JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 21)
If $$\alpha$$ denotes the number of solutions of $$|1-i|^x=2^x$$ and $$\beta=\left(\frac{|z|}{\arg (z)}\right)$$, where $$z=\frac{\pi}{4}(1+i)^4\left[\frac{1-\sqrt{\pi} i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}\right], i=\sqrt{-1}$$, then the distance of the point $$(\alpha, \beta)$$ from the line $$4 x-3 y=7$$ is __________.
Answer
3
Explanation
$$\begin{aligned} & (\sqrt{2})^x=2^x \Rightarrow x=0 \Rightarrow \alpha=1 \\ & z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right] \\ & =-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right) \\ & =2 \pi i \\ & \beta=\frac{2 \pi}{\frac{\pi}{2}}=4 \end{aligned}$$
Distance from $$(1,4)$$ to $$4 x-3 y=7$$
Will be $$\frac{15}{5}=3$$
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