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A curve passes through the point (1,pi/6...

A curve passes through the point `(1,pi/6)` . Let the slope of the curve at each point `(x , y)` be `y/x+sec(y/x),x > 0.` Then the equation of the curve is

A

`sin(y/x)=logx+1/2`

B

`"cosec "(y/x)=logx+2`

C

`sec(2y)/(x)=logx+2`

D

`cos(2y)/(x)=logx+1/2`

Text Solution

Verified by Experts

The correct Answer is:
A
`(dy)/(dx) =y/x+secy/x`.
Let `y=vx`. Then given equation reduces to `(dv)/(secv) = (dv)/x`
or `intcosvdv=int(dx)/x`
or `sinv="ln"x+c` or `sin(y/x)=logx+c`
The curve passes through `(1,pi6)`.
Thus, `sin(y/x)=logx+1/2`.
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Knowledge Check

  • The gradient (siope) of a curve at any point (x,y) is (x^(2)-4)/(x^(2)) . If the curve passes through the point (2 , 7) , then the equation of the curve is ,

    A
    `y = x + (4)/(x) + 3 `
    B
    `y = x + (4)/(x) + 4 `
    C
    `y = x^(2) + 3x + 4 `
    D
    `y = x^(2) - 3x + 6 `

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