Question

Let $C$ be a curve defined by $y=e^{a+b{x}^2}$. The curve $C$ passes through the point $P(1,1)$ and the slope of the tangent at $P$ is $(-2)$. Then the value of $(2a-3b)$ is

Answer 1

$y=e^{a+b{x}^2}$, passes through $(1,1)$
$\Rightarrow 1=e^{a+b}\cdots \left(1\right)$
$\Rightarrow a+b=0\cdots \left(2\right)$
$\Rightarrow \left(\frac{dy}{dx}\right)=e^{a+b{x}^2}\cdot 2bx$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,1)}=-2$
$\Rightarrow e^{a+b(1{)}^2}\cdot 2b\left(1\right)=-2$
Using $\left(1\right),$ we get $b=-1$
Using $\left(2\right),$ we get $a=1$
$\Rightarrow 2a-3b=5$