If the slope of the tangent to the curve at any point P(x,y) { is } y/x - ² y/x , then the equation of a curve passing through ≤ft(1, π/4) is

Correct Answer is : tan⁡(yx)+log⁡x=1 ≤ft(y/x) + x = 1tan(xy)+logx=1

Correct Answer is : tan⁡(yx)+log⁡x=1 ≤ft(y/x) + x = 1tan(xy)+logx=1

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BITSAT 2021 Paper

Section: Mathematics

Question : 118 of 150

Marks: +1, -0

P(x,y) \text{ is } \frac{y}{x} - \cos^2

\frac{y}{x}

\left(1, \frac{\pi}{4}\right)

Solution:

According to the condition,

\frac{dy}{dx} = \frac{y}{x} - \cos^2

\frac{y}{x}

...(i)

This is a homogeneous differential equation Substituting

y = vx

, we get

v + x \frac{dv}{dx} = v - \cos^2 v

⇒

x \frac{dv}{dx} = - \cos^2 v

⇒

\int \sec^2 v dv = - \int \frac{dx}{x}

⇒

\tan v = -\log x + C

⇒

\tan \frac{y}{x} + \log x = C

Substituting

x = 1, y = \frac{\pi}{4}

, we get

C = 1

Thus, we get

\tan \left(\frac{y}{x}\right) + \log x = 1

which is the required solution,

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