A) \[\left[ -\frac{1}{3},1 \right]\]
B) \[\left[ \frac{1}{2},1 \right]\]
C) \[\left[ -\frac{1}{2},\frac{1}{3} \right]\]
D) \[\left[ -1,\frac{1}{3} \right]\]
E) \[\left[ -\frac{1}{3},\frac{1}{2} \right]\]
Correct Answer: E
Solution :
The given function is \[f(x)=\sqrt{1-2x}+2{{\sin }^{-1}}\left( \frac{3x-1}{2} \right)\] For domain of\[\sqrt{1-2x,}1-2x\ge 0\] \[\Rightarrow \] \[1\ge 2x\Rightarrow x\le \frac{1}{2}\] \[\Rightarrow \] \[x\in \left( -\infty ,\frac{1}{2} \right]\] and for domain of\[2{{\sin }^{-1}}\left( \frac{3x-1}{2} \right),\] \[-1\le \frac{3x-1}{2}\le 1\] \[\Rightarrow \] \[-2\le 3x-1\le 2\] \[\Rightarrow \] \[-2+1\le 3x\le 2+1\] \[\Rightarrow \] \[-1\le 3x\le 3\] \[\Rightarrow \]\[-\frac{1}{3}\le x\le 1\] \[\therefore \]Domain of\[f(x)=\left[ -\frac{1}{3},\frac{1}{2} \right]\]
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