1. If A ad B are acute positive agles satisfyig the equatio 3si A si B 1 ad 3si A si B 0, the A B (a) (b) (c) (d) 6. 3 si A + si B = 1 3si A 1 si B 3 si A = cosb Also 3 si A si B = 0 si B = 3 si A Now, cos (A + B) = cosa cos B sia sib = cos A (3 si A) si A ( 3 si A) = 3 si A cos A 3 si A cos A = 0 A + B =. The umber of solutios of cos x 1si x,0 x 3, is (a) 3 (b) (c) (d) oe of these. As (a) Clearly 1 + si x > 0 equatio becomes cos x 1 si x cos x si x = 1 Clearly x = 0 is a solutio x = is a solutio 3π x is a solutio No other solutio is possible
3. The sum of all the solutios of the equatio x 0, 6 1 cos x.cos x.cos x, 3 3 is (a) 5 (b) 30 (c) 0 (d) oe of these. As (b) We have 1 cos xcos si x 3 1 1 cos x (1 cos x) 3 1 cos xcos x cos x ( cos x 3) = 1 cos 3x = 1 3x = x π 3 Where = 0, 1,, 3,, 5, 6, 7, 8, 9. the reqd. sum π (1 3... 9) = 3 π 9(9 1). 30 π. 3. If ta m ta, ( m ) the the differet values of are i (a) A.P. (c) G.P. (b) H.P. (d) o particular sequece. ta m ta m K m K K where K I m These values of are i A.P. with commo differece = m Hece (a) is the correct aswer
5. The set of value of x for which 0, (B) 0, 3 3 si x.cos x cos x.si x, 0 x, is, (D) Noe of these 3 3 si xcos x cos xsi x si x cos x cos x si x 0 1 si x cos x 0 si x 0 x ; I 0 x (give i the optios) 6. The umber of values of K for which the system of equatios K 1 x 8y K ad Kx K 3 y 3K 1 has ifiitely may solutios is 0 (B) 1 (D) Ifiite For ifiitely may solutios, we have k 1 8 k k k 3 3k 1 from first two k k 3 8k k k 3 0 k 1,3 from first ad last, k k 1 0 k 1 0 k 1 3k k 1 k takig the commo solutio we get k = 1 Hece oly oe solutio is possible
7. The umber of differet permutatios of all the letters of the word 'PERMUTATION' such that ay two cosecutive letters i the arragemet are either both vowels or both idetical is 63 6! 5! (B) 57 5! 5! 33 6! 5! (D) 7 7! 5! The letters other tha vowels are : PRMTTN Number of permutatios with o two vowels together is 6!! 7 C 5 5! Further amog these permutatios the umber of cases i which T's are together is 5! 6 C 5 5! So the required umber = 6!! 7 C 5 5! - 5! 6 C 5 5! = 57 (5!) 8. The umber of differet words that ca be formed usig all the letters of the word 'SHASHANK' such that i ay word the vowels are separated by atleast two cosoats, is 700 (B) 1800 900 (D) 600 The letters other tha vowels are SHSHNK which ca be arraged i 6!!! ways Now i its each case, let the first A be placed i the r th gap the the umber of ways to place the d A will be (7 - r - 1). So, the total umber of ways = 5 6! (6 r)!! r 1 = 6! (5 + + 3 + + 1) = 700.!!
9 13 17 9. The sum of the series upto ifiity 3 5..1 5.3. 5..3 1 (B) 9 5 1 5 (D) 5 r 1 Hit : Tr r, r 5 rr ( 1) 5 r( r1) 1 1 r r1 r 5 r( r 1) 5 ( r 1) 5 r r T r 1 1 1 1 1 1 1 3 3 5.1 5. 5. 5.3 5.3 5. 1 5 10. If 15 13 11 9 7 5 3 x x x x x x x x 7 the 16 x is 15 (B) 16 x is less tha 15 16 x greater tha 15 (D) Nothig ca be said about 16 x Hit : 15 13 11 9 7 5 3 x x x x x x x x 8 x( x 1)( x 1)( x 1) 16 8 x 1 ( x 1)( x 1)( x 1)( x 1) 7 1 ( x 1) 7 x 1 x x 16 x 15
11. If a, b, c, d are distict itegers i AP such that 0 (B) 1 d a b c the a + b + c + d is (D) Noe Hit : d a b c a 3 t ( a t) a ( a t) 5t 3(a 1) t 3a a 0 D 0 a 16a 9 0 1 70 1 70 a 3 3 a 1,0 3 a0, t 0, 5 a 1, t 1, 5 t 1 a b c d 1. The least value of sec + cosec + sec + cosec i (0, /) is + (B) - 5 (D) 5 - Sice sec + cosec ad sec + cosec has miimum value at = /, we have miimum value of sec + cosec + sec + cosec = +
13. If cosa = cosb + cos 3 B, sia = sib - si 3 B. The si(a-b) = 1/ (B) 1/3 /3 (D) 1/5 si(a - B) = sia cosb - cosa sib... (i) Substitutig the values of cosa ad sia from the give equatios i (i), he have si(a - B) = si Bcos B... (ii) squarig ad addig give equatio we get cosb = 1/3 si(a - B) = 1/3 1. If x + y + z + w = 5, the the least value of x cot9 + y cot7 + z cot63 + w cot81 is 5/ (B) 5 5 5/ (D) 5 5 We have x y z w ta9 ta 7 ta 63 ta81 5 5 5 5 15. The umber of solutios of [si x] + cos x = 1, i x, [where [x] = greatest iteger fuctio], is 3 (B) 5 (D) 6 [si x] = - 1, 0, 1 If [si x] = - 1 the cosx ot possible If si x 0 the cos x 1 x =,, 3, If si x 1 The cosx 0 5 x Number or solutios = 5 for x,.
16. Let f(x) = cosec x + sec x + cosec x, the miimum value of f(x), for x (0, /), is 1 1 (B) 1 1 1 (D) 1 (B) f(x) = si x cos x = si x si x 1si x cos x si x = 1 si x cos x = si x cos x 1 si x cos x 1 f (x) cos x si x si x cos x 1 f(x) is decreasig for 0 < x < / ad icrease for / < x < / miimum is at x = f mi (x) = f 1 1 17. 1 lim, r r1 where [] deotes the G.I.F, is 1 (B) 0 does ot exist (D) oe of these (B) Clearly r1 1 1 1 r 1 lim 0 r r1
x x 18. The maximum value of f(x) = x 1, (xr), is - (B) -3 - (D) -1 f(x) = 3 - ( x 1) ( x 1) 3 - = -1 19. The umber of all the odd divisor of 3600 is 5 (B) 18 (D) 9 Solutio : (D) 3600 = 3 5 so umber of odd divisors = ( + 1) ( + 1) = 9 0. Let A be the set of -digit umbers a 1 a a 3 a where a 1 > a > a 3 > a, the is equal to 16 (B) 8 10 (D) oe of these Solutio: Ay selectio of four digits from the te digits 0, 1,, 3,..., 9 gives oe such umber. So, the required umber of umbers = 10 C = 10. Hece is the correct aswer.
1. Total umber of ways of selectig two umbers from the set {1,, 3,,., 3} so that their sum is divisible by 3 is equal to: (B) 3 Solutio : (D) 3 Give umbers ca be rearraged as 1 7. 3 3 type 5 8. 3 1 3 1 type 3 6 9. 3 3 type That meas we must take two umbers from last row or oe umber each from first ad secod row. Total ways = C + C 1. C 1 = 1 3 =. A polygo has 65 diagoals. The umber of its sides is 8 (B) 10 11 (D) 13 Solutio : (D) Let umber of sides be, the C = 65 ( 3) = 130 = 13
3. I a chess touramet, all participats were to play oe game with the other. Two players fell ill after havig played 3 games each. If total umber of games played i the touramet is equal to 8, the total umber of participats i the begiig was equal to: Solutio : 10 (B) 15 1 (D) 1 Let there were players i the begiig. Total umber of games to be played was played to C ad each player would have played ( 1) games. Thus C (( 1) + ( 1) 1) + 6 = 8 5 150 = 0 = 15. If objects are arraged i a row, the the umber of ways of selectig three objects so that o two of them are ext to each other is 3 (B) 6 - C 3-3 C 3 + -3 C (D) oe of these. Solutio: Let x 0 be the umber of objects to the left of the first object chose, x 1 the umber of objects betwee the first ad the secod, x the umber of objects betwee the secod ad the third ad x 3 the umber of objects to the right of the third object. We have x 0, x 3 0, x 1, x 1 ad x 0 + x 1 +x + x 3 = 3... (1) The umber of solutios of (1) = coefficiet of y -3 i (1+ y+ y +...)(1+ y + y +...)(y + y + y 3 +...)(y + y + y 3 +...) = coefficiet of y -3 i y ( 1+ y + y +y 3 +...) = coefficiet of y -5 i (1- y) - = coefficiet of y -5 i ( 1+ C 1 y + 5 C y + 6 C 3 y 3 +...) = -5 +3 C -5 = - C 3 3 = 6 also -3 C 3 + -3 C = - C 3. Hece, (B) ad are the correct aswers.
I order to solve the equatio of the form Put a rcos ; b rsi b Paragraph for Questios Nos. 5 to 7 a cos bsi c... 1 r a b, ta equatio (1) becomes a rcos cos rsi si c cos a c b If c a b, the the equatio acos bsi c has o solutio If c a b, the put a c b cos t cos cos t t t 5. The solutio of si x 3 cos x is 5, (B) 1 1 5, 1 1, (D) Noe of these 6 6 1 3 1 si x cos x cosx cos 6 x 6 x 6 5 x or x 1 1
6. kcos x 3si x k 1 is solvable oly if k (, ] (B) k (, ) k cos x 3si x k 1 k 3 k 1 cos x si x k 9 k 9 k 9 k 1, (D) Noe of these k 1 k the cos x ; where cos k 9 k 9 ad si 3 k 9 1 cos x 1 ; k1 1 1 k 9 k1 1 k 1 k 9 k 9 7. If 0 x, ; k 3 3 the umber of solutios of equatio 3 si x cos x si x cos x 8 is 0 (B) 1 (D) 3 si x cos x si x cos x si x cos x si x cos x 8 si x cos x 3 1si x cos x 8 si x cos x1 si x cos x 8 si x cos x 3 8 ; si x cos x six 1, Not possible The give equatio has o. solutio
Paragraph for Questio Nos. 8 to 0 If a sequece or series is ot a direct form of a AP, GP, etc. The its th term ca ot be determied. I such cases, we use the followig steps to fid the th term T of the give sequece. Step I : Fid the differeces betwee the successive terms of the give sequece. If these differeces are i AP, the take T a b c, where a,b,c are costats. Step II : If the successive differeces foud i step I are i GP with commo ratio r, the 1 take T a b cr, where a, b, c are costats. Step III : If the secod successive differeces (Differeces of the differeces) i step I are i 3 AP, the take T a b c d, where a, b, c, d are costats. Step IV : If the secod successive differeces (Differeces of the differeces) i step I are i 1 GP, the take T a b c dr, where a, b, c, d are costats. Now let sequeces : A : 1, 6, 18, 0, 75, 16,.. B : 1, 1, 6, 6, 91, 91,. C : l l, l 3, l 10.. 3 8. If the th term of the sequece A is T a b c d the the value 6a b d is l (B) l 8 (D) Hit : (D) 3 T a b c d T1 a b c d 1 T 8a b c d 6 6a b d 9. For the sequece 1, 1, 6, 6, 91, 91,.. Fid the S 50 where S 50 50 T r r1 5 50 3 1 3075 50 (B) 8 5 3 1 5075 8
5 3 1 175 (D) Noe of these 8 50 Hit : T 5 1 5 9 3 9 5 5 9 S50 1 3... 3 1... 50 50. 50 5 3 1 5 50.51 50. 5 50 15.51 50 3 1 8 5 3 50 1 3075 8 1.. 1 3.....1 30. The sum of the series 1 6 (B) 1 3 1 1 6 (D) 1 1 3 Hit : r r 1 1 r r r1 r1 r1 1 1 1 1 6 1 3 3 1 = 6 1 6