Question 1

Show that the function given by \(f(x) = 3x + 17\) is increasing on \(\mathbb{R}\).

Solution:

Given: \(f(x) = 3x + 17\)

Differentiating with respect to \(x\):

\[ f'(x) = 3 \]

Since \(3 > 0\), it follows that \(f'(x) > 0\) for all \(x \in \mathbb{R}\).

Thus, \(f\) is strictly increasing on \(\mathbb{R}\).

Question 2

Show that the function given by \(f(x) = e^{2x}\) is increasing on \(\mathbb{R}\).

Solution:

Given: \(f(x) = e^{2x}\)

Differentiating with respect to \(x\):

\[ f'(x) = 2e^{2x} \]

We know that the exponential function \(e^{2x}\) is always positive for all real values of \(x\). Therefore, \(2e^{2x} > 0\).

Since \(f'(x) > 0\) for all \(x \in \mathbb{R}\), the function is strictly increasing on \(\mathbb{R}\).

Question 3

Show that the function given by \(f(x) = \sin x\) is (a) increasing in \((0, \pi/2)\) (b) decreasing in \((\pi/2, \pi)\) (c) neither increasing nor decreasing in \((0, \pi)\).

Solution:

Given: \(f(x) = \sin x \implies f'(x) = \cos x\).

(a) For \(x \in (0, \pi/2)\), \(\cos x\) is positive (I quadrant). So \(f'(x) > 0\). Thus, \(f\) is increasing in \((0, \pi/2)\).

(b) For \(x \in (\pi/2, \pi)\), \(\cos x\) is negative (II quadrant). So \(f'(x) < 0\). Thus, \(f\) is decreasing in \((\pi/2, \pi)\).

(c) Since the function increases in \((0, \pi/2)\) and decreases in \((\pi/2, \pi)\), it is neither increasing nor decreasing in \((0, \pi)\).

Question 4

Find the intervals in which the function \(f\) given by \(f(x) = 2x^2 - 3x\) is (a) increasing (b) decreasing.

Solution:

\[ f(x) = 2x^2 - 3x \implies f'(x) = 4x - 3 \]

Setting \(f'(x) = 0 \implies 4x - 3 = 0 \implies x = 3/4\).

The point \(x = 3/4\) divides the real line into intervals \((-\infty, 3/4)\) and \((3/4, \infty)\).

(a) In \((3/4, \infty)\), \(f'(x) > 0\). So \(f\) is increasing in \((3/4, \infty)\).

(b) In \((-\infty, 3/4)\), \(f'(x) < 0\). So \(f\) is decreasing in \((-\infty, 3/4)\).

Question 5

Find the intervals in which the function \(f\) given by \(f(x) = 2x^3 - 3x^2 - 36x + 7\) is (a) increasing (b) decreasing.

Solution:

\[ f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2) \]

Setting \(f'(x) = 0\) gives \(x = -2\) and \(x = 3\).

Intervals: \((-\infty, -2)\), \((-2, 3)\), and \((3, \infty)\).

1. In \((-\infty, -2)\), \(f'(x) = 6(-)(- ) = (+)\). So \(f\) is increasing.

2. In \((-2, 3)\), \(f'(x) = 6(-)(+) = (-)\). So \(f\) is decreasing.

3. In \((3, \infty)\), \(f'(x) = 6(+)(+) = (+)\). So \(f\) is increasing.

Answer: (a) Increasing in \((-\infty, -2) \cup (3, \infty)\); (b) Decreasing in \((-2, 3)\).

Question 6

Find the intervals in which the following functions are strictly increasing or decreasing:

(a) \(x^2 + 2x - 5\)

Solution: \(f'(x) = 2x + 2 = 2(x + 1)\). Critical point \(x = -1\). Increasing on \((-1, \infty)\), decreasing on \((-\infty, -1)\).

(b) \(10 - 6x - 2x^2\)

Solution: \(f'(x) = -6 - 4x = -2(3 + 2x)\). Critical point \(x = -3/2\). Increasing on \((-\infty, -3/2)\), decreasing on \((-3/2, \infty)\).

(c) \(-2x^3 - 9x^2 - 12x + 1\)

Solution: \(f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x + 1)(x + 2)\). Critical points \(-2, -1\). Increasing on \((-2, -1)\), decreasing on \((-\infty, -2) \cup (-1, \infty)\).

Question 7

Show that \(y = \log(1 + x) - \frac{2x}{2 + x}, x > -1\), is an increasing function of \(x\) throughout its domain.

Solution:

\[ \frac{dy}{dx} = \frac{1}{1 + x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2} \]

\[ \frac{dy}{dx} = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \frac{4 + x^2 + 4x - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2} \]

Since \(x > -1\), \((1+x) > 0\). Also \(x^2 \ge 0\) and \((2+x)^2 > 0\). Thus, \(\frac{dy}{dx} \ge 0\).

So, the function is increasing throughout its domain.

Question 8

Find the values of \(x\) for which \(y = [x(x - 2)]^2\) is an increasing function.

Solution:

\[ y = [x^2 - 2x]^2 \implies \frac{dy}{dx} = 2(x^2 - 2x)(2x - 2) = 4x(x - 2)(x - 1) \]

Critical points are \(x = 0, 1, 2\). Analyzing signs:

In \((0, 1)\), \(y' = 4(+)(-)(-) = (+)\). (Increasing)

In \((2, \infty)\), \(y' = 4(+)(+)(+) = (+)\). (Increasing)

Answer: \(x \in (0, 1) \cup (2, \infty)\).

Question 9

Prove that \(y = \frac{4\sin\theta}{2 + \cos\theta} - \theta\) is an increasing function of \(\theta\) in \([0, \pi/2]\).

Solution:

\[ \frac{dy}{d\theta} = \frac{(2 + \cos\theta)(4\cos\theta) - (4\sin\theta)(-\sin\theta)}{(2 + \cos\theta)^2} - 1 \]

\[ \frac{dy}{d\theta} = \frac{8\cos\theta + 4\cos^2\theta + 4\sin^2\theta}{(2 + \cos\theta)^2} - 1 = \frac{8\cos\theta + 4}{(2 + \cos\theta)^2} - 1 \]

\[ \frac{dy}{d\theta} = \frac{8\cos\theta + 4 - (4 + \cos^2\theta + 4\cos\theta)}{(2 + \cos\theta)^2} = \frac{\cos\theta(4 - \cos\theta)}{(2 + \cos\theta)^2} \]

In \([0, \pi/2]\), \(\cos\theta \in [1]\). So \(\cos\theta \ge 0\) and \((4 - \cos\theta) > 0\).

Thus, \(\frac{dy}{d\theta} \ge 0\), implying \(y\) is increasing.

Question 10

Prove that the logarithmic function is increasing on \((0, \infty)\).

Solution:

Let \(f(x) = \log x\). Differentiating gives \(f'(x) = 1/x\).

For \(x \in (0, \infty)\), \(x > 0 \implies 1/x > 0\).

Since \(f'(x) > 0\) for all \(x > 0\), the logarithmic function is strictly increasing on \((0, \infty)\).

Question 11

Prove that the function \(f\) given by \(f(x) = x^2 - x + 1\) is neither strictly increasing nor decreasing on \((-1, 1)\).

Solution:

Given: \(f(x) = x^2 - x + 1\)

Differentiating with respect to \(x\):

\[ f'(x) = 2x - 1 \]

To find the critical point, set \(f'(x) = 0\):

\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

The point \(x = 1/2\) divides the interval \((-1, 1)\) into two sub-intervals: \((-1, 1/2)\) and \((1/2, 1)\).

• In the interval \((-1, 1/2)\), take \(x = 0\): \(f'(0) = 2(0) - 1 = -1 < 0\). Thus, \(f\) is strictly decreasing on \((-1, 1/2)\).
• In the interval \((1/2, 1)\), take \(x = 3/4\): \(f'(3/4) = 2(3/4) - 1 = 1/2 > 0\). Thus, \(f\) is strictly increasing on \((1/2, 1)\).

Since the function is decreasing in one part of the interval and increasing in the other, it is neither strictly increasing nor decreasing on \((-1, 1)\).

Question 12

Which of the following functions are decreasing on \((0, \pi/2)\)?

(A) \(\cos x\) (B) \(\cos 2x\) (C) \(\cos 3x\) (D) \(\tan x\)

Solution:

(A) Let \(f_1(x) = \cos x \implies f_1'(x) = -\sin x\). For \(x \in (0, \pi/2)\), \(\sin x > 0\), so \(f_1'(x) < 0\). Decreasing.

(B) Let \(f_2(x) = \cos 2x \implies f_2'(x) = -2\sin 2x\). For \(x \in (0, \pi/2)\), \(2x \in (0, \pi)\), so \(\sin 2x > 0\), so \(f_2'(x) < 0\). Decreasing.

(C) Let \(f_3(x) = \cos 3x \implies f_3'(x) = -3\sin 3x\). For \(x \in (0, \pi/2)\), \(3x \in (0, 3\pi/2)\). \(\sin 3x\) is positive in \((0, \pi/3)\) and negative in \((\pi/3, \pi/2)\). Thus, it is not decreasing on the whole interval.

(D) Let \(f_4(x) = \tan x \implies f_4'(x) = \sec^2 x > 0\). Increasing.

Both (A) and (B) are decreasing on the given interval.

Question 13

On which of the following intervals is the function \(f\) given by \(f(x) = x^{100} + \sin x - 1\) decreasing?

(A) \((0, 1)\) (B) \((\pi/2, \pi)\) (C) \((0, \pi/2)\) (D) None of these

Solution:

\[ f(x) = x^{100} + \sin x - 1 \implies f'(x) = 100x^{99} + \cos x \]

(A) In \((0, 1)\), \(x > 0 \implies 100x^{99} > 0\). Also \(\cos x > 0\) for \(x \in (0, 1)\). So \(f'(x) > 0\). Increasing.

(B) In \((\pi/2, \pi)\), \(x > 1.57 \implies 100x^{99}\) is a very large positive value. \(\cos x\) is between \(0\) and \(-1\). Thus, \(f'(x) > 0\). Increasing.

(C) In \((0, \pi/2)\), both \(100x^{99}\) and \(\cos x\) are positive. So \(f'(x) > 0\). Increasing.

The correct option is (D) None of these.

Question 14

For what values of \(a\) the function \(f\) given by \(f(x) = x^2 + ax + 1\) is increasing on \([1, 2]\)?

Solution:

\[ f(x) = x^2 + ax + 1 \implies f'(x) = 2x + a \]

For \(f\) to be increasing on \([1, 2]\), we must have \(f'(x) \ge 0\) for all \(x \in [1, 2]\).

\[ 2x + a \ge 0 \implies a \ge -2x \]

Since the minimum value of \(2x + a\) occurs at the smallest value of \(x\) in the interval (which is \(x = 1\)), we have:

\[ 2(1) + a \ge 0 \implies a \ge -2 \]

Thus, the function is increasing on \([1, 2]\) for all \(a \in [-2, \infty)\).

Question 15

Let \(I\) be any interval disjoint from \([-1, 1]\). Prove that the function \(f\) given by \(f(x) = x + \frac{1}{x}\) is increasing on \(I\).

Solution:

\[ f(x) = x + \frac{1}{x} \implies f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} \]

For the function to be increasing, \(f'(x) > 0\):

\[ \frac{x^2 - 1}{x^2} > 0 \]

Since \(x^2 > 0\) for \(x \neq 0\), we need \(x^2 - 1 > 0\):

\[ x^2 > 1 \implies |x| > 1 \]

This means \(x \in (-\infty, -1) \cup (1, \infty)\). This set represents all intervals disjoint from \([-1, 1]\). Hence, the function is increasing on \(I\).

Question 16

Prove that the function \(f\) given by \(f(x) = \log \sin x\) is increasing on \((0, \pi/2)\) and decreasing on \((\pi/2, \pi)\).

Solution:

\[ f(x) = \log \sin x \implies f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \]

• In the interval \((0, \pi/2)\), \(\cot x > 0\) (as it is the first quadrant). Since \(f'(x) > 0\), the function is strictly increasing.
• In the interval \((\pi/2, \pi)\), \(\cot x < 0\) (as it is the second quadrant). Since \(f'(x) < 0\), the function is strictly decreasing.

Question 17

Prove that the function \(f\) given by \(f(x) = \log |\cos x|\) is decreasing on \((0, \pi/2)\) and increasing on \((3\pi/2, 2\pi)\).

Solution:

\[ f(x) = \log |\cos x| \implies f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \]

• In the interval \((0, \pi/2)\), \(\tan x > 0\). Thus, \(f'(x) = -\tan x < 0\). The function is decreasing.
• In the interval \((3\pi/2, 2\pi)\), \(\tan x < 0\) (fourth quadrant). Thus, \(f'(x) = -\tan x > 0\). The function is increasing.

Question 18

Prove that the function given by \(f(x) = x^3 - 3x^2 + 3x - 100\) is increasing in \(\mathbb{R}\).

Solution:

\[ f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2 \]

Since \((x - 1)^2 \ge 0\) for all real \(x\), \(f'(x) \ge 0\) for all \(x \in \mathbb{R}\).

Thus, \(f\) is increasing on \(\mathbb{R}\).

Question 19

The interval in which \(y = x^2 e^{-x}\) is increasing is:

Solution:

\[ \frac{dy}{dx} = 2x e^{-x} + x^2 (-e^{-x}) = x e^{-x} (2 - x) \]

For increasing, \(\frac{dy}{dx} > 0\). Since \(e^{-x} > 0\), we need \(x(2 - x) > 0\).

This holds when \(x \in (0, 2)\).

Correct option is (D).