Suppose $$f(x+y)=f(x)$$⋅$$f(y)$$, $$f(x)=1+xg(x)$$ where x→$$0$$ $$Ltg(x)=1$$ then the value of log $$f(8)$$ is $$___$$

$$f(x+y)=f(x)f(y)f(0)=f(0)f(0)$$⇒$$f(0)=0$$ or $$f(0)=1And$$ $$f(x+0)=f(x)f(0)If$$ $$f(0)=0$$, then $$f(x)=f(x+0)=f(x)0=0f(x)=0$$∀xLtx→$$0f(x)=1+Ltx$$→$$0x$$⋅$$g(x)=1+0=1$$ contradicts $$f(x)=0$$∴$$f(0)=1limh$$→$$0f(x+h)$$−$$f(x)h=limh$$→$$0$$⁡$$f(x)f(h)$$−$$f(x)h=limh$$→$$0$$⁡$$f(x)f(h)$$−$$1hf$$'$$x=f(x)limh$$→$$0f(h)$$−$$1h=fxlimh$$→$$01+hg(h)-1h=fxlimh$$→$$0hg(h)h=fxlimh$$→$$0g(h)=f(x)$$ $$f1(x)=f(x)$$⇒$$f(x)=c$$.ex⇒$$f(0)=c=1$$ $$f(x)=ex$$ Log $$f(x)=x$$ Log $$f(8)=8$$

Questions

Suppose $f (x + y) = f (x) \cdot f (y), f (x) = 1 + xg (x)$ where $_{x \to 0}^{L t} g (x) = 1$ then the value of log f(8) is ___

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detailed solution

$\begin{array}{l} f (x + y) = f (x) f (y) \\ f (0) = f (0) f (0) \\ \Rightarrow f (0) = 0 or f (0) = 1 \end{array}$

And $f (x + 0) = f (x) f (0)$

If $f (0) = 0, then f (x) = f (x + 0) = f (x) 0 = 0$

$f (x) = 0 \forall x$

$\underset{x \to 0}{Lt} f (x) = 1 + \underset{x \to 0}{Lt} x \cdot g (x) = 1 + 0 = 1 contradicts f (x) = 0$

$∴ f (0) = 1$

$\begin{array}{l} \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{f (x) f (h) - f (x)}{h} \\ = \lim_{h \to 0} f (x) \frac{f (h) - 1}{h} \\ f' (x) = f (x) \lim_{h \to 0} \frac{f (h) - 1}{h} = f (x) \lim_{h \to 0} \frac{1 + hg (h) - 1}{h} = f (x) \lim_{h \to 0} \frac{hg (h)}{h} = f (x) \lim_{h \to 0} g (h) = f (x) \end{array}$

$\begin{matrix} f^{1} (x) = f (x) \\ \Rightarrow f (x) = c . e^{x} \Rightarrow f (0) = c = 1 \\ f (x) = e^{x} \\ Log f (x) = x \\ Log f (8) = 8 \end{matrix}$