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)

Options

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`

Answer

A

Explanation

`(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`.