Question

If `=(e^y-x)^(-1),` where `y(0)=0` , then `y` is expressed explicitly as (a) `( b ) (c) (d)1/( e )2( f ) (g)1n(( h ) (i)1+( j ) x^(( k )2( l ))( m ) (n))( o )` (p) (b) `( q ) (r)1n(( s ) (t)1+( u ) x^(( v )2( w ))( x ) (y))( z )` (aa) (c) `( d ) (e)1n(( f ) (g) x+sqrt(( h ) (i)1+( j ) x^(( k )2( l ))( m ) (n))( o ) (p))( q )` (r) (d) `( s ) (t)1n(( u ) (v) x+sqrt(( w ) (x)1-( y ) x^(( z )2( a a ))( b b ) (cc))( d d ) (ee))( f f )` (gg)
A. `1/2log_(e)(1+x^(2))`
B. `log_(e)(1+x^(2))`
C. `log_(e)(x+sqrt(1+x^(2)))=c`
D. None of these

Answer 1

Correct Answer - C
We have `(dy)/(dx) = (e^(y)-x)^(-1) rArr (dx)/(dy)=e^(y)-x`
`rArr (dx)/(dy)+x=e^(y)`
`rArr x=e^(y)/2+Ce^(-y)`
As `y(0)=0, C=-1/2` ,
`therefore x=e^(y)/2-1/2e^(-y)`
`rArr e^(2y)-2xe^(y)-1=0`
`rArr 2e^(y)=2x+-sqrt(4x^(2)+4)`.
But `e^(y)=x-sqrt(x^(2)+1)` (Rejected)
Hence `y=log_(e)(x+sqrt(x^(2)+1))`