Let f: R → R be defined as f(x) = \ cases - 55x, & ifx 4 cases ) Let A = x ∈ R: f is increasing . Then A is equal to:

Correct answer: B. (- 5, - 4) (4, ∞) f (x) = \ cases - 55x, & ifx 4 cases ) f ' (x) = \ cases - 55, & ifx 4 cases ) f ^ ' (x) = \ cases - 55; & x 4 cases ) f (x) is increasing ⇒ f ^ ' (x) > 0 So, x ∈ (- 5, - 4) (4, ∞)

JEE Main 202124 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2021 · 24 February · Evening Shift · Mathematics · Application of Derivatives

Let f: R → R be defined as f(x) = \ cases - 55x, & ifx 4 cases )...

Let $f: R \to R$ be defined as

$f\left(x\right) = \left\{$ \begin{cases}- 55x, & ifx 4\end{cases} $\right)</p><p>Let$ A = { x \in R: f $is increasing}. Then$ A is equal to:

Options

A $\left(- 5, \infty\right)$
B $\left(- 5, - 4\right) \cup \left(4, \infty\right)$
C $\left(- \infty , - 5\right) \cup \left(- 4, \infty\right)$
D $\left(- \infty , - 5\right) \cup \left(4, \infty\right)$

Correct answer

B. $\left(- 5, - 4\right) \cup \left(4, \infty\right)$

Step-by-step solution

$f$ \left(x\right) = \left\{ $\begin{cases}- 55x, & ifx 4\end{cases}$ \right)

$f '$ \left(x\right) = \left\{ $\begin{cases}- 55, & ifx 4\end{cases}$ \right)

${f}^{'}\left(x\right) = \left\{$ \begin{cases}- 55; & x 4\end{cases} $\right)</p><p>$ f $\left(x\right)$ is increasing

$⇒ {f}^{'}\left(x\right) > 0$

So, $x \in \left(- 5, - 4\right) \cup \left(4, \infty\right)$

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