Application of Derivatives 2 Question 16
16. Let $f$ and $g$ be increasing and decreasing functions respectively from $[0, \infty)$ to $[0, \infty)$ and $h(x)=f{g(x)}$. If $h(0)=0$, then $h(x)-h(1)$ is
$(1987,2 M)$
(a) always negative
(b) always positive
(c) strictly increasing
(d) None of these
Objective Questions II
(One or more than one correct option)
Show Answer
Answer:
Correct Answer: 16. (c)
Solution:
- Let $F(x)=h(x)-h(1)=f{g(x)}-f{g(1)}$
$$ \begin{aligned} \therefore \quad F^{\prime}(x) & =f^{\prime}{g(x)} \cdot g^{\prime}(x) \\ & =(+)(\rightarrow)=-ve \end{aligned} $$
[since, $f(x)$ is an increasing function $f^{\prime}(g(x))$ is + ve and $g(x)$ is decreasing function $g^{\prime}(f(x))$ is $\rightarrow$ ve ]
Since, $\quad f^{\prime}(x)$ is -ve.
$\therefore f(x)$ is decreasing function.
When
$0 \leq x<1$
$\Rightarrow$
$$ h(x)-h(1)=+ve $$
When
$x \geq 1$,
$\Rightarrow$
$$ h(x)-h(1)=-ve $$
Hence, for $x>0$,
$h(x)-h(1)$ is neither always positive nor always negative, so it is not strictly increasing throughout.
Therefore, option (d) is the answer.
