If x sqrt{3} and frac{x^2+1}{(x^2+2)(x^2+3) | vedclass

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

Question

(D) Let $u = x^{-2}$. Since $x > \sqrt{3}$,we have $x^2 > 3$,so $u = \frac{1}{x^2} < \frac{1}{3}$.We rewrite the expression as $\frac{x^2+1}{(x^2+2)(x

Vedclass · Accepted Answer

$-46$

Vedclass · Answer

(D) Let $u = x^{-2}$. Since $x > \sqrt{3}$,we have $x^2 > 3$,so $u = \frac{1}{x^2} We rewrite the expression as $\frac{x^2+1}{(x^2+2)(x^2+3)}$.Dividing numerator and denominator by $x^4$,we get $\frac{x^{-2} + x^{-4}}{(1 + 2x^{-2})(1 + 3x^{-2})} = (u + u^2)(1 + 2u)^{-1}(1 + 3u)^{-1}$.Using the binomial expansion $(1+z)^{-1} = 1 - z + z^2 - z^3 + \dots$,we have:$(1+2u)^{-1} = 1 - 2u + 4u^2 - 8u^3 + \dots$$(1+3u)^{-1} = 1 - 3u + 9u^2 - 27u^3 + \dots$Multiplying these: $(1+2u)^{-1}(1+3u)^{-1} = (1 - 2u + 4u^2 - 8u^3 + \dots)(1 - 3u + 9u^2 - 27u^3 + \dots) = 1 - 5u + 19u^2 - 65u^3 + \dots$Now,$(u + u^2)(1 - 5u + 19u^2 - 65u^3 + \dots) = u - 5u^2 + 19u^3 - 65u^4 + u^2 - 5u^3 + 19u^4 - 65u^5 + \dots$$= u - 4u^2 + 14u^3 - 46u^4 + \dots$The term $x^{-8}$ corresponds to $u^4$. The coefficient is $-46$.

If $x > \sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x^{-2}$,then the coefficient of $x^{-8}$ is

Similar Questions

$1+\frac{4}{15}+\frac{4 \times 10}{15 \times 30}+\frac{4 \times 10 \times 16}{15 \times 30 \times 45}+\ldots \quad \infty=$

For $0 < x < 1$,the expansion of $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ is

For $x=\frac{5}{7}$,if $t_k$ is the first negative term in the expansion of $(1+x)^{7/5}$,then $t_1+t_2+\ldots+t_k=$

If the $(r + 1)^{th}$ term is the first negative term in the expansion of $(1 + x)^{7/2}$,then the value of $r$ is

If $|x| < \frac{1}{2}$,then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{(1-2x)^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module