A) 4/3 sq. units
B) 7/3 sq. units
C) 7/6 sq. units
D) None of these
Correct Answer: C
Solution :
[c] Given \[y={{x}^{2}}+x+1={{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}\] \[\Rightarrow y-\frac{3}{4}={{\left( x+\frac{1}{2} \right)}^{2}}\] This is a parabola with vertex at \[\left( -\frac{1}{2},\frac{3}{4} \right)\]and the curve is concave upwards. \[y={{x}^{2}}+x+1\] or \[\frac{dy}{dx}=2x+1\]or \[{{\left( \frac{dy}{dx} \right)}_{(1.3)}}=3\] Equation of the tangent at \[A(1,3)\] is \[y=3x\] Required (shaded) area = Area ABDMN \[-\] Area ONA Now, area ABDMN\[=\int\limits_{-1}^{1}{({{x}^{2}}+x+1)dx}\] \[=2\int\limits_{0}^{1}{({{x}^{2}}+1)=\frac{8}{3}}\] Area of ONA\[=\frac{1}{2}\times 1\times 3=\frac{3}{2}.\] \[\therefore \] Required area \[=\frac{8}{3}-\frac{3}{2}\] \[=\frac{16-9}{6}=\frac{7}{6}\]sq. units.You need to login to perform this action.
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