Question

Let f and g be increasing and decreasing functions respectively from $(0,\mathrm{\infty })$ to $(0,\mathrm{\infty })$ and let h(x) = f[g(x)]. If h(0)= 0, then h(x) – h(1) is
[a] always zero
[b] always negative
[c] always positive
[d] strictly increasing
[e] None of these.

Answer 1

Hint: Use the fact that if f(x) is an increasing function then for all $x_1>x_2$, we have $f\left(x_1\right)\ge f\left(x_2\right)$. Similarly, if $f\left(x\right)$ is a decreasing function, then for all $x_1>x_2$, we have $f\left(x_1\right)\le f\left(x_2\right)$. Hence choose $x_1,x_2\in (0,\mathrm{\infty })$ and check whether $h\left(x_1\right)-1-(h\left(x_2\right)-1)\ge 0$ or $\le 0$ and hence determine the nature of $h\left(x\right)-h\left(1\right)$.

Complete step-by-step answer:
We know that if $f\left(x\right)$ is increasing in the interval $I$, then for all $x_1,x_2\in I$, we have $x_1>x_2\Rightarrow f\left(x_1\right)\ge f\left(x_2\right)$ and if f(x) is decreasing in the interval $I$, then for all $x_1,x_2\in I$, we have $x_1>x_2\Rightarrow f\left(x_1\right)\le f\left(x_2\right)$
Now consider $x_1,x_2\in (0,\mathrm{\infty })$, we have since g(x) is a decreasing function
$g\left(x_1\right)\le g\left(x_2\right)$
Now since $g\left(x_1\right),g\left(x_2\right)\in (0,\mathrm{\infty })$ because codomain of g(x) is $(0,\mathrm{\infty })$ and since f(x) is increasing in $(0,\mathrm{\infty })$, we have
$f\left(g\left(x_1\right)\right)\le f\left(g\left(x_2\right)\right)$
Hence we have $h\left(x_1\right)\le h\left(x_2\right)$
Hence h(x) is a decreasing function.
Since in the interval $(0,\mathrm{\infty })$, there exist $x_1<1$ and $x_2>1$, we have $h\left(x_1\right)-h\left(1\right)\ge 0$ and $h\left(x_2\right)-1\le 0$
Hence h(x) – h(1) is both positive as well as negative in the interval $(0,\mathrm{\infty })$
Hence none of the options is correct.

Note: Do not try proving that h(x) is a decreasing function by differentiating both sides and showing h’(x) is non-positive. This method is incorrect as it justifies h(x) being decreasing only if f(x) and g(x) are differentiable and not in general.