Exams › JEE Advanced › Maths
Correct answer: 10
Write f(x) = 1+x+2x²+4x⁴. We need [x⁷⁸] f(x)²⁰. Note that 4x⁴ = 4x⁴ and 2x². Let's think combinatorially: each factor contributes one of {1, x, 2x², 4x⁴}. If in 20 factors, a factors contribute x, b factors contribute 2x², c factors contribute 4x⁴, and d=20-a-b-c contribute 1: exponent condition a+2b+4c=78, a+b+c<=20. Coefficient = C(20;a,b,c,d) * 1^a * 2^b * 4^c = 20!/(a!b!c!d!) * 2^b * 2^(2c). We want to write this as lambda*2⁴⁰: so we need 2^(b+2c)=2⁴⁰ from those terms, meaning b+2c=40 always (constrained). From a+2b+4c=78 and b+2c=40: a+2(b+2c)-... let s=b+2c. If b+2c=40 then 2b+4c=2s=80, so a+80-2b+... wait: a+2b+4c=78, and b+2c=40=>2b+4c=80-2(4c-... Hmm: 2b+4c = 2(b+2c)=80. But a+2b+4c=a+80... but a+80=78 => a=-2, impossible. So b+2c is NOT always 40. We need the total coefficient of x⁷⁸ and express as lambda*2⁴⁰. Sum over all valid (a,b,c): [20!/(a!b!c!(20-a-b-c)!)]*2^(b+2c) where a+2b+4c=78, a+b+c<=20. Max power = 20*4=80, so we need a+2b+4c=78. With c<=20, b<=20, a<=20: c_max: if c=19,a=0,b=0: 4*19=76<78. c=19,b=1:76+2=78,a=0: valid. c=18: 4*18=72, remaining=6=2b+a, with b+a<=2. b=3 impossible (b<=2). b=2,a=2: 4+2=6 works, a+b+c=2+2+18=22>20: invalid. b=2,a=0: 4+0=4≠6. b=1,a=4: 2+4=6, sum=1+4+18=23>20: invalid. Hmm. c=19,b=0,a=2: 76+0+2=78, d=20-2-0-19=-1: invalid. c=19,b=1,a=0: 76+2=78, d=0. Sum=20. Valid! Coeff=20!/(0!1!19!0!)*2^(1+38)=20*2³⁹. c=20: 4*20=80>78, need a+2b=−2: impossible. c=18: 4*18=72, need a+2b=6, a+b+c<=20=>a+b<=2. Max a+2b with a+b<=2: a=0,b=2:4, or a=2,b=0:2, or a=1,b=1:3. None reach 6. c=17: 4*17=68, need a+2b=10, a+b<=3. Max a+2b=2+1*2=... a+b<=3: max 2b with b<=3: b=3,a=0:6<10. Not enough. c<=16 won't help since deficit grows. So only case: c=19,b=1,a=0,d=0. Coeff = C(20,19)*C(1,1)*2¹*(2²)¹⁹ = wait. Coeff = [20!/(0!*1!*19!*0!)]*2¹*4¹⁹ = 20*2*2³⁸ = 20*2³⁹ = 10*2⁴⁰. So lambda=10.