If the slope of the tangent drawn to the curv | vedclass

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(A) Given the curve $y = e^{a+bx^2}$.First, find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = \frac{d}{dx}(e^

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$5$

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(A) Given the curve $y = e^{a+bx^2}$.First, find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = \frac{d}{dx}(e^{a+bx^2}) = e^{a+bx^2} \cdot \frac{d}{dx}(a+bx^2) = e^{a+bx^2} \cdot (2bx) = 2bxy$.At the point $P(1,1)$, the slope is $-2$:$\left. \frac{dy}{dx} \right|_{(1,1)} = 2b(1)(1) = -2$.$2b = -2 \implies b = -1$.Since the point $P(1,1)$ lies on the curve, it must satisfy the equation:$1 = e^{a+b(1)^2} \implies 1 = e^{a+b}$.Taking the natural logarithm on both sides:$\ln(1) = a+b \implies 0 = a+b$.Substituting $b = -1$:$0 = a - 1 \implies a = 1$.Finally, calculate $2a - 3b$:$2a - 3b = 2(1) - 3(-1) = 2 + 3 = 5$.

If the slope of the tangent drawn to the curve $y=e^{a+bx^2}$ at the point $P(1,1)$ is $-2$, then the value of $2a-3b$ is

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