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

Mathematics (Maths) Class 11

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
AD
2p
?
?? ………..(i) 
Similarly, 
? ? q1
b
AD
2q
?
?? ……….(ii) 
and 
? ? r1
c
AD
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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