If the gradient of the tangent at any point (x, y) of a curve which passes through the point ≤ft(1,π/4) is ≤ft\{y/x - ² ≤ft(y/x)\} , then equation of the curve is

Correct Answer is : y = x cot⁡−1^{-1}cot−1 (log⁡e_{e}loge ex)

Correct Answer is : y = x cot⁡−1^{-1}cot−1 (log⁡e_{e}loge ex)

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VITEEE 2013 Paper

Section: Mathematics

Question : 120 of 120

Marks: +1, -0

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

\left\{\frac{y}{x} - \sin^2 \left(\frac{y}{x}\right)\right\}

, then equation of the curve is

Solution:

\frac{dy}{dx}

\frac{y}{x}

\sin^2

(y/x)

Put y = vx ⇒

\frac{dy}{dx}

= v + x

\frac{dv}{dx}

v + x

\frac{dv}{dx}

= v -

\sin^2

⇒ -

\csc^2

v dv =

\frac{dx}{x}

Integrating both sides, we get

\int \csc^2

v dv = ∫

\frac{dx}{x}

⇒ cot v = log x + C

cot

\frac{y}{x}

= log x + C

Curve passes through the point

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

∴ C = 1

⇒ cot

\frac{y}{x}

= log x +

\log_{e} e

⇒ cot

\frac{y}{x}

= log xe

⇒ y = x

\cot^{-1}

(

\log_{e}

ex)

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