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Monotonicity

Let y = f x b...

Question

Let $y = f (x)$ be a real valued function satisfying $x \frac{d y}{d x} = x^{2} + y - 2, f (1) = 1$ ,, then which of the following option(s) is/are correct?

f(x) is minimum at x = 1

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f(x) is maximum at x = 1

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f(3) = 5

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f(2) = 3

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Solution

The correct options are
A f(x) is minimum at x = 1

C f(3) = 5

$\frac{x d y - y d x}{x^{2}} = (\frac{x^{2} - 2}{x^{2}}) d x$

$\int d (\frac{y}{x}) = \int (1 - \frac{2}{x^{2}}) d x$

$\Rightarrow y = x^{2} - 2 x + 2 [∵ f (1) = 1]$

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Let y=f(x) be a real valued function satisfying xdydx=x2+y−2,f(1)=1,, then which of the following option(s) is/are correct?

The correct options are A f(x) is minimum at x = 1 C f(3) = 5 xdy−ydxx2=(x2−2x2)dx ∫d(yx)=∫(1−2x2)dx ⇒y=x2−2x+2[∵f(1)=1]

Let $y = f (x)$ be a real valued function satisfying $x \frac{d y}{d x} = x^{2} + y - 2, f (1) = 1$ ,, then which of the following option(s) is/are correct?

The correct options are
A f(x) is minimum at x = 1

C f(3) = 5

$\frac{x d y - y d x}{x^{2}} = (\frac{x^{2} - 2}{x^{2}}) d x$

$\int d (\frac{y}{x}) = \int (1 - \frac{2}{x^{2}}) d x$

$\Rightarrow y = x^{2} - 2 x + 2 [∵ f (1) = 1]$