A) There exists a point where \[f(x)\] has a maximum value
B) There exists a point where \[f(x)\] has a minimum value
C) There exists a point where \[f(x)\] has neither maximum nor minimum value
D) All of the above
Correct Answer: D
Solution :
Given, \[f(x)={{(x-1)}^{2}}{{(x+1)}^{3}}\] \[f'(x)={{(x-1)}^{2}}3{{(x+1)}^{2}}+2(x-1){{(x+1)}^{3}}\] \[=(x-1){{(x+1)}^{2}}[3(x-1)+2(x+1)]\] \[=(x-1){{(x+1)}^{2}}[3x-3+2x+2]\] \[\Rightarrow \] \[f'(x)=(x-1)\,{{(x+1)}^{2}}\,(5x-1)\] ?..(i) For maxima or minima, \[f'(x)=0\] \[\Rightarrow \] \[(x-1){{(x+1)}^{2}}(5x-1)=0\] \[\Rightarrow \] \[x=-1,\,\,\,1,\frac{1}{5}\] Again, differentiating Eq. (i) . r. t, x. we get \[f''(x)=(x-1)(5x-1)\frac{d}{dx}{{(x+1)}^{2}}\] \[+{{(x+1)}^{2}}\frac{d}{dx}(x-1)(5x-1)\] \[=(5{{x}^{2}}-6x+1)2(x+1)+{{(x+1)}^{2}}(10x-6)\] \[=2[5{{x}^{3}}-{{x}^{2}}-5x+1]+(5{{x}^{3}}+7{{x}^{2}}-x-3)\,(2)\]\[\Rightarrow \] \[f''(x)=2[10{{x}^{3}}+6{{x}^{2}}-6x-2]\] At \[x=-1,\,f''(x)=0\] At \[x=\frac{1}{5},\,\,f''(x)<0\] and at \[x=1,\,\,f''(x)>0\] So, at \[x=-1,\] \[f(x)\] has neither maximum not minimum value At \[x=\frac{1}{5},\,\,f(x)\] has maximum value At \[x=1,\,\,\,f(x)\] has minimum value
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