Application of Derivatives 2 Question 19
19. If $h(x)=f(x)-f(x)^{2}+f(x)^{3}$ for every real number $x$. Then,
(1998, 2M)
(a) $h$ is increasing, whenever $f$ is increasing
(b) $h$ is increasing, whenever $f$ is decreasing
(c) $h$ is decreasing, whenever $f$ is decreasing
(d) Nothing can be said in general
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Answer:
Correct Answer: 19. (b)
Solution:
- Given, $h(x)=f(x)-f(x)^{2}+f(x)^{3}$
On differentiating w.r.t. $x$, we get
$$ \begin{aligned} h^{\prime}(x)=f^{\prime}(x)- & 2 f(x) \cdot f^{\prime}(x)+3 f^{2}(x) \cdot f^{\prime}(x) \\ & =f^{\prime}(x)\left[1-2 f(x)+3 f^{2}(x)\right] \\ & =3 f^{\prime}(x) \quad(f(x))^{2}-\frac{2}{3} f(x)+\frac{1}{3} \\ & =3 f^{\prime}(x) \quad f(x)-\frac{1}{3}+\frac{1}{3}-\frac{1}{9} \\ & =3 f^{\prime}(x) \quad f(x)-\frac{1}{3}^{2}+\frac{3-1}{9} \\ & =3 f^{\prime}(x) \quad f(x)-\frac{1}{3}^{2}+\frac{2}{9} \end{aligned} $$
NOTE $h^{\prime}(x)<0$, if $f^{\prime}(x)<0$ and $h^{\prime}(x)>0$, if $f^{\prime}(x)>0$
| $\Rightarrow$ | $f^{\prime}(x)=\frac{1}{1+x}$ | $-\frac{x}{1+x}$ |
|---|---|---|
| $\Rightarrow$ | $f^{\prime}(x)>0$ | |
| when | $-1<x<0$ | |
| and | $f^{\prime}(x)<0$ | |
| when | $x>0$ |
Therefore, $h(x)$ is an increasing function, if $f(x)$ is increasing function and $h(x)$ is decreasing function, if $f(x)$ is decreasing function.
Therefore, options (a) and (c) are correct answers.
