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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) `( b ) (c)sin(( d ) (e) (f) y/( g ) x (h) (i) (j))=logx+( k )1/( l )2( m ) (n) (o)` (p) (q) `( r ) (s)"c o s e c"(( t ) (u) (v) y/( w ) x (x) (y) (z))=logx+2( a a )` (bb) (cc) `( d d ) (ee)sec(( f f ) (gg) (hh)(( i i )2y)/( j j ) x (kk) (ll) (mm))=logx+2( n n )` (oo) (pp) `( q q ) (rr)cos(( s s ) (tt) (uu)(( v v )2y)/( w w ) x (xx) (yy) (zz))=logx+( a a a )1/( b b b )2( c c c ) (ddd) (eee)` (fff)

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

  • 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)),xgt0. 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)`

  • 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 gt 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))= log x+(1)/(2)`

  • If a curve passes through the point (1,(pi)/(4)) and its slope (dy)/(dx) at any point (x,y) is given by (dy)/(dx) = (y)/(x) - cos^(2) ((y)/(x)), then the equation of the curve is

    A
    `y = tan^(-1) {log(( e )/(x))}`
    B
    `y = x tan^(-1) { log ((x)/( e))}`
    C
    `y = x tan^(-1) {log (( e)/(x))}`
    D
    None of these

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