A) 1 sq unit
B) \[\frac{91}{30}sq\text{ }unit\]
C) \[\frac{30}{9}sq\text{ }unit\]
D) \[4sq\text{ }unit\]
E) \[\frac{30}{91}sq\text{ }unit\]
Correct Answer: B
Solution :
The equation of given curve is \[y={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+3\] On differentiating w. r. t.\[x,\]we get \[\frac{dy}{dx}=4{{x}^{3}}-6{{x}^{2}}+2x\] Again differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=12{{x}^{2}}-12x+2\] Put \[\frac{dy}{dx}=0\]for maxima or minima \[\Rightarrow \] \[2x(2{{x}^{2}}-3x+1)=0\] \[\Rightarrow \] \[x=0,1,\frac{1}{2}\] \[\therefore \] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=0}}=2\] \[\therefore \]Function is minimum at\[x=0\] and \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=1}}=12-12+2=2\] \[\therefore \]Function is minimum at\[x=\frac{1}{2}\]. also \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\frac{1}{2}}}=-1<0\] \[\therefore \]Function is maximum at\[x=\frac{1}{2}\]. \[\therefore \]Required area \[=\int_{0}^{1}{({{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+3})dx\] \[=\left[ \frac{{{x}^{5}}}{5}-\frac{2{{x}^{4}}}{4}+\frac{{{x}^{3}}}{3}+3x \right]_{0}^{1}\] \[=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=\frac{91}{30}\,sq\,unit\]
You need to login to perform this action.
You will be redirected in
3 sec
