If n is the degree of the polynomial, [ 1… - JEE Main 2018 Mathematics PYQ

JEE Main 2018 — Mathematics Algebra

2018mcqhard

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

Official previous-year question

Held on 15 Apr 2018 · Verified 6 Jul 2026.

Options

A
$\left(12,(20)^4\right)$
B
$\left(8,5(10)^4\right)$
C
$\left(24,(10)^8\right)$
D
$\left(12,8(10)^4\right)$

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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 $\begin{aligned} &x^{12}=\left[{ }^8 \mathrm{C}_0 5^4+{ }^8 \mathrm{C}_2 5^4+{ }^8 \mathrm{C}_4 5^4+{ }^8 \mathrm{C}_6 5^4+{ }^8 \mathrm{C}_8 5^4\right] \\ &=5^4 \times \frac{2^8}{2}=5^4 \times 2^4 \times \frac{2^2}{2} \\ &=10^4 \times 2^3=8(10)^4 \end{aligned}$

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