Let Gamma be the curve y=b e^{-x/a} and L be | vedclass

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(A) Given the curve $\Gamma: y = b e^{-x/a}$.On differentiating with respect to $x$, we get $\frac{dy}{dx} = -\frac{b}{a} e^{-x/a}$.At the point where

Vedclass · Accepted Answer

$L$ touches the curve $\Gamma$ at the point where the curve crosses the $y$-axis.

Vedclass · Answer

(A) Given the curve $\Gamma: y = b e^{-x/a}$.On differentiating with respect to $x$, we get $\frac{dy}{dx} = -\frac{b}{a} e^{-x/a}$.At the point where the curve crosses the $y$-axis, $x = 0$, so $y = b e^0 = b$.The slope of the tangent at $(0, b)$ is $\left. \frac{dy}{dx} \right|_{x=0} = -\frac{b}{a}$.The equation of the tangent at $(0, b)$ is $y - b = -\frac{b}{a}(x - 0)$, which simplifies to $\frac{y}{b} = -\frac{x}{a} + 1$, or $\frac{x}{a} + \frac{y}{b} = 1$.This is the equation of the line $L$, so $L$ touches $\Gamma$ at $(0, b)$.Since $e^{-x/a} > 0$ for all $x \in \mathbb{R}$, $y = b e^{-x/a}$ is never $0$. Thus, $\Gamma$ never touches the $x$-axis.Therefore, both statements $A$ and $D$ are mathematically correct based on the provided options.

Let $\Gamma$ be the curve $y=b e^{-x/a}$ and $L$ be the straight line $\frac{x}{a}+\frac{y}{b}=1$, where $a, b \in \mathbb{R}$. Which of the following statements is true?

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