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# Sample Solution Paper 6 - Math, Class 11 JEE Notes | EduRev

## JEE : Sample Solution Paper 6 - Math, Class 11 JEE Notes | EduRev

``` Page 1

CBSE XI | Mathematics
Sample Paper – 6 Solution

CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution

SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x

2.  He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?

OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?

4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4

SECTION – B

5. n(A) = 200, n(B) = 300  and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300

6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??

OR

Page 2

CBSE XI | Mathematics
Sample Paper – 6 Solution

CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution

SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x

2.  He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?

OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?

4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4

SECTION – B

5. n(A) = 200, n(B) = 300  and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300

6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??

OR

CBSE XI | Mathematics
Sample Paper – 6 Solution

f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1

7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?

OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40

8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}

9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?

22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?

? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?

22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?

= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?

Page 3

CBSE XI | Mathematics
Sample Paper – 6 Solution

CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution

SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x

2.  He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?

OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?

4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4

SECTION – B

5. n(A) = 200, n(B) = 300  and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300

6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??

OR

CBSE XI | Mathematics
Sample Paper – 6 Solution

f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1

7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?

OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40

8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}

9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?

22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?

? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?

22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?

= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?

CBSE XI | Mathematics
Sample Paper – 6 Solution

OR

tan
4
? + tan
2
? = (tan
2
?)
2
+ tan
2
?
= (sec
2
? - 1)
2
+ sec
2
? - 1
= sec
4
? - 2sec
2
? + 1 + sec
2
? - 1
= sec
4
? - sec
2
?

10.  We observe that 9 is an odd number which is not prime. Similarly, 21, 25 etc are odd
integers which are not prime.

11.  a, b, c are in AP
b
2
= ac
(b
2
)
n
= (ac)
n

b
2n
= a
n
c
n

log b
2n
= log a
n
c
n
log (b
n
)
2
= log a
n
+ log c
n
2log b
n
= log a
n
+ log c
n

log a
n
, log b
n
and log c
n
are in AP

12.  y
2
= 12x comparing with y
2
= 4ax,we get a = 3
The focal distance of any point (x, y) on y
2
= 4ax is x + a.
The focal distance is x + 3.
?x + 3 = 4
?x = 1
Hence, the abscissa of the given point is 1.

SECTION – C

13.  tan (pcos ?) = cot (psin ?)
? ?
? ?
? ?
? ?
sin cos cos sin
cos cos sin sin
? ? ? ?
?
? ? ? ?

sin(pcos ?) sin(psin ?)  = cos (psin ?) cos (pcos ?)
cos (psin ?) cos (pcos ?) - sin(pcos ?) sin(psin ?) = 0
cos (pcos ? + psin ?) = 0
pcos ? + psin ? =
2
?
?
cos ? + sin ? =
1
2
?
1 1 1
cos sin
2 2 2 2
? ? ? ? ?
Page 4

CBSE XI | Mathematics
Sample Paper – 6 Solution

CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution

SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x

2.  He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?

OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?

4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4

SECTION – B

5. n(A) = 200, n(B) = 300  and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300

6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??

OR

CBSE XI | Mathematics
Sample Paper – 6 Solution

f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1

7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?

OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40

8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}

9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?

22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?

? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?

22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?

= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?

CBSE XI | Mathematics
Sample Paper – 6 Solution

OR

tan
4
? + tan
2
? = (tan
2
?)
2
+ tan
2
?
= (sec
2
? - 1)
2
+ sec
2
? - 1
= sec
4
? - 2sec
2
? + 1 + sec
2
? - 1
= sec
4
? - sec
2
?

10.  We observe that 9 is an odd number which is not prime. Similarly, 21, 25 etc are odd
integers which are not prime.

11.  a, b, c are in AP
b
2
= ac
(b
2
)
n
= (ac)
n

b
2n
= a
n
c
n

log b
2n
= log a
n
c
n
log (b
n
)
2
= log a
n
+ log c
n
2log b
n
= log a
n
+ log c
n

log a
n
, log b
n
and log c
n
are in AP

12.  y
2
= 12x comparing with y
2
= 4ax,we get a = 3
The focal distance of any point (x, y) on y
2
= 4ax is x + a.
The focal distance is x + 3.
?x + 3 = 4
?x = 1
Hence, the abscissa of the given point is 1.

SECTION – C

13.  tan (pcos ?) = cot (psin ?)
? ?
? ?
? ?
? ?
sin cos cos sin
cos cos sin sin
? ? ? ?
?
? ? ? ?

sin(pcos ?) sin(psin ?)  = cos (psin ?) cos (pcos ?)
cos (psin ?) cos (pcos ?) - sin(pcos ?) sin(psin ?) = 0
cos (pcos ? + psin ?) = 0
pcos ? + psin ? =
2
?
?
cos ? + sin ? =
1
2
?
1 1 1
cos sin
2 2 2 2
? ? ? ? ?

CBSE XI | Mathematics
Sample Paper – 6 Solution

1
cos cos sin sin
44
22
??
? ? ? ? ?
1
cos
4
22
? ??
? ? ? ?
??
??

14.  For any a ?R we have 1 + a
2
> 0 for (a, a) ?R1
(a, a) ?R1 for all a ?R
Let (a, b) ?R1 then,
1 + ab > 0
? 1 + ba > 0
? (b, a) ?R1
(a, b) ?R1 ? (b, a) ?R1 for all a, b ?R

15. Let
9
?
??
Let C = cos cos2 cos3 cos4 ? ? ? ?
S = sin sin2 sin3 sin4 ? ? ? ?
C × S = ? ? ? ? ? ? ? ? sin cos sin2 cos2 sin3 cos3 sin4 cos4 ? ? ? ? ? ? ? ?
=
1 1 1 1
sin2 sin4 sin6 sin8
2 2 2 2
? ? ? ?
But
9
?
?? ? 9 ? ? ?
? ? ? sin8 sin sin ? ? ? ? ? ? ? and ? ? sin6 sin 3 sin3 ? ? ? ? ? ? ?
C × S =
44
11
sin2 sin4 sin3 sin S
22
? ? ? ? ? ? ?
C =
4
1
2
? S ? 0

16. Let
2
, ? ? ? ? ? ? then
2 3 2
, 1, 1 ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
2 3 2 4 2 3
2 2 2 3 4 3
2 2 2
22
33
xyz a b a b a b
a b a b a b
a b a ab ab b
a b a ab ab b 1,
a b a ab b
a b a ab b
xyz a b
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
??
??

Page 5

CBSE XI | Mathematics
Sample Paper – 6 Solution

CBSE Board
Class XI Mathematics
Sample Paper – 6 Solution

SECTION – A
1. cos[sin v x]’ = -sin (sin v x) × cos v x × 1/2v x

2.  He is not bald and he is not tall.
3.
2
2
6 6 i 6i
6i i 1
i i i i
? ? ? ? ? ?
? ? ?

OR
z = sin ? – i cos ?
Comparing with a + bi we get a = sin ? and b = - cos ?
2 2 2 2
z a b sin cos 1 ? ? ? ? ? ? ?

4. Sample space S = {HH, HT, TH, TT} i.e. total number of cases = 4
Favourable cases for two heads are {HH}.
Required probability =
1
4

SECTION – B

5. n(A) = 200, n(B) = 300  and n(A ? B) = 100
n(A’ ? B’) = n(A ? B)’
= 700 - n(A ? B)
= 700 – [n(A) + n(B) – n(A n B)]
= 700 – (200 + 300 – 100)
= 700 – 400
= 300

6. f(x) = vx
f(25) = 5, f(16) = 4 and f(1) = 1
Hence,
? ?
? ? ? ?
f 25
5
1
f 16 f 1 4 1
??
??

OR

CBSE XI | Mathematics
Sample Paper – 6 Solution

f(x) = x
2
– 1 and g(x) = vx
f ° g(x) = f[g(x)] = f(vx) = (vx)
2
– 1 = x – 1

7. f(x) = |x – 3|
f(x) is defined for all x ? R. Therefore, domain f = R
|x – 3| > 0 for all x ? R.
0 x 3 ? ? ? ? for all x ? R.
? ? ? ? f x 0, ?? for all x ? R.
Range of f is
? ? 0, ?

OR
n(A) = 50, n(A ? B) = 60 and n(A n B) = 10
n(A ? B) = n(A) + n(B) – n(An B)
60 = 50 + n(B) – 10
n(B) = 60 – 50 + 10 = 20
n(A – B) = n(A) – n(An B)
= 50 – 10 = 40

8. A × B = {(6, 3), (6, 5), (8, 3), (8, 5)}
A × A = {(6, 6), (6, 8), (8, 6), (8, 8)}

9. LHS =
cos A sin A cos A sin A
sin A cosA
1 tan A 1 cot A
11
cosA sin A
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
??
? ? ? ? ? ? ? ?
??
? ? ? ?
? ? ? ?

22
cos A sin A
cosA sin A sin A cosA
? ? ? ?
??
? ? ? ?
? ? ? ?
??
? ? ? ?

? ?
22
cos A sin A
cosA sin A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ?
? ? ? ?

22
cos A sin A
cos A sin A
1
cos A sin A
?
?
?
?
?

= RHS
cos A sin A 1
1 tan A 1 cot A cosA sin A
? ? ? ?
??
? ? ? ?
? ? ?
? ? ? ?

CBSE XI | Mathematics
Sample Paper – 6 Solution

OR

tan
4
? + tan
2
? = (tan
2
?)
2
+ tan
2
?
= (sec
2
? - 1)
2
+ sec
2
? - 1
= sec
4
? - 2sec
2
? + 1 + sec
2
? - 1
= sec
4
? - sec
2
?

10.  We observe that 9 is an odd number which is not prime. Similarly, 21, 25 etc are odd
integers which are not prime.

11.  a, b, c are in AP
b
2
= ac
(b
2
)
n
= (ac)
n

b
2n
= a
n
c
n

log b
2n
= log a
n
c
n
log (b
n
)
2
= log a
n
+ log c
n
2log b
n
= log a
n
+ log c
n

log a
n
, log b
n
and log c
n
are in AP

12.  y
2
= 12x comparing with y
2
= 4ax,we get a = 3
The focal distance of any point (x, y) on y
2
= 4ax is x + a.
The focal distance is x + 3.
?x + 3 = 4
?x = 1
Hence, the abscissa of the given point is 1.

SECTION – C

13.  tan (pcos ?) = cot (psin ?)
? ?
? ?
? ?
? ?
sin cos cos sin
cos cos sin sin
? ? ? ?
?
? ? ? ?

sin(pcos ?) sin(psin ?)  = cos (psin ?) cos (pcos ?)
cos (psin ?) cos (pcos ?) - sin(pcos ?) sin(psin ?) = 0
cos (pcos ? + psin ?) = 0
pcos ? + psin ? =
2
?
?
cos ? + sin ? =
1
2
?
1 1 1
cos sin
2 2 2 2
? ? ? ? ?

CBSE XI | Mathematics
Sample Paper – 6 Solution

1
cos cos sin sin
44
22
??
? ? ? ? ?
1
cos
4
22
? ??
? ? ? ?
??
??

14.  For any a ?R we have 1 + a
2
> 0 for (a, a) ?R1
(a, a) ?R1 for all a ?R
Let (a, b) ?R1 then,
1 + ab > 0
? 1 + ba > 0
? (b, a) ?R1
(a, b) ?R1 ? (b, a) ?R1 for all a, b ?R

15. Let
9
?
??
Let C = cos cos2 cos3 cos4 ? ? ? ?
S = sin sin2 sin3 sin4 ? ? ? ?
C × S = ? ? ? ? ? ? ? ? sin cos sin2 cos2 sin3 cos3 sin4 cos4 ? ? ? ? ? ? ? ?
=
1 1 1 1
sin2 sin4 sin6 sin8
2 2 2 2
? ? ? ?
But
9
?
?? ? 9 ? ? ?
? ? ? sin8 sin sin ? ? ? ? ? ? ? and ? ? sin6 sin 3 sin3 ? ? ? ? ? ? ?
C × S =
44
11
sin2 sin4 sin3 sin S
22
? ? ? ? ? ? ?
C =
4
1
2
? S ? 0

16. Let
2
, ? ? ? ? ? ? then
2 3 2
, 1, 1 ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
2 3 2 4 2 3
2 2 2 3 4 3
2 2 2
22
33
xyz a b a b a b
a b a b a b
a b a ab ab b
a b a ab ab b 1,
a b a ab b
a b a ab b
xyz a b
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ?
??
?? ? ? ? ?
??
??

CBSE XI | Mathematics
Sample Paper – 6 Solution

17. Let A denote the event : both the balls are of the same colour.
B : both are white
C : both are red
A = B ? C ? D
P(A) = P (B ? C ? D)
= P(B) + P(C) + P(D)                      B, C and D are mutually exclusive.
P(B) =
? ?
? ?
7
2
16
2
nB
C 21
n S C 120
??
P(C) =
? ?
? ?
5
2
16
2
nC
C 10
n S C 120
??
P(D) =
? ?
? ?
4
2
16
2
nD
C 6
n S C 120
??
P(A) =
21 10 6 37
120 120 120 120
? ? ?
Required probability = 1 – P(A) =
83
120

18. Let A be the first term and D the common difference. It is given that Sp = a
? ?
p
2A p 1 D a
2
?? ? ? ?
??

? ? p1
a
2p
?
?? ………..(i)
Similarly,
? ? q1
b
2q
?
?? ……….(ii)
and
? ? r1
c
2r
?
?? ……….(iii)
Multiplying by (i) by (q – r), (ii) by (r – p) and (iii) by (p – q) and adding
? ? ? ? ? ? ? ? ? ? ? ? ? ?
D
A q r r p p q p 1 q r q 1 r p r 1 p q
2
?? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
??

? ? ? ? ? ? ? ?
D a b c
A 0 0 q r r p p q 0
2 p q r
? ? ? ? ? ? ? ? ?

```
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## Mathematics (Maths) Class 11

158 videos|187 docs|161 tests

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