If f(x) = 1/(1-x), then for x > 1, f(x) is: Option 1: decreasing Option 2: constant Option 3: increasing Option 4: neither decreasing nor increasing

Answer: increasing. Solution: f'(x) = d/dx(1-x)^(-1) = 1/((1-x)^2). Since (1-x)^2 > 0 for x ≠ 1, we have f'(x) > 0 for x > 1. Hence f is increasing.

Calculus

Application of Derivatives

Medium

common

If $f(x) = \frac{1}{1-x}$ , then for $x > 1, f(x)$ is:

decreasing

constant

increasing

neither decreasing nor increasing

✅ Correct Option: 3

$f'(x) = \frac{d}{dx}(1-x)^{-1} = \frac{1}{(1-x)^2}$ . Since $(1-x)^2 > 0$ for $x \neq 1$ , we have $f'(x) > 0$ for $x > 1$ . Hence $f$ is increasing.

2023: 30 May Shift 3

The interval in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly decreasing is :

2025: 16 May Shift 1

If the interval in which f(x) = $\frac{x}{4}$ + $\frac{4}{x}$ , x ≠ 0 is strictly increasing is (-∞, a) ∪ (b, ∞), then

2025: 15 May Shift 1

If $f(x) = a \log_e|x| + bx^2 + x$ has critical points at $x = -2$ and $x = 1$ , then

2025: 13 May Shift 1

Consider the function $f(x) = x^3 - 3x$ . Then

Match List-I with List-II

List-I	List-II
(A) Point of local Maxima	(I) 1
(B) Point of local Minima	(II) -1
(C) Local maximum value	(III) 2
(D) Local minimum value	(IV) -2

Choose the correct answer from the options given below:

2025: 3 June Shift 1

The edge of a cube is increasing at a rate of 7cm/s. The rate of change of area of the cube when edge of the cube is 3cm is:

2025: 22 May Shift 2

The curve $x = y^2$ and $xy = k$ cut orthogonally, then $k^2$ is equal to:

If f(x)=11−xf(x) = \frac{1}{1-x}f(x)=1−x1​, then for x>1,f(x)x > 1, f(x)x>1,f(x) is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $f(x) = \frac{1}{1-x}$ , then for $x > 1, f(x)$ is: