A) \[-1/4\]
B) \[0\]
C) \[1/4\]
D) \[1/2\]
Correct Answer: B
Solution :
Given, \[y={{x}^{40}}-{{x}^{20}}\] \[\therefore \] \[\frac{dy}{dx}=40{{x}^{39}}-20{{x}^{19}}\] \[\frac{dy}{dx}=20{{x}^{19}}(2{{x}^{20}}-1)\] Now, put \[\frac{dy}{dx}=0\] ie, \[20{{x}^{19}}(2{{x}^{20}}-1)=0\] \[\Rightarrow \] \[x=0\] or \[{{x}^{20}}=1/2\] \[{{y}_{x=0}}=0\] and \[{{y}_{x=1}}=0\] Now, \[y={{x}^{40}}-{{x}^{20}}={{({{x}^{20}})}^{2}}-{{x}^{20}}\] \[{{y}_{{{x}^{20}}=\frac{1}{2}}}={{\left( \frac{1}{2} \right)}^{2}}-\frac{1}{2}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\] \[\therefore \] Absolute maximum value of y is 0.
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