If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is {{{x^2} - 4x + y + 8} \over {x - 2}}, then this curve also passes through the point :

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Answer: B. (5, 5) Given y (0) = 0 & {{dy} \over {dx}} = {{{{(x - 2)}^2} + y + 4} \over {x - 2}} \Rightarrow {{dy} \over {dx}} - {y \over {x - 2}} = (x - 2) + {4 \over {x - 2}} \Rightarrow I.F. = {e^{ - \int {{1 \over {x - 2}}dx} }} = {1 \over {x - 2}} Solution of D.E. \Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} ight)} \,.\,dx \Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C Now, at x = 0, y = 0 \Rightarrow C = -2 	herefore y = x (x - 2) - 4 - 2 (x - 2) \Rightarrow y = x2 - 4x This curve passes through (5, 5)

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $${{{x^2} - 4x + y + 8} \over {x - 2}}$$, then this curve also passes through the point :

Answer

Explanation

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