SATHEE: Application of Derivatives 2 Question 4

Application of Derivatives 2 Question 4

####4. Let $f : [0, 2] \to R$ be a twice differentiable function such that $f^{''} (x) > 0$ , for all $x \in (0, 2)$ . If $φ (x) = f (x) + f (2 - x)$ , then $φ$ is

(2019 Main, 8 April I)

(a) increasing on $(0, 1)$ and decreasing on $(1, 2)$

(b) decreasing on $(0, 2)$

(d) increasing on $(0, 2)$

Show Answer

Answer:

(c)

Solution:

Given, $φ (x) = f (x) + f (2 - x), \forall x \in (0, 2)$

$\Rightarrow φ^{'} (x) = f^{'} (x) - f^{'} (2 - x)$

Also, we have $f^{''} (x) > 0 \forall x \in (0, 2)$

$\Rightarrow f^{'} (x)$ is a strictly increasing function

$\forall x \in (0, 2)$ . Now, for $φ (x)$ to be increasing,

$\begin{array}{lcr} φ^{'} (x) \geq 0 \\ \Rightarrow & f^{'} (x) - f^{'} (2 - x) \geq 0 & [using Eq. (i)] \\ \Rightarrow & f^{'} (x) \geq f^{'} (2 - x) \Rightarrow x > 2 - x \\ \Rightarrow & [∵ f^{'} is a strictly increasing function] \\ 2 x > 2 \Rightarrow \end{array}$

Thus, $φ (x)$ is increasing on $(1, 2)$ .

Similarly, for $φ (x)$ to be decreasing,

$\Rightarrow f^{'} (x) - f^{'} (2 - x) \leq 0$

[using Eq. (i)]

$\Rightarrow$	$f^{'} (x) \leq f^{'} (2 - x)$
$\Rightarrow$	$x < 2 - x [∵ f^{'}$ is a strictly increasing
function $]$
$\Rightarrow$	$2 x < 2$
$\Rightarrow$	$x < 1$

Application of Derivatives 2 Question 4

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Application of Derivatives 2 Question 4

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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