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 |
\[x-3=-{{a}^{2}}{{x}^{2}}+5\,\,ax-4\] |
\[{{a}^{2}}{{x}^{2}}+(1-5\,a)\,x+1=0\] |
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 ] |
\[\therefore \,\,\,S=1\] \[\therefore \,\,\,12\,S=12\] |
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