The coefficient of x^2 in the expansion of th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6$.We need the coefficient of $x^2$ in $(2 - x^2)f(x)$.This is equal to $2 	imes (	ext{coefficient of

Vedclass · Accepted Answer

$106$

Vedclass · Answer

(A) Let $f(x) = (1 + 2x + 3x^2)^6 + (1 - 4x^2)^6$.We need the coefficient of $x^2$ in $(2 - x^2)f(x)$.This is equal to $2 	imes (	ext{coefficient of } x^2 	ext{ in } f(x)) - 1 	imes (	ext{constant term in } f(x))$.First,find the constant term in $f(x)$:Constant term $= (1 + 2(0) + 3(0)^2)^6 + (1 - 4(0)^2)^6 = 1^6 + 1^6 = 2$.Next,find the coefficient of $x^2$ in $f(x)$:For $(1 + 2x + 3x^2)^6$,using the multinomial expansion or binomial expansion $(1 + (2x + 3x^2))^6 = 1 + 6(2x + 3x^2) + \binom{6}{2}(2x)^2 + \dots = 1 + 12x + 18x^2 + 15(4x^2) + \dots = 1 + 12x + 78x^2 + \dots$The coefficient of $x^2$ is $78$.For $(1 - 4x^2)^6$,the expansion is $1 + 6(-4x^2) + \dots = 1 - 24x^2 + \dots$The coefficient of $x^2$ is $-24$.Thus,the coefficient of $x^2$ in $f(x)$ is $78 - 24 = 54$.Finally,the coefficient of $x^2$ in $(2 - x^2)f(x)$ is $2(54) - 1(2) = 108 - 2 = 106$.

The coefficient of $x^2$ in the expansion of the product $(2 - x^2)((1 + 2x + 3x^2)^6 + (1 - 4x^2)^6)$ is

Similar Questions

$x^5 + 10x^4a + 40x^3a^2 + 80x^2a^3 + 80xa^4 + 32a^5 = $

If $k$ is a positive integer and $10^k$ is a divisor of the number $9^{11}+11^9$,then the greatest value of $k$ is

For $|x| < \frac{1}{\sqrt{2}}$, the coefficient of $x$ in the expansion of $\frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}$ is

The sum of the coefficients of all odd degree terms in the expansion of $(x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$,where $x > 1$,is:

Using the Binomial Theorem,determine which number is larger: $(1.1)^{10000}$ or $1000$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module