Functions
Onto concave function — inequalities with inverse
MJAT_TS4_P2
Grade 12
Question:
If $f:[0,1]\to[0,1]$ is an onto function such that $f'(x)>0$ and $f''(x)<0$, then:
A) $\dfrac{f(x)}{x}$ is a decreasing function $\forall x\in(0,1]$
B) $\dfrac{f(x)}{x}$ is an increasing function $\forall x\in(0,1]$
C) $f(x)\cdot f^{-1}(x)\leq x^2$ for some $x\in(0,1)$
D) $f^{-1}(x)\leq x\leq f(x)$ for all $x\in[0,1]$
Step-by-Step Solution
Key Concept: Since $f$ is concave ($f''<0$), strictly increasing ($f'>0$), onto $[0,1]$ with $f(0)=0$: by concavity, $f(x)/x$ is decreasing (A ✓). Also, for concave $f$: $f(x)\geq x$ for $x\in[0,1]$ (since the chord from $(0,0)$ to $(1,1)$ lies below the curve for concave $f$... wait actually concave means above).
A ✓ ($f(x)/x$ decreasing for concave $f$ through origin). D ✓ ($f^{-1}(x)\leq x\leq f(x)$). Answer: A, D.
Correct Answer: AD