Let f:R−{0}→R be a function satisfying f(yx)=f(y)f(x) for all x,y,f(y)=0. If f′(1)=2024, then
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Solution
f(yx)=f(y)f(x)f′(1)=2024f(1)=1 Partially differentiating w. r. t. x f′(yx)⋅y1=f(y)1f′(x)y→xf′(1)⋅x1=f(x)f′(x)2024f(x)=xf′(x)⇒xf′(x)−2024f(x)=0