1. Class : XI (J-Batch)
Time : 3 hour Max. Marks : 228
INSTRUCTIONS
1. The question paper contain 2-parts. Part-A contains 54 objective question, Part-B contains 2 "Match the
Column" questions.All questions are compulsory.
PART
-A
(i) Q.1 to Q.45 have only one correct alternative and carry 3 marks each.
There is NEGATIVE marking and 1 mark will be deducted for each wrong answer.
(ii) Q.46 to Q.54 have More than one are correct alternative and carry 5 marks each.
There is NO NEGATIVE marking. Marks will be awarded only if all the correct alternatives are
selected.
PART
-B
(iii) Q.1 to Q.9 are "Match the Column" type which may have one or more than one matching options and
carry 8 marks for each question. 2 marks will be awarded for each correct match within a question.
There is NO NEGATIVE marking.
2. Indicate the correct answer for each question by filling appropriate bubble in your answer sheet.
3. Use only HB pencil for darkening the bubble.
4. Use of Calculator, Log Table, Slide Rule and Mobile is not allowed.
5. Theanswer(s) ofthequestions must bemarkedbyshadingthecirclesagainst thequestionbydarkHB pencilonly.
USEFUL DATA
Atomic weights:Al = 27, Mg = 24, Cu = 63.5, Mn = 55, Cl = 35.5, O= 16, H = 1,P= 31,Ag =108, N= 14,
Li = 7, I = 127, Cr = 52, K=39, S = 32, Na = 23, C = 12, Br = 80, Fe = 56, Ca = 40, Zn = 65.4, Pt = 195,
REVIEW TEST-4
PAPER CODE : A
PART-A
For example if only 'B' choice is
correct then, the correct method for
filling the bubble is
A B C D
For example if only 'B & D' choices
are correct then, the correct method
for filling the bubble is
A B C D
the wrong method for filling the
bubble are
The answer of the questions inwrong
or any other manner will be treated
as wrong.
PART-B
For example if Correct match for
(A) is P, Q; for (B) is P, R;
for (C) is P and for (D) is S then the
correct method for filling the bubble
is
P Q R S
(A)
(B)
(C)
(D)

2. XI(J-BATCH) MATHEMATICS REVIEWTEST-4
Select the correct alternative. (Only one is correct) [45×3 = 135]
There is NEGATIVE marking and 1 mark will be deducted for each wrong answer.
Q.1s&p Consider the sequence 8A+ 2B, 6A + B, 4A, 2A– B, ........ Which term of this sequence will have
a coefficient of A which is twice the coefficient of B?
(A) 10th (B) 14th (C) 17th (D*) none
[Sol. coefficient of A in nth term = 8 + (n – 1)(– 2) [11th, 12th and 13th]
= 10 – 2n
coefficient of B in nth term = 2 + (n – 1)(– 1)
= 3 – n
10 – 2n = 2(3 – n)  10 = 6
which is absurd  none ]
Q.2s&p Find the sum of the infinite series .....
63
1
45
1
30
1
18
1
9
1





(A*)
3
1
(B)
4
1
(C)
5
1
(D)
3
2
[Hint: Tn = 









 ........
21
1
15
1
10
1
6
1
3
1
3
1
= 








 ........
6
·
5
1
5
·
4
1
4
·
3
1
3
·
2
1
3
2
hence Tn using method of diff; Tn =
)
2
n
)(
1
n
(
1
3
2


= 







 2
n
1
1
n
1
3
2
 Sn =
2
1
·
3
2
=
3
1
Ans. ]
Q.3ph-2 Number of degrees in the smallest positive angle x such that
8 sin x cos5x – 8 sin5x cos x = 1, is
(A) 5° (B*) 7.5° (C) 10° (D) 15°
[Sol. 2 sin x cos x[4 cos4x – 4 sin4x] = 1
(sin 2x)[2(cos2x + sin2x)] [2 cos2x – 2 sin2x] = 1
(sin 2x) 2 · 2 cos 2x = 1
2 sin 4x = 1
sin 4x =
2
1
 4x = 30°  x = 7.5° Ans. ]
Q.4log There exist positive integers A, B and C with no common factors greater than 1, such that
A log2005 + B log2002 = C. The sum A + B + C equals
(A) 5 (B*) 6 (C) 7 (D) 8
[Sol. A log2005 + B log2002 = C
200
log
2
log
B
200
log
5
log
A
 = C
A log 5 + B log 2 = C log 200 = C log(52 23) = 2C log 5 + 3 C log 2
hence, A = 2C
B = 3C
for no common factor greater than 1, C = 1
 A = 2; B = 3  A + B + C = 6 Ans. ]

3. Q.5ph-3 A triangle with sides 5, 12 and 13 has both inscribed and circumscribed circles. The distance
between the centres of these circles is
(A) 2 (B)
2
5
(C) 65 (D*)
2
65
[Sol. r =
s

=
)
13
12
5
(
2
2
·
)
5
·
12
(


=
30
60
= 2
 d2 =
2
2
2
5






 + (4)2
=
4
1
+ 16 =
2
65
Ans. ]
Q.6QE The graph of a certain cubic polynomial is as shown. If the polynomial
can be written in the form f (x) = x3 + ax2 + bx + c, then
(A) c = 0 (B) c < 0
(C*) c > 0 (D) c = – 1
[Sol. From graph we see that this cubic has 2 + ve and one – ve root
 product of roots is – ve and product of roots – c
 – c is + ve ]
Q.7ph-3 In an isosceles triangleABC, AB =AC, BAC = 108° and
AD trisects BAC and BD > DC. The ratio
DC
BD
is
(A*)
2
1
5 
(B)
2
3
(C) 1
5  (D) 2
[Sol. In  ABC (
y
x
= ?)


108
sin
y
x
=

36
sin
x
x
y
x 
=


36
sin
108
sin
....(1)
1 +
x
y
=


36
sin
72
sin
= 2 cos 36°
x
y
=
2
1
5 
– 1 =
2
1
5 
y
x
=
)
1
5
)(
1
5
(
)
1
5
(
2



=
2
1
5 
Ans. ]

4. Q.8QE If a function f (x) = ax3 + bx2 + cx + d where a, b, c and d are integers and a > 0 is such that
f 




 
18
sin = 0. Then the smallest possible value of f (1) is
(A) 1 (B) 2 (C*) 3 (D) 4
[Sol. sin
18

= sin 10°, sin 30° =
2
1
[for 11th, 12th and 13th]
also sin 30° = 3 sin 10° – 4 sin310°
2
1
= 3 sin 10° – 4 sin310°
8 sin310° + 0 sin210° – 6 sin 10° + 1 = 0 ....(1)
given, f (sin 10°) = 0
a sin310° + b sin210° + c sin 10° + d = 0 ....(2)
comparing (1) and (2)
a = 8, b = 0, c = – 6, d = 1
hence f (1) = a + b + c + d
f (1) = 3 Ans. ]
Q.9ph-3 The sides of a triangle are 6 and 8 and the angle  between these sides varies such that 0° <  < 90°.
The length of 3rd side x is
(A) 2 < x < 14 (B) 0 < x < 10 (C*) 2 < x < 10 (D) 0 < x < 14
[Sol. x2 = b2 + c2 – 2bc cos 
x2 = 64 + 36 – 96 cos 
= 100 – 96 cos 
and   (0, 90°)
cos   (0, 1)
 x2|max = 100 – 0  x = 10
x2|min = 100 – 96 = 4  x = 2
 2 < x < 10 Ans. ]
Q.10s&p If a, b, c are in G.P. then
a
b
1

,
b
2
1
,
c
b
1

are in
(A*) A.P. (B) G.P. (C) H.P. (D) none
[Sol. Let a = x; b = xr; c = xr2
hence the number are
)
1
r
(
x
1

, r
x
2
1
,
)
r
1
(
xr
1

now,
xr
2
2
= r
x
1
)
1
r
(
x
1

+
)
r
1
(
xr
1

=
)
1
r
(
x
1
 






r
1
1 = r
x
1
hence
a
b
1

,
b
2
1
,
c
b
1

are in A.P.]
.]

5. Q.11s&p How many terms are there in the G.P. 5, 20, 80, .........20480.
(A) 6 (B) 5 (C*) 7 (D) 8
[Sol. 20480 = 5 · 4n – 1
4096 = 4n – 1  212 = 22(n – 1)  2n – 2 = 12  n = 7 Ans. ]
Q.12s&p The sum of the first 14 terms of the sequence
x
1
1

+
x
1
1

+
x
1
1

+ ....... is
(A*)
x
1
)
x
11
2
(
7


(B)
x
1
)
x
7
1
(
7


(C)
)
x
1
)(
x
1
)(
x
1
(
14



(D) none
[Sol. Sequence is A.P. with d =
x
1
x

as
x
1
1

–
x
1
1

=
x
1
1

–
x
1
x
1


=
x
1
x

 S14 =












x
1
x
13
x
1
)
x
1
(
2
7 = 









x
1
x
11
2
7 Ans. ]
Q.13logIf x, y > 0, logyx + logxy =
3
10
and xy = 144, then arithmetic mean of x and y is
(A) 24 (B) 36 (C) 2
12 (D*) 3
13
[Hint: logyx = t  t = 3 or 1/3]
Q.14ph-3 A circle of radius R is circumscribed about a right triangle ABC. If r is the radius of incircle
inscribed in triangle then the area of the triangle is
(A) r(2r + R) (B*) r(r + 2R) (C) R(r + 2R) (D) R(2r + R)
[Sol. 2 = ar + br + 2Rr
2 = r[(a + b) + 2R]
also r = (s – c)tan
2
C
(take C = 90°)
2r = (2s – 2c)
2r = a + b – c = a + b – 2R
a + b = 2(r + R)
 2 = 2r(r + 2R)   = r(r + 2R) Ans. ]
Q.15general The simplest form of
a
1
1
1
a
1


 is
(A) a for a  1 (B*) a for a  0 and a  1
(C) – a for a  0 and a  1 (D) 1 for a  1
[Sol.
a
1
1
1
a
1


 = 1 +
1
a
1
)
a
1
(
a



=
a
a
a
a 2




= a  a  0, 1 ]

6. Select the correct alternatives. (More than one are correct) [9 × 5 = 45]
Q.16 Which of the following equations can have sec2 and cosec2 as its roots (  R)?
(A) x2 – 3x + 3 = 0 (B*) x2 – 6x + 6 = 0
(C*) x2 – 9x + 9 = 0 (D) x2 – 2x + 2 = 0
Q.17 The equation 1
x
10 2
|
2
x
| 
 = x
3
|
2
x
|  has
(A) 3 integral solutions (B*) 4 real solutions
(C*) 1 prime solution (D*) no irrational solution
[Sol. x – 2  0  x  2
Case-I: x – 2 > 0  x > 2
(10x2 – 1) log(x – 2) = 3x log (x – 2)
log(x – 2)(10x2 – 3x – 1) = 0
x – 2 = 1 or 10x2 – 5x + 2x – 1 = 0  (5x + 1)(2x – 1) = 0
x = 3 x = – 1/5 (reject), x = 1/2 (reject)
Case-II: x – 2 < 0  x < 2
(10x2 – 1) log(2 – x) = 3x log(2 – x)
log(2 – x)(10x2 – 3x – 1) = 0
2 – x = 1 or x = – 1/5, x = 1/2
x = 1  x 






 3
,
1
,
2
1
,
5
1
]
Q.18 Which of the following statements hold good?
(A*) If M is the maximum and m is the minimum value of y = 3 sin2x + 3 sin x · cos x + 7 cos2x
then the mean of M and m is 5.
(B) The value of cosec

18
– 3 sec

18
is a rational which is not integral.
(C*) If x lies in the third quadrant, then the expression x
2
sin
x
sin
4 2
4
 + 4 cos2 







2
x
4
is
independent of x.
(D) There are exactly 2 values of  in [0, 2] which satisfy 4 cos2 2 2 cos  1 = 0.
[Sol. (A)829/ph-1 3 sin2 x + 3 sin x . cos x + 7 cos2 x
=
2
)
x
2
cos
1
(
7
2
x
2
sin
3
2
)
x
2
cos
1
(
3 



y =
2
x
2
cos
4
x
2
sin
3 
+ 5
ymax = 5/2 + 5 = 7.5
ymin = –5/2+5 = 2.5
(B)84/ph-1
18
/
cos
3
18
/
sin
1



=
 
2
9
sin
18
sin
2
3
18
cos
2
1
2






 


=
9
sin
18
sin
6
cos
18
cos
6
sin
4






 




= 4 Ans

7. (C)824/ph-1
 
2
2
2
2
2
x
4
cos
4
x
2
sin
x
sin
4


































x
2
cos
1
2
x
2
sin
4
)
x
2
cos
1
(
4 2
2
)
x
sin
1
(
2
x
2
cos
2
1
1 



x
sin
2
2
)
x
sin
2
(
2 2


x
sin
2
2
|
x
sin
|
2 
 = – 2 sin x + 2 + 2 sin x = 2 Ans
(D)25/ph-1 4 cos2 – 2 2 cos – 1 = 0
cos =
8
16
8
2
2 

=
4
6
2 
cos =
4
2
6 
12
23
12
2
;
12









cos =
4
2
6 
 = – 






 
2
2
1
3
cos = cos(–5/12) ; cos(+5/12)
 = 7/12 ; 17/12]
PART
-B
MATCH THE COLUMN [6 × 8 = 48]
INSTRUCTIONS:
Column-I and column-II contains four entries each. Entries of column-I are to be matched with
some entries of column-II. One or more than one entries of column-I may have the matching with
the same entries of column-II and one entry of column-I may have one or more than one matching
with entries of column-II.
Column-I Column-II
Q.1 (A) In a triangle ABC, AB = 3
2 , BC = 6
2 , AC > 6, (P) 60°
and area of the triangle ABC is 6
3 .  B equals (Q) 90°
(B) In a triangle ABC is b = 3 , c = 1 and A = 30° (R) 120°
then angle B equals
(C) In a  ABC if (a + b + c)(b + c – a) = 3bc (S) 75°
then  A equals
(D) Area of a triangle ABC is 6 sq. units. If the radii of its
excircles are 2, 3 and 6 then largest angle of the triangle is
[Ans. (A) R; (B) R; (C) P; (D) Q]

8. [Hint: (A)  =
2
1
ac sinB
6
3 =
2
1
· 3
2 · 6
2 sin B
sin B =
2
3
hence B = 120° (as b > 6)
if B = 60° then b will be less than 6. Hence (A)  Q
(B) b = 3 , c = 1 and A = 30°
cos 30° =
3
2
a
3
1 2


2
3
=
3
2
a
4 2

 3 = 4 – a2  a2 = 1  a = 1
 cos B =
1
1
2
3
1
1




 cos B =
2
3
2 
= –
2
1
 B = 120°
(C) (b + c)2 – a2 = 3bc
b2 + c2 – a2 = bc
bc
2
a
c
b 2
2
2


=
2
1
cos A =
2
1
  A = 60° Ans.
(D)  = 6 sq. units
r1 = 2; r2 = 3; r3 = 6
r1 =
a
s 

; r2 =
b
s 

; r3 =
c
s 

 s – a =
2
6
= 3; s – b = 2; s – c = 1
 3s – (a + b + c) = 3 + 2 + 1
3s – 2s = s – 6
 6 – a = 3 6 – b = 2 6 – c = 1
 a = 3 b = 4 c = 5
 largest  is 90° ]

9. Column-I Column-II
Q.2 (A) The sequence a, b, 10, c, d is an arithmetic progression. (P) 10
The value of a + b + c + d
(B) The sides of right triangle form a three term geometric (Q) 20
sequence. The shortest side has length 2. The length
of the hypotenuse is of the form b
a  where a  N (R) 26
and b is a surd, then a2 + b2 equals
(C) The sum of first three consecutive numbers of an (S) 40
infinite G..P. is 70, if the two extremes be multipled
each by 4, and the mean by 5, the products are in A.P.
The first term of the G.P. is
(D) The diagonals of a parallelogram have a measure of
4 and 6 metres. They cut off forming an angle of 60°.
If the perimeter of the parallelogram is  
b
a
2 
where a, b  N then (a + b) equals
[Ans. (A) S; (B) R; (C) S; (D) R]
[Hint: (A) b + c = a + d = 2 · 10
 a + b + c + d = 40
(B) (ar2)2 = a2 + a2r2 where a = 2
 r4 = 1 + r2
r4 – r2 – 1 = 0
let r2 = t
t2 – t – 1 = 0
t =
2
5
1

2
5
1
(
2
5
1
reject)
r2 =
2
5
1
 hypotenuse is 2 × 






 
2
5
1
= 1 + 5
comparing with b
a 
a = 1, b = 5
 a2 + b2 = 1 + 25 = 26 Ans.
(C) a, ar, ar2  G.P. | r | < 1
a + ar + ar2 = 70
 10ar = 4a + 4ar2
10r = 4 + 4r2
2r2 – 5r + 2 = 0
2r2 – 4r – r + 2 = 0
(2r – 1)(r – 2) = 0
 r = 2 (reject) or r = 1/2
for r = 1/2

10. a +
2
a
+
4
a
= 70
a +
4
a
3
= 70 
4
a
7
= 70  a = 40
 series is 40, 20, 10
 first term of G.P. is 40 Ans.
(D) Using cosine rule
a2 = 9 + 4 – 2 · 2 · 3 · 






2
1
= 13 + 6 = 19
a2 = 19  a = 19
|||ly b2 = 9 + 4 – 2 · 2 · 3 · 





2
1
b2 = 7  b = 7
 P =  
7
19
2   a + b = 26 Ans. ]