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Question

Let f$$:\mathbb{R}\rightarrow \mathbb{R}$$ and $$g:\mathbb{R}\rightarrow \mathbb{R}$$ be two non-constant differentiable functions. If $$f'(x)=(e^{(f(x)-g(x))})g'(x)$$ for all $$x\epsilon \mathbb{R}$$, and $$f(1)=g(2)=1$$, then which of the following statement(s) is (are) TRUE?


A
f(2)<1loge2
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B
f(2)>1loge2
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C
g(1)>1loge2
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D
g(1)<1loge2
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Solution

The correct options are
B $$g(1) > 1 -log_e 2$$
D $$f(2) > 1-log_e 2$$
$$f'(x)=e^{(f(x)-g(x))}g'(x)\forall x\epsilon \mathbb{R}$$

$$\Rightarrow e^{-f(x)}\cdot f'(x)-e^{-g(x)}g'(x)=0$$

$$\Rightarrow \displaystyle\int(e^{-f(x)}f'(x)-e^{-g(x)}\cdot g'(x))dx=C$$

$$\Rightarrow -e^{-f(x)}+e^{-g(x)}=C$$

$$\Rightarrow -e^{-f(1)}+e^{-g(1)}=-e^{-f(2)}+e^{-g(2)}$$

$$\Rightarrow -\displaystyle\dfrac{1}{e}+e^{-g(1)}=-e^{-f(2)}+\dfrac{1}{e}$$

$$\Rightarrow e^{-f(2)}+e^{-g(1)}=\dfrac{2}{e}$$

$$\therefore$$ $$e^{-f(2)} < \dfrac{2}{e}$$ and $$e^{-g(1)} < \dfrac{2}{e}$$

$$\Rightarrow -f(2) < log_{e}2 -1$$ and $$-g(1) < log_{e}2-1$$

$$\Rightarrow f(2) > 1 -log_{e}2$$  and  $$g(1) > 1-log_{e}2$$.

Mathematics

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