Application of Derivatives 2 Question 4

####4. 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:

  1. 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) $

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 $]$
$\Rightarrow$ $2 x<2$
$\Rightarrow$ $x<1$


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