For every real number $$x \neq-1$$, let $$\mathrm{f}(x)=\frac{x}{x+1}$$. Write $$\mathrm{f}_1(x)=\mathrm{f}(x)$$ & for $$\mathrm{n} \geq 2, \mathrm{f}_{\mathrm{n}}(x)=\mathrm{f}\left(\mathrm{f}_{\mathrm{n}-1}(x)\right)$$. Then $$\mathrm{f}_1(-2) \cdot \mathrm{f}_2(-2) \ldots . . \mathrm{f}_{\mathrm{n}}(-2)$$ must be
If $$\mathrm{U}_{\mathrm{n}}(\mathrm{n}=1,2)$$ denotes the $$\mathrm{n}^{\text {th }}$$ derivative $$(\mathrm{n}=1,2)$$ of $$\mathrm{U}(x)=\frac{\mathrm{L} x+\mathrm{M}}{x^2-2 \mathrm{~B} x+\mathrm{C}}$$ (L, M, B, C are constants), then $$\mathrm{PU}_2+\mathrm{QU}_1+\mathrm{RU}=0$$, holds for
The equation $$2^x+5^x=3^x+4^x$$ has
Consider the function $$\mathrm{f}(x)=(x-2) \log _{\mathrm{e}} x$$. Then the equation $$x \log _{\mathrm{e}} x=2-x$$