Binomial Theorem PDF Past Papers

Binomial Theorem Question1 [27-Jan-2024 Shift 1] Options: A. B. C. D. Answer: A Solution: ------------------------------------------------------------------------------------------------- Question2 If A denotes the sum of all the coefficients in the expansion of (1 − 3x + 10x2)n and B den...

Binomial Theorem Question1 [27-Jan-2024 Shift 1] Options: A. B. C. D. Answer: A Solution: ------------------------------------------------------------------------------------------------- Question2 If A denotes the sum of all the coefficients in the expansion of (1 − 3x + 10x2)n and B denotes the sum of all the coefficients in the expansion of (1 + x2)n, then : [27-Jan-2024 Shift 1] Options: A. A = B3 B. 3A = B C. B = A3 D. A = 3B Answer: A Solution: ------------------------------------------------------------------------------------------------- Question3 The coefficient of x2012 in the expansion of (1 − x)2008 (1 + x + x2)2007 is equal to [27-Jan-2024 Shift 2] Answer: 0 Solution: ------------------------------------------------------------------------------------------------- Question4 [29-Jan-2024 Shift 1] Answer: 2041 Solution: ------------------------------------------------------------------------------------------------- Question5 [29-Jan-2024 Shift 2] Answer: 1 Solution: ------------------------------------------------------------------------------------------------- Question6 Number of integral terms in the expansion of is equal to [30-Jan-2024 Shift 1] Answer: 138 Solution: ------------------------------------------------------------------------------------------------- Question7 Suppose 2 − p, p, 2 − α, α are the coefficient of four consecutive terms in the expansion of (1 + x)n. Then the value of p2 − α2 + 6α + 2p equals [30-Jan-2024 Shift 2] Options: A. 4 B. 10 C. 8 D. 2 Answer: D Solution: ------------------------------------------------------------------------------------------------- Question8 [30-Jan-2024 Shift 2] Answer: 10 Solution: Question9 In the expansion of (1 + x)(1 − x2) the sum of the coefficient of x3 and x−13 is equal to [31-Jan-2024 Shift 1] Options: Answer: 118 Solution: Question10 [31-Jan-2024 Shift 2] Options: A. 380 B. 376 C. 384 D. 372 Answer: D Solution: Question11 [31-Jan-2024 Shift 2] Answer: 25 Solution: ------------------------------------------------------------------------------------------------- Question12 If the Coefficient of x30 in the expansion of (1 + 1/x)6(1 + x2)7(1 − x3)8; x ≠ 0 is α, then |α| equals______ [1-Feb-2024 Shift 1] Answer: 678 Solution: Coeff. = 7 + (15 × 21) + (15 × 35) + (35) − (6 × 8) − (20 × 7 × 8) − (6 × 21 × 8) + (15 × 28) + (7 × 28) = −678 = α |α| = 678 Question13 Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of (n/m)1/3 is: [1-Feb-2024 Shift 2] Options: A. 4/9 B. 1/9 C. 1/4 D. 9/4 Answer: D Solution: ------------------------------------------------------------------------------------------------- Question14 22 22 The value ∑ r=0 Cr23Cr is [24-Jan-2023 Shift 1] Options: 45 A. C23 44 B. C23 45 C. C24 44 D. C22 Answer: A Solution: Solution: 22 22 22 ∑ Cr ⋅ 23Cr = ∑ 22 Cr ⋅ 23C23 − r r=0 r=0 45 = C23 ------------------------------------------------------------------------------------------------- Question15 2023 Suppose ∑ r=0 r2 2023 Cr = 2023 × α × 22022. Then the value of α is [24-Jan-2023 Shift 1] Answer: 1012 Solution: using result n ∑ r2nCr = n(n + 1) ⋅ 2n − 2 r=0 2023 Then ∑ r2 2023 Cr = 2023 × 2024 × 22021 r=0 = 2023 × α × 22022 So, α = 1012 ------------------------------------------------------------------------------------------------- Question16 30 2 30 2 30 2 30 2 α60! If ( C1) + 2( C2) + 3( C3) +....... + 30( C30) = , then α is (30!)2 equal to [24-Jan-2023 Shift 2] Options: A. 30 B. 60 C. 15 D. 10 Answer: C Solution: Solution: 30 2 30 2 30 2 30 2 S = 0 ⋅ ( C0) + 1 ⋅ ( C1) + 2 ⋅ ( C2) +......⋅ + 30 ⋅ ( C30) 30 2 30 2 30 2 30 2 S = 30 ⋅ ( C0) + 29 ⋅ ( C1) + 28 ⋅ ( C2) +...... + 0 ⋅ ( C0) 30 2 30 2 30 2 2S = 30 ⋅ ( C0 + C1 +......⋅ + ⋅ C30 ) 60 60! S = 15 ⋅ C30 = 15⋅ 2 (30!) 15 ⋅ 10! α ⋅ 60! 2 = 2 (30!) (30!) ⇒ α = 15 ------------------------------------------------------------------------------------------------- Question17 Let the sum of the coefficients of the first three terms in the expansion ( ) n 3 4 of x− 2 , x ≠ 0, n ∈ N, be 376. Then the coefficient of x is___ x [24-Jan-2023 Shift 2] Answer: 405 Solution: Given Binomial x 3 n 2 ( x− , x ≠ 0, n ∈ N, ) Sum of coefficients of first three terms n n n 2 C0 − C1 ⋅ 3 + C23 = 376 2 ⇒ 3n − 5n − 250 = 0 ⇒ (n − 10)(3n + 25) = 0 ⇒ n = 10 ( ) r 10 10 − r −3 Now general term Crx 2 x 10 10 − r r −2r = Crx (−3) ⋅ x 10 r 10 − 3r = Cr(−3) ⋅ x 4 Coefficient of x ⇒ 10 − 3r = 4 ⇒r=2 10 2 C2(−3) = 405 ------------------------------------------------------------------------------------------------- Question18 If ar is the coefficient of x10 − r in the Binomial expansion of (1 + x)10, ( ) 2 10 3 ar then r=1 ∑ r ar − 1 is equal to [25-Jan-2023 Shift 1] Options: A. 4895 B. 1210 C. 5445 D. 3025 Answer: B Solution: Solution: ar = 10C10 − r = 10 Cr ( ) 10 2 10 3 Cr ⇒ ∑ r 10 r=1 Cr − 1 10 ( 11r− r ) 2 3 = ∑ r r=1 10 = ∑ r(11 − r)2 r=1 10 3 2 = ∑ (121r + r − 22r ) = 1210 r=1 ------------------------------------------------------------------------------------------------- Question19 The constant term in the expansion of ( ) 5 1 2 2x + + 3x is ______. x7 [25-Jan-2023 Shift 1] Answer: 1080 Solution: Solution: n −7 n 2 n3 5!(2x) 1(x ) 2(3x ) General term is ∑ n1!n2!n3! For constant term, n1 + 2n3 = 7n2 &n1 + n2 + n3 = 5 Only possibility n1 = 1, n2 = 1, n3 = 3 ⇒ constant term = 1080 ------------------------------------------------------------------------------------------------- Question20 6 51 − k ∑ k=0 C3 is equal to [25-Jan-2023 Shift 2] Options: 51 45 A. C4 − C4 51 45 B. C3 − C3 52 45 C. C4 − C4 52 45 D. C3 − C3 Answer: C Solution: Solution: 6 51 − k ∑ C3 k=0 51 50 49 45 = C3 + C3 + C3 +... + C3 45 46 51 = C3 + C3 +.... + C3 45 45 46 51 45 = C4 + C3 + C3 +.... + C3 − C4 n n n+1 ( Cr + Cr − 1 = Cr) 52 45 = C4 − C4 ------------------------------------------------------------------------------------------------- Question21 2023 The remainder when (2023) is divided by 35 is ________. [25-Jan-2023 Shift 2] Answer: 7 Solution: (2023)2023 = (2030 − 7)2023 = (35K − 7)2023 = 2023C0(35K)2023(−7)0 + 2023C1(35K)2022(−7)+..... +....... + 2023C2023(−7)2023 = 35N − 72023.  Now, − 72023 = −7 × 72022 = −7(72)1011 1011 = −7(50 − 1) 1011 1011 1011 1010 1011 = −7( C050 − C1(50) +...... C1011) = −7(5λ − 1) = −35λ + 7 2023 ∴ when (2023) is divided by 35 remainder is 7 ------------------------------------------------------------------------------------------------- Question22 ( ) 11 9 3 1 −9 If the co-efficient of x in αx + βx and the co-efficient of x in ( αx − ) 11 1 2 are equal, then (αβ) is equal to _______. βx3 [29-Jan-2023 Shift 1] Answer: 1 Solution: 5 Coefficient of x in 9 ( αx + 31 βx )= 11 α C6 ⋅ 6 β ∵ Both are equal 5 6 11 α 11 α ∴ ⋅ 6 =− ⋅ 5 C6 β C5 β 1 ⇒ = −α β ⇒ αβ = −1 2 ⇒ (αβ) = 1 ------------------------------------------------------------------------------------------------- Question23 Let the coefficients of three consecutive terms in the binomial n expansion of (1 + 2x) be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of these three terms, is _______. [29-Jan-2023 Shift 1] Answer: 1120 Solution: tr + 1 = nCr(2x)r  n r−1 Cr − 1(2) 2 ⇒ n r = Cr(2) 5 n! (r − 1)!(n − r + 1)! 2 ⇒ = n!(2) 5 r!(n − r)! r 4 ⇒ = ⇒ 5r = 4n − 4r + 4 n−r+1 5 ⇒ 9r = 4(n + 1)... (1) n r Cr(2) 5 ⇒ n r+1 = C (2) 8 r+1 n! r!(n − r)! 5 r+1 5 ⇒ = ⇒ = n! 4 n−r 4 (r + 1)!(n − r − 1)! ⇒ 4r + 4 = 5n − 5r ⇒ 5n − 4 = 9r... (2) From (1) and (2) ⇒4n + 4 = 5n − 4 ⇒ n = 8 (1) ⇒r = 4 so, coefficient of middle term is 8 4 8×7×6×5 C42 = 16× = 16 × 70 = 1120 4×3×2×1 ------------------------------------------------------------------------------------------------- Question24 Let K be the sum of the coefficients of the odd powers of x in the 99 expansion of (1 + x). Let a be the middle term in the expansion of ( ) 200 200 C99K 2l m 1 2+. If a = n , where m and n are odd numbers, then the √2 ordered pair (ℓ, n) is equal to : [29-Jan-2023 Shift 2] Options: A. (50, 51) B. (51, 99) C. (50, 101) D. (51, 101) Answer: C Solution: In the expansion of 99 2 99 (1 + x) = C0 + C1x + C2x +.... + C99x 98 K = C1 + C3 +.... + C99 = 2 ( 2 + √21 ) 200 a⇒ Middle in the expansion of ( √21 ) 100 200 100 T 200 = C100(2) +1 2 200 50 = C100 ⋅ 2 200 98 C99 × 2 100 48 So, 200 50 = ×2 C100 × 2 101  25 50 m ′ So, ×2 = 2 101 n ∴m, n are odd so (ℓ, n) become (50, 101) Ans. ------------------------------------------------------------------------------------------------- Question25 ( ) 15 15 3 1 If the coefficient of x in the expansion of ax + 1 is equal to bx 3 ( ) 1 15 −15 3 1 the coefficient of x in the expansion of ax − 3 , where a and b bx are positive real numbers, then for each such ordered pair (a, b) : [30-Jan-2023 Shift 1] Options: A. a = b B. ab = 1 C. a = 3b D. ab = 3 Answer: B Solution: Solution: Option (2) ( ax ) 15 Coefficient Of x15 in 3 + 1 1∕3 bx ( bx1 ) r 15 3 15 − r T r+1 = Cr(ax ) 1∕3 r 45 − 3r − = 15 3 10r 30 = 3 r=9 15 15 6 −9 Coefficient of x = C9a b ( ax − bx1 ) 15 Coefficient of x−15 in 1∕3 3 ( − bx1 ) r 15 1 ∕ 3 15 − r T r+1 = Cr(ax ) 3 r 5− − 3r = −15 3 10r = 20 3 r=6 15 9 −6 Coefficient = C6a × b 9 6 a a ⇒ = 9 b6 b 3 3 ⇒ a b = 1 ⇒ ab = 1 ------------------------------------------------------------------------------------------------- Question26 The coefficient of x301 in (1 + x)500 + x(1 + x)499 + x2(1 + x)498 +..... + x500 is: [30-Jan-2023 Shift 1] Options: 501 A. C302 500 B. C301 500 C. C300 501 D. C200 Answer: D Solution: Solution: 500 499 2 498 500 (1 + x) + x(1 + x) + x (1 + x) +.... + x { ( 1 +x x ) } 501 1− = (1 + x)500 ⋅ x 1− 1+x 501 501 ((1 + x) −x ) = (1 + x)500 ⋅ (1 + x) (1 + x)501 501 501 = (1 + x) −x 301 501 501 Coefficient of x in (1 + x) −x is given by 501 501 C301 = C200 ------------------------------------------------------------------------------------------------- Question27 13 9 Let x = (8√3 + 13) and y = (7√2 + 9). If [t] denotes the greatest integer ≤t, then [30-Jan-2023 Shift 2] Options: A. [x] + [y] is even B. [x] is odd but [y] is even C. [x] is even but [y] is odd D. [x] and [y] are both odd Answer: A Solution: Solution: 13 13 13 12 1 x = (8√3 + 13) = C0 ⋅ (8√3 ) + C1(8√3 ) (13) +...  ′ 13 13 13 13 12 1 x = (8√3 − 13) = C (8√3 ) 0 − C1(8√3 ) (13) +... ′ 13 12 1 13 10 3 x − x = 2[ C1 ⋅ (8√3 ) (13) + C3(8√3 ) ⋅ (13)...] ′ therefore, x − x is even integer, hence [x] is even Now, y = (7√2 + 9)9 = 9C0(7√2 )9 + 9C1(7√2 )8(9)1 + 9C2(7√2 )7(9)2....... y′ = (7√2 − 9)9 = 9C0(7√2 )9 − 9C1(7√2 )8(9)1 + 9C2(7√2 )7(9)2....... y − y′ = 2[9C1(7√2 )8(9)1 + 9C3(7√2 )6(9)3 +...] y − y′ = Even integer, hence [y] is even ------------------------------------------------------------------------------------------------- Question28 th th 50 root of a number x is 12 and 50 root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _______. [30-Jan-2023 Shift 2] Answer: 23 Solution: x + y = 1250 + 1850 = (150 − 6)25 + (325 − 1)25 = 25K − (625 + 1) = 25K − ((5 + 1)25 + 1) = 25K1 − 2 Remainder = 23 ------------------------------------------------------------------------------------------------- Question29 Let α > 0, be the smallest number such that the expansion of ( ) 2 30 3 2 −α x + 3 has a term βx , β ∈ N. x Then α is equal to _______. [31-Jan-2023 Shift 1] Answer: 2 Solution: ( ) r 30 2 ∕ 3 30 − r 2 Tr + 1 = Cr(x ) 3 x 60 − 11r 30 3 = Cr ⋅ 2 ⋅ x  60 − 11r 60 < 0 ⇒ 11r > 60 ⇒ r > ⇒r=6 3 11 T7 = 30C6 ⋅ 26x−2 30 We have also observed β = C6(2)6 is a natural number. ∴α = 2 ------------------------------------------------------------------------------------------------- Question30 The remainder on dividing 599 by 11 is ________. [31-Jan-2023 Shift 1] Answer: 9 Solution: 599 = 54 ⋅ 595 = 62519 = 62519 = 625[3124 + 1]19 = 625[11k × 19 + 1] = 625 × 11k × 19 + 625 = 11k1 + 616 + 9 = 11(k2) + 9 Remainder = 9 ------------------------------------------------------------------------------------------------- Question31 ( ) 9 The Coefficient of x−6, in the expansion of 4x 5 + 5 , is _______ 2x2 [31-Jan-2023 Shift 2] Answer: 5040 Solution: ( ) 4x 9 5 + 5 2x 2 ( ) ( ) 4x 9−r r 9 5 Now, Tr + 1 = Cr ⋅ 5 2x 2 4 9−r ( ) ( 52 ) ⋅ x r 9 9 − 3r = Cr ⋅ 5 −6 Coefficient of x i.e. 9 − 3r = −6 ⇒ r = 5 ------------------------------------------------------------------------------------------------- Question32 ( ) 9 5 x2 4 If the constant term in the binomial expansion of 2 − is −84 xℓ and the Coefficient of x−3ℓ is 2αβ, where β < 0 is an odd number, Then |αℓ − β| is equal to _________ [31-Jan-2023 Shift 2] Answer: 98 Solution: ( ) 5 9 In, x2 − 4 ℓ 2 x 5∕2 9−r ( ) (x ) r 9 −4 T r + 1 = Cr 9−r ℓ 2 x 9 45 − 5r − r r Cr r = (−1) 9 − r 4 x 2 2 2 = 45 − 5r − 21r = 0 45 r=... (1) 5 + 21 9 r Cr r Now, according to the question, (−1) 9−r 4 = −84 2 r9 3r − 9 = (−1) Cr2 = 21 × 4 9 Only natural value of r possible if 3r − 9 = 0 r = 3 and C3 = 84 ∴1 = 5 from equation (1) 45 5r − − lr −31 Now, coefficient of x =x 2 2 at 1 = 5, gives r=5 5 9 4 α ∴ c5(−1) 4 = 2 × β 2 7 = −63 × 2 ⇒ α = 7, β = −63 ∴ value of | αℓ − β | = 98 ------------------------------------------------------------------------------------------------- Question33 1 1 1 1 1 The value of 1!50! + 3!48! + 5!46! +.... + 49!2! + 51!1! is [1-Feb-2023 Shift 1] Options: 250 A. 50! 50 2 B. 51!  251 C. 51! 51 2 D. 50! Answer: B Solution: Solution: 26 26 1 1 ∑ = ∑ 51C(2r − 1) r = 1 (2r − 1)!(51 − (2r − 1))! r=1 51! 1 51 1 = { C1 + 51C3 +.... + 51C51} = (250) 51! 51! ------------------------------------------------------------------------------------------------- Question34 200 200 The remainder when 19 + 23 is divided by 49 , is _______. [1-Feb-2023 Shift 1] Answer: 29 Solution: (21 + 2)200 + (21 − 2)200 ⇒ 2[100 C021200 + 200C221198 ⋅ 22 +..... + 200C198. 212 ⋅ 2198 + 2200 ] ⇒ 2[49I1 + 2200] = 49I1 + 2201 Now , 2201 = (8)67 = (1 + 7)67 = 49I2 + 67C067C1 ⋅ 7 = 49I2 + 470 = 49I2 + 49 × 9 + 29 ------------------------------------------------------------------------------------------------- Question35 Let the sixth term in the binomial expansion of ( √ 2log (10 − 3x) + 5 √ 2(x − 2)log 3 ) m, in the increasing powers of 2(x − 2)log 3, 2 2 2 be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is ________. [1-Feb-2023 Shift 2] Answer: 4 Solution:  m−5 T6 = C5(10 − 3 ) 2 ⋅ (3x − 2) = 21 m x m C1, mC2, mC3 are in A.P. 2. mC2 = mC1 + mC3 Solving for m, we get m = 2 (rejected), 7 Put in equation (1) 3x 21. (10 − 3x) = 21 9 3x = 30, 32 x = 0, 2 Sum of the squares of all possible values of x = 4 ------------------------------------------------------------------------------------------------- Question36 ( ) 2 22 3 α If the term without x in the expansion of x + is 7315 , then |α| x3 is equal to _______. [1-Feb-2023 Shift 2] Answer: 1 Solution: 2 C ⋅ (x 3 ) 22 − r 22 r −3r Tr + 1 = r ⋅ (α) , x 44 2r − − 3r 22 r = Cr ⋅ x 3 3 (α) 44 11r = 3 3 r=4 22 4 C4 ⋅ α = 7315 22 × 21 × 20 × 19 4 ⋅ α = 7315 24 α=1 ------------------------------------------------------------------------------------------------- Question37 If the ratio of the fifth term from the beginning to the fifth term from ( ) a 4 1 the end in the expansion of √2 + 4 is √6 : 1, then the third term √3 from the beginning is : [6-Apr-2023 shift 1] Options: A. 30√2 B. 60√2 C. 30√3 D. 60√3 Answer: D Solution: Solution: 1 ( (2) 4 ) n−4 ( ) n 1 4 C4 ⋅ 1 T5 34 √6 = = T5 ′ 1 1 ( ) 4 ( ) n 1 n−4 C4 24 1 34 n−8 1 −4 − 4 + n 2 4 ⋅ 34 ( ) = √6 n−8 n−8 2 4 ⋅ 3 4 = √6 n−8 1 = ⇒ n − 8 = 2 ⇒ n = 10 4 2 1 8 ( ) ( ) 1 2 T3 = 10C2 2 4 1 34 1 − 10.9 1 = 10C2 ⋅ 22 ⋅ 3 2 = ⋅ 4⋅ = 60√3 2 √3 ------------------------------------------------------------------------------------------------- Question38 2n If C3 : nC3 : 10 : 1, then the ratio (n2 + 3n) : (n2 − 3n + 4) is : [6-Apr-2023 shift 1] Options: A. 27 : 11 B. 35 : 16 C. 2 : 1 D. 65 : 37 Answer: C Solution: Solution: 2n C3 2n!(n − 3)! n = 10 ⇒ = 10 C3 (2n − 3)!n!  2n(2n − 1)(2n − 2) = 10 n(n − 1)(n − 2) 4(2n − 1) = 10 ⇒ 8n − 4 = 10n − 20 n−2 2n = 16 n2 + 3n Now n2 − 3n + 4 64 + 24 88 = = =2 64 − 24 + 4 44 ------------------------------------------------------------------------------------------------- Question39 ( ) 15 18 4 1 The coefficient of x in the expansion of x − is ______. x3 [6-Apr-2023 shift 1] Answer: 5005 Solution: ( ) 15 4 1 x − x3 ( ) r 15 4 15 − r −1 T r+1 = Cr(x ) 3 x 60 − 7r = 18 r=6 18 15 Hence coeff. of x = C6 = 5005 ------------------------------------------------------------------------------------------------- Question40 ( ) ( ) 11 11 7 2 1 −7 1 If the coefficients of x in ax + 2bx and x in ax − 2 are 3bx equal, then : [6-Apr-2023 shift 2] Options: A. 64 ab = 243 B. 32 ab = 729 C. 729ab = 32 D. 243 ab = 64 Answer: C Solution: Solution: (ax2 + 1 11 2bx ) 11 × 2 − 7 r= =5 3 ( 2b1 ) 5 7 11 6 Coefficient of x is = C5(a) ( ax − 3bx1 ) 11 2 11 × 1 − (−7) r= =6 3 Coefficient of x−7 is = 11 C6 ⋅ a5 36b6 ∵ 11 C5(a6) ( 2 1b ) 5 5 = 11C6 ⋅ a5 6 6 3 b 5 2 ⇒ ab = 6 3 ⇒ 729 ab = 32 Ans. Opiton 3 ------------------------------------------------------------------------------------------------- Question41 Among the statements : 2022 2022 (S1) : 2023 − 1999 is divisible by 8 n (S2) : 13(13) − 11n − 13 is divisible by 144 for infinitely many n ∈ ℕ [6-Apr-2023 shift 2] Options: A. only (S2) is correct B. only (S1) is correct C. both (S1) and (S2) are incorrect D. both (S1) and (S2) are correct Answer: D Solution: Solution: ∵xn − yn = (x − y)[xn − 1 + xn − 2y + xn − 3y2 +...... + yn − 1] xn − yn is divisible by x − y Stat 1 → (2023)2022 − (1999)2022 (2023) − (1999) = 24 2022 2022 Stat 2 → (2023) − (1999) is divisible by 8 13(1 + 12)n − 11n − 13 13[1 + nC1, (12) + nC2(12)2 +...] − 11n − 13 ⇒ (156n − 11n) + 13 ⋅ nC2(12)2 + 13 ⋅ nC3(12)3 +... ⇒ 145n + 13 ⋅ nC2(12)2 + 13 ⋅ nC3(12)3 +... If (n = 144m, m ∈ N) then it is divisible by 144 for infinite values of n. Ans. Option 4 ------------------------------------------------------------------------------------------------- Question42 Let (t) denote the greatest integer ≤t, If the constant term in the ( ) 7 expansion of 3x2 − 1 5 is α, then [α] is equal to _______. 2x [8-Apr-2023 shift 1] Answer: 1275 Solution: Solution: ( ) 7 3x2 − 1 5 2x ( ) r Tr + 1 = 7Cr(3x2)7 − r − 1 5 2x 14 − 2r − 5r = 14 − 7r = 0 ∴r=2 ∴ T3 = 7C2 ⋅ 35 − 1 2 2 ( = 4) 21 × 243 = 1275.75 ∴ [α] = 1275 ------------------------------------------------------------------------------------------------- Question43 3 n If aa is the greatest term in the sequence an = 4 , n = 1, 2, 3,..., n + 147 then a is equal to ______. [8-Apr-2023 shift 1] Options: A. Answer: 5 Solution: 3 x f (x) = 4 x + 147 4 2 3 3 ′ (x + 147)3x − x (4x ) f (x) = 4 2 (x + 147) 6 2 6 3x + 147 × 3x − 4x 2 4 = = x (44 − x ) +ve ′ 6 2 f (x) = 0 at x = 147 × 3x 2 4 x = 0, x = 147 × 3 2 x = 0, x = ±√147 × 3 2 x = ±21 x = ±√21 fmax at f (4) or f (5) 64 125 f (4) = ≃ 0.158 f(5) = ≃ 0.161 403 772 ∴a=5 ------------------------------------------------------------------------------------------------- Question44 The largest natural number n such that 3n divides 66 ! is _______. [8-Apr-2023 shift 1] Answer: 31 Solution: Solution: [66 3 +] [66 9 + ] [ 66 27 ] 22 + 7 + 2 = 31 ------------------------------------------------------------------------------------------------- Question45 The absolute difference of the coefficients of x10 and x7 in the expansion ( ) 11 of 2x2 + 1 2x is equal to [8-Apr-2023 shift 2] Options: A. 103 − 10 B. 113 − 11 C. 123 − 12 D. 133 − 13 Answer: C Solution: Solution:  ( 2x1 ) r 11 Tr + 1 = Cr(2x2)11 − r = 11Cr211 − 2rx22 − 3r 22 − 3r = 10 and 22 − 3r = 7 r = 4 and r = 5 10 11 3 Coefficient of x = C4 ⋅ 2 7 11 1 Coefficient of x = C5 ⋅ 2 11 3 11 difference = C4 ⋅ 2 − C5 ⋅ 2 11 × 10 × 9 × 8 11 × 10 × 9 × 8 × 7 = ×8− ×2 24 120 = 11 × 10 × 3 × 8 − 11 × 3 × 4 × 7 = 11 × 3 × 4 × (20 − 7) = 11 × 12 × 13 = 12(12 − 1)(12 + 1) = 12(122 − 1) = 123 − 12( Option 3) ------------------------------------------------------------------------------------------------- Question46 190 190 190 190 25 − 19 −8 +2 is divisible by [8-Apr-2023 shift 2] Options: A. 34 but not by 14 B. 14 but not by 34 C. Both 14 and 34 D. Neither 14 nor 34 Answer: A Solution: Solution: 25190 − 8190 is divisible by 25 − 8 = 17 19190 − 2190 is divisible by 19 − 2 = 17 190 190 25 − 19 is divisible by 25 − 19 = 6 190 190 8 −2 is divisible by 8 − 2 = 6 L.C.M. of 1746 = 34 ∴ divisible by 34 but not by 14 ------------------------------------------------------------------------------------------------- Question47 ( ) 13 If the coefficient of x7 in ax − 1 2 and the coefficient of x−5 in bx ( ) 13 1 4 4 ax + are equal, then a b is equal to : bx2 [10-Apr-2023 shift 1] Options: A. 22 B. 44 C. 11 D. 33 Answer: A Solution: Solution: ( ) 13 ax − 1 2 bx We have, Tr + 1 = nCr(p)n − r(q)r ( ) r Tr + 1 = Cr(ax)13 − r − 1 2 13 bx 13 = Cr(a) 13 − r − b ( 1 r 13 − r (x) ) ⋅ (x)−2r = 13Cr(a)13 − r − b ( 1 r 13 − 3r (x) )... (1) Coefficient of x7 ⇒ 13 − 3r = 7 r=2 r in equation (1) 13 T3 = C2(a) 13 − 2 − b (x)( 1 2 13 − 6 ) = 13C2(a)11 1 2 7 b ( ) (x) 11 (a) Coefficient of x7 is 13C2 2 b Now, ax + ( 1 13 bx2 ) Tr + 1 = 13Cr(ax)13 − r 1 r bx2 ( ) ( ) 1 r = 13Cr(a)13 − r (x)13 − r(x)−2r b = 13Cr(a)13 − r b (x) ( ) 1 r 13 − 3r... (2) Coefficient of x−5 ⇒ 13 − 3r = −5 r=6 r in equation 13 T7 = C6(a) 13 − 6 b ( ) 1 6 13 − 18 (x) ( ) 1 6 T7 = 13C6(a)7 (x)−5 b Coefficient of x−5 is 13C6(a)7 1 6 b ( ) ATQ Coefficient of x7 = coefficient of x−5 T3 = T7 13 C2 ( ) a11 b 2 = 13 C6(a)7 ( b1 ) 6 13 C6 a4 ⋅ b4 = 13 C2 13 × 12 × 11 × 10 × 9 × 8 × 1 = = 22 13 × 12 × 6 × 5 × 4 × 3 ------------------------------------------------------------------------------------------------- Question48  7 3 10 The coefficient of x in (1 − x + 2x ) is _______. [10-Apr-2023 shift 1] Answer: 960 Solution: 3 10 (1 − x + 2x ) 10! b 3 c Tn= (−2x) (x ) a!b!c! 10! b b + 3c = (−2) x a!b!c! ⇒ b + 3c = 7, a + b + c = 10 7 10! 7 10! 4 ∴ Coefficient of x = (−1) + (−1) (2) 3!7!0! 5!4!1! 10! 1 2 + (−1) (2) 7!1!2! = −120 + 2520 − 1440 = 960 ------------------------------------------------------------------------------------------------- Question49 Let the number (22)2022 + (2022)22 leave the remainder α when divided by 3 and β when divided by 7. Then ( α2. +β2 ) is equal to [10-Apr-2023 shift 2] Options: A. 13 B. 20 C. 10 D. 5 Answer: D Solution: Solution: 2022 22 (22) + (2022) divided byy 3 (21 + 1)2022 + (2022)22 = 3k + 1 (α = 1) Divided by 7 (21 + 1)2022 + (2023 − 1)22 7k + 1 + 1 (β = 2) 7k + 2 2 2 So α + β ⇒ 5 ------------------------------------------------------------------------------------------------- Question50 If the coefficients of x and x2 in (1 + x)p(1 − x)q are 4 and -5 respectively, then 2p + 3q is equal to [10-Apr-2023 shift 2] Options: A. 60 B. 63 C. 66 D. 69 Answer: B Solution: Solution: (1 + x)p(1 − x)q (1 + px + p(p − 1) 2 2! x +... ) (1 − qx + q(q − 1) 2 2! x −... ) p−q=4 p(p − 1) q(q − 1) + − pq = −5 2 2 2 2 p + q − p − q − 2 pq = −10 2 2 (q + 4) + q − (q + 4) − q − 2(4 + q)q = −10 q + 8q + 16 − q2 − q − 4 − q − 8q − 2q2 = −10 2 − 2q = −22 q = 11 p = 15 2(15) + 3(11) 30 + 33 = 63 ------------------------------------------------------------------------------------------------- Question51 (3 ) 1 1 680 2 4 The number of integral terms in the expansion of +5 is equal to : [11-Apr-2023 shift 1] Answer: 171 Solution: Solution: 1 1 (32 +54) 680 The number of integral term in the expression of is equal to  1 1 ( ) (5 4 ) 680 − r r 680 General term = Cr 3 2 680 − r r = 680C 3 r 2 54 r Values' s of r, where goes to integer 4 r = 0, 4, 8, 12,...... 680 680 − r All value of r are accepted for as well so 2 No of integral terms = 171. ------------------------------------------------------------------------------------------------- Question52 2 7 The mean of the coefficients of x, x ,... x in the binomial expansion of (2 + x)9 is ________. [11-Apr-2023 shift 1] Answer: 2736 Solution: Solution: Coefficient of x = 9C128 Coef. x2 = 9C227 Coef. x7 = 9C7 ⋅ 22 9 C1 ⋅ 28 + 9C2 ⋅ 27... + 9C7 ⋅ 22 Mean = 7 (1 + 2)9 − 9C0 ⋅ 29 − 9C8 ⋅ 21 − 9C9 = 7 39 − 29 − 18 − 1 = 7 19152 = = 2736 7 ------------------------------------------------------------------------------------------------- Question53 th If the 1011 term from the end in the binominal expansion of ( ) 2022 4x 5 th 5 − 2x is 1024 times 1011 term from the beginning, the |x| is equal to [11-Apr-2023 shift 2] Options: A. 8 B. 12 C. 10 D. 15 Answer: C Solution: Solution: T1011 from beginning = T1010 + 1 ( 4x5 ) ( −5 2x ) 1012 1010 2022 = C1010 T1011 from end ( −5 2x ) ( 4x5 ) 1012 1010 2022 = C1010 ( −5 2x ) ( 4x5 ) 1012 1010 2022 Given: = C 1010 ( −52x ) ( 4x5 ) 1010 1012 = 210 ⋅ 2022C1010 ( −5 2x ) ( 4x5 ) 2 2 = 210 4 54 x = 216 |x |= 5 16 ------------------------------------------------------------------------------------------------- Question54 The sum of the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 2, which are in the ratio 1 : 3 : 5, is equal to [11-Apr-2023 shift 2] Options: A. 63 B. 92 C. 25 D. 41 Answer: A Solution: Solution: n+2 cr − 1 : n + 2cr : n + 2cr + 1 : : 1 : 3 : 5 (n + 2)! r!(n + 2 − r)! 1 × = (r − 1)!(n − r + 3)! (n + 2)! 3 r 1 = ⇒ n − r + 3 = 3r (n − r + 3) 3 n = 4r − 3 − 0 (n + 1)! (r + 1)!(n − r + 1)! 3 × = r!(n + 2 − r)! (n + 2)! 5 r+1 3 = n+2−r 5 8r − 1 = 3n... (2) By equation 1 and 2 8r − 1 = 4r − 3 n = 4r − 3 3 r=2n=4×2−3 n=5 Sum: 7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63 ------------------------------------------------------------------------------------------------- Question55 n n 1n n If n +1 1 Cn + n1 Cn − 1 +... + 2 C1 + C0 = 1023 10 then n is equal to [12-Apr-2023 shift 1] Options: A. 7 B. 9 C. 6 D. 8 Answer: B Solution: Solution: n n Cr 1 n+1 ∑ = ∑ C T=0 r+1 n + 1 r = 0n r + 1 1 n+1 1023 = (2 − 1) = n+1 10 n + 1 = 10 ⇒ n = 9 ------------------------------------------------------------------------------------------------- Question56 The sum, of the coefficients of the first 50 terms in the binomial expansion of (1 − x)100, is equal to [12-Apr-2023 shift 1] Options: A. − 101C50 99 B. C49 101 C. C50 D. − 99C49 Answer: D Solution: Solution: (1 − x)100 = C0 − C1x + C2x2 − C3x3 +...... C99x99 + C100x100 ⇒ C0 − C1 + C2 − C3 +... − C99 + C100 = 0 1 C0 − C1 + C2 +... C99 = − 100C50 2 1 100! 1 100 × 99! − =− × = − 99C49 2 50!50! 2 50!50! ------------------------------------------------------------------------------------------------- Question57 2022 4 Fractional part of the number is 15 equal to [13-Apr-2023 shift 1] Options: A. 4 15 8 B. 15 1 C. 15 D. 14 15 Answer: C Solution: Solution: { 42022 15 }={ 24044 15 }={ (1 + 15)1011 15 } = 1 15 ------------------------------------------------------------------------------------------------- Question58 Let α be the constant term in the binomial expansion of ( ) n 6 √x − 3 , n ≤ 15. If the sum of the coefficients of the remaining x2 terms in the expansion is 649 and the coefficient of x−n is λα, then λ is equal to _______. [13-Apr-2023 shift 1] Answer: 36 Solution:  n−k −3 k n k Tk + 1 = Ck(x) 2 (−6) (x) 2 n−k 3 − k=0 2 2 n − 4k = 0 n ( (−5)n − nC n (−6) 4 = 649 ) By observation (625 + 24 = 649), we get n = 4 ∵n = 4&k = 1 Required is coefficient of ( ) 6 4 x−4 is √4 − 3 x2 4 3 C1(−6) By calculating we will get λ = 36 ------------------------------------------------------------------------------------------------- Question59 x Let for x ∈ R, S0(x) = x, Sk(x) = Ckx + k ∫0 Sk − 1(t) dt where 1 C0 = 1, Ck = 1 − ∫0 Sk − 1(x) dx, k = 1, 2, 3,... Then S2(3) + 6C3 is equal to _______. [13-Apr-2023 shift 1] Answer: 18 Solution: Solution: x Given, Sk(x) = Ckx + k ∫ Sk − 1(t) dt 0 put k = 2 and x = 3 3 S2(3) = C2(3) + 2 ∫ S1(t) dt 0 x Also, S1(x) = C1(x) + ∫ S0(t) dt...... (1) 0 2 x = C1x + 2 ( C t + t2 ) dt 3 2 S2(3) = 3C2 + 2 ∫ 1 0 = 3C2 + 9C1 + 9 Also, 1 1 C1 = 1 − ∫ S0(x) dx = 0 2 1 C2 = 1 − ∫ S1(x) dx = 0 0 1 C3 = 1 − ∫ S2(x) dx 0 (C x+C x x3 ) dx = 34 1 2 =1−∫ 2 1 + 0 3 x S2(x) = C2x + 2 ∫ S1(t) dt 0 3 2 x = C2x + C1x + 3  ⇒ S2(3) + 6C3 = 6C3 + 3C2 + 9C1 + 9 = 18 ------------------------------------------------------------------------------------------------- Question60 ( ) 5 5 3 1 The coefficient of x in the expansion of 2x − is 3x2 [13-Apr-2023 shift 2] Options: A. 80 9 B. 8 C. 9 D. 26 3 Answer: A Solution: Solution: ( 2x − 3x1 ) 5 3 general term for 2 C ( − 1 ) (2x ) r 5 3 5−r Tr + 1 = r 2 3x 5 r −r 5 − r 15 − 5r Cr(−1) 3 2 ⋅x 15 − 5r = 5 ⇒ r = 2 5 5 2 −2 3 Coeff. Of x = C2(−1) 3 2 1 = 10× × 8 9 80 = 9 ------------------------------------------------------------------------------------------------- Question61 The remainder, when 7103 is divided by 17 , is ______. [13-Apr-2023 shift 2] Answer: 12 Solution: 103 102 7 = 7.7 2 51 = 7(7 ) 51 51 = 7(51 − 2) → remainder = 7(−2)  − 7(23)(16)12 = −56(17 − 1)12 → remainder = −56(−1)12 Remainder = −56 + 17k = −56 + 68 = 12 Question62 20 2 10 i Let (a + bx + cx ) = ∑ i=0 pix , a, b, c ∈ N. If p1 = 20 and p2 = 210, then 2(a + b + c) is equal to [15-Apr-2023 shift 1] Options: A. 8 B. 12 C. 6 D. 15 Answer: B Solution: Solution: 20 (a + bx + cx 2)10 = ∑ pixi i=0 Coefficient of x1 = 20 10! 20 = × a9 × b1 9!1! 9 a ⋅b=2 a = 2, b = 2 Coefficient of x2 = 210 10! 10! 210 = × a9 × c1 + × a8b2 9!1! 8!2! 210 = 10. c + 45 × 4 10c = 30 c=3 2(a + b = c) = 12 ------------------------------------------------------------------------------------------------- Question63 The remainder when 32022 is divided by 5 is : [24-Jun-2022-Shift-1] Options: A. 1 B. 2 C. 3 D. 4 Answer: D Solution: ------------------------------------------------------------------------------------------------- Question64 2 3 2021 The remainder on dividing 1 + 3 + 3 + 3 +..... + 3 by 50 is__ [24-Jun-2022-Shift-2] Answer: 4 Solution: ------------------------------------------------------------------------------------------------- Question65 Let Cr denote the binomial coefficient of xr in the expansion of (1 + x)10. If for α, β ∈ R, C1 + 3.2C2 + 5.3C3 +...... upto 10 terms = α × 211 2β − 1 (C 0 + C1 2 + C2 3 +.... upto 10 terms ) then the value of α + β is equal to [25-Jun-2022-Shift-1] Answer: 286 Solution: Question66 101 The coefficient of x in the expression 500 499 (5 + x) + x(5 + x) + x2(5 + x)498 +... + x500, x > 0, is [25-Jun-2022-Shift-2] Options: 501 A. C101(5)399 501 B. C101(5)400 501 C. C100(5)400 500 D. C101(5)399 Answer: A Solution: Given, (5 + x)500 + x(5 + x)499 + x2(5 + x)498 +...... x500 This is a G.P. with first term (5 + x)500 x(5 + x)499 x Common ratio = 500 = and 501 terms present. (5 + x) 5 + x (( ) ) x 501 (5 + x)500 −1 5+x ∴ Sum = x −1 5+x = (5 + x)500 ( x501 − (5 + x)501 (5 + x)501 ) x−5−x 5+x 501 501 x − (5 + x) 5+x = −5 5+x 1 = ((5 + x)501 − x501) 5 Coefficient of x101 in (5 + x)501 is = 501C101 ⋅ 5400 1 500 501 101 1 501 400 ∴ln ((5 + x) − x ) coefficient of x is = ⋅ C101 ⋅ 5 5 5 = 501C101 ⋅ 5399 Question67 If the sum of the co-efficient of all the positive even powers of x in the ( ) 10 3 3 10 9 binomial expansion of 2x + x is 5 − β.3 , then β is equal to____ [25-Jun-2022-Shift-2] Answer: 83 Solution: Given, Binomial Expansion ( 3 2x + 3 10 x ) General term 10 T r + 1 = Cr ⋅ (2x ) 3 10 − r ⋅ 3 r x ( ) 10 10 − r r 30 − 3r −r = Cr ⋅ 2 ⋅3 ⋅x ⋅x 10 10 − r r 30 − 4r = Cr ⋅ 2 ⋅3 ⋅x For positive even power of x, 30 − 4r should be even and positive. For r = 0, 30 − 4 × 0 = 30 (even and positive) For r = 1, 30 − 4 × 1 = 26 (even and positive) For r = 2, 30 − 4 × 2 = 22 (even and positive) For r = 3, 30 − 4 × 3 = 18 (even and positive) For r = 4, 30 − 4 × 4 = 14 (even and positive) For r = 5, 30 − 4 × 5 = 10 (even and positive) For r = 6, 30 − 4 × 6 = 6 (even and positive) For r = 7, 30 − 4 × 7 = 2 (even and positive) For r = 8, 30 − 4 × 8 = −2 (even but not positive) So, for r = 1, 2, 3, 4, 5, 6 and 7 we can get positive even power of x. ∴ Sum of coefficient for positive even power of 10 10 0 10 9 1 10 8 2 10 7 3 10 6 4 10 5 5 10 4 6 10 3 7 x = C0 ⋅ 2 ⋅ 3 + C1.2 ⋅ 3 + C2 ⋅ 2 ⋅ 3 + C3 ⋅ 2 ⋅ 3 + C4 ⋅ 2 ⋅ 3 + C5 ⋅ 2 ⋅ 3 + C6 ⋅ 2 ⋅ 3 + C7 ⋅ 2 ⋅ 3 10 10 0 10 9 1 10 0 10 10 2 8 10 9 10 0 10 = C10 ⋅ 2 ⋅3 + C1.2 ⋅ 3 +...... + C10 ⋅ 2 ⋅ 3 − [ C8 ⋅ 2 ⋅ 3 + C9 ⋅ 2 ⋅ 3 + C10 ⋅ 2 ⋅ 3 ] 10 8 9 10 = (2 + 3) − [45.4.3 + 10.2.3 + 1.1.3 ] 10 9 9 9 = 5 − [60 × 3 + 20 ⋅ 3 + 3 ⋅ 3 ] 10 9 = 5 − (60 + 20 + 3)3 10 9 = 5 − 83.3 ∴β = 83 Question68 The remainder when (2021)2023 is divided by 7 is : [26-Jun-2022-Shift-1] Options: A. 1 B.2 C.5 D.6 Answer: C Solution: Question69 40 41 42 60 m 60 If ( C0) + ( C1) + ( C2) +...... + ( C20) = n C20, m and n are coprime, then m + n is equal to____ [26-Jun-2022-Shift-2] Answer: 102 Solution: Question70 ( ) 60 10 √x √5 If the coefficient of x in the binomial expansion of 1 + 1 is 54 x3 k 5. l , where I, k ∈ N and I is co-prime to 5 , then k is equal to [27-Jun-2022-Shift-1] Answer: 5 Solution: Given Binomial Expansion ( ) √x √x 60 = 1 + 1 54 53 ∴ General term ( ) ( ) 60 − r r 60 x1 ∕ 2 51 ∕ 2 T r+1 = Cr ⋅ ⋅ 51 ∕ 4 x1 ∕ 3 60 ( r − 15 + 2r ) ( 30 − 2r − 3r ) = Cr ⋅ 5 4 ⋅x ( 3r −4 60 ) ( 1806− 5r ) = 60Cr ⋅ 5 ⋅x For x10 term, 180 − 5r = 10 6 ⇒5r = 120 ⇒r = 24 ∴ Coefficient of x10 = 60 C24 ⋅ 5 ( 3 × 244 − 60 ) 60 3 = C24 ⋅ 5 60! = ⋅ 53 24!36! It is given that, 60! ⋅ 53 = 5k ⋅ l...... (1) 24!36! Also given that, I is coprime to 5 means I can't be multiple of 5. So we have to find all the factors of 5 in 60!, 24 ! and 36 ! [Note : Formula for exponent or degree of prime number in n!. Exponent of p in n! = n p⌈ ⌋ ⌈ ⌋ [ ] + n + p2 n +..... until 0 comes p3 here p is a prime number. ] ∴ Exponent of 5 in 60 ! = ⌈ 60 5 ⌋ ⌈+ 60 52 ⌋ ⌈ ⌋ + 60 53 +...... = 12 + 2 + 0 +.... = 14 Exponent of 5 in 24 ! = ⌈ 24 5 ⌋ ⌈ + 24 52 + 24 ⌋ ⌈ ⌋ 53 +...... = 4 + 0 + 0...... =4 Exponent of 5 in 36 ! = ⌈ 36 5 ⌋ ⌈ + 36 52 + 36 ⌋ ⌈ ⌋ 53 +...... = 7 + 1 + 0...... =8 ∴ From equation (1), exponent of 5 overall 514 4 8 ⋅ 53 = 5k 5 ⋅5 ⇒55 = 5k ⇒k = 5 Question71 If the sum of the coefficients of all the positive powers of x, in the ( ) 7 n 2 Binomial expansion of x + is 939 , then the sum of all the x5 possible integral values of n is____ [27-Jun-2022-Shift-2] Answer: 57 Solution: Solution: Given, Binomial expression is ( = x + 5 n x 2 7 ) ∴ General term T r + 1 = 7Cr ⋅ (xn)7 − r ⋅ 2 r x5 ( ) 7 7n − nr − 5r r = Cr ⋅ x ⋅2 For positive power of x, 7n − nr − 5r > 0 ⇒7n > r(n + 5) 7n ⇒r < n+5 As r represent term of binomial expression so r is always integer. Given that sum of coefficient is 939. When r = 0, sum of coefficient = 7C0 ⋅ 20 = 1 when r = 1, sum of coefficient = 7C0 ⋅ 20 + 7C1 ⋅ 21 = 1 + 14 = 15 when r = 2, sum of coefficient = 7C0 ⋅ 20 + 7C1 ⋅ 21 + 7C2 ⋅ 22 = 1 + 14 + 84 = 99 when r = 3, sum of coefficient = 7C0 ⋅ 20 + 7C1 ⋅ 21 + 7C2 ⋅ 22 + 7C3 ⋅ 23 = 1 + 14 + 84 + 280 = 379 when r = 4, sum of coefficient = 7C0 ⋅ 20 + 7C1 ⋅ 21 + 7C2 ⋅ 22 + 7C3 ⋅ 23 + 7C4 ⋅ 24 = 1 + 14 + 84 + 280 + 560 = 939 7n To get value of r = 4, value of should be between 4 and 5. n+5 7n ∴4 < 20 20 ⇒n > 3 ⇒n > 6.66 and 7n < 5n + 25 ⇒2n < 25 ⇒n < 12.5 ∴6.66 < n < 12.5 ∴ Possible integer values of n = 7, 8, 9, 10, 11, 12 ∴ Sum of values of n = 7 + 8 + 9 + 10 + 11 + 12 = 57 ------------------------------------------------------------------------------------------------- Question72 If 31 30 31 31 31 31 α(60!) ∑ k=1 ( Ck)( Ck − 1) − ∑ ( C )( C k=1 k k − 1) = (30!)(31!) where α ∈ R, then the value of 16α is equal to [28-Jun-2022-Shift-1] Options: A. 1411 B. 1320 C. 1615 D. 1855 Answer: A Solution: Solution: Given, 31 30 ∑ (31Ck)(31Ck − 1) − ∑ (30Ck)(30Ck − 1) = α(60!) k=1 k=1 (30!)(31!) Now, 31 ∑ (31Ck)(31Ck − 1) k=1 = (31C1 ⋅ 31C0 + 31C2 ⋅ 31C1 + 31C3 ⋅ 31C2 +...... + 31C31 ⋅ 31C30) = (31C0 ⋅ 31C31 − 1 + 31C1 ⋅ 31C31 − 2 +...... + 31C30 ⋅ 31C31 − 31) n [using Cr = nCn − r ] = (31C0 ⋅ 31C30 + 31C1 ⋅ 31C29 +...... + 31C30 ⋅ 31C0) 62 = C30 30 Now, ∑ 30Ck ⋅ 30Ck − 1 k=1 = (30C1 ⋅ 30C0 + 30C2 ⋅ 30C1 +...... + 30C30 ⋅ 30C29) = (30C0 ⋅ 30C29 + 30C1 ⋅ 30C28 +... + 30C29 ⋅ 30C0) 60 = C29 α(60!) ∴ 60C30 − 60C29 = 30!31! 62.61.60! 60! α(60!) ⇒ − = 30!32! 29!31! 30!31! 62.61.60! 60! α(60!) ⇒ − = 30!32! 30! 30!31! ⋅ 31! 30 ⇒ 60! 30!31! 62.61 32 (− 30 = 30!31!) α(60!) 62.61 ⇒α = − 30 32 62.61 − 30 × 32 ⇒16α = 2 2822 ⇒16α = = 1411 2 ------------------------------------------------------------------------------------------------- Question73 The number of positive integers k such that the constant term in the ( ) 12 3 3 8 binomial expansion of 2x + , x ≠ 0 is 2. I, where I is an odd xk integer, is___ [28-Jun-2022-Shift-1] Answer: 2 Solution: ( 2x ) 12 3 3 + xk ( x3 ) 12 − r 12 tr + 1 = Cr(2x3)r k x3r − (12 − r)k → constant ∴3r − 12k + rk = 0 3r ⇒k = 12 − r ∴ possible values of r are 3, 6, 8, 9, 10 and corresponding values of k are 1, 3, 6, 9, 15 12 Now Cr = 220, 924, 495, 220, 66 8 ∴ possible values of k for which we will get 2 are 3,6 ------------------------------------------------------------------------------------------------- Question74 The term independent of x in the expansion of ( ) 11 2 3 5 3 1 (1 − x + 3x ) 2 x − , x ≠ 0 is : 5x2 [28-Jun-2022-Shift-2] Options: 7 A. 40 33 B. 200 39 C. 200 11 D. 50 Answer: B Solution: Solution: ( 52x ) 11 General term of Binomial expansion 3 − 1 is 5x2 ( 52 x ) ⋅ ( − 5x1 ) 11 − r r 11 3 T r+1 = Cr ⋅ 2 C ⋅ ( ) ⋅ (− ) ⋅x 5 1 11 − r r 11 33 − 5r = r 2 5 In the term, (1 − x2 + 3x3) 1 11 5x2 ( 52x 3 − ) Term independent of x is when (1) 33 − 5r = 0 33 ⇒r = ∉ integer 5 (2) 33 − 5r = −2 ⇒5r = 35 ⇒r = 7 ∈ integer (3) 33 − 5r = −3 ⇒5r = 36 36 ⇒r = ∉ integer 5 ∴ Only for r = 7 independent of x term possible. ∴ Independent of x term  ( ( 52 ) ⋅ ( − 15 ) ) 4 7 11 =− C7 =− ( 11 ⋅ 10 ⋅ 9 ⋅ 8 54 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 24 ⋅ − 17 5 ) = 11 ⋅ 10 ⋅ 3 24 ⋅ 53 = 11 3 ⋅3 2 2 ⋅5 33 = 200 ------------------------------------------------------------------------------------------------- Question75 If the constant term in the expansion of ( ) 10 3x3 − 2x2 + 5 is 2k. I, where I is an odd integer, then the value of k x5 is equal to: [29-Jun-2022-Shift-1] Options: A. 6 B. 7 C. 8 D. 9 Answer: D Solution: Solution: Note : Multinomial Theorem : The general term of (x1 + x2 +... + xn)n the expansion is n! n n n x 1x 2... xn n n1!n2!... nn! 1 2 where n1 + n2 +..... + nn = n Given, ( ) 10 3x2 − 2x2 + 55 x 8 7 10 (3x − 2x + 5) = 50 x ( ) 10 Now constant term in 3x3 − 2x2 + 55 = x50 term in (3x8 − 2x7 + 5)10 x General term in (3x8 − 2x7 + 5)10 is 10! n n n = (3x8) 1(−2x7) 2(5) 3 n1!n2!n3! 10! 1 n 3 8n + 7n2 = (3)n (−2) 2(5)n ⋅ x 1 n1!n2!n3! 8n + 7n ∴ Coefficient of x 1 2 is 10! n n n = (3) (−2) (5) 3 1 2 n1!n2!n3! where n1 + n2 + n3 = 0 For coefficient of x50 : 8n1 + 7n2 = 50 ∴ Possible values of n1, n2 and n3 are  50 ∴ Coefficient of x 10! = (3)1(−2)6(5)3 1!6!3! 10 × 9 × 8 × 7 = × 3 × 53 × 26 6 = 5 × 3 × 8 × 7 × 3 × 53 × 26 = 7 × 54 × 32 × 29 = 2k ⋅ l ∴l = 7 × 54 × 32 = An odd integer and 2k = 29 ⇒k = 9 ------------------------------------------------------------------------------------------------- Question76 Let n ≥ 5 be an integer. If 9n − 8n − 1 = 64α and 6n − 5n − 1 = 25β, then α − β is equal to [29-Jun-2022-Shift-2] Options: A. 1 + nC2(8 − 5) + nC3(82 − 52) +..... + nCn(8n − 1 − 5n − 1) B. 1 + nC3(8 − 5) + nC4(82 − 52) +..... + nCn(8n − 2 − 5n − 2) C. nC3(8 − 5) + nC4(82 − 52) +....... + nCn(8n − 2 − 5n − 2) D. nC4(8 − 5) + nC5(82 − 52) +...... + nCn(8n − 3 − 5n − 3) Answer: C Solution: Solution: Given, 9n − 8n − 1 = 64α (1 + 8)n − 8n − 1 ⇒α = 64 (nC0 ⋅ 1 + nC1 ⋅ 81 + nC2 ⋅ 82 +...... + nCn ⋅ 8n) − 8n − 1 = 82 1 + 8n + nC2 ⋅ 82 +.... + nCn ⋅ 8n − 8n − 1 = 2 8 n C2 ⋅ 82 + nC3 ⋅ 83 +.... + nCn ⋅ 8n = 82 = C2 + C3 ⋅ 8 + C4 ⋅ 82 +... nCn ⋅ 8n − 2 n n n Also given, 6n − 5n − 1 = 25β (1 + 5)n − 5n − 1 ⇒β = 25 n C0 ⋅ 1 + nC1 ⋅ 5 + nC2 ⋅ 52 +..... + nCn ⋅ 5n − 5n − 1 = 52 n 2 n 3 n 2 1 + 5n + C2 ⋅ 5 + C3 ⋅ 5 +.... + Cn ⋅ 5 − 5n − 1 = 2 5  n C2 ⋅ 52 + nC3 ⋅ 53 + nC4 ⋅ 54 +.... + nCn ⋅ 5n = 52 = C2 + C3 ⋅ 5 + C4 ⋅ 52 +....... + nCn ⋅ 5n − 2 n n n ∴α − β = (nC2 + nC3 ⋅ 8 + nC4 ⋅ 82 +.... + nCn ⋅ 8n − 2) − ( nC2 + nC3 ⋅ 5 + nC4 ⋅ 52 +... + nCn ⋅ 5n − 2. = nC3 ⋅ (8 − 5) + nC4 ⋅ (82 − 52) +... + nCn(8n − 2 − 5n − 2) ------------------------------------------------------------------------------------------------- Question77 Let the coefficients of x−1 and x−3 in the expansion of ( ) 1 15 5 1 2x − 1 , x > 0, be m and n respectively. If r is a positive integer x5 such that mn2 = 15 Cr ⋅ 2r, then the value of r is equal to___ [29-Jun-2022-Shift-2] Answer: 5 Solution: Solution: 15 − 2r Tr + 1 = (−1)r ⋅ 15Cr ⋅ 215 − r × 5 m = 15C1025 n = −1 so mn2 = 15C525 ------------------------------------------------------------------------------------------------- Question78 If the maximum value of the term independent of t in the expansion of ( ) 15 1 1 2 5 (1 − x) 10 t x + t , x ≥ sl ant0, is K , then 8K is equal to [25-Jul-2022-Shift-1] Answer: 6006 Solution:  ( ) 1 1 15 10 General term of t x 5 + (1 − x) 2 is t ( ) 1 1 r ( ) 15 − r 15 2 (1 − x) 10 T r + 1 = Cr ⋅ t x5 ⋅ t 15 − r r = 15Cr ⋅ t30 − 2r ⋅ x 5 ⋅ (1 − x) 10 ⋅ t−r 15 − r r = 15C ⋅ t30 − 3r ⋅ x 5 r ⋅ (1 − x) 10 Term will be independent of t when 30 − 3r = 0 ⇒ r = 10 ∴T 10 + 1 = T 11 will be independent of t 15 − 10 10 ∴T 11 = 15C10 ⋅ x 5 ⋅ (1 − x) 10 = 15C10 ⋅ x1 ⋅ (1 − x)1 T11 will be maximum when x(1 − x) is maximum. 2 Let f (x) = x(1 − x) = x − x ′ f (x) is maximum or minimum when f (x) = 0 ′ ∴f (x) = 1 − 2x For maximum ∕ minimum f ′(x) = 0 ∴1 − 2x = 0 1 ⇒x = 2 Now, f ′′(x) = −2 < 0 1 ∴ At x = , f (x) maximum 2 ∴ Maximum value of T 11 is = 15 1 C10 ⋅ 2 1− 1 2 ( ) 15 1 = C10 ⋅ 4 1 Given K = 15C10 ⋅ 4 Now, 8K = 2(15C10) = 6006 ------------------------------------------------------------------------------------------------- Question79 The remainder when (11)1011 + (1011)11 is divided by 9 is [25-Jul-2022-Shift-2] Options: A. 1 B. 4 C. 6 D. 8 Answer: D Solution: Solution: ((11)1011 + (1011)11 ) = Re ( 2 ) 1011 + 311 Re 9 9 ( ) 1011 2 For Re 9 21011 = (9 − 1)337 = 337 C09337(−1)0 + 337C19336(−1)1 + 337C29335(−1)2 +...... + 337C33790(−1)337 So, remainder is 8 and Re ( 311 9 =0 ) So, remainder is 8 ------------------------------------------------------------------------------------------------- Question80 If the coefficients of x and x2 in the expansion of (1 + x)p(1 − x)q, p, q ≤ 15, are −3 and −5 respectively, then the coefficient of x3 is equal to _________. [26-Jul-2022-Shift-1] Answer: 23 Solution: Solution: Coefficient of x in (1 + x)p(1 − x)q p C0qC1 + pC1qC0 = −3 ⇒ p − q = −3 Coefficient of x2 in (1 + x)p(1 − x)q p C0qC2 − pC1qC1 + pC2qC0 = −5 q(q − 1) p(q − 1) − pq + = −5 2 2 q2 − q (q − 3)(q − 4) − (q − 3)q + = −5 2 2 ⇒q = 11, p = 8 Coefficient of x3 in (1 + x)8(1 − x)11 = − 11C3 + 8C111C2 − 8C211C1 + 8C3 = 23 ------------------------------------------------------------------------------------------------- Question81 n n ∑ i, j = 0 CinCj is equal to i≠j [26-Jul-2022-Shift-2] Options: A. 22n − 2nCn B. 22n − 1 − 2n − 1Cn − 1 C. 22n − 1 2 2n Cn 2n − 1 D. 2 + 2n − 1Cn Answer: B Solution: Solution: n n n n n n ∑ Ci Cj = ∑ Ci nCj − ∑ n Ci n Cj i, j = 0, i ≠ j i, j = 0 i=j n n n n n n = ∑ Ci ∑ Cj − ∑ CiCi j=0 j=0 i=0 n n 2n =2 ⋅2 − Cn 2n 2n =2 − Cn ------------------------------------------------------------------------------------------------- Question82 The remainder when (2021)2022 + (2022)2021 is divided by 7 is [27-Jul-2022-Shift-1] Options: A. 0 B. 1 C. 2 D. 6 Answer: A Solution: Solution: (2021)2022 + (2022)2021 = (7k − 2)2022 + (7k1 − 1)2021 = [(7k − 2)3]674 + (7k1)2021 − 2021(7k1)2020 +... − 1 = (7k2 − 1)674 + (7m − 1) = (7n + 1) + (7m − 1) = 7(m + n) (multiple of 7) ∴ Remainder = 0 ------------------------------------------------------------------------------------------------- Question83 Let for the 9 th term in the binomial expansion of (3 + 6x)n, in the increasing powers of 6x, to be the greatest for x = 32 , the least value of n 6 3 is n0. If k is the ratio of the coefficient of x to the coefficient of x , then k + n0 is equal to: [27-Jul-2022-Shift-2] Answer: 24 Solution: Solution: (3 + 6x)n = 3n(1 + 2x)n If T 9 is numerically greatest term ∴T 8 ≤ T 9 ≤ T 10 n C73n − 7(6x)7 ≤ nC83n − 8(6x)8 ≥ nC93n − 9(6x)9 n! n! n! ⇒ 9≤ 3 ⋅ (6x) ≥ (6x)2 (n − 7)!7! (n − 8)!8! (n − 9)!9! ⇒ 9 ≤ 18 3 2 ≥ ( ) 36 9 (n − 7)(n − 8) (n − 8)8 9.8 4 72 ≤ 27(n − 7) and 27 ≥ 9(n − 8) 29 ≤ n and n ≤ 11 3 ∴n0 = 10 For (3 + 6x)10 T r + 1 = 10Cr 310 − r(6x)r For coeff. of x6 r = 6 ⇒ 10C634 ⋅ 66 For coeff. of x3 r = 3 ⇒ 10C337 ⋅ 63 10 C6 34 ⋅ 66 10!7!3! ∴k = 10 ⋅ 7 3= ⋅8 C3 3 ⋅ 6 6!4!10! ⇒k = 14 ∴k + n0 = 24 ------------------------------------------------------------------------------------------------- Question84 The remainder when 72022 + 320222 is divided by 5 is : [28-Jul-2022-Shift-1] Options: A. 0 B. 2 C. 3 D. 4 Answer: C Solution: Solution: 72022 + 32022 1011 1011 = (49) + (9) 1011 1011 = (50 − 1) + (10 − 1) = 5λ − 1 + 5k − 1 = 5m − 2 Remainder = 5 − 2 = 3 Question85 Let the coefficients of the middle terms in the expansion of ( ) (1− ) 4 6 1 2 β + βx , (1 − 3βx) and 2 x , β > 0, respectively form the first √6 three terms of an A.P. If d is the common difference of this A.P. , then 50 − 2d2 is equal to _________. β [28-Jul-2022-Shift-2] Answer: 57 Solution: Solution: ( −β 2 ) 1 2 2 3 Coefficients of middle terms of given expansions are 4C2 β , C1(−3β), 6C3 form an A.P. 6 3 5β ∴2.2(−3β) = β2 − 2 ⇒−24 = 2β − 5β2 ⇒5β2 − 2β − 24 = 0 ⇒5β2 − 12β + 10β − 24 = 0 ⇒β(5β − 12) + 2(5β − 12) = 0 12 β= 5 d = −6β − β2 (−6β − β2) 12 ∴50 − 2d2 = 50 − 2 = 50 + + 2 = 57 β β2 β ------------------------------------------------------------------------------------------------- Question86 If 1 + (2 + 49C1 + 49C2 +... + 49C49)(50C2 + 50C4 +... + 50C50) is equal to n 2 ⋅ m, where m is odd, then n + m is equal to _______. [28-Jul-2022-Shift-2] Answer: 99 Solution: Solution: l = 1 + (1 + 49C0 + 49C1 +... + 49C49)(50C2 + 50C4 +... + 50C50) 49 As C0 + 49C1 +... + 49C49 = 249 50 and C0 + 50C2 +... + 50C50 = 249 ⇒ 50C2 + 50C4 +... + 50C50 = 249 − 1 ∴l = 1 + (249 + 1)(249 − 1) = 298 ∴m = 1 and n = 98 m + n = 99 ------------------------------------------------------------------------------------------------- Question87 Let the ratio of the fifth term from the beginning to the fifth term from ( ) n 4 1 the end in the binomial expansion of √2 + 4 , in the increasing √3 1 4 α powers of 4 be √6 : 1. If the sixth term from the beginning is 4 , √3 √3 then α is equal to ________. [29-Jul-2022-Shift-1] Answer: 84 Solution: Solution: 1 −1 ( ) (3 ) n−4 4 Fifth term from beginning = nC4 2 4 4 1 −1 ( ) (3 ) 3 n−4 Fifth term from end = (n − 5 + 1)th term from begin = nCn − 4 2 4 4 n−4 n 1 C42 4 ⋅ 3−1 Given =64 4 −( n−4 n ) Cn − 32 4 ⋅ 3 4 n−8 1 ⇒6 4 =64 n−8 1 ⇒ = ⇒n=9 4 4 1 4 −1 5 T 6 = T 5 + 1 = 9C5 2 4 ( 3 4 ) ( ) 9 C5 ⋅ 2 84 α = 1 = = 1 1 34⋅3 34 34 ⇒α = 84 ------------------------------------------------------------------------------------------------- Question88 10 If k=1 ∑ K 2 ( 10C ) 2 = 22000L, then L is equal to _______. K [29-Jul-2022-Shift-2] Answer: 221 Solution: Solution: Given, 10 ∑ k2(10Ck)2 = 2200L k=1 10 ⇒ ∑ (k ⋅ 10Ck)2 = 22000L k=1 10 ( k⋅ 10k ⋅ ) 2 9 ⇒ ∑ Ck − 1 = 22000L k=1 10 ⇒ ∑ (10 ⋅ 9Ck − 1)2 = 22000L k=1 10 ⇒100 ⋅ ∑ (9Ck − 1)2 = 22000L k=1 ⇒100((9C0)2 + (9C1)2 +... + (9C9)2) = 22000L ⇒100(18C9) = 22000L [Note : (nC1)2 + (nC2)2 +... + (nCn)2 = 2n Cn ] 18! ⇒100× = 22000L 9!9! ⇒L = 221 ------------------------------------------------------------------------------------------------- Question89 If n ≥ 2 is a positive integer, then the sum of the series n+1 C2 + 2(2C2 + 3C2 + 4C2 +... + nC2) is [2021, 24 Feb. Shift-II] Options: A. n(n − 1)(2n 6 + 1) B. n(n + 1)(2n + 1) 6 C. n(2n + 1)(3n 6 + 1) n(n + 1)2(n + 2) D. 12 Answer: B Solution: Solution: Given, n ≥ 2 Let S = 2C2 + 3C2 +... + nC2 = n+1 C3 n+1 Now, C2 + 2 × ( C2 + C2 +... + nC2) 2 3 n+1 = C2 + 2 × n + 1C3 n+1 =( C2 + n + 1C3) + n + 1C3 (n + 2)! (n + 1)! = n + 2C3 + n + 1C3 = + 3!(n − 1)! 3!(n − 2)! (n + 2)(n + 1)n(n − 1)! = 3 × 2 × 1 × (n − 1)! (n + 1) × n × (n − 1) × (n − 2)! + 3 × 2 × 1 × (n − 2)! n(n + 1) = [n + 2 + n − 1] 6 n(n + 1)(2n + 1) = 6 ------------------------------------------------------------------------------------------------- Question90 ( ) { n n Cr if n ≥ r ≥ 0 For integers n and r, let = The maximum value r 0 otherwise. of k for which the sum, k ∑ i=0 ( )( ) 10 i k−i 5 k+1 + i =∑ 0 ( )( 12 i k+1−i 13 ) exists, is equal to [2021, 24 Feb Shift-II] Answer: 1 Solution: Solution: ( ) { n n Cr if n≥r≥0 Given, = r 0 otherwise. ( )( ) ( )( k 10 15 k+1 12 and ∑ i=0 i k−i + ∑ i=0 i 13