Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Calculus

Question:

Determine the intervals in which the function $f(x) = 3x^2 - 4x^3 – 12x^2 + 5$ are strictly increasing or strictly decreasing.

Options:

Strictly increasing on $[−1,0]∪[2,∞)$ and strictly decreasing on $(−∞,−1]∪[0,2]$.

Strictly decreasing on $[−1,0]∪[2,∞)$ and strictly increasing on $(−∞,−1]∪[0,2]$.

Strictly increasing on $(−∞,−1]∪[2,∞)$ and strictly decreasing on $[−1,2]$.

Strictly decreasing on $(−∞,−1)∪(2,∞)$ and strictly increasing on $(−1,2)$.

Correct Answer:

Strictly increasing on $[−1,0]∪[2,∞)$ and strictly decreasing on $(−∞,−1]∪[0,2]$.

Explanation:

The correct answer is Option (1) → Strictly increasing on $[−1,0]∪[2,∞)$ and strictly decreasing on $(−∞,−1]∪[0,2]$.

Given $f(x) = 3x^4-4x^3-12x^2 +5, D_f= R$.

Differentiating w.r.t. x, we get

$f'(x)=3.4x^3-4.3x^2-12.2x$

$= 12x(x^2-x-2)$

$= 12(x+1)x(x-2)$

Now $f'(x) > 0$ iff $12 (x + 1) x (x-2) > 0$

$⇒ (x+1)x(x-2) > 0$

$⇒x∈ (-1, 0) ∪ (2,∞)$

⇒ f(x) is strictly increasing in $[-1, 0] ∪ [2, ∞)$.

And $f'(x) < 0$ iff $12(x + 1) x (x-2) <0$

$⇒ (x+1)x(x-2) <0$

$⇒x∈ (-∞,-1) ∪ (0,2)$

⇒ f(x) is strictly decreasing in $(-∞, -1] ∪ [0,2]$.