2012 Njc Prelim H2 Math |top|

Differentiate $y = (x-1) - 3(x+1)^-1$. $$ \fracdydx = 1 - 3(-1)(x+1)^-2 = 1 + \frac3(x+1)^2 $$ Set $\fracdydx = 0$: $$ 1 + \frac3(x+1)^2 = 0 \implies \frac3(x+1)^2 = -1 $$ Since $(x+1)^2 \ge 0$ and $3 > 0$, the LHS is always positive. There are no real stationary points . The curve is strictly increasing everywhere it is defined.

NJC set a three-part question linking parametric differentiation to Maclaurin’s series. 2012 njc prelim h2 math

You can find full solutions and step-by-step workings on the following platforms: Course Hero (Paper 1) Course Hero (Paper 2) Scribd (Full Solutions Archive) step-by-step breakdown Differentiate $y = (x-1) - 3(x+1)^-1$