Let f:R→R be defined by fx=−55x, if x 4, let A=x∈R,fxis increasing then A is

f1(x)=-55x 0∀x∈(-5,-4)=6x2-6x-36 x>4 ⇒f1(x)>0 ∀x∈(4,∞) Therefore, the function is increasing in the interval −5,−4∪4,∞

Let f:R→R be defined by fx=−55x, if x 4, let A=x∈R,fxis increasing then A is

f1(x)=-55x 0∀x∈(-5,-4)=6x2-6x-36 x>4 ⇒f1(x)>0 ∀x∈(4,∞) Therefore, the function is increasing in the interval −5,−4∪4,∞

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Let $f : R \to R$ be defined by $f (x) = \{\begin{cases} \begin{matrix} - 55 x, & i f x < - 5 \end{matrix} \\ \begin{matrix} 2 x^{3} - 3 x^{2} - 120 x, & i f - 5 \leq x \leq 4 \end{matrix} \\ \begin{matrix} 2 x^{3} - 3 x^{2} - 36 x - 336, & i f x > 4, \end{matrix} \end{cases}$
let $A = \{x \in R, f (x) i s i n c r e a \sin g\}$ then $A$ is

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$(- \infty, - 5) \cup (4, \infty)$

$(- \infty, - 5) \cup (- 4, \infty)$

$(- 5, \infty)$

$(- 5, - 4) \cup (4, \infty)$

answer is D.

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Detailed Solution

$\begin{array}{ccc} f^{1} (x) = - 55 & x < - 5 \\ = 6 x^{2} - 6 x - 120 - 5 \leq x \leq 4 & \Rightarrow f^{1} (x) > 0 & \forall x \in (- 5, - 4) \\ = 6 x^{2} - 6 x - 36 x > 4 & \Rightarrow f^{1} (x) > 0 \forall x \in (4, \infty) \end{array}$

Therefore, the function is increasing in the interval $(- 5, - 4) \cup (4, \infty)$

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