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Page 1
8
th
January 2020 (Shift 1), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. The maximum values of ??
19
?? , ??
20
?? , ??
21
?? are ?? ,?? ,?? respectively. Then, the relation
between ?? ,?? ,?? is
a.
?? 22
=
?? 42
=
?? 11
b.
?? 11
=
?? 22
=
?? 42
c.
?? 22
=
?? 11
=
?? 42
d.
?? 21
=
?? 11
=
?? 22
Answer: ( ?? )
Solution:
We know that, ??
?? ?? is maximum when ?? = {
?? 2
, ?? is even
?? +1
2
or
?? -1
2
,?? is odd
Therefore, max( ??
19
?? )= ??
19
9
= ??
max( ??
20
?? )= ??
20
10
= ??
max( ??
21
?? )= ??
21
11
= ??
?
?? ??
19
9
=
?? 20
10
× ??
19
9
=
?? 21
11
×
20
10
× ??
19
9
?? 1
=
?? 2
=
?? 42
11
?
?? 11
=
?? 22
=
?? 42
.
2. Let ?? ( ?? )=
1
3
,?? ( ?? )=
1
6
where ?? and ?? are independent events, then
a. ?? (
?? ?? )=
2
3
b. ?? (
?? ?? '
)=
5
6
c. ?? (
?? ?? '
)=
1
3
d. ?? (
?? ?? )=
1
6
Answer: ( ?? )
Solution:
If ?? and ?? are independent events, then
?? (
?? ?? )=
?? ( ?? n ?? )
?? ( ?? )
=
?? ( ?? ) ?? ( ?? )
?? ( ?? )
= ?? ( ?? )
Page 2
8
th
January 2020 (Shift 1), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. The maximum values of ??
19
?? , ??
20
?? , ??
21
?? are ?? ,?? ,?? respectively. Then, the relation
between ?? ,?? ,?? is
a.
?? 22
=
?? 42
=
?? 11
b.
?? 11
=
?? 22
=
?? 42
c.
?? 22
=
?? 11
=
?? 42
d.
?? 21
=
?? 11
=
?? 22
Answer: ( ?? )
Solution:
We know that, ??
?? ?? is maximum when ?? = {
?? 2
, ?? is even
?? +1
2
or
?? -1
2
,?? is odd
Therefore, max( ??
19
?? )= ??
19
9
= ??
max( ??
20
?? )= ??
20
10
= ??
max( ??
21
?? )= ??
21
11
= ??
?
?? ??
19
9
=
?? 20
10
× ??
19
9
=
?? 21
11
×
20
10
× ??
19
9
?? 1
=
?? 2
=
?? 42
11
?
?? 11
=
?? 22
=
?? 42
.
2. Let ?? ( ?? )=
1
3
,?? ( ?? )=
1
6
where ?? and ?? are independent events, then
a. ?? (
?? ?? )=
2
3
b. ?? (
?? ?? '
)=
5
6
c. ?? (
?? ?? '
)=
1
3
d. ?? (
?? ?? )=
1
6
Answer: ( ?? )
Solution:
If ?? and ?? are independent events, then
?? (
?? ?? )=
?? ( ?? n ?? )
?? ( ?? )
=
?? ( ?? ) ?? ( ?? )
?? ( ?? )
= ?? ( ?? )
8
th
January 2020 (Shift 1), Mathematics Page | 2
Therefore, ?? (
?? ?? )= ?? ( ?? )=
1
3
? ?? (
?? ?? '
)= ?? ( ?? )=
1
3
.
3. If ?? ( ?? ) =
8
2?? -8
-2?? 8
2?? +8
-2?? , then inverse of ?? ( ?? ) is
a.
1
2
log
8
(
1+?? 1-?? )
b.
1
2
log
8
(
1-?? 1+?? )
c.
1
4
log
8
(
1-?? 1+?? ) d.
1
4
log
8
(
1+?? 1-?? )
Answer: ( ?? )
Solution:
?? ( ?? )=
8
2?? - 8
-2?? 8
2?? + 8
-2?? =
8
4?? - 1
8
4?? + 1
Put ?? =
8
4?? -1
8
4?? +1
Applying componendo-dividendo on both sides
?? + 1
?? - 1
=
2× 8
4?? -2
?? +1
?? -1
= -8
4?? ? 8
4?? =
1+?? 1-??
? ?? =
1
4
log
8
(
1+?? 1-?? )
?? -1
( ?? )=
1
4
log
8
(
1+?? 1-?? ) .
4. Roots of the equation ?? 2
+ ???? + 45 = 0,?? ? ?? lies on the curve |?? + 1| = 2v10, where ??
is a complex number, then
a. ?? 2
+ ?? = 12 b. ?? 2
- ?? = 36
c. ?? 2
- ?? = 30 d. ?? 2
+ ?? = 30
Answer: ( ?? )
Solution:
Given ?? 2
+ ???? + 45 = 0,?? ? ?? , let roots of the equation be ?? ± ????
Then, sum of roots = 2?? = -??
Product of roots = ?? 2
+ ?? 2
= 45
As ?? ± ???? lies on |?? + 1| = 2v10, we get
( ?? + 1)
2
+ ?? 2
= 40
? ?? 2
+ ?? 2
+ 2?? + 1 = 40
Page 3
8
th
January 2020 (Shift 1), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. The maximum values of ??
19
?? , ??
20
?? , ??
21
?? are ?? ,?? ,?? respectively. Then, the relation
between ?? ,?? ,?? is
a.
?? 22
=
?? 42
=
?? 11
b.
?? 11
=
?? 22
=
?? 42
c.
?? 22
=
?? 11
=
?? 42
d.
?? 21
=
?? 11
=
?? 22
Answer: ( ?? )
Solution:
We know that, ??
?? ?? is maximum when ?? = {
?? 2
, ?? is even
?? +1
2
or
?? -1
2
,?? is odd
Therefore, max( ??
19
?? )= ??
19
9
= ??
max( ??
20
?? )= ??
20
10
= ??
max( ??
21
?? )= ??
21
11
= ??
?
?? ??
19
9
=
?? 20
10
× ??
19
9
=
?? 21
11
×
20
10
× ??
19
9
?? 1
=
?? 2
=
?? 42
11
?
?? 11
=
?? 22
=
?? 42
.
2. Let ?? ( ?? )=
1
3
,?? ( ?? )=
1
6
where ?? and ?? are independent events, then
a. ?? (
?? ?? )=
2
3
b. ?? (
?? ?? '
)=
5
6
c. ?? (
?? ?? '
)=
1
3
d. ?? (
?? ?? )=
1
6
Answer: ( ?? )
Solution:
If ?? and ?? are independent events, then
?? (
?? ?? )=
?? ( ?? n ?? )
?? ( ?? )
=
?? ( ?? ) ?? ( ?? )
?? ( ?? )
= ?? ( ?? )
8
th
January 2020 (Shift 1), Mathematics Page | 2
Therefore, ?? (
?? ?? )= ?? ( ?? )=
1
3
? ?? (
?? ?? '
)= ?? ( ?? )=
1
3
.
3. If ?? ( ?? ) =
8
2?? -8
-2?? 8
2?? +8
-2?? , then inverse of ?? ( ?? ) is
a.
1
2
log
8
(
1+?? 1-?? )
b.
1
2
log
8
(
1-?? 1+?? )
c.
1
4
log
8
(
1-?? 1+?? ) d.
1
4
log
8
(
1+?? 1-?? )
Answer: ( ?? )
Solution:
?? ( ?? )=
8
2?? - 8
-2?? 8
2?? + 8
-2?? =
8
4?? - 1
8
4?? + 1
Put ?? =
8
4?? -1
8
4?? +1
Applying componendo-dividendo on both sides
?? + 1
?? - 1
=
2× 8
4?? -2
?? +1
?? -1
= -8
4?? ? 8
4?? =
1+?? 1-??
? ?? =
1
4
log
8
(
1+?? 1-?? )
?? -1
( ?? )=
1
4
log
8
(
1+?? 1-?? ) .
4. Roots of the equation ?? 2
+ ???? + 45 = 0,?? ? ?? lies on the curve |?? + 1| = 2v10, where ??
is a complex number, then
a. ?? 2
+ ?? = 12 b. ?? 2
- ?? = 36
c. ?? 2
- ?? = 30 d. ?? 2
+ ?? = 30
Answer: ( ?? )
Solution:
Given ?? 2
+ ???? + 45 = 0,?? ? ?? , let roots of the equation be ?? ± ????
Then, sum of roots = 2?? = -??
Product of roots = ?? 2
+ ?? 2
= 45
As ?? ± ???? lies on |?? + 1| = 2v10, we get
( ?? + 1)
2
+ ?? 2
= 40
? ?? 2
+ ?? 2
+ 2?? + 1 = 40
8
th
January 2020 (Shift 1), Mathematics Page | 3
? 45- ?? + 1 = 40
? ?? = 6
? ?? 2
- ?? = 30.
5. Rolle’s theorem is applicable on ?? ( ?? )= ln(
?? 2
+?? 7?? ) in [3,4]. The value of ?? '' ( ?? ) is equal
to
a.
1
12
b.
-1
12
c.
-1
6
d.
1
6
Answer: ( ?? )
Solution:
Rolle’s theorem is applicable on ?? ( ?? ) in [3,4]
? ?? ( 3)= ?? ( 4)
? ln(
9 + ?? 21
)= ln(
16+ ?? 28
)
?
9+ ?? 21
=
16+ ?? 28
? 36+ 4?? = 48+ 3??
? ?? = 12
Now,
?? ( ?? )= ln(
?? 2
+ 12
7?? )? ?? '
( ?? )=
7?? ?? 2
+ 12
×
7?? × 2?? - ( ?? 2
+ 12)× 7
( 7?? )
2
?? '
( ?? )=
?? 2
- 12
?? ( ?? 2
+ 12)
?? '
( ?? )= 0 ? ?? = 2v3
?? ''
( ?? )=
-?? 4
+ 48?? 2
+ 144
?? 2
( ?? 2
+ 12)
2
?? ''
( ?? )=
1
12
.
6. Let ?? ( ?? )= ?? cos
-1
( sin ( -|?? |) ),?? ? ( -
?? 2
,
?? 2
) , then
a. ?? '
( 0)= -
?? 2
b. ?? '
( ?? ) is not defined at ?? = 0
c. ?? '
( ?? ) is decreasing in ( -
?? 2
, 0) and ?? '
( ?? ) is decreasing in ( 0,
?? 2
)
d. ?? '
( ?? ) is increasing in ( -
?? 2
, 0) and ?? '( ?? ) is increasing in ( 0,
?? 2
)
Page 4
8
th
January 2020 (Shift 1), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. The maximum values of ??
19
?? , ??
20
?? , ??
21
?? are ?? ,?? ,?? respectively. Then, the relation
between ?? ,?? ,?? is
a.
?? 22
=
?? 42
=
?? 11
b.
?? 11
=
?? 22
=
?? 42
c.
?? 22
=
?? 11
=
?? 42
d.
?? 21
=
?? 11
=
?? 22
Answer: ( ?? )
Solution:
We know that, ??
?? ?? is maximum when ?? = {
?? 2
, ?? is even
?? +1
2
or
?? -1
2
,?? is odd
Therefore, max( ??
19
?? )= ??
19
9
= ??
max( ??
20
?? )= ??
20
10
= ??
max( ??
21
?? )= ??
21
11
= ??
?
?? ??
19
9
=
?? 20
10
× ??
19
9
=
?? 21
11
×
20
10
× ??
19
9
?? 1
=
?? 2
=
?? 42
11
?
?? 11
=
?? 22
=
?? 42
.
2. Let ?? ( ?? )=
1
3
,?? ( ?? )=
1
6
where ?? and ?? are independent events, then
a. ?? (
?? ?? )=
2
3
b. ?? (
?? ?? '
)=
5
6
c. ?? (
?? ?? '
)=
1
3
d. ?? (
?? ?? )=
1
6
Answer: ( ?? )
Solution:
If ?? and ?? are independent events, then
?? (
?? ?? )=
?? ( ?? n ?? )
?? ( ?? )
=
?? ( ?? ) ?? ( ?? )
?? ( ?? )
= ?? ( ?? )
8
th
January 2020 (Shift 1), Mathematics Page | 2
Therefore, ?? (
?? ?? )= ?? ( ?? )=
1
3
? ?? (
?? ?? '
)= ?? ( ?? )=
1
3
.
3. If ?? ( ?? ) =
8
2?? -8
-2?? 8
2?? +8
-2?? , then inverse of ?? ( ?? ) is
a.
1
2
log
8
(
1+?? 1-?? )
b.
1
2
log
8
(
1-?? 1+?? )
c.
1
4
log
8
(
1-?? 1+?? ) d.
1
4
log
8
(
1+?? 1-?? )
Answer: ( ?? )
Solution:
?? ( ?? )=
8
2?? - 8
-2?? 8
2?? + 8
-2?? =
8
4?? - 1
8
4?? + 1
Put ?? =
8
4?? -1
8
4?? +1
Applying componendo-dividendo on both sides
?? + 1
?? - 1
=
2× 8
4?? -2
?? +1
?? -1
= -8
4?? ? 8
4?? =
1+?? 1-??
? ?? =
1
4
log
8
(
1+?? 1-?? )
?? -1
( ?? )=
1
4
log
8
(
1+?? 1-?? ) .
4. Roots of the equation ?? 2
+ ???? + 45 = 0,?? ? ?? lies on the curve |?? + 1| = 2v10, where ??
is a complex number, then
a. ?? 2
+ ?? = 12 b. ?? 2
- ?? = 36
c. ?? 2
- ?? = 30 d. ?? 2
+ ?? = 30
Answer: ( ?? )
Solution:
Given ?? 2
+ ???? + 45 = 0,?? ? ?? , let roots of the equation be ?? ± ????
Then, sum of roots = 2?? = -??
Product of roots = ?? 2
+ ?? 2
= 45
As ?? ± ???? lies on |?? + 1| = 2v10, we get
( ?? + 1)
2
+ ?? 2
= 40
? ?? 2
+ ?? 2
+ 2?? + 1 = 40
8
th
January 2020 (Shift 1), Mathematics Page | 3
? 45- ?? + 1 = 40
? ?? = 6
? ?? 2
- ?? = 30.
5. Rolle’s theorem is applicable on ?? ( ?? )= ln(
?? 2
+?? 7?? ) in [3,4]. The value of ?? '' ( ?? ) is equal
to
a.
1
12
b.
-1
12
c.
-1
6
d.
1
6
Answer: ( ?? )
Solution:
Rolle’s theorem is applicable on ?? ( ?? ) in [3,4]
? ?? ( 3)= ?? ( 4)
? ln(
9 + ?? 21
)= ln(
16+ ?? 28
)
?
9+ ?? 21
=
16+ ?? 28
? 36+ 4?? = 48+ 3??
? ?? = 12
Now,
?? ( ?? )= ln(
?? 2
+ 12
7?? )? ?? '
( ?? )=
7?? ?? 2
+ 12
×
7?? × 2?? - ( ?? 2
+ 12)× 7
( 7?? )
2
?? '
( ?? )=
?? 2
- 12
?? ( ?? 2
+ 12)
?? '
( ?? )= 0 ? ?? = 2v3
?? ''
( ?? )=
-?? 4
+ 48?? 2
+ 144
?? 2
( ?? 2
+ 12)
2
?? ''
( ?? )=
1
12
.
6. Let ?? ( ?? )= ?? cos
-1
( sin ( -|?? |) ),?? ? ( -
?? 2
,
?? 2
) , then
a. ?? '
( 0)= -
?? 2
b. ?? '
( ?? ) is not defined at ?? = 0
c. ?? '
( ?? ) is decreasing in ( -
?? 2
, 0) and ?? '
( ?? ) is decreasing in ( 0,
?? 2
)
d. ?? '
( ?? ) is increasing in ( -
?? 2
, 0) and ?? '( ?? ) is increasing in ( 0,
?? 2
)
8
th
January 2020 (Shift 1), Mathematics Page | 4
Answer: ( ?? )
Solution:
?? ( ?? )= ?? cos
-1
( sin ( -|?? |) )
? ?? ( ?? )= ?? cos
-1
( -sin|?? |)
? ?? ( ?? )= ?? [?? - cos
-1
( sin|?? |) ]
? ?? ( ?? )= ?? [?? - (
?? 2
- sin
-1
( sin|?? |) ) ]
? ?? ( ?? )= ?? (
?? 2
+ |?? |)
? ?? ( ?? )= {
?? (
?? 2
+ ?? ), ?? = 0
?? (
?? 2
- ?? ), ?? < 0
? ?? '( ?? )= {
(
?? 2
+ 2?? ), ?? = 0
(
?? 2
- 2?? ), ?? < 0
Therefore, ?? '( ?? ) is decreasing ( -
?? 2
,0) and increasing in ( 0,
?? 2
) .
7. Ellipse 2?? 2
+ ?? 2
= 1 and ?? = ???? meet at a point ?? in the first quadrant. Normal to the
ellipse at ?? meets ?? -axis at ( -
1
3v2
,0) and ?? -axis at ( 0,?? ) , then |?? | is
a.
2
3
b.
2v2
3
c.
v2
3
d.
2
v3
Answer: ( ?? )
Solution:
Page 5
8
th
January 2020 (Shift 1), Mathematics Page | 1
Date: 8
th
January 2020 (Shift 1)
Time: 9:30 A.M. to 12:30 P.M.
Subject: Mathematics
1. The maximum values of ??
19
?? , ??
20
?? , ??
21
?? are ?? ,?? ,?? respectively. Then, the relation
between ?? ,?? ,?? is
a.
?? 22
=
?? 42
=
?? 11
b.
?? 11
=
?? 22
=
?? 42
c.
?? 22
=
?? 11
=
?? 42
d.
?? 21
=
?? 11
=
?? 22
Answer: ( ?? )
Solution:
We know that, ??
?? ?? is maximum when ?? = {
?? 2
, ?? is even
?? +1
2
or
?? -1
2
,?? is odd
Therefore, max( ??
19
?? )= ??
19
9
= ??
max( ??
20
?? )= ??
20
10
= ??
max( ??
21
?? )= ??
21
11
= ??
?
?? ??
19
9
=
?? 20
10
× ??
19
9
=
?? 21
11
×
20
10
× ??
19
9
?? 1
=
?? 2
=
?? 42
11
?
?? 11
=
?? 22
=
?? 42
.
2. Let ?? ( ?? )=
1
3
,?? ( ?? )=
1
6
where ?? and ?? are independent events, then
a. ?? (
?? ?? )=
2
3
b. ?? (
?? ?? '
)=
5
6
c. ?? (
?? ?? '
)=
1
3
d. ?? (
?? ?? )=
1
6
Answer: ( ?? )
Solution:
If ?? and ?? are independent events, then
?? (
?? ?? )=
?? ( ?? n ?? )
?? ( ?? )
=
?? ( ?? ) ?? ( ?? )
?? ( ?? )
= ?? ( ?? )
8
th
January 2020 (Shift 1), Mathematics Page | 2
Therefore, ?? (
?? ?? )= ?? ( ?? )=
1
3
? ?? (
?? ?? '
)= ?? ( ?? )=
1
3
.
3. If ?? ( ?? ) =
8
2?? -8
-2?? 8
2?? +8
-2?? , then inverse of ?? ( ?? ) is
a.
1
2
log
8
(
1+?? 1-?? )
b.
1
2
log
8
(
1-?? 1+?? )
c.
1
4
log
8
(
1-?? 1+?? ) d.
1
4
log
8
(
1+?? 1-?? )
Answer: ( ?? )
Solution:
?? ( ?? )=
8
2?? - 8
-2?? 8
2?? + 8
-2?? =
8
4?? - 1
8
4?? + 1
Put ?? =
8
4?? -1
8
4?? +1
Applying componendo-dividendo on both sides
?? + 1
?? - 1
=
2× 8
4?? -2
?? +1
?? -1
= -8
4?? ? 8
4?? =
1+?? 1-??
? ?? =
1
4
log
8
(
1+?? 1-?? )
?? -1
( ?? )=
1
4
log
8
(
1+?? 1-?? ) .
4. Roots of the equation ?? 2
+ ???? + 45 = 0,?? ? ?? lies on the curve |?? + 1| = 2v10, where ??
is a complex number, then
a. ?? 2
+ ?? = 12 b. ?? 2
- ?? = 36
c. ?? 2
- ?? = 30 d. ?? 2
+ ?? = 30
Answer: ( ?? )
Solution:
Given ?? 2
+ ???? + 45 = 0,?? ? ?? , let roots of the equation be ?? ± ????
Then, sum of roots = 2?? = -??
Product of roots = ?? 2
+ ?? 2
= 45
As ?? ± ???? lies on |?? + 1| = 2v10, we get
( ?? + 1)
2
+ ?? 2
= 40
? ?? 2
+ ?? 2
+ 2?? + 1 = 40
8
th
January 2020 (Shift 1), Mathematics Page | 3
? 45- ?? + 1 = 40
? ?? = 6
? ?? 2
- ?? = 30.
5. Rolle’s theorem is applicable on ?? ( ?? )= ln(
?? 2
+?? 7?? ) in [3,4]. The value of ?? '' ( ?? ) is equal
to
a.
1
12
b.
-1
12
c.
-1
6
d.
1
6
Answer: ( ?? )
Solution:
Rolle’s theorem is applicable on ?? ( ?? ) in [3,4]
? ?? ( 3)= ?? ( 4)
? ln(
9 + ?? 21
)= ln(
16+ ?? 28
)
?
9+ ?? 21
=
16+ ?? 28
? 36+ 4?? = 48+ 3??
? ?? = 12
Now,
?? ( ?? )= ln(
?? 2
+ 12
7?? )? ?? '
( ?? )=
7?? ?? 2
+ 12
×
7?? × 2?? - ( ?? 2
+ 12)× 7
( 7?? )
2
?? '
( ?? )=
?? 2
- 12
?? ( ?? 2
+ 12)
?? '
( ?? )= 0 ? ?? = 2v3
?? ''
( ?? )=
-?? 4
+ 48?? 2
+ 144
?? 2
( ?? 2
+ 12)
2
?? ''
( ?? )=
1
12
.
6. Let ?? ( ?? )= ?? cos
-1
( sin ( -|?? |) ),?? ? ( -
?? 2
,
?? 2
) , then
a. ?? '
( 0)= -
?? 2
b. ?? '
( ?? ) is not defined at ?? = 0
c. ?? '
( ?? ) is decreasing in ( -
?? 2
, 0) and ?? '
( ?? ) is decreasing in ( 0,
?? 2
)
d. ?? '
( ?? ) is increasing in ( -
?? 2
, 0) and ?? '( ?? ) is increasing in ( 0,
?? 2
)
8
th
January 2020 (Shift 1), Mathematics Page | 4
Answer: ( ?? )
Solution:
?? ( ?? )= ?? cos
-1
( sin ( -|?? |) )
? ?? ( ?? )= ?? cos
-1
( -sin|?? |)
? ?? ( ?? )= ?? [?? - cos
-1
( sin|?? |) ]
? ?? ( ?? )= ?? [?? - (
?? 2
- sin
-1
( sin|?? |) ) ]
? ?? ( ?? )= ?? (
?? 2
+ |?? |)
? ?? ( ?? )= {
?? (
?? 2
+ ?? ), ?? = 0
?? (
?? 2
- ?? ), ?? < 0
? ?? '( ?? )= {
(
?? 2
+ 2?? ), ?? = 0
(
?? 2
- 2?? ), ?? < 0
Therefore, ?? '( ?? ) is decreasing ( -
?? 2
,0) and increasing in ( 0,
?? 2
) .
7. Ellipse 2?? 2
+ ?? 2
= 1 and ?? = ???? meet at a point ?? in the first quadrant. Normal to the
ellipse at ?? meets ?? -axis at ( -
1
3v2
,0) and ?? -axis at ( 0,?? ) , then |?? | is
a.
2
3
b.
2v2
3
c.
v2
3
d.
2
v3
Answer: ( ?? )
Solution:
8
th
January 2020 (Shift 1), Mathematics Page | 5
Let ?? = ( ?? 1
,?? 1
)
2?? 2
+ ?? 2
= 1 is given equation of ellipse.
? 4?? + 2?? ?? '
= 0
? ?? '
|
( ?? 1
,?? 1
)
= -
2?? 1
?? 1
Therefore, slope of normal at ?? ( ?? 1
,?? 1
) is
?? 1
2?? 1
Equation of normal at ?? ( ?? 1
,?? 1
) is
( ?? - ?? 1
)=
?? 1
2?? 1
( ?? - ?? 1
)
It passes through ( -
1
3v2
,0)
? -?? 1
=
?? 1
2?? 1
( -
1
3v2
- ?? 1
)
? ?? 1
=
1
3v2
? ?? 1
=
2v2
3
as ?? lies in first quadrant
Since ( 0, ?? ) lies on the normal of the ellipse at point ?? , hence we get
?? =
?? 1
2
=
v2
3
8. If ?????? is a triangle whose vertices are ?? ( 1, -1) , ?? ( 0,2) , ?? ( ?? '
, ?? ') and area of ??????? is
5, and ?? ( ?? '
, ?? ') lies on 3?? + ?? - 4?? = 0, then
a. ?? = 3 b. ?? = 4
c. ?? = -3 d. ?? = 2
Answer: ( ?? )
Solution:
Area of triangle is
?? =
1
2
|
0 2 1
1 -1 1
?? '
?? '
1
| = ±5
-2( 1- ?? '
)+ ( ?? '
+ ?? '
)= ±10
-2+ 2?? '
+ ?? '
+ ?? '
= ±10
3?? '
+ ?? '
= 12 or 3?? '
+ ?? '
= -8
? ?? = 3 or -2
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FAQs on JEE Main 2020 January 8 Question Paper Shift 1 - JEE Main & Advanced Previous Year Papers
| 1. What is the exam pattern of JEE Main 2020? |  |
Ans. JEE Main 2020 consists of multiple-choice questions (MCQs) and numerical value-based questions. The exam is conducted in online mode and includes three sections - Physics, Chemistry, and Mathematics.
| 2. How many times is JEE Main conducted in a year? | |
Ans. JEE Main is conducted twice a year - in January and April. Candidates have the option to appear for both exams and the best of the two scores are considered for admission.
| 3. What is the marking scheme for JEE Main 2020? | |
Ans. For MCQs, candidates are awarded 4 marks for a correct answer and 1 mark is deducted for an incorrect answer. For numerical value-based questions, there is no negative marking.
| 4. How can I prepare effectively for JEE Main 2020? | |
Ans. To prepare for JEE Main 2020, it is important to create a study schedule, focus on understanding concepts, practice previous years' question papers, and take regular mock tests to assess your preparation.
| 5. What are the eligibility criteria for JEE Main 2020? | |
Ans. To be eligible for JEE Main 2020, candidates must have passed their Class 12 examination or equivalent with Physics, Chemistry, and Mathematics as compulsory subjects. Additionally, there are age and attempts limit criteria that candidates need to fulfill.
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