is
A)
B)
C)
D)
Correct Answer: B
Solution :
We have, \[\operatorname{f}(x)\,\,=\,\,exp\,\left( \sqrt{5x-3-2{{x}^{2}}} \right)\] i.e., \[f(x)={{e}^{\sqrt{5x-3-2{{x}^{2}}}}}\] For Domain of \[\operatorname{f}(x),\,\,5x-3-2{{x}^{2}}\,\,should\,\,be+ve.\] i.e., \[\sqrt{5x-3-2{{x}^{2}}}\ge 0\] \[=\,\,\,2{{x}^{2}}-5x+3\le 0\] (By taking -ve sign common) \[\Rightarrow \,\,\,\,2x\left( x-1 \right)-3\left( x-1 \right)\le 0\] \[\Rightarrow \,\,(2x-3)(x-1)\le 0\] \[\Rightarrow \,\,\,2x-3\le 0\,\,or\,\,x-1\ge 0\] \[\Rightarrow \,\,\,x\le \frac{3}{2}\,\,or\,\,x\ge 0\] \[\therefore \,\,\,1\le \,\,x<\frac{3}{2}\,\,i.e.,\,\,x\in \,\,\left[ 1,\,\,\frac{3}{2} \right]\] Hence, domain of the given function is \[\left[ 1,\,\,\frac{3}{2} \right]\]
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