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Page 1
9
th
January 2020 (Shift 1), Mathematics Page | 1
Date of Exam: 9
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is
a.
1
12?? b.
1
18??
c.
1
9?? d.
1
36?? Answer: (?? )
Solution:
Let thickness of ice be ?? cm.
Therefore, net radius of sphere =(10+?? ) cm
Volume of sphere ?? =
4
3
?? (10+?? )
3
?
????
????
=4?? (10+?? )
2
????
????
At ?? =5,
????
????
=50 cm
3
/min
?50=4?? ×225×
????
????
????
????
=
1
18?? cm/min
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is
a. 1 b. 2
c. 3 d. 4
Answer: (?? )
Solution:
?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0
??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0
?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0
?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0
Let ?? ?? +
1
?? ?? =??
Then, ?? 2
+?? -6=0
Page 2
9
th
January 2020 (Shift 1), Mathematics Page | 1
Date of Exam: 9
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is
a.
1
12?? b.
1
18??
c.
1
9?? d.
1
36?? Answer: (?? )
Solution:
Let thickness of ice be ?? cm.
Therefore, net radius of sphere =(10+?? ) cm
Volume of sphere ?? =
4
3
?? (10+?? )
3
?
????
????
=4?? (10+?? )
2
????
????
At ?? =5,
????
????
=50 cm
3
/min
?50=4?? ×225×
????
????
????
????
=
1
18?? cm/min
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is
a. 1 b. 2
c. 3 d. 4
Answer: (?? )
Solution:
?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0
??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0
?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0
?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0
Let ?? ?? +
1
?? ?? =??
Then, ?? 2
+?? -6=0
9
th
January 2020 (Shift 1), Mathematics Page | 2
??? =2,-3
?? ?-3 as ?? >0 (??? ?? >0)
??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0
Hence, only one real solution is possible.
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is
a.
?? -1
4
b.
?? +1
4
c.
?? +1
2
d. 0
Answer: (?? )
Solution:
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? )
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
)
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
]
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
]
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)]
?? '(?? )=
?? 4
+
?? 2
??? (?? )=
?? 4
?? +
?? 2
4
+??
?? (0)=0??? =0
?? (1)=
?? 4
+
1
4
=
?? +1
4
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is
a. 2 b. 4
c. 8 d. 6
Answer: (?? )
Page 3
9
th
January 2020 (Shift 1), Mathematics Page | 1
Date of Exam: 9
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is
a.
1
12?? b.
1
18??
c.
1
9?? d.
1
36?? Answer: (?? )
Solution:
Let thickness of ice be ?? cm.
Therefore, net radius of sphere =(10+?? ) cm
Volume of sphere ?? =
4
3
?? (10+?? )
3
?
????
????
=4?? (10+?? )
2
????
????
At ?? =5,
????
????
=50 cm
3
/min
?50=4?? ×225×
????
????
????
????
=
1
18?? cm/min
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is
a. 1 b. 2
c. 3 d. 4
Answer: (?? )
Solution:
?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0
??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0
?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0
?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0
Let ?? ?? +
1
?? ?? =??
Then, ?? 2
+?? -6=0
9
th
January 2020 (Shift 1), Mathematics Page | 2
??? =2,-3
?? ?-3 as ?? >0 (??? ?? >0)
??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0
Hence, only one real solution is possible.
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is
a.
?? -1
4
b.
?? +1
4
c.
?? +1
2
d. 0
Answer: (?? )
Solution:
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? )
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
)
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
]
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
]
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)]
?? '(?? )=
?? 4
+
?? 2
??? (?? )=
?? 4
?? +
?? 2
4
+??
?? (0)=0??? =0
?? (1)=
?? 4
+
1
4
=
?? +1
4
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is
a. 2 b. 4
c. 8 d. 6
Answer: (?? )
9
th
January 2020 (Shift 1), Mathematics Page | 3
Solution:
log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ]
?log1
2
|sin?? ||cos?? |=2
?|sin?? cos?? |=
1
4
?sin2?? =±
1
2
? We have 8 solutions for ?? ?[0,2?? ]
5. If ?? 1
and ?? 2
are the eccentricities of
?? 2
18
+
?? 2
4
=1 and
?? 2
9
-
?? 2
4
=1, respectively. If the
points (?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? . Then the value of ?? is
a. 16 b. 14
c. 15 d. 17
Answer: (?? )
Solution:
e
1
=
v
1-
4
18
=
v7
3
& ?? 2
=
v
1+
4
9
=
v13
3
?(?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=??
?15?? 1
2
+3?? 2
2
=??
?15×
7
9
+3×
13
9
=?? ??? =16
Page 4
9
th
January 2020 (Shift 1), Mathematics Page | 1
Date of Exam: 9
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is
a.
1
12?? b.
1
18??
c.
1
9?? d.
1
36?? Answer: (?? )
Solution:
Let thickness of ice be ?? cm.
Therefore, net radius of sphere =(10+?? ) cm
Volume of sphere ?? =
4
3
?? (10+?? )
3
?
????
????
=4?? (10+?? )
2
????
????
At ?? =5,
????
????
=50 cm
3
/min
?50=4?? ×225×
????
????
????
????
=
1
18?? cm/min
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is
a. 1 b. 2
c. 3 d. 4
Answer: (?? )
Solution:
?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0
??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0
?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0
?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0
Let ?? ?? +
1
?? ?? =??
Then, ?? 2
+?? -6=0
9
th
January 2020 (Shift 1), Mathematics Page | 2
??? =2,-3
?? ?-3 as ?? >0 (??? ?? >0)
??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0
Hence, only one real solution is possible.
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is
a.
?? -1
4
b.
?? +1
4
c.
?? +1
2
d. 0
Answer: (?? )
Solution:
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? )
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
)
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
]
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
]
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)]
?? '(?? )=
?? 4
+
?? 2
??? (?? )=
?? 4
?? +
?? 2
4
+??
?? (0)=0??? =0
?? (1)=
?? 4
+
1
4
=
?? +1
4
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is
a. 2 b. 4
c. 8 d. 6
Answer: (?? )
9
th
January 2020 (Shift 1), Mathematics Page | 3
Solution:
log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ]
?log1
2
|sin?? ||cos?? |=2
?|sin?? cos?? |=
1
4
?sin2?? =±
1
2
? We have 8 solutions for ?? ?[0,2?? ]
5. If ?? 1
and ?? 2
are the eccentricities of
?? 2
18
+
?? 2
4
=1 and
?? 2
9
-
?? 2
4
=1, respectively. If the
points (?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? . Then the value of ?? is
a. 16 b. 14
c. 15 d. 17
Answer: (?? )
Solution:
e
1
=
v
1-
4
18
=
v7
3
& ?? 2
=
v
1+
4
9
=
v13
3
?(?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=??
?15?? 1
2
+3?? 2
2
=??
?15×
7
9
+3×
13
9
=?? ??? =16
9
th
January 2020 (Shift 1), Mathematics Page | 4
6. The value of ?
????
(?? -3)
6
7
×(?? +4)
8
7
is –
a. 7(
?? -3
?? +4
)
1
7
+?? b. 7(
?? -3
?? +4
)
6
7
+??
c. (
?? -3
?? +4
)
1
7
+?? d. 7(
?? +4
?? -3
)
6
7
+??
Answer: (?? )
Solution:
?? =?
????
(?? -3)
6
7
×(?? +4)
8
7
??? =?
(?? +4)
6
7
????
(?? -3)
6
7
×(?? +4)
2
=?(
?? -3
?? +4
)
-
6
7
×
????
(?? +4)
2
Put
?? -3
?? +4
=?? ????? =7(
1
(?? +4)
2
)????
??? =
??? -
6
7
7
???? =?? 1
7
+?? =(
?? -3
?? +4
)
1
7
+??
7. If |
?? -?? ?? +2?? | =1 ,|?? |=
5
2
then the value of |?? +3?? | is
a. v10 b. v5
c.
7
2
d. v3
Answer: (?? )
Solution:
If |
?? -?? ?? +2?? | =1 & |?? |=
5
2
?|?? -?? |=|?? +2?? |
??? 2
+(?? -1)
2
=?? 2
+(?? +2)
2
??? -1=±(?? +2)
??? -1=-?? -2
??? =-
1
2
?|?? |=
5
2
??? 2
+?? 2
=
25
4
??? 2
+
1
4
=
25
4
??? =±v6
Page 5
9
th
January 2020 (Shift 1), Mathematics Page | 1
Date of Exam: 9
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. A sphere of 10 cm radius has a uniform thickness of ice around it. If the ice is melting at
the rate of 50 cm
3
/min when thickness is 5 cm, then the rate of change of thickness is
a.
1
12?? b.
1
18??
c.
1
9?? d.
1
36?? Answer: (?? )
Solution:
Let thickness of ice be ?? cm.
Therefore, net radius of sphere =(10+?? ) cm
Volume of sphere ?? =
4
3
?? (10+?? )
3
?
????
????
=4?? (10+?? )
2
????
????
At ?? =5,
????
????
=50 cm
3
/min
?50=4?? ×225×
????
????
????
????
=
1
18?? cm/min
2. The number of real roots of ?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0 is
a. 1 b. 2
c. 3 d. 4
Answer: (?? )
Solution:
?? 4?? +?? 3?? -4?? 2?? +?? ?? +1=0
??? 2?? +?? ?? -4+
1
?? ?? +
1
?? 2?? =0
?(?? 2?? +
1
?? 2?? )+(?? ?? +
1
?? ?? )-4=0
?(?? ?? +
1
?? ?? )
2
-2+(?? ?? +
1
?? ?? )-4=0
Let ?? ?? +
1
?? ?? =??
Then, ?? 2
+?? -6=0
9
th
January 2020 (Shift 1), Mathematics Page | 2
??? =2,-3
?? ?-3 as ?? >0 (??? ?? >0)
??? ?? +
1
?? ?? =2?(?? ?? -1)
2
=0??? ?? =1??? =0
Hence, only one real solution is possible.
3. If ?? '(?? )=tan
-1
(sec?? +tan?? ),?? ?(-
?? 2
,
?? 2
) and ?? (0)=0, then the value of ?? (1) is
a.
?? -1
4
b.
?? +1
4
c.
?? +1
2
d. 0
Answer: (?? )
Solution:
?? '(?? )=tan
-1
(sec?? +tan?? )=tan
-1
(
1+sin?? cos?? )
?? '(?? )=tan
-1
(
sin
2
?? 2
+cos
2
?? 2
+2sin
?? 2
cos
?? 2
cos
2
?? 2
-sin
2
?? 2
)
?? '(?? )=tan
-1
[
(cos
?? 2
+sin
?? 2
)
2
(cos
?? 2
+sin
?? 2
)(cos
?? 2
-sin
?? 2
)
]
?? '(?? )=tan
-1
[
1+tan
?? 2
1-tan
?? 2
]
?? '(?? )=tan
-1
[tan(
?? 4
+
?? 2
)]
?? '(?? )=
?? 4
+
?? 2
??? (?? )=
?? 4
?? +
?? 2
4
+??
?? (0)=0??? =0
?? (1)=
?? 4
+
1
4
=
?? +1
4
4. The number of solutions of log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ] is
a. 2 b. 4
c. 8 d. 6
Answer: (?? )
9
th
January 2020 (Shift 1), Mathematics Page | 3
Solution:
log1
2
|sin?? | =2-log1
2
|cos?? |,?? ?[0,2?? ]
?log1
2
|sin?? ||cos?? |=2
?|sin?? cos?? |=
1
4
?sin2?? =±
1
2
? We have 8 solutions for ?? ?[0,2?? ]
5. If ?? 1
and ?? 2
are the eccentricities of
?? 2
18
+
?? 2
4
=1 and
?? 2
9
-
?? 2
4
=1, respectively. If the
points (?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=?? . Then the value of ?? is
a. 16 b. 14
c. 15 d. 17
Answer: (?? )
Solution:
e
1
=
v
1-
4
18
=
v7
3
& ?? 2
=
v
1+
4
9
=
v13
3
?(?? 1
,?? 2
) lies on the ellipse 15?? 2
+3?? 2
=??
?15?? 1
2
+3?? 2
2
=??
?15×
7
9
+3×
13
9
=?? ??? =16
9
th
January 2020 (Shift 1), Mathematics Page | 4
6. The value of ?
????
(?? -3)
6
7
×(?? +4)
8
7
is –
a. 7(
?? -3
?? +4
)
1
7
+?? b. 7(
?? -3
?? +4
)
6
7
+??
c. (
?? -3
?? +4
)
1
7
+?? d. 7(
?? +4
?? -3
)
6
7
+??
Answer: (?? )
Solution:
?? =?
????
(?? -3)
6
7
×(?? +4)
8
7
??? =?
(?? +4)
6
7
????
(?? -3)
6
7
×(?? +4)
2
=?(
?? -3
?? +4
)
-
6
7
×
????
(?? +4)
2
Put
?? -3
?? +4
=?? ????? =7(
1
(?? +4)
2
)????
??? =
??? -
6
7
7
???? =?? 1
7
+?? =(
?? -3
?? +4
)
1
7
+??
7. If |
?? -?? ?? +2?? | =1 ,|?? |=
5
2
then the value of |?? +3?? | is
a. v10 b. v5
c.
7
2
d. v3
Answer: (?? )
Solution:
If |
?? -?? ?? +2?? | =1 & |?? |=
5
2
?|?? -?? |=|?? +2?? |
??? 2
+(?? -1)
2
=?? 2
+(?? +2)
2
??? -1=±(?? +2)
??? -1=-?? -2
??? =-
1
2
?|?? |=
5
2
??? 2
+?? 2
=
25
4
??? 2
+
1
4
=
25
4
??? =±v6
9
th
January 2020 (Shift 1), Mathematics Page | 5
|?? +3?? |=v?? 2
+(?? +3)
2
? |?? +3?? |=
7
2
8. The value of 2
1
4
×4
1
16
×8
1
48
…8 is
a. 2 b. 1
c. v2
d. 2
1
4
Answer: (?? )
Solution:
2
1
4
×4
1
16
×8
1
48
…8= 2
1
4
×2
2
16
×2
4
48
…8
?2
1
4
×2
1
8
×2
1
16
…8= 2
1
4
+
1
8
+
1
16
+?8
? 2
(
1
4
1-
1
2
)
=v2
9. The value of cos
3
?? 8
cos
3?? 8
+sin
3
?? 8
sin
3?? 8
is –
a.
1
2
b. -
1
2
c.
1
v2
d.
1
2v2
Answer: (?? )
Solution:
cos
3
?? 8
cos
3?? 8
+sin
3
?? 8
sin
3?? 8
=cos
3
?? 8
[4cos
3
?? 8
-3cos
?? 8
]+sin
3
?? 8
[3sin
?? 8
-4sin
3
?? 8
]
=4[cos
6
?? 8
-sin
6
?? 8
]+3[sin
4
?? 8
-cos
4
?? 8
]
=4[cos
2
?? 8
-sin
2
?? 8
][cos
4
?? 8
+sin
4
?? 8
+cos
2
?? 8
sin
2
?? 8
]-3[cos
2
?? 8
-sin
2
?? 8
]
=[cos
2
?? 8
-sin
2
?? 8
][4(1-cos
2
?? 8
sin
2
?? 8
)-3]
=cos
?? 4
[1-sin
2
?? 4
]=
1
2v2
10. The value of ?
?? sin
8
?? sin
8
?? +cos
8
?? ???? 2?? 0
is
a. 4?? 2
b. 2?? 2
c. ?? 2
d. 3?? 2
Answer: (?? )
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