JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 6)
Let us consider a curve, y = f(x) passing through the point ($$-$$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
$${x^2} + 2xf(x) - 12 = 0$$
$${x^3} + xf(x) + 12 = 0$$
$${x^3} - 3xf(x) - 4 = 0$$
$${x^2} + 2xf(x) + 4 = 0$$
Explanation
$$y + {{xdy} \over {dx}} = {x^2}$$ (given)
$$ \Rightarrow {{dy} \over {dx}} + {y \over x} = x$$
If $${e^{\int {{1 \over x}dx} }} = x$$
Solution of DE
$$ \Rightarrow y\,.\,x = \int {x\,.\,x\,dx} $$
$$ \Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$$
Passes through ($$-$$2, 2), so
$$-$$12 = $$-$$ 8 + c $$\Rightarrow$$ c = $$-$$ 4
$$\therefore$$ 3xy = x3 $$-$$ 4
i.e. 3x . f(x) = x3 $$-$$ 4
$$ \Rightarrow {{dy} \over {dx}} + {y \over x} = x$$
If $${e^{\int {{1 \over x}dx} }} = x$$
Solution of DE
$$ \Rightarrow y\,.\,x = \int {x\,.\,x\,dx} $$
$$ \Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$$
Passes through ($$-$$2, 2), so
$$-$$12 = $$-$$ 8 + c $$\Rightarrow$$ c = $$-$$ 4
$$\therefore$$ 3xy = x3 $$-$$ 4
i.e. 3x . f(x) = x3 $$-$$ 4
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