The degree of the polynomial $\left(x+\sqrt{x^4-1}\right)^9+\left(x-\sqrt{x^4-1}\right)^9$ is

The degree of the polynomial $\left(x+\sqrt{x^4-1}\right)^9+\left(x-\sqrt{x^4-1}\right)^9$ is
  1. $14$
  2. $15$
  3. $16$
  4. $17$

Solution

$\begin{aligned} & \text {}\left(x+\sqrt{x^4-1}\right)^9+\left(x-\sqrt{x^4-1}\right)^9 \\ & \text { Let } y=\sqrt{x^4-1} \Rightarrow y^2=x^4-1 \\ & =(x+y)^9+(x-y)^9 \\ & =2\left[{ }^9 C_1 x^1 y^8+{ }^9 C_3 x^3 y^6+{ }^9 C_5 x^5 y^4+{ }^9 C_7 x^7 \cdot y^2+{ }^9 C_9 x^7 y^0\right] \\ & =2\left[{ }^9 C_1 x^1\left(x^4-1\right)^4+{ }^9 C_3 x^3\left(x^4-1\right)^3+{ }^9 C_4 x^5\left(x^4-1\right)^2\right. \\ & \left.\quad+{ }^9 C_7 x^7\left(x^4-1\right)+{ }^9 C_9 x^7\right] \\ & \therefore \text { Degree of polynomial }=1+4 \times 4=17\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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