If n is the degree of the polynomial,left[ {f | vedclass

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

Question

(D) Let $P(x) = \left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}\right]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}\right]^{8}$.Rationalizi

Vedclass · Accepted Answer

$\left( {12,8{{\left( {10} \right)}^4}} \right)$

Vedclass · Answer

(D) Let $P(x) = \left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}ight]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}ight]^{8}$.Rationalizing the terms,we get:$P(x) = \left[\frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{2}ight]^{8} + \left[\frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{2}ight]^{8}$.$P(x) = \frac{1}{2^8} \left[ (\sqrt{5x^3+1} + \sqrt{5x^3-1})^8 + (\sqrt{5x^3+1} - \sqrt{5x^3-1})^8 ight]$.Using the expansion $(a+b)^8 + (a-b)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} a^{8-k} b^k$,we have:$P(x) = \frac{2}{2^8} \left[ \binom{8}{0} (5x^3+1)^4 + \binom{8}{2} (5x^3+1)^3(5x^3-1) + \binom{8}{4} (5x^3+1)^2(5x^3-1)^2 + \binom{8}{6} (5x^3+1)(5x^3-1)^3 + \binom{8}{8} (5x^3-1)^4 ight]$.The highest power of $x$ is $x^{3 	imes 4} = x^{12}$,so $n = 12$.The coefficient of $x^{12}$ is given by $\frac{2}{2^8} 	imes 5^4 	imes \left[ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} ight]$.Since $\sum_{k 	ext{ even}} \binom{8}{k} = 2^{8-1} = 2^7$,we have:$m = \frac{2}{2^8} 	imes 5^4 	imes 2^7 = \frac{2^8}{2^8} 	imes 5^4 	imes 2^3 = 8 	imes 10^4$.Thus,$(n, m) = (12, 8(10)^4)$.

If $n$ is the degree of the polynomial,$\left[ {\frac{1}{{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8 + \left[ {\frac{1}{{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8$ and $m$ is the coefficient of $x^{12}$ in it,then the ordered pair $(n, m)$ is equal to

Similar Questions

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}, x \neq 1,$ is equal to ....... .

Find the mean of the values $0, 1, 2, \dots, n$ with respective weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$.

If the $7^{th}$ term from the beginning in the binomial expansion of ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln x} \right)^9}$ for $x > 0$ is equal to $729$,then the possible value of $x$ is:

The positive value of $\lambda$ for which the coefficient of $x^2$ in the expression $x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10}$ is $720$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module