InterviewSolution
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InterviewSolution
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`sin^(-1){sin(2x^(2)+4)/(x^(2)+1)=ltpi-3` ifA. `x in [-1,0]`B. `x in [0,1]`C. `x in (-1,1)`D. `x in (1,oo)` |
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Answer» We have `(2x^(2)+4)/(x^(2)+1)=(2(x^(2)+1)+2)/(x^(2)+1)=2+(2)/(x^(2)+1)` `therefore 2lt(2x^(2)+4)/(x^(2)+1)le 4 forall x in R` `rarr pi-4ltpi-(2x^(2)+45)/(x^(2)+1)ltpi-2 forall x in R` `rarr -(pi)/(2)ltpi-(2x^(2)+4)/(x^(2)+1 lt(pi)/(2) forall x in R` `=pi-(2x^(2)+4)/(x^(2)+1)` Hence `sin^(-1){sin(2x^(2)+4)/(x^(2)+1)}ltpi-3` `rarr pi-(2x^(2)+4)/(x^(2)+1)ltpi-3` `rarr (2x^(2)+4)/(x^(2)+1)gt3` `rarr 2+(2)/(x^(2)+1)rarr2gt3gtx^(2)+1rarrx^(2)1lt0rarrx in (-1,1)` |
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