A) One
B) Two
C) Zero
D) Infinite
Correct Answer: C
Solution :
[c] as \[0\le \sqrt{{{x}^{2}}-3x+2}<\infty \] so \[0\le ta{{n}^{\text{-}1}}\sqrt{{{x}^{2}}-3x+2}<\pi /2\] .. ... (i) and \[0\le \sqrt{4x-{{x}^{2}}-3}<\infty \] so \[0\le co{{s}^{-1}}\sqrt{4x-{{x}^{2}}-3}\le \pi /2\] ..... (ii) so from (i) + (ii) \[0<ta{{n}^{-1}}\sqrt{{{x}^{2}}-3x+2}+co{{s}^{-1}}\sqrt{4x-{{x}^{2}}-3}<\pi \] so no solution
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