A curve $C$ has equation $y={{\mathrm{e}}^{\frac{{{x}^{2}}-4}{12}}}$, where $x0$. The

Question

A curve $C$ has equation $y={{\mathrm{e}}^{\frac{{{x}^{2}}-4}{12}}}$, where $x>0$. The curve $C$ undergoes the following sequence of transformations: $\mathrm{A}$: Stretch by factor $\frac{1}{2}$ parallel to the $x$-axis. $\mathrm{B}$: Translate by $1$ unit in the negative $y$-direction. $\mathrm{C}$: Reflection in the line $y=x$. Find the equation of the new curve in the form $y=\mathrm{q}\left( x \right)$ and state the domain of $\mathrm{q}$.[5]

Accepted Answer

$y=\mathrm{q}(x)=\sqrt{3\ln(x+1)+1}$ $D_{\mathrm{q}}=(\mathrm{e}^{-1/3}-1,\infty)$

2022 TMJC P1 Q7

2022 TMJC P1 Q7

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