Show that $x^2 - x + 1 0$ for all real values of $x$.[1] Using the result in part (i),

Question

Show that $x^2 - x + 1 > 0$ for all real values of $x$.[1] Using the result in part (i), solve the inequality $\frac{1}{x-1} + \frac{3}{x+1} \leq 2 - 2x$. [4] Hence, solve the inequality $\frac{1}{\ln x - 1} + \frac{3}{\ln x + 1} \leq 2 - \ln x^2$, giving your answer in exact form.[2]

Accepted Answer

$x<-1$ or $0\le x<1$; $0<x<e^{-1}$ or $1\le x<e$

2025 SAJC Promo Q2

2025 SAJC Promo Q2

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