K is a constant. If the
functionf(x)is aconstantfunction, thevalueofk is
(A) – 1 (B*) – 1/2 (C) 0 (D) 1/2 (E) 1
[Sol. Your first impulse, like mine, might be to complete the square to get the expression
(cos2x +sin2x)2 somewhereinthe problem. Whilethis works, it iseasier to substitute intwo values for
x andforcethentobe equal,sincethefunction isto beconstant.Anytwo values will work,but wewant
touseones forwhich weknowthesineandcosine.All multiples of/2resultin1+K, soweneedto put
in one value that is not a multiple of /2.Try/4.This gives us
cos4(/4) + Kcos2(/2) + sin4(/4) = 1/4 + K · 0 + 1/4 = 1/2. So now we know that
1 + K = 1/2, so K = – 1/2. This does in fact give us the function
f (x) = (cos2x + sin2x)2 – 1/2 = 1/2 ]

Subjective:
Q.839/6 Prove that : sin
7
2
+ sin
7
4
– sin
7
6
= 4 sin
7

sin
7
3
sin
7
5
[Sol. RHS = 4 sin
7

sin
7
2
sin
7
4





 





7
4
sin
7
3
sin
and
7
2
sin
7
5
sin
g
sin
u
2 sin
7






 

7
4
sin
7
2
sin
2 or 2 sin
7






 


7
6
cos
7
2
cos = 




 



7
cos
7
2
cos
7
sin
2
sin
7
2
+ 2 sin
7

cos
7
2
 sin
7
2
+ sin
7
3
– sin
7

sin
7
2
+ sin
7
4
– sin
7
6
= L.H. S. proved ]
Q.927/6 Find the two smallest positive values of x for which sin x° = sin (xc) [Ans.



180
180
or



180
360
]
[Sol. Let x be the number
now  radian = 180 degree
1 radian = 






180
degree
x radian =

x
180
degree
now 






x
180
sin = sin x  x = 0 (not possible)
or 






x
180
sin = sin(180 – x) ;

x
180
= 180 – x








1
180
x = 180  x =



180
180
Ans.
or 






x
180
sin = sin(360 + x) ;

x
180
= 360 + x








1
180
x = 360  x =



180
360
Ans. ]
Q.10 If one root of the quadratic equation x2 + mx – 24 = 0 is twice a root of the equation
x2 –(m + 1)x + m = 0 thenfind thevalue of m.
[Ans. m = – 2, 2, 10]
[Sol. 2nd equationgiven
x = m or x = 1
now put x = 1 and then x = m to determine the value of m. ]

Q.1166/6 For  = 1°, prove that
2 sin2 + 4 sin4 + 6 sin6 + ......... + 180 sin180 = 90 cot
[Sol. Let LHS =S, now multiplyingbothsides bysin
sin · S = 2 sin2 sin + 4 sin4 sin + 6 sin6 sin + ....... + 180 sin180 sin
T1 = (cos – cos3)
T2 = 2(cos3 – cos5)
T3 = 3(cos5 – cos7)
T4 = 4(cos7 – cos9)
 
= 89(cos177 – cos179)
T90 = 90(cos179 – cos181)
————————————
S sin = cos + cos3 + cos5 + ....... + cos(177) + cos(179) – 90 cos(181)
put  = 1°
S·sin1° = 










 1
cos
90
179
cos
177
cos
.......
5
cos
3
cos
1
cos
terms
90










 










 

S·sin1° = 

 


 



 


 



 


 

vanish
vanish
vanish
91
cos
89
cos
........
177
cos
3
cos
179
cos
1
cos 











+ 90 cos1°
 hence S sin 1° = 90 cos 1°  S = cot1° ]
*Q.12 If y=
8
x
2
x
3
x
2
x
2
2




then findthe interval in which ycan lie for everyx  R wherever defined. [6]
[Sol. y =
)
2
x
)(
4
x
(
)
1
x
)(
3
x
(




[Ans. y )
,
1
(
9
4
, 








 ]
x2(y – 1) + 2x(y – 1) – 8y + 3 = 0,
 x  R
(y – 1)2 – (y – 1)(3 – 8y)  0
or (y – 1)(9y – 4)  0
 y  1 or y 
9
4
for y = 1, x2 + 2x – 8 = x2 + 2x – 3
 – 8 = – 3 (notpossible)
 y  1
|||ly y =
9
4
; we get,
x2 + 2x + 1 = 0
x = – 1
 y  )
,
1
(
9
4
, 








 ]