RIGHTARROW f'(x)\] \[=\frac{2}{x-2}-2x+4\] \[\Rightarrow f'(x)=2\left[ \frac{1-{{(x-2)}^{2}}}{x-2} \right]=-2\frac{(x-1)(x-3)}{x-2}\] \[\Rightarrow f'(x)=\frac{2(x-1)(x-3)(x-2)}{{{(x-2)}^{2}}}\] \[\THEREFORE f'(x)>0\Rightarrow -2(x-1)(x-3)(x-2)>0\] \[\Rightarrow (x-1)(x-2)(x-3)<0\Rightarrow x\in (-\infty ,1)\cup (2,3)\] THUS, f(x) is INCREASING on \[(-\infty ,1)\cup (2,3)\]. Clearly, it includes answer [b] and (c).

"> RIGHTARROW f'(x)\] \[=\frac{2}{x-2}-2x+4\] \[\Rightarrow f'(x)=2\left[ \frac{1-{{(x-2)}^{2}}}{x-2} \right]=-2\frac{(x-1)(x-3)}{x-2}\] \[\Rightarrow f'(x)=\frac{2(x-1)(x-3)(x-2)}{{{(x-2)}^{2}}}\] \[\THEREFORE f'(x)>0\Rightarrow -2(x-1)(x-3)(x-2)>0\] \[\Rightarrow (x-1)(x-2)(x-3)<0\Rightarrow x\in (-\infty ,1)\cup (2,3)\] THUS, f(x) is INCREASING on \[(-\infty ,1)\cup (2,3)\]. Clearly, it includes answer [b] and (c).

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The function \[f(x)=2\,\,\log (x-2)-{{x}^{2}}+4x+1\]increases on the interval

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 11 months ago

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[b] \[f(x)=2\log (x-2)-{{x}^{2}}+4x+1\RIGHTARROW f'(x)\] \[=\frac{2}{x-2}-2x+4\] \[\Rightarrow f'(x)=2\left[ \frac{1-{{(x-2)}^{2}}}{x-2} \right]=-2\frac{(x-1)(x-3)}{x-2}\] \[\Rightarrow f'(x)=\frac{2(x-1)(x-3)(x-2)}{{{(x-2)}^{2}}}\] \[\THEREFORE f'(x)>0\Rightarrow -2(x-1)(x-3)(x-2)>0\] \[\Rightarrow (x-1)(x-2)(x-3)<0\Rightarrow x\in (-\infty ,1)\cup (2,3)\] THUS, f(x) is INCREASING on \[(-\infty ,1)\cup (2,3)\]. Clearly, it includes answer [b] and (c).

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