Let f: R -> R be defined as

$f (x) = t {\begin{matrix} - 55 x, & i f x < - 5 \\ 2 x^{3} - 3 x^{2} - 120 x, & i f - 5 \leq x \leq 4 \\ 2 x^{3} - 3 x^{2} - 36 x - 336, & i f x > 4, \end{matrix}$

Let A = ${x \in R : f i s i n c r e a s i n g} .$ Then A is equal to:

Option 1 -

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

Option 2 -

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

Option 3 -

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

Option 4 -

$(- 5, \infty)$

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

let f : R -> R

$f' (x) = {\begin{cases} - 55, x < - 5 \\ 6 x^{2} - 6 x - 120, - 5 < x < 4 \\ 6 x^{2} - 6 x - 36, x > 4 \end{cases}$

f’ (x) increasing in x $\in (- 5, - 4) \cup (4, \infty)$

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Let f: R -> R be defined as

$f (x) = t {\begin{matrix} - 55 x, & i f x < - 5 \\ 2 x^{3} - 3 x^{2} - 120 x, & i f - 5 \leq x \leq 4 \\ 2 x^{3} - 3 x^{2} - 36 x - 336, & i f x > 4, \end{matrix}$

Let A = ${x \in R : f i s i n c r e a s i n g} .$ Then A is equal to:

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Let f: R -> R be defined as f ( x ) = t { − 5 5 x , i f x < − 5 2 x 3 − 3 x 2 − 1 2 0 x , i f − 5 ≤ x ≤ 4 2 x 3 − 3 x 2 − 3 6 x − 3 3 6 , i f x > 4 , Let A = { x ∈ R : f i s i n c r e a s i n g } . Then A is equal to:

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Let f: R -> R be defined as

$f (x) = t {\begin{matrix} - 55 x, & i f x < - 5 \\ 2 x^{3} - 3 x^{2} - 120 x, & i f - 5 \leq x \leq 4 \\ 2 x^{3} - 3 x^{2} - 36 x - 336, & i f x > 4, \end{matrix}$

Let A = ${x \in R : f i s i n c r e a s i n g} .$ Then A is equal to: