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JEE MainMathematicsToughFunctions

If [x] denotes the greatest integer [IMAGE] x,
then the system of linear equations

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if θ∈(π2,2π3)\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right)
⇒ cosθ∈(−12,0)⇒−cosθ∈(0,1/2)cos\theta \in \left( - \frac{1}{2},0 \right) \Rightarrow - cos\theta \in (0,1/2)
sinθ∈(32,1)&cotθ∈(−13,0)sin\theta \in \left( \frac{\sqrt{3}}{2},1 \right)\& cot\theta \in \left( - \frac{1}{\sqrt{3}},0 \right)
If
θ∈(π,7π6)⇒cosθ∈(−1,−32)⇒−cosθ∈(32,1)\theta \in \left( \pi,\frac{7\pi}{6} \right) \Rightarrow cos\theta \in \left( - 1,\frac{- \sqrt{3}}{2} \right) \Rightarrow - cos\theta \in \left( \frac{\sqrt{3}}{2},1 \right)
sinθ∈(−12,0)&cotθ∈(3,∞)sin\theta \in \left( - \frac{1}{2},0 \right)\& cot\theta \in (\sqrt{3},\infty)
Hence in
θ∈(π2,2π3)⇒[sinθ]=0;[−cosθ]=0;[cotθ]=−1\theta \in \left( \frac{\pi}{2},\frac{2\pi}{3} \right) \Rightarrow \lbrack sin\theta\rbrack = 0;\lbrack - cos\theta\rbrack = 0;\lbrack cot\theta\rbrack = - 1
Hence 0=0&−x+y=0Have infinitely solutionsHence\ 0 = 0\& - x + y = 0Have\ infinitely\ solutions
$\begin{matrix}
\ \\
\ \
\end{matrix}in\ \theta \in \left( \pi,\frac{7\pi}{6} \right) \Rightarrow \lbrack sin\theta\rbrack = - 1;\lbrack - cos\theta\rbrack = 0;\lbrack cot\theta\rbrack = 1,2,3,\ldots\ldots\ldots\ldots\ldots$
Hence −x−0.y=0⇒x=0- x - 0.y = 0 \Rightarrow x = 0
[cotθ]x+y=0⇒1⋅x+y=0 or 2⋅x+y=0 or \lbrack cot\theta\rbrack x + y = 0 \Rightarrow 1 \cdot x + y = 0\ or\ 2 \cdot x + y = 0\ or\ .........
each line well cut x=0 at exactly one point. Hence unique solution.

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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