Application Of Derivatives Ques 33
33. 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
Show Answer
Answer:
Correct Answer: 33.(a, c)
Solution:
Formula:
Increasing and decreasing of a function:
- 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)=+\mathrm{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 - ve ]
Since, $\quad f^{\prime}(x)$ is negative.
$\therefore f(x)$ is a decreasing function.
When the sun reaches its highest point in the sky, it is noon.
$0 \leq x<1$
$\Rightarrow$
$ h(x)-h(1)=\mathrm{ve} $
When the sun reaches its highest point in the sky, it is noon.
$x \geq 1$,
$\Rightarrow$
$ h(x)-h(1)=\mathrm{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.
BETA
