Binomial Theorem - JEE Main 2024 April Question Bank - MathonGo.pdf

Binomial Theorem JEE Main 2024 April Question Bank Questions...

Binomial Theorem JEE Main 2024 April Question Bank Questions MathonGo Q1 - 2024 (04 Apr Shift 1) 15 The sum of all rational terms in the expansion of (2 5 + 5 3 ) 1 1 is equal to : (1) 3133 (2) 931 (3) 6131 (4) 633 Q2 - 2024 (04 Apr Shift 1) 2C 3C 4C 2 2 2 Let a = 1 + 3! + 4! +5! + …, 1 C +1 C 2 C +2 C +2 C 3C 3C 3C 3C 0 1 0 1 2 0+ 1+ 2+ 3 b=1+ 1! + 2! + 3! +… 2b Then a2 is equal to Q3 - 2024 (04 Apr Shift 2) If the coefficients of x4 , x5 and x6 in the expansion of (1 + x)n are in the arithmetic progression, then the maximum value of n is: (1) 7 (2) 21 (3) 28 (4) 14 Q4 - 2024 (05 Apr Shift 1) 9 If the constant term in the expansion of (1 + 2x − 3x3 ) ( 32 x2 − 1 3x ) is p, then 108p is equal to Q5 - 2024 (05 Apr Shift 2) 12 If the constant term in the expansion of ( ) 5 √ 3 2x x + 3 , x ≠ 0, is α × 28 × √ 5 3, then 25α is equal to : √5 (1) 724 #MathBoleTohMathonGo www.mathongo.com Binomial Theorem JEE Main 2024 April Question Bank Questions MathonGo (2) 742 (3) 639 (4) 693 Q6 - 2024 (06 Apr Shift 1) 10 If the second, third and fourth terms in the expansion of (x + y)n are 135,30 and 3 , respectively, then 6 (n3 + x2 + y) is equal to _______ Q7 - 2024 (08 Apr Shift 1) Let α = ∑r=0 (4r2 + 2r + 1) n Cr and β = (∑r=0 ) nC n n 1 2α r+1 r + n+1. If 140 < β < 281, then the value of n is _______ Q8 - 2024 (08 Apr Shift 2) 10 If the term independent of x in the expansion of (√ax2 + 1 ) is 105 , then a2 is equal to : 2x3 (1) 2 (2) 4 (3) 6 (4) 9 Q9 - 2024 (09 Apr Shift 1) The coefficient of x70 in x2 (1 + x)98 + x3 (1 + x)97 + x4 (1 + x)96 + … + x54 (1 + x)46 is 99 Cp − 46 Cq. Then a possible value of p + q is : (1) 55 (2) 83 (3) 61 (4) 68 #MathBoleTohMathonGo www.mathongo.com Binomial Theorem JEE Main 2024 April Question Bank Questions MathonGo Q10 - 2024 (09 Apr Shift 1) The remainder when 4282024 is divided by 21 is__________ Q11 - 2024 (09 Apr Shift 2) 9 The sum of the coefficient of x2/3 and x−2/5 in the binomial expansion of (x2/3 + 12 x−2/5 ) is (1) 21/4 (2) 63/16 (3) 19/4 (4) 69/16 #MathBoleTohMathonGo www.mathongo.com Binomial Theorem JEE Main 2024 April Question Bank Questions MathonGo Answer Key Q1 (1) Q2 (8) Q3 (4) Q4 (54) Q5 (4) Q6 (806) Q7 (5) Q8 (2) Q9 (2) Q10 (1) Q11 (1) #MathBoleTohMathonGo www.mathongo.com Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo Q1 15−r r 1 Tr+1 = 15 Cr (5 3 ) ( ) 1 25 r 15−r = 15 Cr 5 ⋅ 2 3 5 R = 3λ, 15μ ⇒ r = 0, 15 2 rational terms ⇒ 15 C0 23 + 15 C15 (5)5 = 8 + 3125 = 3133 Q2 (1 + x) (1 + x)2 (1 + x)3 f(x) = 1 + + + + ….. 1! 2! 3! e(1+x) 1 (1 + x) (1 + x)2 (1 + x)2 = +1+ + + 1+x 1+x 2! 3! 4! 2 3 C2 C2 coef x2 in RHS : 1 + + +…=a 3 4 coeff. x2 in L.H.S. e (1 + x + x2 2! ) … (1 − x + x2 2! … …) e is e − e + 2! =a 2 22 23 b=1+ + + + … … = e2 1! 2! 3! 2b =8 a2 Q3 Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo Coeff. of x4 = n C4 Coeff. of x5 = n C5 Coeff. of x6 = n C6 n C4 , n C5 , n C6 …. AP 2C5 = n C4 + n C6 n n n n−r+1 2 = n 4 + n 6 {n r = } C C C C5 C5 Cr r 5 n−5 2= + n−4 6 12(n − 4) = 30 + n2 − 9n + 20 n2 − 21n + 98 = 0 (n − 14)(n − 7) = 0 nmax = 14 nmin = 7 Q4 3 2 1 9 (1 + 2x − 3x ) ( x − 3 ) 2 3x 3 2 1 9 General term m( x − ) 2 3x 39−2r = Cr ⋅ 9−r (−1)r ⋅ x18−3r 9 2 1 7 Put r = 6 to get coeff. of x0 = 9 C6 ⋅ ⋅ x0 = 18 x0 63 −5 Put r = 7 to get coeff. of x−3 = 9 Cr ⋅ 3 2 (−1)7 ⋅ x−3 2 1 −1 −3 = −9 C7 ⋅ 5 2 ⋅ x−3 = x 3 ⋅2 27 7 1 −3 (1 + 2x − 3x3 ) ( x0 − x ) 18 27 7 3 7 1 7+2 9 1 + = + = = = 18 27 18 9 18 18 2 1 ∴ 108 ⋅ = 54 2 Q5 Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo 12−r r Cr ( 31/5 ) 2x Tr+1 = 12 ( ) x 51/3 12−r 12 C (2)r (x)2r−12 r (3) 5 Tr+1 = (5)r/3 r=6 12 C6 (3)6/5 (2)6 9 × 11 × 7 T7 = =( ) 28 ⋅ 31/5 5 2 25 25α = 693 Q6 n C1 xn−1 y = 135....(i) n C2 xn−2 y 2 = 30....(ii) n C3 xn−3 y3 = 10 3....(iii) (i) By ( ii) n C1 x 9 =... (iv) nC y 2 2 ( ii) By ( iii) nC 2 x nC y = 9....(v) 3 Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo ( iv ) By (v) n n C1 C3 1 = nC nC 2 2 2 2 2n (n − 1)(n − 2) n(n − 1) n(n − 1) = 6 2 2 4n − 8 = 3n − 3 ⇒n=5 put in (v) x =9 y x = 9y put in (i) C1 x4 ( ) = 135 5 x 9 5 x = 27 × 9 1 ⇒ x = 3, y = 3 3 2 6 (n + x + y) 1 = 6 (125 + 9 + ) 3 = 806 Q7 Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo n α = ∑ (4r2 + 2r + 1) ⋅ n Cr r=0 n n n α = 4 ∑ r2 ⋅ Cr−1 + 2 ∑ r ⋅ ⋅ n−1 Cr− + ∑ n Cr n n−1 n ⋅ r=0 r r=0 r r=0 n n n + 4n ∑ n−1 Cr−1 + 2n ∑ n−1 Cr−1 + ∑ n Cr r=0 r=0 r=0 α = 4n(n − 1) ⋅ 2 + 4n ⋅ 2 + 2n ⋅ 2 n−2 n−1 n−1 + 2n n−2 α = 2 [4n(n − 1) + 8n + 4n + 4] α = 2n−2 [4n2 + 8n + 4] α = 2n(n + 1)2 n Cr n 1 β=∑ + r=0 r+1 n+1 n C n+1 Cr+1 1 =∑ + r=0 n+1 n+1 1 = (1 + n+1 C1 + …. +n+1 Cn+1 ) n+1 2n+1 = n+1 2α 2n+1 (n + 1)2 = ⋅ (n + 1) = (n + 1)3 β 2 n+1 140 < (n + 1)3 < 281 n = 4 ⇒ (n + 1)3 = 125 n = 5 ⇒ (n + 1)3 = 216 n = 6 ⇒ (n + 1)3 = 343 ∴n=5 Q8 10 (√ax2 + 1 2x3 ) r ( 2x1 3 ) 10−r General term = 10 Cr (√ax2 ) 20 − 2r − 3r = 0 r=4 10 1 C4 a 3 ⋅ = 105 16 a3 = 8 a2 = 4 SECTION - B Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo Q9 x2 (1 + x)98 + x3 (1 + x97 ) + x4 (1 + x)96 + … … x54 (1 + x)46 Coeff. of x70 : 98 C68 + 97 C67 + 96 C66 + … … … 47 C17 + 46 C16 = 46 C30 + 47 C30 + … … ….. 98 C30 = (46 C31 + 46 C30 ) + 47 C30 + … … …. 98 C30 − 46 C31 = 47 C31 + 47 C30 + … … ….. 98 C30 − 46 C31 …… = 99 C31 − 46 C31 = 99 Cp − 46 Cq Possible values of (p + q) are 62, 83, 99, 46 ⇒ p + q = 83 Q10 (428)2024 = (420 + 8)2024 = (21 × 20 + 8)2024 = 21 m + 82024 1012 Now 82024 = (82 ) = (64)1012 = (63 + 1)1012 = (21 × 3 + 1)1012 = 2 ln +1 ⇒ Remainder is 1. Q11 r ( x−2/5 ) 9−r Tr+1 = Cr (x 9 2/3 ) 2 1 r (6 2r 2r ) = 9 Cr ( ) (r) 3 5 2 2r 2r 2 for coefficient of x2/3 , put 6 − 3 − 5 = 3 ⇒r=5 5 ∴ Coefficient of x2/3 is = 9 C5 ( 15 ) 2r 2r For coefficient of x−2/5 , put 6 − 3 − 5 = − 25 ⇒r=6 Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App Binomial Theorem JEE Main 2024 April Question Bank Solutions MathonGo 6 Coefficient of x−2/5 is 9 C6 ( 12 ) 5 6 Sum = 9 C5 ( 12 ) + 9 C6 ( 12 ) = 21 4 Do you want to practice these PYQs along with PYQs of JEE Main from 2002 till 2024? Click here to download MARKS App