The interval(s), where the function f(x) = cases (1-e^x)/(e^2x-1) & : x ≠ 0 \\ (-1)/2 & : x = 0 cases is increasing, is/ are: Option 1: (1, ∞) Option 2: (-∞, 1) Option 3: (-∞, ∞) Option 4: (-∞, ∞) - \0\

Answer: (-∞, ∞). Solution: Given: f(x) = cases (1-e^x)/(e^2x-1) & : x ≠ 0 \\ (-1)/2 & : x = 0 cases For x ≠ 0: Numerator: 1-e^x = -(e^x - 1) Denominator: e^2x-1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1) Therefore: f(x) = (-(e^x-1))/((e^x-1)(e^x+1)) f(x) = (-1)/(e^x+1) for x ≠ 0 At x = 0, the given value is f(0) = -1/2 The limit as x 0: _(x 0) (-1)/(e^x+1) = (-1)/(e^0+1) = (-1)/2 Since _(x 0) f(x) = f(0) = -1/2, the function is continuous at x = 0. Therefore f(x) = (-1)/(e^x+1) for all x R Rewriting: f(x) = -(e^x+1)^(-1) Using the chain rule: f'(x) = -(-1)(e^x+1)^(-2) * e^x f'(x) = e^x/((e^x+1)^2) For f'(x) = e^x/((e^x+1)^2): Numerator: e^x > 0 for all x R Denominator: (e^x+1)^2 > 0 for all x R Therefore f'(x) > 0 for all x R The function is increasing on (-∞, ∞)

Calculus

Application of Derivatives

Medium

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The interval(s), where the function $f(x) = \begin{cases} \frac{1-e^x}{e^{2x}-1} & : x \neq 0 \\ \frac{-1}{2} & : x = 0 \end{cases}$ is increasing, is/ are:

$(1, \infty)$

$(-\infty, 1)$

$(-\infty, \infty)$

$(-\infty, \infty) - \{0\}$

✅ Correct Option: 3

Given: $f(x) = \begin{cases} \frac{1-e^x}{e^{2x}-1} & : x \neq 0 \\ \frac{-1}{2} & : x = 0 \end{cases}$

For $x \neq 0$ :

Numerator: $1-e^x = -(e^x - 1)$

Denominator: $e^{2x}-1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1)$

Therefore:

$f(x) = \frac{-(e^x-1)}{(e^x-1)(e^x+1)}$

$f(x) = \frac{-1}{e^x+1}$ for $x \neq 0$

At $x = 0$ , the given value is $f(0) = -\frac{1}{2}$

The limit as $x \to 0$ :

$\lim_{x \to 0} \frac{-1}{e^x+1}$

$= \frac{-1}{e^0+1}$

$= \frac{-1}{2}$

Since $\lim_{x \to 0} f(x) = f(0) = -\frac{1}{2}$ , the function is continuous at $x = 0$ .

Therefore $f(x) = \frac{-1}{e^x+1}$ for all $x \in \mathbb{R}$

Rewriting: $f(x) = -(e^x+1)^{-1}$

Using the chain rule:

$f'(x) = -(-1)(e^x+1)^{-2} \cdot e^x$

$f'(x) = \frac{e^x}{(e^x+1)^2}$

For $f'(x) = \frac{e^x}{(e^x+1)^2}$ :

Numerator: $e^x > 0$ for all $x \in \mathbb{R}$

Denominator: $(e^x+1)^2 > 0$ for all $x \in \mathbb{R}$

Therefore $f'(x) > 0$ for all $x \in \mathbb{R}$

The function is increasing on $(-\infty, \infty)$

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The function $f(x) = -x^3 + 12x^2 - 36x + 21$ is

A. Increasing on $(-\infty, 2)$

B. Decreasing on $(-\infty, 2)$

C. Increasing on $(2, 6)$

D. Decreasing on $(6, \infty)$

Choose the correct answer from the options given below:

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Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) The minimum value of } f(x) = (2x - 1)^2 + 3 & \text{(I) } 4 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) The maximum value of } f(x) = -|x + 1| + 4 & \text{(II) } 10 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) The minimum value of } f(x) = \sin(2x) + 6 & \text{(III) } 3 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) The maximum value of } f(x) = -(x - 1)^2 + 10 & \text{(IV) } 5 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

The interval(s), where the function f(x)={1−exe2x−1:x≠0−12:x=0f(x) = \begin{cases} \frac{1-e^x}{e^{2x}-1} & : x \neq 0 \\ \frac{-1}{2} & : x = 0 \end{cases}f(x)={e2x−11−ex​2−1​​:x=0:x=0​ is increasing, is/ are:

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CUET Mathematics Past Year Questions (Topic-Wise)

The interval(s), where the function $f(x) = \begin{cases} \frac{1-e^x}{e^{2x}-1} & : x \neq 0 \\ \frac{-1}{2} & : x = 0 \end{cases}$ is increasing, is/ are: