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`        Please solve with steps.,........................................`
4 months ago

```							let the 2 fns be y= f(x) and y= g(x). now consider a point with abscissa x. then eqn of tangent for f and g would be:Y – f(x)= f’(x)(X – x)and Y – g(x)= g’(x)(X – x). here X and Y represent the axes.now, since they intersect at y axis, X=0 so thatY= f(x) – xf’(x)= g(x) – xg’(x).....(1)similarly the eqn of normals to f and g would be:Y – f(x)= (–1/f’(x))(X – x)and Y – g(x)= (–1/g’(x))(X – x). these intesect on x axis so that Y=0so X= x + f(x)*f’(x)= x + g(x)*g’(x)or f(x)*f’(x)= g(x)*g’(x).....(2)from (2), we can sayf(x)*df(x)/dx= g(x)*dg(x)/dxintegrating both sides, ∫f(x)df(x)= ∫g(x)dg(x)f^2(x)/2= g^2(x)/2 + Cor f^2(x)= g^2(x) + K......(3) where K= 2C is a constant.(1) can be written as f(x) – g(x)= x(f’(x) – g’(x))f(x) – g(x)= xd(f(x) – g(x))/dxor dx/x= d(f(x) – g(x))/f(x) – g(x)integrateln|x|= ln|f(x) – g(x)| + Craise to power of ex= A(f(x) – g(x)), ….(4) A= ±e^C being a constant.(3) and (4) can be solved for f and g by noting that (3) can be factorised as (f – g)(f+g)= K and then putting f – g= x/A from (4) and finally adding these eqns.we shall obtain f(x)= (KA/x + x/A)/2 and g(x)= (KA/x – x/A)/2 if we let 2A= B (another constant) and K= 4M, thenf(x)= MB/x + x/B and g(x)= MB/x – x/B here M and B are constants and obviously B cant be zero.kindly approve :)
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4 months ago
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