Question

Let $f$ and $g$ be non-increasing and non-decreasing functions respectively from $\left[ {0,\infty } \right]$ to $\left[ {0,\infty } \right]$ and $h\left( x \right) = f\left( {g\left( x \right)} \right)$, $h\left( 0 \right) = 0$, then in $\left[ {0,\infty } \right]$, $h\left( x \right) - h\left( 1 \right)$ is
A. $< 0$
B. $> 0$
C. $= 0$
D. None of these

Answer 1

Here, $f$and $g$ be non-increasing and non-decreasing functions. Hence choose $x1,x2 \in \left[ {0,\infty } \right]$and check whether $h\left( {x1} \right) - 1 - \left( {h\left( {x2} \right) - 1} \right) \geqslant 0$ or $h\left( {x1} \right) - 1 - \left( {h\left( {x2} \right) - 1} \right) \leqslant 0$and hence based on this determine the nature of $h\left( x \right) - h\left( 1 \right)$.

Complete answer:
As per given $f\left( x \right)$ is an increasing function then for all $x1 > x2$, it implies to the function that$f\left( {x1} \right) \geqslant f\left( {x2} \right)$. Similarly, if $f\left( x \right)$ is a decreasing function, then for all $x1 > x2$, we have $f\left( {x1} \right) \leqslant f\left( {x2} \right)$.
We know that if $f\left( x \right)$ is an increasing function in the interval $I$, then for all $x1,x2 \in I$, we have
$x1 > x2 \Rightarrow f\left( {x1} \right) \geqslant f\left( {x2} \right)$.
If $f\left( x \right)$ is a decreasing function in the interval $I$, then for all $x1,x2 \in I$, we have
$x1 > x2 \Rightarrow f\left( {x1} \right) \leqslant f\left( {x2} \right)$.
Now let us consider
$x1,x2 \in \left[ {0,\infty } \right]$, since we have $g\left( x \right)$ is a decreasing function
$g\left( {x1} \right) \leqslant g\left( {x2} \right)$
Now since $g\left( {x1} \right),g\left( {x2} \right) \in \left( {0,\infty } \right)$ because codomain of $g\left( x \right)$is $\left[ {0,\infty } \right]$ and since $f\left( x \right)$ is an increasing function in $\left[ {0,\infty } \right]$, we have
$f\left( {g\left( {x1} \right)} \right) \leqslant f\left( {g\left( {x2} \right)} \right)$
Hence, we have
$h\left( {x1} \right) \leqslant h\left( {x2} \right)$
Here $h\left( x \right)$ is a decreasing function.
Since in the interval $\left[ {0,\infty } \right]$, there exist $x1 < 1$ and $x2 > 1$, we have
$h\left( {x1} \right) - h\left( 1 \right) \geqslant 0$ and
$h\left( {x2} \right) - 1 \leqslant 0$
Hence h (x1)- h (1) is both positive and negative in the interval $\left[ {0,\infty } \right]$.

Therefore, none of the option is correct, hence option D is the answer.

Additional Information:
A (strictly) increasing function f is one where $x1 < x2 \Rightarrow f(x1) < f(x2)$and a non-decreasing function f is one where $x1 < x2 \Rightarrow f(x1) \leqslant f(x2)$.
The dual terms are (strictly) decreasing and non-increasing (reverse the direction of the inequalities), respectively.

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) both are differentiable and not in general.