Question

Let $f: R \rightarrow R$ be defined as, $f(x) = \begin{cases} -55x, & \text{if } x < -5 \\ 2x^3 -3x^2 -120x, & \text{if } 5 \le x \le 4 \\ 2x^3 -3x^2-36x-336& \text{if } x > 4, \end{cases} $ Let $A=\{ x \in R : f$ is increasing $\} .$ Then $A$ is equal to

• A. $(-\infty,-5) \cup(4, \infty)$
• B. $(-5, \infty)$
• C. $(-\infty,-5) \cup(-4, \infty)$
• D. (-5,-4)$\cup(4, \infty)$

Answer 1

The Correct Option is D

$f'(x) = \begin{cases} -55; & x < -5 \\ 6(x - 5)(x+4); & -5 < x <4\\ 6(x-3)(x+2); & x > 4 \end{cases} $ $f ( x )$ is increasing in $x \in(-5,-4) \cup(4, \infty)$

Answer 2

The correct answer is (D) : (-5,-4)\(\cup(4, \infty)\)
For f'(x) ≥ 0
Case I :- -5 < x < 4
Then f'(x)  ≥ 0 ≥ 0  implies 
6x² - 6x - 120 ≥  0 
⇒ x² - x - 20  ≥  0
⇒ (x - 5) (x + 4) ≥  0
⇒ x ≤  - 4 or x  ≥  5 
But -5 <x <4 
∴ -5 < x < 4
Hence, function f(x) increases in domain x ∈ (-5, -4)
Case II :- x > 4
Then f'(x)  ≥ ≥  0 implies 
6x² - 6x - 36  ≥ 0 
⇒ x² - x - 6  ≥ 0 
⇒ (x - 3) (x + 2) ≥ 0
⇒ x ≤  - 2 or x ≥ 3 
But x > 4 
Hence, f(x) increases in (4, ∞ )
From both cases, we can say that f(x) is increases in (-5, -4) U (4, ∞ ). 
A = {x|x ∈ (-5, -4) U (4, ∞ )}