Question

The equation of the curve in the form y = f(x) if the cuve passese through the point (1, 0) and Find f’(x)
= 2x-1 is
(a) y = x²-x
(b) x= y²-y
(c) y = x²
(d) none of these

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

The equation of the curve in the form y = f(x) if the curve passes through the point (1, 0) and f ' (x) = 2x-1 is

(a) y = x²-x

(b) x= y²-y

(c) y = x²

(d) None of these

EVALUATION

Here it is given that the curve is of the form y = f(x) and the curve passes through the point (1, 0) and f ' (x) = 2x-1

So by the given condition

$\displaystyle \sf{ \frac{dy}{dx} = 2x - 1}$

$\implies \sf{dy = (2x - 1)dx}$

Integrating both sides we get

$\displaystyle \sf{ \int dy = \int \: (2x - 1)dx}$

$\displaystyle \implies \sf{ y = 2. \frac{ {x}^{2} }{2} - x + c}$

$\displaystyle \implies \sf{ y = {x}^{2} - x + c}$

Where C is integration constant

Now the curve passes through the point (1, 0)

So we have

$\displaystyle \sf{ 0 = {(1)}^{2} - 1 + c}$

$\displaystyle \implies \sf{ c = 0}$

Hence the required equation of the curve is

$\displaystyle \sf{ y = {x}^{2} - x }$

FINAL ANSWER

Hence the correct option is

$\boxed{ \: \displaystyle \sf{ (a) \: \: \: \: \: y = {x}^{2} - x } \: \: \: }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. d²y/dx² + 2dy/dx + 2y = sin hx

include complete solution ( Complementary Function + Particular Integral)

https://brainly.in/question/28609831

2. Solve the differential equation :-

(cosx.cosy - cotx) dx - (sinx.siny) dy=0

https://brainly.in/question/23945760

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

The equation of the curve in the form y = f(x) if the curve passes through the point (1, 0) and f ' (x) = 2x-1 is

(a) y = x²-x

(b) x= y²-y

(c) y = x²

(d) None of these

EVALUATION

Here it is given that the curve is of the form y = f(x) and the curve passes through the point (1, 0) and f ' (x) = 2x-1

So by the given condition

$\sf \: \dfrac{dy}{df} = 2x−1 \\ </p><p>\implies \sf{dy = (2x - 1) dx} ⟹dy=(2x−1)dx</p><p> \: Integrating \: both \: sides \: we \: get</p><p>\displaystyle \sf{ \int dy = \int \: (2x - 1)dx}∫dy=∫(2x−1)dx</p><p>\displaystyle \implies \sf{ y = 2. \frac{ {x}^{2} }{2} - x + c}⟹y=2.2x2−x+c</p><p>\displaystyle \implies \sf{ y = {x}^{2} - x + c}⟹y=x2−x+c</p><p>$

Where C is integration constant

Now the curve passes through the point (1, 0)

So we have

$\displaystyle \sf{ 0 = {(1)}^{2} - 1 + c}$

$\sf \: ⟹c=0$

Hence the required equation of the curve is

$\displaystyle \sf{ y = {x}^{2} - x }y=x </p><p>2</p><p> −x$

FINAL ANSWER

Hence the correct option is

$\boxed{ \: \displaystyle \sf{ (a) \: \: \: \: \: y = {x}^{2} - x } \: \: \: } </p><p> \sf \\ \\ \sf(a)y=x </p><p>2</p><p> −x</p><p>$

━━━━━━━━━━━━━━━━