A) 12
B) 15
C) 17
D) 19
Correct Answer: A
Solution :
Slope of the tangent \[{{\left. \frac{dy}{dx}=\frac{1}{{{(1-x)}^{2}}}\, \right|}_{x\,=\,2}}=1\] |
Equation of tangent is \[y+1=1\,(x-2)\] |
i.e. \[y=x-3\] |
parabola \[y=-{{a}^{2}}{{x}^{2}}+5ax-4\] |
Solving the equations of tangent and the parabola |
\[x-3=-{{a}^{2}}{{x}^{2}}+5ax-4\] |
\[{{a}^{2}}{{x}^{2}}+(1-5a)\,x+1=0\] |
Since x is real \[\therefore \] \[{{(1-5a)}^{2}}-4{{a}^{2}}\ge 0\] |
\[\Rightarrow \]\[a\le \frac{1}{7}\] or \[a\ge \frac{1}{3}\] |
Sum of roots \[=\frac{5a-1}{{{a}^{2}}}=2\times 2\] |
\[4{{a}^{2}}-5a+1=0\] |
\[a=1,\]\[\frac{1}{4}\] [\[a=\frac{1}{4}\]is rejected] |
\[\therefore \] \[S=1\] \[\therefore \]\[12S=12\] |
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