SAMPLE QUESTION PAPER–6
Section-A
1. We have R = {(x, y) : (x – y) is odd natural number x A, y B}. 1 R = {5, 4} 2. f (x) = 1 5x . It is defined when 5 – x > 0 x < 5. Domain of f = {x : x R , x < 5} Range = (0, ). 1 3. 0 tan 8 lim sin 3 x x x = x x x x x x x 0 tan 8 8 8 lim sin 3 3 3 = 8. 3 1 4. cosec (– 1410°) = cosec (– 360° × 4 + 30) = cosec 30° = 2. 1 5. 2 2 x = 2 x = 4 + 2 = 6 and 3 2 y = – 5 y – 3 = – 10 y = – 10 + 3 = – 7 Co-ordinates of B are (6, – 7). 1
6. If two lines do not intersect in plane, then they are parallel. 1
Section - B
7. L.H.S. = cosA cosB cosC
a b c 1
=
2 2 2 2 2 2 2 2 2
2 2 2
b c a a c b a b c abc abc abc
2 = 2 2 2 2 2 2 2 2 2 2 b c a a c b a b c abc 1 = 2 2 2 2 a b c abc = R.H.S. Hence proved. 1 OR
Given that, cot = 1
2 and sec = 5 3
1
tan = 2 and tan =
2 5 1 3 O (2, –5) B (x, y) A (–2, –3)
AglaSem Schools
www.schools.aglasem.com
tan = 25 9 9 tan = + 4 3 [ B 2, 2
tan ( +) = tan tan
1 tan tan 1 = 4 2 3 4 1 2 3 = 2 3 11 3 1 tan ( + ) = 2 11 8. L.H.S. = c o t 71 2 = 1 c os 7 2 1 s in 7 2 1 = 1 1 2 cos 7 . cos 7 2 2 1 1 2 sin 7 . cos 7 2 2 = 2 1 2 cos 7 2 sin 15 = 1 cos 15 sin 1 5 1 = 1 cos (45 30 ) sin (45 30)
= 1 cos 45 cos 30 sin 45 sin 30 sin 45 . cos 30 cos 45 sin 30
= 1 3 1 1 1 2 2 2 2 1 3 1 1 2 2 2 2 1 = ( 3 1) 1 2 2 3 1 2 2 = 2 2 3 1 3 1 = (2 2 3 1) ( 3 1) ( 3 1) ( 3 1) = 2 2 2 6 3 3 2 2 3 1 ( 3) 1
AglaSem Schools
www.schools.aglasem.com
= 2 6 2 3 2 2 4 2 = 6 3 22 = 6 3 2 4 = R.H.S. Hence proved. 1 9. We have 2kx – 5k = (x – 3).x2 (k – 3).x2 – 2kx + 5k = 0 = (–2 k)2 – 4 (k – 3). 5k 1 = 4k2 – 20k (k – 3) = 4k (k – 5k + 15) = 4k (– 4k + 15) = 4k (4k – 15) 1
For real and unequal root
> 0 1 – 4k (4k – 15) > 0 4k (4k – 15) < 0 0 < k < 15 4 1 OR (3 5)(3 5) ( 3 2) ( 3 2) i i i i = 2 9 5 2 2 2 i i i 2 = i i2 7 2 2 = 7 2 2 i = 0 7 2 2 i 2
10. Veena visits to four cities A, B, C and D.
Total arrangements in which Veena can visit = 4 ! = 24 1 (i) She visits A before B and B before C
X = {ABCD, DABC, ABDC, ADBC}
P(X) = 4
24= 1
6 1
(ii) Y = Order of cities A first and B last
Y = {ACDB, ADCB} P(Y) = 2 24= 1 . 12 2
11. We have a pack of 52 playing cards
Total ways of selecting 4 cards =52C 4 = 52 51 50 49 4 3 2 1 1
AglaSem Schools
www.schools.aglasem.com
A = Event of getting 1 card from each suit P(A) = 13 13 13 13 1 1 1 1 52 51 50 49 4 3 2 1 C C C C 1 = 13 13 13 13 4 3 2 52 51 50 49 P(A) = 2197 20825 2 12. P(n) : 4n + 15n – 1 is divisible by 9 at n = 1, P(1) : 4 + 15 – 1 = 18 is divisible by 9 So, P(1) is true 1 Let P(k) is true. i.e., 4k + 15k–1 is divisible by 9 4k +15k–1 = 9, N. 4k = 9 – 15k + 1 ....(i) 1
Now to show P(k +1) is true.
i.e., 4k+1 + 15(k + 1) – 1 is divisible by 9.
Takeing L.H.S.= 4k+1 + 15(k +1) – 1 = 4k.4 + 15k + 15 – 1
Using equation (i), we get 1
4 (9 – 15k+1) + 15k + 14
= 36 – 60k + 4 + 15k + 14 = 36 – 45k + 18
= 9(4 – 5k + 2)
= 9, = 4 – 5k + 2) 1
P(k + 1) is true, when P(k) is true. So, P(n) is true for all n N.
OR P(n) 3·51 5·71 7·91 ....(2n 1)(21 n 3) = n n 3(2 3) Put n = 1, P(1) : 1 3·5= 1 3·5 So, P(1) is true. 1 Let P(k) is true. P(k) : k k 1 1 1 .... 3·55·7 (2 1)(2 3) = k k 3(2 3) ....(i)
Now prove that P(k + 1) is true.
i.e., k k k k 1 1 1 1 .... 3·5 5·7 (2 1)(2 3) (2 3)(2 5) = ( 1) 3(2 5) k k
AglaSem Schools
www.schools.aglasem.com
Take L.H.S.
k k k k
1 1 1 1
....
3·5 5·7 (2 1)(2 3) (2 3)(2 5) 1
Using equation (i), we get
= k k k k 1 3(2 3) (2 3) (2 5) 1 = (2 5) 3 3(2 3) (2 5) k k k k = 2 2 5 3 2(2 3)(2 5) k k k k = (2 3) ( 1) 3(2 3) (2 5) k k k k = 1 3(2 5) k k = R.H.S. 1
Hence, P(k + 1) is true, when P(k) it true. P(n) is true for all n N.
13. f(x) = 2 2 1 x x , x R ½ For domain, as x2 > 0 x2 + 1 > 0. ½
f(x) is defined for all x R.
Domain of f = R.
For range, let f(x) = y
y = 2 2 1 x x 1 x2 = 1 y y ½ x = 1 y y 1 1 – y > 0 y < 1 and y > 0. Range : 0 < y < 1. ½ 14. f (x) = sin (5x – 8) f (x) = 0 ( ) ( ) lim h f x h f x h = h x h x h 0 sin {5 ( ) 8} sin (5 8) lim 1 = h x h x x h x h 0 2 5 5 8 5 8 5 5 8 5 8
lim cos ·sin
2 2 = h h h x h 0 1 5 5
2 lim cos 5 8 ·sin
2 2 1
AglaSem Schools
www.schools.aglasem.com
= h h h x h 0 5 sin 5 2 5 lim cos (5 8 5 2 2 1 = 5.cos (5x – 8). h h h 0 5 sin 2 lim 5 2 = 5 cos (5x – 8), 0 sin lim 1 x x x 1 15. f(x) = 1 , [ ] x R x x As, 0 < (x – [x]) < 1, x R 1 But when x Z x – [x] = 0 0 < (x – [x]) < 1, x R – Z 1 Domain of f = R – Z. For range, as 0 < x – [x] < 1, x R – Z 0 < x[ ]x < 1 1 1 < 1 x x < 1 < f(x) < Range = (1, ). 1
16. Let two positive integers be x and y. According to question, a = 2 x y ....(i) and x, b, c, y are in G.P. 1
Let r be common ratio.
b = xr, c = xr2, y = xr3 r = y x 1 / 3 b = x y x 1 / 3 and c= x y x 2 / 3 1 b3 = 3 , x y x c 3 = 3 2 2 x y x b3 + c3 = 3 3 2 x y x y x x = x2y + xy2 1
AglaSem Schools
www.schools.aglasem.com
b3 + c3 = xy (x + y)
Using (i), x + y = 2a
and xy = bc
b3c3 2abc 1 17. A (1, 4, 2), B (2, – 3, 4), and C (–2, 1, 2) are co-ordinates of vertices of ABC, 1
AB2 = (2 –1)2 + (– 3 – 4)2 + (4 – 2)2 = 1 + 49 + 4 = 54 1 BC2 = (– 2 – 2)2 + (1 + 3)2 + (2 – 4)2 BC2 = 16 + 16 + 4 = 36 and AC2 = (– 2 – 1)2 + (1 – 4)2 + (2 – 2)2 = 9 + 9 = 18 BC2 + AC2 = 36 + 18 = 54 = AB2 2
ABC is a right triangle.
OR
A (–1, 3), B (2, – 1) and C (0, 0) are the vertices of ABC. P is concurrent point of altitude, so
it is orthocentre. Now, slope of BC = 1 2 1 slope of AD = 2 Equation of AD is y – 3 = 2 (x + 1) 2x – y + 5 = 0 ....(i) 1 Now, slope of AC = 3 1 = – 3 slope of BE = 1 3 Equation of BE is y + 1 = 1(x 2) 3 3y + 3 = x – 2 x – 3y = 5 ....(ii) 1 On solving equations (i) and (ii), we get
x = – 4, and y = – 3. Co-ordinates of orthocentre = P (– 4, – 3). 1 18. a 0 30 2 15 3 9 6 18 30 60 3 45 2 4 6 12 60 90 5 75 1 1 5 5 90 120 10 105 0 0 0 0 120 150 3 135 1 1 3 3 150 180 5 165 2 4 10 20 180 210 2 195 3 9 6 18 2 2 105 Class 30 Total 30 2 76 i i i i i i i i i x f x u u f u f u A (1, 4, 2) C (– 2, 1, 2) B (2, – 3, 4) A (–1, 3) C (0, 0) E P (x, y) B (2, – 1)
AglaSem Schools
www.schools.aglasem.com
Variance = 2 = 2 2 h N [N. fiui 2 – (f i ui) 2] Here h = 30, N = 30. 2 2 =
2 2 2 30 30 76 (2) 30 = [2280 – 4] 2 2276 2 19. We have word DAUGHTER which consists 3 vowels and 5 consonants. Number of words consisting 2 vowels and 3 consonants = 3C 2 × 5C 3 × 5 ! 2 = 3 2 2 1 × 5 4 3 120 3 2 1 = 3600. 2
Section - C
20. L.H.S. = cos 4 cos 2 cos3
sin 4 sin 2 sin 3
x x x x x x = x x x x x x x x x x 4 2 4 2
2 cos ·cos cos 3
2 2
4 2 4 2
2 sin ·cos sin 3
2 2 2 =
2 cos 3 . cos cos 3 2 sin 3 . cos sin 3
x x x x x x 2 = cos 3 (2 cos 1) sin 3 (2 cos 1) x x x x 2
= cos 3x = R.H.S. Hence proved.
21. A = Number of people who read English news-paper 1
B = Number of people who read Hindi news-paper 1
According to question, 1
n(A) = 2250, n(B) = 1750 and n(A B) = 875.
n (A B) = n(A) + n(B) – n (A B)
= 2250 + 1750 – 875
= 400 – 875 = 3125 1
Number of people who neither read Hindi nor English = n(U) – n (A B)
= 5000 – 3125 = 1875. 1
Number of people who reads only English news-paper = 2250 – 875
= 1375. 1
22. Equations of two straight roads are
2x – 3y – 4 = 0 ...(i) and 3x + 4y – 5 = 0 ...(ii) 1 A B Hindi English n(U)=5000 n A B =875
AglaSem Schools
www.schools.aglasem.com
Family is standing at the point P.
On solving equations (i) and (ii), we get
x = 31 71, y = 2 17 P 31, 2 17 17 1 Because family is at P 31, 2 17 17 want to reach the path 6x – 7y + 8 = 0 ...(iii) ½ in least time.
So they have to walk along PN.
Now, slope of path (iii) = 6
7 1 Slope of PN = 7 6 Equation of PN = 2 17 y = 7 31 6 x 17 17 2 17 y = 7 17 31 6 17 x 1 102y + 12 = – 119x + 217 119x + 102y = 217 – 12 ½ 119x + 102y= 205.
So, he should follow the path given by 119x + 102y = 205. 1 Because of unavailability of road transport, they have less time to reach to station, that is why they have decided to took the shortest path.
23. The expansion is (1 + x)n
Tr + 1 =nC
r x r
Coefficients of (r – 1)th, rth and (r + 1)th terms are nCr–1, nCr and nCr+1 1 According to question, nC r–1 : nC r : nC r+1 = 45 : 120 : 210 n r n r C C 1 = 45 120 1 n r n r r n r n ( )! !( ) ! ( 1)!( 1)! ! = 45 120 3x – 11r + 3 = 0 ....(i) 1 Also, n r n r C C1 = 120 210 n r n r r n r n ! ( 1) !( 1) ! !( ) ! ! = 4 7 1 4n – 11r – 7 = 0 ....(ii) 1
On solving (i) and (ii), we get
n = 10 and r = 3 n10 1 P(x, y) N 3x + 4y – 5 = 0 6x – 7y + 8 = 0 2x – 3y – 4 = 0
AglaSem Schools
www.schools.aglasem.com
OR
(52)4 = (50 + 2)4 1
Using Binomial theorem, we get
= (50)4 + 4C1 503 × 2 + 4C2 × 502 × 22 + 4C3 × 50 × 23 + 4C4 ×24 2 = 6250000 + 8 × 125000 + 24 × 2500 + 32 × 50 + 16 1
= 6250000 + 1000000 + 60000 + 1600 + 16 1
= 73,11,616 2
24. Given inequations are
3x + 2y < 60 ....(i) x + 3y < 30, x > 0, y > 0 ....(ii) Take 3x + 3y = 60 At x = 0, y = 20 1 and y = 0, x = 20 At (0, 0); 0 + 0 < 60, (True) i.e., (0, 0) is included. 1 Now take x + 3y = 30 At x = 0, y = 10 and y = 0, x = 30. At (0, 0); 0 + 0 < 30, (True)
So, origin is also included. 2
So, shaded region OABC is the solution region.
25. L.H.S. = n n n n 2 2 2 2 2 2 1·2 2·3 .... ( 1) 1 ·2 2 ·3 .... ( 1) = n n n n 2 2 [ ( 1) ] [ ( 1)] 1 = n n n n n 3 2 3 2 [ 2 ] [ ] = n n n n n 3 2 3 2 2 1 = n n n n n n n n n n n n 2 2 2 2 ( 1) ( 1)(2 1) ( 1) 2 4 6 2 ( 1) ( 1)(2 1) 4 6 = n n n n n n n n n n n n 2 2 2 2 3 ( 1) 4 ( 1)(2 1) 6 ( 1) 12 3 ( 1) 2 ( 1)(2 1) 12 1 = n n n n n n n n n n 2 2 ( 1)(3 3 8 4 6) ( 1)(3 3 4 2) 1 = n n n n 2 2 3 11 10 3 7 6 1 40 30 20 3x + 3 y = 60 10 0 20 30 40 A(20, 5) B(15, 5) C(0, 10) 10 x + 3y = 30 X X' Y' Y 2
AglaSem Schools
www.schools.aglasem.com
= n n n n (3 5)( 2) (3 1)( 2) = n n 3 5 3 1 = R.H.S. Hence proved. 1 26. Given that,
(a + ib) (c + id) (e + if) (g + ih) = A + iB 1
(a – ib) (c – id) (e – if) (g – (ih) = A – iB 2
(A + iB) (A – iB) = (a + ib) (a – ib) (c + id)
(c – id) (e + if) (e – if) (g + ih) (g – ih) 2
A2 + B2= (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2). 1 OR We have 2 5 4 z i z i
= 5, where z is only complex number. 2 2 5 4 z i z i = 25 1 2 2 5 4 x iy i x iy i = 25 1 2 ( 2) ( 1) ( 5) ( 4) x i y x i y = 25 As, |z|2 = zz 1 ( 2) ( 1) ( 2) ( 1) ( 5) ( 4) ( 5) ( 4) x i y x i y x i y x i y = 25