Let f : R rightarrow R and g : R rightarrow R | vedclass

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(D) Given $f^{\prime}(x) = e^{f(x)} e^{-g(x)} g^{\prime}(x)$.Dividing by $e^{f(x)}$,we get $e^{-f(x)} f^{\prime}(x) = e^{-g(x)} g^{\prime}(x)$.Integra

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$B, C$

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(D) Given $f^{\prime}(x) = e^{f(x)} e^{-g(x)} g^{\prime}(x)$.Dividing by $e^{f(x)}$,we get $e^{-f(x)} f^{\prime}(x) = e^{-g(x)} g^{\prime}(x)$.Integrating both sides with respect to $x$,we get $\int e^{-f(x)} f^{\prime}(x) dx = \int e^{-g(x)} g^{\prime}(x) dx$.This yields $-e^{-f(x)} = -e^{-g(x)} + C$,or $e^{-g(x)} - e^{-f(x)} = C$.Using the condition $f(1) = 1$ and $g(2) = 1$,we evaluate the constant $C$ at different points.Since $e^{-g(x)} - e^{-f(x)} = C$ holds for all $x$,we have $e^{-g(1)} - e^{-f(1)} = e^{-g(2)} - e^{-f(2)}$.Substituting the given values $f(1) = 1$ and $g(2) = 1$,we get $e^{-g(1)} - e^{-1} = e^{-1} - e^{-f(2)}$.Rearranging gives $e^{-f(2)} + e^{-g(1)} = 2e^{-1} = \frac{2}{e}$.Since $e^{-f(2)} > 0$ and $e^{-g(1)} > 0$,we must have $e^{-f(2)} Taking the natural logarithm on both sides,$-f(2)  1 - \ln(2)$.Similarly,$-g(1)  1 - \ln(2)$.Thus,statements $(B)$ and $(C)$ are true.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

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

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