A) (1, 2)
B) (2, 3)
C) (1/2, 3)
D) (2, 4)
Correct Answer: B
Solution :
[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).
You need to login to perform this action.
You will be redirected in
3 sec
