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If the slope of the tangent drawn to the curve $y=e^{a+b x^2}$ at the point $\mathrm{P}(1,1)$ is -2 , then the value of $2 \mathrm{a}-3 \mathrm{~b}$ is
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$5$
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x} e^{a+b x^2} \\ & \quad=2 b x \cdot e^{a+b x^2}=2 b x \cdot y \\ & \left.\frac{d y}{d x}\right|_{(1,1)}=2 b(1) \cdot(1)=-2 \\ & \Rightarrow \quad 2 b=-2 \Rightarrow b=-1 \\ & \because \quad(1,1) \text { lie on } y=e^{a+b x^2} \\ & \therefore \quad 1=e_a^{-1} \\ & \therefore \quad a-1=0 \Rightarrow a=1 \\ & \therefore \quad 2 a-3 b=2(1)-3(-1)=5 .\end{aligned}$
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