Question

The slope of the tangent at `(x , y)` to a curve passing through `(1,pi/4)` is given by `y/x-cos^2(y/x),` then the equation of the curve is (a) `( b ) (c) y=( d ) (e)tan^(( f ) (g)-1( h ))( i )(( j ) (k)log(( l ) (m) (n) e/( o ) x (p) (q) (r))( s ))( t )` (u) (v) `( w ) (x) y=x (y) (z)tan^(( a a ) (bb)-1( c c ))( d d )(( e e ) (ff)log(( g g ) (hh) (ii) x/( j j ) e (kk) (ll) (mm))( n n ))( o o )` (pp) (qq) `( r r ) (ss) y=x (tt) (uu)tan^(( v v ) (ww)-1( x x ))( y y )(( z z ) (aaa)log(( b b b ) (ccc) (ddd) e/( e e e ) x (fff) (ggg) (hhh))( i i i ))( j j j )` (kkk) (d) none of these
A. `y=tan^(-1)log(e/x)`
B. `y=xtan^(-1)log(x/e)`
C. `y=xtan^(-1)log(e/x)`
D. None of these

Answer 1

Correct Answer - C
We have `(dy)/(dx) = y/x-cosp^(2)(y/x)`
Putting `y=vx` so that `(dy)/(dx) = v+x(dv)/(dx)`, we get
`v+x(dv)/(dx) =v-cos^(2)v`
or `(dv)/(cos^(2)v)=-(dv)/(x)`
or `sec^(2)udu=-1/xdx`
On integration, we get
`tanu=-logx+logC`
or `tan(y/x)=-logx+logC`
This passes through `(1,pi//4)`. Therefore, `1=logC`.
So, `tan(y/2)=-logx+1=-logx+log_(e)`
or `y=xtan^(-1)log(ex)`