The interval in which the function f(x)=x^4-x^3/3 is increasing, is Option 1: (-∞, 1/4) Option 2: (0, 1/4) Option 3: (1/4, ∞) Option 4: (-∞, -1/4)

Answer: (1/4, ∞). Solution: f'(x) = 4x^3 - x^2 = x^2(4x-1). Since x^2≥ 0, the sign of f' depends on 4x-1. So f'(x)>0 when x>1/4, and the function increases on (1/4,∞).

Calculus

Application of Derivatives

Easy

applied

The interval in which the function $f(x)=x^4-\dfrac{x^3}{3}$ is increasing, is

$(-\infty, 1/4)$

$(0, 1/4)$

$(1/4, \infty)$

$(-\infty, -1/4)$

✅ Correct Option: 3

$f'(x) = 4x^3 - x^2 = x^2(4x-1)$ .

Since $x^2\ge 0$ , the sign of $f'$ depends on $4x-1$ .

So $f'(x)>0$ when $x>\dfrac{1}{4}$ , and the function increases on $\left(\dfrac{1}{4},\infty\right)$ .

2026: 29 May Shift 2

The interval in which the function $f(x)=x^3-12x^2+36x+20$ is increasing, is

2026: 19 May Shift 2

Function $f(x) = x + \frac{1}{x}, x \neq 0$ is increasing in the interval

2026: 19 May Shift 1

The maximum area of a rectangle inscribed in a circle of radius $3\sqrt{2}$ cm, is:

2025: 16 May Shift 1

Match List-I with List-II

Consider the function f(x) = 2x³ - 21x² + 36x + 80, x∈[0, 6]. Then

List-I	List-II
(A) one of its critical points is at x =	(I) -28
(B) Its absolute maximum value is	(II) -42
(C) Its absolute minimum value is	(III) 97
(D) Its second derivative at x = 0 is	(IV) 6

Choose the correct answer from the options given below:

2025: 16 May Shift 1

The point on the curve y = (x - 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4) is:

2026: 21 May Shift 1

The maximum and the minimum value of the function $f(x) = -||x+3|-5| + 2$ are :

The interval in which the function f(x)=x4−x33f(x)=x^4-\dfrac{x^3}{3}f(x)=x4−3x3​ is increasing, is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The interval in which the function $f(x)=x^4-\dfrac{x^3}{3}$ is increasing, is