A) \[\left( \frac{6}{5},\,\frac{36}{25} \right)\]
B) \[\left( \frac{2}{3},\,\frac{4}{9} \right)\]
C) \[\left( \frac{1}{3},\,\frac{1}{9} \right)\]
D) \[\left( \frac{1}{5},\,\frac{1}{25} \right)\]
Correct Answer: A
Solution :
Let \[{{F}_{1}}(x)={{y}_{1}}=\int_{2}^{x}{(2t-5)dt}\]and \[{{F}_{2}}(x)={{y}_{2}}=\int_{0}^{x}{2t\,\,dt}\] Now point of intersection means those point at which \[{{y}_{1}}={{y}_{2}}=y\Rightarrow {{y}_{1}}={{x}^{2}}-5x+6\]and \[{{y}_{2}}={{x}^{2}}\]. On solving, we get \[{{x}^{2}}={{x}^{2}}-5x+6\Rightarrow x=\frac{6}{5}\] and \[y={{x}^{2}}=\frac{36}{25}\]. Thus point of intersection is \[\left( \frac{6}{5},\frac{36}{25} \right)\].
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