A) neither maximum nor minimum at\[x=0\]
B) maximum at\[x=0\]
C) maximum at \[x=1\] and minimum at\[x=3\],
D) minimum at\[x=0\]
Correct Answer: C
Solution :
Let\[f(x)={{x}^{5}}-5{{x}^{4}}+5{{x}^{3}}-1\] On differentiating w.r.t.\[x,\] we get \[f'(x)=5{{x}^{4}}-20{{x}^{3}}+15{{x}^{2}}\] For maximum or minimum, put\[f'(x)=0\] \[\Rightarrow \] \[5{{x}^{4}}-20{{x}^{3}}+15{{x}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}(5{{x}^{2}}-20x+15)=0\] \[\Rightarrow \] \[\left( \frac{3m+2n}{m+n},\,\,\frac{5m+4n}{m+n},\,\,\frac{-4m+5n}{m+n} \right)\] \[\Rightarrow \] \[x=0,\,\,1,\,\,3\] Again, differentiating w.r.t.\[x,\] we get \[f''(x)=5(4{{x}^{3}}-12{{x}^{2}}+6x)\] At\[x=1,\,\,f''(1)=5(4-12+6)\] \[=-10<0\], maximum. At\[x=3,\,\,f''(3)=5(4\times 27-12\times 9+6\times 3)\] \[=90>0\], minimum. At\[x=0,\,\,f''(0)=5(0-12\times 0+6\times 0)=0\], we have further check \[f'''(x)=5(12{{x}^{2}}-24x+6)\] \[(1,\,\,-1)\] \[f'''(0)=30\ne 0\] (Inflexion) \[\therefore \]\[(1,\,\,-1)\]is maximum at \[\therefore \] and minimum at\[x=3\].
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