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(C) Given the curve equation is $e^y = 1 + x^2$.Differentiating both sides with respect to $x$,we get:$e^y \frac{dy}{dx} = 2x$Thus,the slope of the ta

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$|m| \leq 1$

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(C) Given the curve equation is $e^y = 1 + x^2$.Differentiating both sides with respect to $x$,we get:$e^y \frac{dy}{dx} = 2x$Thus,the slope of the tangent $m$ is given by:$m = \frac{dy}{dx} = \frac{2x}{e^y} = \frac{2x}{1 + x^2}$We need to find the range of $|m| = \left| \frac{2x}{1 + x^2} \right|$.We know that for any real number $x$,$(|x| - 1)^2 \geq 0$.$x^2 - 2|x| + 1 \geq 0$$x^2 + 1 \geq 2|x|$Dividing both sides by $x^2 + 1$ (which is always positive):$1 \geq \frac{2|x|}{x^2 + 1}$Therefore,$|m| = \left| \frac{2x}{1 + x^2} \right| \leq 1$.

If the slope of a tangent to the curve $e^y = 1 + x^2$ is $m$,then

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