InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 theequation 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)` |
|
Answer» Correct Answer - A It is given that `(dy)/(dx)=(y)/(x)+sec ((y)/(x)),x gt0" …(i)"` Let `y=vx.` Then, `(dy)/(dx)=v+x(dv)/(x)` Substituting these values in (i), we obtain `v+x(dv)/(dx)=v+sec v` `rArr" "x(dv)/(dx)=sec v` `rArr" "cos v dv =(1)/(x)dx` Integrating both sides, we get `sin v+log x+C` `rArr" "sin((y)/(x))=logx+C" ...(ii)"` `therefore" "sin.(pi)/(6)=log1+CrArr C=(1)/(2)` Putting `C=(1)/(2)` in (ii), we obtain the equation of the curve as `sin((y)/(x))=logx+(1)/(2)` |
|