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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:

Option 1 -

$x^{3} + x f (x) + 12 = 0$

Option 2 -

$x^{3} - 3 x f (x) - 4 = 0$

Option 3 -

$x^{2} + 2 x f (x) + 4 = 0$

Option 4 -

$x^{2} + 2 x f (x) - 12 = 0$

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Given curve is f(x) + xf’(x) = x² i.e. y + x $\frac{d y}{d x} = x^{2}$

where P = $\frac{1}{x}, Q = x$

$I . F . = e^{\int P d x} = e^{\int \frac{1}{x} d x} = e^{l n x} = x$

Solution be y.x = $\int x . x d x$

$x y = \frac{x^{3}}{3} + c . . . . . . . . . (i)$

(i) passes through (-2, 2) then $- 4 = \frac{- 8}{3} + c$

$c = \frac{- 4}{3}$

(i) -> 3xy = x³ – 4

or x³ – 3xf(x) – 4 = 0

Given curve is f(x) + xf’(x) = x2 i.e. y + x <math> <mrow> <mfrac> <mrow> <mi>d</mi> <mi>y</mi> </mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> where P = <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> <mo>,</mo> <mi>Q</mi> <mo>=</mo> <mi>x</mi> </mrow> </math>  <math> <mrow> <mi>I</mi> <mo>.</mo> <mi>F</mi> <mo>.</mo> <mo>=</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mstyle displaystyle="true"> <mrow> <mo>∫</mo> <mrow> <mi>P</mi> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mstyle displaystyle="true"> <mrow> <mo>∫</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mi>l</mi> <mi>n</mi> <mi>x</mi> </mrow> </msup> <mo>=</mo> <mi>x</mi> </mrow> </math> Solution be y.x = <math> <mrow> <mstyle displaystyle="true"> <mrow> <mo>∫</mo> <mrow> <mi>x</mi> <mo>.</mo> <mi>x</mi> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> </mrow> </math>  <math> <mrow> <mi>x</mi> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mi>c</mi> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mrow> <mo>(</mo> <mrow> <mi>i</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> (i) passes through (-2, 2) then <math> <mrow> <mo>−</mo> <mn>4</mn> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>8</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mi>c</mi> </mrow> </math>  <math> <mrow> <mi>c</mi> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>4</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>        (i) -> 3xy = x3 – 4or x3 – 3xf(x) – 4 = 0

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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:

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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:

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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: