Mathematics › Binomial Theorem › Terms of Binomial Expansion
If $n$ is the degree of the polynomial, $ \left[\frac{1}{\sqrt{5 x^3+1}-\sqrt{5…
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 $\mathrm{m}$ is the coefficient of $\mathrm{x}^{\mathrm{n}}$ in it, then the ordered pair $(\mathrm{n}, \mathrm{m})$ is equal to
$\left(12,(20)^4\right)$
$\left(8,5(10)^4\right)$
$\left(24,(10)^8\right)$
$\left(12,8(10)^4\right)$ Solution
$
\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
$
After rationalise the polynomial we get
$
\begin{aligned}
&{\left[\frac{1}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}} \times \frac{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}{\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}} \times \frac{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}\right]^8
\end{aligned}
$
$
\left[\frac{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}{\left(5 x^3+1\right)-\left(5 x^3-1\right)}\right]^8+\left[\frac{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}{\left(5 x^3+1\right)-\left(5 x^3-1\right)}\right]^8
$
$
\begin{aligned}
&=\frac{1}{2^8}\left[\left(\sqrt{5 x^3+1}+\sqrt{5 x^3-1}\right)^8+\left(\sqrt{5 x^3+1}-\sqrt{5 x^3-1}\right)^8\right] \\
&\left.=\frac{1}{2^8}\left[\begin{array}{l}
\left.{ }^8 C_0\left(\sqrt{5 x^3+1}\right)^8+{ }^8 C_4\left(\sqrt{5 x^3+1}\right)^6\left(\sqrt{5 x^3-1}\right)^2\right)^4\left(\sqrt{5 x^3-1}\right)^4+ \\
{ }^8 C_6\left(\sqrt { 5 x ^ { 3 } + 1 } { } ^ { 2 } \left(\sqrt{\left.5 x^3-\right)^1}{ }^6+{ }^8 C_8\left(\sqrt{5 x^3-1}{ }^8\right.\right.\right.
\end{array}\right]\right) \\
&\frac{1}{2^8}\left[\begin{array}{l}
{ }^8 C_0\left(5 x^3+1\right)^4+{ }^8 C_2\left(5 x^3+1\right)^3\left(5 x^3-1\right)+8_{C_4} \\
\left(5 x^3+1\right)^2\left(5 x^3-1\right)^2+ \\
{ }^8 C_6\left(5 x^3+1\right)\left(5 x^3-1\right)^3+8_{C_8}\left(5 x^3-1\right)^4
\end{array}\right] \\
&
\end{aligned}
$
So, the degree of polynomial is 12 ,
Now, coefficient of
$
\begin{aligned}
&x^{12}=\left[{ }^8 \mathrm{C}_0 5^4+{ }^8 \mathrm{C}_2 5^4+{ }^8 \mathrm{C}_4 5^4+{ }^8 \mathrm{C}_6 5^4+{ }^8 \mathrm{C}_8 5^4\right] \\
&=5^4 \times \frac{2^8}{2}=5^4 \times 2^4 \times \frac{2^2}{2} \\
&=10^4 \times 2^3=8(10)^4
\end{aligned}
$
Asked in: JEE Main 2018 (15 Apr Shift 1 Online)
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