A) \[[-(\pi +2),\,0]\]
B) \[[0,4+\pi ]\]
C) \[[2,3]\]
D) \[(0,\,2+\pi ]\]
Correct Answer: B
Solution :
We have \[f:[0,\,1]\to R\] \[f(x)={{x}^{3}}-{{x}^{2}}+4x+2{{\sin }^{-1}}x\] Now \[f'(x)=3{{x}^{2}}-2x+4+\frac{2}{\sqrt{1-{{x}^{2}}}}\] For \[x\,[0,\,\,1],\,\,\,f'\,(x)>0\] Hence, it is a increasing function at \[x=0,\,f(0)=0-0+4(0)+2si{{n}^{-1}}(0)=0\] at \[x=1,f(1)=1-1+4(1)+2{{\sin }^{-1}}(1)\] \[=4+2\left( \frac{\pi }{2} \right)=4+\pi \] \[\therefore \] Range of \[f(x)\,[0,\,4+\pi ]\]
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