JEE Main 2018 — Binomial Theorem Question with Solution
From: JEE Main 2018 (Online) 16th April Morning Slot
Question
The coefficient of x2 in the expansion of the product
(2x2) .((1 + 2x + 3x2)6 + (1 4x2)6) is :
(2x2) .((1 + 2x + 3x2)6 + (1 4x2)6) is :
Choose an option
Show full solutionCorrect option: B
Correct answer
B106
Step-by-step explanation
Given,
(2 x2) . (1 + 2x + 3x2) 6 + (1 4x2)6)
Let, a = ((1 + 2x + 3x2)6 + (1 4x2)6)
Given statement becomes,
(2 x2) . (a)
= 2a x2 (a)
Here coefficients of x2 is
= 2 (coefficient of x2 in a ) 1 (constant in a)
(1 + 2x + 3x2) 6 = 6C0 + 6C1 (2x + 3x2) + 6C2 (2x + 3x2)2
+ . . . . . .+ (2x + 3x2)6
(1 4x2)6 = 6C0 6C1 (4x2) + 6C2 (4x2)2
+ . . . . .+ (4x2)6
Coefficient of x2 in (1 + 2x + 3x2)6
= 6C1 3 + 6C2 4
= 18 + 60
Coefficient of x2 in (1 4x2)6
= 6C1 4
= 24
Coefficient of x2 in ((1 + 2x + 3x2)6 + (1 4x2)6)
= 60 + 18 24
= 54
Constant term in (1 + 2x + 3x2)6 = 6C0 = 1
Constant term in (1 4x)6 = 6C0 = 1
Constant term in ((1 + 2x + 3x2)6 + (1 4x)6)
= 1 + 1 = 2
Coefficient of x2 in (2 x2) ((1 + 2x + 3x2)6 + (1 4x2)6)
= 2 (54) 1 (2)
= 108 2
= 106
(2 x2) . (1 + 2x + 3x2) 6 + (1 4x2)6)
Let, a = ((1 + 2x + 3x2)6 + (1 4x2)6)
Given statement becomes,
(2 x2) . (a)
= 2a x2 (a)
Here coefficients of x2 is
= 2 (coefficient of x2 in a ) 1 (constant in a)
(1 + 2x + 3x2) 6 = 6C0 + 6C1 (2x + 3x2) + 6C2 (2x + 3x2)2
+ . . . . . .+ (2x + 3x2)6
(1 4x2)6 = 6C0 6C1 (4x2) + 6C2 (4x2)2
+ . . . . .+ (4x2)6
Coefficient of x2 in (1 + 2x + 3x2)6
= 6C1 3 + 6C2 4
= 18 + 60
Coefficient of x2 in (1 4x2)6
= 6C1 4
= 24
Coefficient of x2 in ((1 + 2x + 3x2)6 + (1 4x2)6)
= 60 + 18 24
= 54
Constant term in (1 + 2x + 3x2)6 = 6C0 = 1
Constant term in (1 4x)6 = 6C0 = 1
Constant term in ((1 + 2x + 3x2)6 + (1 4x)6)
= 1 + 1 = 2
Coefficient of x2 in (2 x2) ((1 + 2x + 3x2)6 + (1 4x2)6)
= 2 (54) 1 (2)
= 108 2
= 106
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