A) ? 37, ? 9
B) 10, 0
C) It has 2 min. and 1 max. values
D) It has 2 max. and 1 min. values
Correct Answer: A
Solution :
\[y={{x}^{5}}-5{{x}^{4}}+5{{x}^{3}}-10\] \ \[\frac{dy}{dx}=5{{x}^{4}}-20{{x}^{3}}+15{{x}^{2}}\]\[=\,\,5{{x}^{2}}({{x}^{2}}-4x+3)\] \[=\,\,5{{x}^{2}}(x-3)\,(x-1)\] \[\frac{dy}{dx}=0\], gives \[x=0,\,1,\,3\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=20{{x}^{3}}-60{{x}^{2}}+30x\] = \[10x(2{{x}^{2}}-6x+3)\] and \[\frac{{{d}^{3}}y}{d{{x}^{3}}}=10(6{{x}^{2}}-12x+3)\] For \[x=0\]: \[\frac{dy}{dx}=0,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,\,\frac{{{d}^{3}}y}{d{{x}^{3}}}\ne 0\] \ Neither minimum nor maximum For\[x=1,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}=-10=\text{negative}\]. \ Maximum value \[{{y}_{\text{max}\text{.}}}=-9\] For \[x=3,\,\frac{{{d}^{2}}y}{d{{x}^{2}}}=90=\text{positive}\] \ Minimum value \[{{y}_{\text{min}\text{.}}}=-37\].
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