question_answer

The curve\[y=a{{x}^{2}}+bx+c\]passes through the point (1, 2) and its tangent at origin is the line\[y=x\]. The area bounded by the curve, the ordinate of the curve at minima and the tangent line is

A)  \[\frac{1}{24}\]                               

B)  \[\frac{1}{12}\]

C)  \[\frac{1}{8}\]                                  

D)  \[\frac{1}{6}\]

Correct Answer: A

Solution :

 Curve passes through the point\[(1,\,\,2).\] \[\therefore \]  \[2=a+b+c\]                                       ?(i) Also curve passes through the point\[(0,0)\]. \[\therefore \]  \[c=0\Rightarrow a+b=2\] Now, \[{{\left. \frac{dy}{dx} \right|}_{(0,0)}}=2a(0)+b=1\] \[\therefore \]  \[b=1;\,a=1\] Hence, the curve is\[y={{x}^{2}}+x={{(x+1\text{/}2)}^{2}}-1\text{/}4\] \[\therefore \] \[x=-1/2\]is point of minima. \[\therefore \]  Required area \[A=\int\limits_{-1/2}^{0}{({{x}^{2}}+x-x)\,dx}\]      \[=\int\limits_{-1/2}^{0}{({{x}^{2}})dx=1/24\,\text{sq}\text{.}\,\text{units}}\]