Question

Let f be a differentiable function satisfying (ay)=xf(y)+ y f (x)-2x-2y+2Vx,y>0. If S'(1) =1 then fle')-(x) - 2 A) Lim =-1 x0+ x sle")= {x} - 2 - = 1 B) Lim xo+ x2 e C) minimum value of f(x) is 2 D) minimum value of f (x) occurs at x = e [Note: {k} denotes fractional part function of k.)​

Answer 1

Answer:

Putting y=0 in the given equation, we have f(x)=f(x)+f(0)-1⇒f(0)=

f′(x)=h→0limhf(x+h)−f(x)=h→0limhf(x)+f(h)+2hx−1−f(x)

   =h→0limhf(h)−1+2x=h→0limhf(h)−f(0)+2x

   =f′(0)+2x=dxd(xf′(0)+x2)

f(x)=xf′(0)+x2+c

but f(0)=1 so c=1. Thus

f(x)=xsinϕ+x2+1

Also  we can write f(x)=(x+2sinϕ)2+4cos

Step-by-step explanation:

please mark me brainly list