JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 6)
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 :
(4, 4)
(5, 5)
(5, 4)
(4, 5)
Explanation
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.
$$ \Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} \right)} \,.\,dx$$
$$ \Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C$$
Now, at x = 0, y = 0 $$ \Rightarrow $$ C = $$-$$2
$$ \therefore $$ y = x (x $$-$$ 2) $$-$$ 4 $$-$$ 2 (x $$-$$ 2)
$$ \Rightarrow $$ y = x2 $$-$$ 4x
This curve passes through (5, 5)
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.
$$ \Rightarrow y.{1 \over {x - 2}} = \int {{1 \over {x - 2}}\left( {(x - 2) + {4 \over {x - 2}}} \right)} \,.\,dx$$
$$ \Rightarrow {y \over {x - 2}} = x - {4 \over {x - 2}} + C$$
Now, at x = 0, y = 0 $$ \Rightarrow $$ C = $$-$$2
$$ \therefore $$ y = x (x $$-$$ 2) $$-$$ 4 $$-$$ 2 (x $$-$$ 2)
$$ \Rightarrow $$ y = x2 $$-$$ 4x
This curve passes through (5, 5)
Comments (0)
