InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The function `y=f(x)`is thesolution of the differential equation `(dy)/(dx)+(x y)/(x^2-1)=(x^4+2x)/(sqrt(1-x^2))`in `(-1,1)`satisfying`f(0)=0.`Then `int_((sqrt(3))/2)^((sqrt(3))/2)f(x)dx`is(a)`( b ) (c) (d)pi/( e )3( f ) (g)-( h )(( i )sqrt(( j )3( k ))( l ))/( m )2( n ) (o) (p)`(q)(b) `( r ) (s) (t)pi/( u )3( v ) (w)-( x )(( y )sqrt(( z )3( a a ))( b b ))/( c c )4( d d ) (ee) (ff)`(gg)(c)`( d ) (e) (f)pi/( g )6( h ) (i)-( j )(( k )sqrt(( l )3( m ))( n ))/( o )4( p ) (q) (r)`(s)(d) `( t ) (u) (v)pi/( w )6( x ) (y)-( z )(( a a )sqrt(( b b )3( c c ))( d d ))/( e e )2( f f ) (gg) (hh)`(ii)A. `pi/3-sqrt(3)/2`B. `pi/3-sqrt(3)/4`C. `pi/6-sqrt(3)/4`D. `pi/6-sqrt(3)/2` |
|
Answer» Correct Answer - B `(dy)/(dx) + x/(x^(2)-1)y=(x^(4)+2x)/sqrt(1-x^(2))` This is a linear differential equation. `I.F. =e^(intx/(x^(2)-1)dx)` `e^(1/2"ln "|x^(2)-1|)=sqrt(1-x^(2))` (`therefore x in (-1,1))` Therefore, solution is: `ysqrt(1-x^(2))=int(x(x^(3)+2))/sqrt(1-x^(2)).sqrt(1-x^(2))dx` or `ysqrt(1-x^(2))=int(x^(4)+2x)dx=x^(5)/5+x^(2)+C` Since, `f(0)=0,C=0` `therefore f(x) sqrt(1-x^(2))=x^(5)/5+x^(2)` `rArr f(x) = x^(5)/(5sqrt(1-x^(2))+x^(2)/sqrt(1-x^(2))` `therefore int_(-sqrt(3//2))^(sqrt(3)/2) f(x)dx= int_(sqrt(3)/2)^(sqrt(3)/2) (x^(5)/(5sqrt(1-x^(2))+x^(2)/sqrt(1-x^(2))))dx` `=2int_(0)^(sqrt(3)//2)(x^(2)/sqrt(1-x^(2)))dx` `rArr 2int_(0)^(pi//3) sin^(2)theta(d)theta` `=2int_(0)^(pi//3)(1-cos2theta)/(2)(d)theta` `=2[theta/2-(sin2theta)/4]_(0)^(pi//3)` `=2(pi/6) -2(sqrt(3))/8 = pi/3-sqrt(3)/4` |
|