The number of distinct solutions of the equations $x^{11}-x^7+x^4-1=0$ is

The number of distinct solutions of the equations $x^{11}-x^7+x^4-1=0$ is
  1. 9
  2. 11
  3. 10
  4. 8

Solution

Given equation is $ \begin{aligned} & x^{11}-x^7+x^4-1=0 \\ & =x^7\left(x^4-1\right)+1\left(x^4-1\right)=0 \\ & =\left(x^4-1\right)\left(x^7+1\right)=0 \end{aligned} $ Case (i): $x^4-1=0$ $ \begin{aligned} & \text { or } \mathrm{x}^4=1 \\ & \Rightarrow \mathrm{x}^4=(\cos \mathrm{O}+\mathrm{i} \sin \mathrm{o}) \\ & \mathrm{x}^4=(\cos 2 \mathrm{k} \pi+i \sin 2 \mathrm{k} \pi) \\ & \mathrm{x}=(\cos 2 \mathrm{k} \pi+i \sin 2 \mathrm{k} \pi)^{1 / 4} \\ & \mathrm{~K}=0,1,2,3 \end{aligned} $ Case (ii): $x^7+1=0$ $ \begin{aligned} & \Rightarrow \mathrm{x}^7=-1=\cos \pi+\mathrm{i} \sin \pi \\ & \Rightarrow \mathrm{x}^7=\cos (2 \mathrm{k} \pi+\pi)+\mathrm{i} \sin (2 \mathrm{k} \pi+\pi) \\ & \Rightarrow \mathrm{x}=(\cos (2 \mathrm{k} \pi+\pi)+\mathrm{i} \sin (2 \mathrm{k} \pi+\pi))^{1 / 7} \\ & \Rightarrow \mathrm{x}=\mathrm{c} \text { is }(2 \mathrm{k}+1)^{\pi / 7} \\ & \Rightarrow \mathrm{k}=0,1,2,3,4,5,6 \end{aligned} $ now values of $x$ are $\mathrm{c}$ is $\mathrm{o}=1, \mathrm{c}$ is $\frac{3 \neq}{2}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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