Question

Let f and g be increasing and decreasing functions respectively from $\left( 0,\infty \right)$ to $\left( 0,\infty \right)$ 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 ${{x}_{1}},{{x}_{2}}\in \left( 0,\infty \right)$ and check whether $h\left( {{x}_{1}} \right)-1-\left( h\left( {{x}_{2}} \right)-1 \right)\ge 0$ 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 ${{x}_{1}},{{x}_{2}}\in \left( 0,\infty \right)$, we have since g(x) is a decreasing function
$g\left( {{x}_{1}} \right)\le g\left( {{x}_{2}} \right)$
Now since $g\left( {{x}_{1}} \right),g\left( {{x}_{2}} \right)\in \left( 0,\infty \right)$ because codomain of g(x) is $\left( 0,\infty \right)$ and since f(x) is increasing in $\left( 0,\infty \right)$, 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 $\left( 0,\infty \right)$, 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 $\left( 0,\infty \right)$
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.