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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(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.

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) $TRUE$?

Similar Questions

If $x^y \cdot y^x = 16$,then $\frac{dy}{dx}$ at $(2, 2)$ is

If ${x^y} = {e^{x - y}}$,then $\frac{dy}{dx} = $

If $\sin (x+y)+\cos (x+y)=\sin \left[\cos ^{-1}\left(\frac{1}{3}\right)\right]$,then $\frac{d y}{d x}=$

If $\tan y = \cot \left(\frac{\pi}{4} - x\right)$, then $\frac{dy}{dx} = $

If $x = y\sqrt{1 - y^2}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module