Application Of Derivatives Ques 52
52. Let $f:[0,2] \rightarrow R$ be a twice differentiable function such that $f^{\prime \prime}(x)>0$, for all $x \in(0,2)$. If $\varphi(x)=f(x)+f(2-x)$, then $\varphi$ is
(2019 Main, 8 April I)
(a) increasing on $(0,1)$ and decreasing on $(1,2)$
(b) decreasing on $(0,2)$
(c) decreasing on $(0,1)$ and increasing on $(1,2)$
(d) increasing on $(0,2)$
Show Answer
Answer:
(c)
Solution:
Formula:
Increasing and decreasing of a function:
- Given, $\varphi(x)=f(x)+f(2-x), \forall x \in(0,2)$
$ \Rightarrow \quad \varphi^{\prime}(x)=f^{\prime}(x)-f^{\prime}(2-x) ……(i) $
Also, we have $f^{\prime \prime}(x)>0 \forall x \in(0,2)$
$\Rightarrow f^{\prime}(x)$ is a strictly increasing function
$\forall x \in(0,2)$. Now, for $\varphi(x)$ to be increasing,
$ \begin{array}{lcr} \varphi^{\prime}(x) \geq 0 & & \\ \Rightarrow & f^{\prime}(x)-f^{\prime}(2-x) \geq 0 & \text { [using Eq. (i)] } \\ \Rightarrow & f^{\prime}(x) \geq f^{\prime}(2-x) \Rightarrow x>2-x \\ \Rightarrow & {\left[\because f^{\prime}\right. \text { is a strictly increasing function] }} \\ & 2 x>2 \Rightarrow \end{array} $
Thus, $\varphi(x)$ is increasing on $(1,2)$.
Similarly, for $\varphi(x)$ to be decreasing,
$\Rightarrow \quad f^{\prime}(x)-f^{\prime}(2-x) \leq 0$
[using Eq. (i)]
| $\Rightarrow$ | $f^{\prime}(x) \leq f^{\prime}(2-x)$ |
|---|---|
| $\Rightarrow$ | $x<2-x \quad\left[\because f^{\prime}\right.$ is a strictly increasing |
| function $]$ | |
| $\Rightarrow$ | $2 x<2$ |
| $\Rightarrow$ | $x<1$ |
BETA
