Let function y=f(x) satisfies the differential equation x2dydx=y2e1x(x≠0) and limx→0f(x)=1 . Identity the correct statements?

Range of f ( x ) is ( 0 , 1 ) - 1 2 , f ( x ) is bounded , ∫ 0 e f x d x > ∫ 0 1 f x d x

Let function y=f(x) satisfies the differential equation x2dydx=y2e1x(x≠0) and limx→0f(x)=1 . Identity the correct statements?

d y y 2 = e 1 x x 2 d x ⇒ - 1 y = - ∫ e t d t + c p u t 1 x = t ⇒ - 1 y = - e 1 x + c ∴ lim x → 0 f ( x ) = 1 ⇒ - 1 = 0 + c ⇒ c = - 1 ∴ y = 1 1 + e 1 x lim x → ∞ 1 1 + e 1 x = 1 2

Range of f ( x ) is ( 0 , 1 ) - 1 2

∫ 0 e f x d x > ∫ 0 1 f x d x

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$Let function y = f (x) satisfies the differential equation x^{2} \frac{d y}{d x} = y^{2} e^{\frac{1}{x}} (x \neq 0) and$ $\lim_{x \to 0} f (x) = 1 . Identity the correct statements?$

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$Range of f (x) is (0, 1) - \{\frac{1}{2}\}$

$f (x) is bounded$

$\lim_{x \to \pm \infty} f (x) = 1$

$\int_{0}^{e} f (x) d x > \int_{0}^{1} f (x) d x$

answer is A, B, D.

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Detailed Solution

$\frac{d y}{y^{2}} = \frac{e^{\frac{1}{x}}}{x^{2}} d x$

$\Rightarrow - \frac{1}{y} = - \int e^{t} d t + c [p u t \frac{1}{x} = t]$

$\Rightarrow - \frac{1}{y} = - e^{\frac{1}{x}} + c$

$∴ \lim_{x \to 0} f (x) = 1 \Rightarrow - 1 = 0 + c \Rightarrow c = - 1$

$∴ y = \frac{1}{1 + e^{\frac{1}{x}}}$

$\lim_{x \to \infty} \frac{1}{1 + e^{\frac{1}{x}}} = \frac{1}{2}$

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