Question
If $$m$$ be the slope of a tangent to the curve $${e^y} = 1 + {x^2}$$ then :
A.
$$\left| m \right| > 1$$
B.
$$m < 1$$
C.
$$\left| m \right| < 1$$
D.
$$\left| m \right| \leqslant 1$$
Answer :
$$\left| m \right| \leqslant 1$$
Solution :
$$\eqalign{
& {\text{Differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}x,{\text{ }}{e^y}{\text{.}}\frac{{dy}}{{dx}}{\text{ }} = 2x \cr
& {\text{or,}}\,\,\frac{{dy}}{{dx}} = \frac{{2x}}{{1 + {x^2}}}\,\,\,\,\left( {\because {e^y} = 1 + {x^2}} \right) \cr
& \therefore m = \frac{{2x}}{{1 + {x^2}}}{\text{ or }}\left| m \right| = \frac{{2\left| x \right|}}{{1 + {{\left| x \right|}^2}}} \cr
& {\text{But }}1 + {\left| x \right|^2} - 2\left| x \right| = \left( {1 - {{\left| x \right|}^2}} \right) \geqslant 0 \cr
& \therefore 1 + {\left| x \right|^2} \geqslant 2\left| x \right| \cr
& \therefore \left| m \right| \leqslant 1 \cr} $$