QPaperGen

Explanation: The area of a circle, denoted by \(A\), with a radius \(r\), is given by the formula:\[A = \pi r^2\]To find the rate of change of the area with respect to its radius, we differentiate the area formula with respect to \(r\):\[\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r\]Now, let's calculate this rate of change for specific values of the radius:

\[\]
1. When \(r = 3\) cm: Substitute \(r=3\) into the derivative: \[\frac{dA}{dr} = 2\pi (3) = 6\pi\] Therefore, when the radius is 3 cm, the area of the circle is changing at a rate of \(6\pi\) cm\(^2\)/s.
\[\]
2. When \(r = 4\) cm: Substitute \(r=4\) into the derivative: \[\frac{dA}{dr} = 2\pi (4) = 8\pi\] Therefore, when the radius is 4 cm, the area of the circle is changing at a rate of \(8\pi\) cm\(^2\)/s.

Explanation: Let the length of a side of the cube be \(x\), its volume be \(V\), and its surface area be \(S\). We know that \(x\) is a function of time \(t\). The formulas for the volume and surface area of a cube are:

\[\] Volume: \(V = x^3\) Surface Area: \(S = 6x^2\)\[\]

It is given that the rate of change of the volume with respect to time is 8 cm\(^3\)/s, i.e.:\[ \frac{dV}{dt} = 8 \text{ cm}^3/\text{s} \]To relate the rate of change of volume to the rate of change of the side length, we use the chain rule on the volume formula:\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} \]\[ 8 = 3x^2 \cdot \frac{dx}{dt} \]From this, we can express the rate of change of the side length, \(\frac{dx}{dt}\):\[ \frac{dx}{dt} = \frac{8}{3x^2} \quad \text{(1)} \]Now, we need to find the rate of change of the surface area with respect to time, \(\frac{dS}{dt}\). We apply the chain rule to the surface area formula:\[ \frac{dS}{dt} = \frac{d}{dt}(6x^2) = \frac{d}{dx}(6x^2) \cdot \frac{dx}{dt} \]\[ \frac{dS}{dt} = 12x \cdot \frac{dx}{dt} \]Substitute the expression for \(\frac{dx}{dt}\) from equation (1) into this equation:\[ \frac{dS}{dt} = 12x \cdot \left(\frac{8}{3x^2}\right) \]\[ \frac{dS}{dt} = \frac{96x}{3x^2} = \frac{32}{x} \text{ cm}^2/\text{s} \]Finally, we calculate the rate of change of the surface area when the length of the edge of the cube is 12 cm.When \(x = 12\) cm:\[ \frac{dS}{dt} = \frac{32}{12} = \frac{8}{3} \text{ cm}^2/\text{s} \]Therefore, if the length of the edge of the cube is 12 cm, its surface area is increasing at the rate of \(\frac{8}{3}\) cm\(^2\)/s.

Explanation: We know that \(A = \pi {r^2}\) \[\therefore \dfrac{{dA}}{{dt}} = \dfrac{d}{{dr}}\left( {\pi {r^2}} \right)\dfrac{{dr}}{{dt}} = 2\pi r\dfrac{{dr}}{{dt}}\] Given that \(\dfrac{{dr}}{{dt}} = 3\;{\text{cm}}/{\text{s}}\) \[\therefore \dfrac{{dA}}{{dt}} = 2\pi r(3) = 6\pi r\] So, when \(r = 10\;{\text{cm}}\), \(\dfrac{{dA}}{{dt}} = 6\pi (10) = 60\pi {\text{c}}{{\text{m}}^2}/{\text{s}}\)

Explanation: Let the side length of the cube be \(x\), and its volume be \(V\). \[\]
Since \(V = x^3\), we differentiate with respect to time \(t\):\[\]

\[\frac{dV}{dt} = \frac{d}{dt}(x^3) = \frac{d}{dx}(x^3) \cdot \frac{dx}{dt} = 3x^2 \cdot \frac{dx}{dt}\]

Given \(\frac{dx}{dt} = 3\,\text{cm/s}\), we substitute:

\[\frac{dV}{dt} = 3x^2 \cdot 3 = 9x^2\]

When \(x = 10\,\text{cm}\), \[\]we find:

\(\frac{dV}{dt} = 9(10)^2 = 900\,\text{cm}^3/\text{s}\) \[\]

Hence, the volume of the cube is increasing at the rate of \(900\,\text{cm}^3/\text{s}\) when the edge is 10
cm long.

Explanation: We are asked to find how fast the area enclosed by the wave is increasing when the radius is \(8\,\text{cm}\).

We know that the area of a circle is given by \(A = \pi r^2\).

\[\]Differentiating both sides with respect to time \(t\):\[\]

\[\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = \frac{d}{dr}(\pi r^2) \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt}\]

Given \(\frac{dr}{dt} = 5\,\text{cm/s}\), we substitute the values:

\(\frac{dA}{dt} = 2\pi (8)(5) = 80\pi\,\text{cm}^2/\text{s}\)

Explanation: We know that \(C = 2\pi r\).
Differentiating with respect to time \(t\):

\[\frac{dC}{dt} = \frac{d}{dr}(2\pi r) \cdot \frac{dr}{dt} = 2\pi \cdot \frac{dr}{dt}\]

Given \(\frac{dr}{dt} = 0.7,\text{cm/s}\): \[\]

\(\frac{dC}{dt} = 2\pi (0.7) = 1.4\pi,\text{cm/s}\)

Explanation: (a) Perimeter:
The perimeter is \(P = 2(x + y)\)

\[\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2,\text{cm/min}\]

(b) Area:
The area is \(A = xy\)

\[\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt} = (-5)(6) + (8)(4) = -30 + 32 = 2,\text{cm}^2/\text{min}\]

Explanation: Volume of a sphere: \(V = \frac{4}{3}\pi r^3\) \[\]

Differentiating:

\[\frac{dV}{dt} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\]

Given \(\frac{dV}{dt} = 900\): This implies

\[900 = 4\pi (15)^2 \cdot \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{900}{4\pi (225)} = \frac{1}{\pi} \text{cm/s}\]

Explanation: Volume of a sphere: \(V = \frac{4}{3}\pi r^3\)\[\]

Differentiating with respect to \(r\):

\[\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2\]

When \(r = 10,\text{cm}\):

\[\frac{dV}{dr} = 4\pi (10)^2 = 400\pi \quad\text{cm}^3/\text{cm}\]

Explanation:
Let \(x\) be the horizontal distance from the wall, and \(y\) be the height on the wall. \[\]
Using Pythagoras: \(x^2 + y^2 = 25\)\[\]

Differentiate:

\[\frac{dy}{dt} = \frac{d}{dt}\left(\sqrt{25 - x^2}\right) = \frac{-x}{\sqrt{25 - x^2}} \cdot \frac{dx}{dt}\]

Given \(\frac{dx}{dt} = 2,\text{cm/s} = 0.02,\text{m/s}\) and \(x = 4\):

\[\frac{dy}{dt} = \frac{-4}{\sqrt{25 - 16}} \cdot 0.02 = \frac{-4}{3} \cdot 0.02 = -\frac{0.08}{3} \approx -0.0267,\text{m/s}\]

So, the top of the ladder is sliding down at approximately \(0.0267,\text{m/s}\).

Explanation: We start with the equation of the curve:
\[6y = x^3 + 2\]

Differentiating both sides with respect to time \(t\) gives:
\[6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}\]

Dividing both sides by 3:
\[2\frac{dy}{dt} = x^2 \frac{dx}{dt}\]

According to the problem, the \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate, i.e.,
\[\frac{dy}{dt} = 8\frac{dx}{dt}\]

Substituting this into the equation:
\[2(8\frac{dx}{dt}) = x^2 \frac{dx}{dt}\]
\[\Rightarrow 16\frac{dx}{dt} = x^2 \frac{dx}{dt}\]

Assuming \(\frac{dx}{dt} \ne 0\), we cancel it out. So :
\[x^2 = 16 \Rightarrow x = \pm 4\]

Now, we substitute these \(x\) values back into the original equation to find \(y\):
\[\]

When \(x = 4\):
\[y = \frac{4^3 + 2}{6} = \frac{64 + 2}{6} = \frac{66}{6} = 11\]

When \(x = -4\):
\[y = \frac{(-4)^3 + 2}{6} = \frac{-64 + 2}{6} = \frac{-62}{6} = -\frac{31}{3}\]

Final Answer:
The required points on the curve are \((4,\ 11)\) and \(\left(-4,\ -\frac{31}{3}\right)\)

Explanation: We are given that the radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s. We want to find the rate at which the volume of the bubble is increasing when the radius is 1 cm.

Assuming the bubble is spherical, the volume is given by: \[V = \frac{4}{3}\pi r^3\]

Differentiating both sides with respect to time \(t\): \[\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\]

Given: \(\frac{dr}{dt} = \frac{1}{2}\) cm/s

At \(r = 1\) cm: \[\frac{dV}{dt} = 4\pi (1)^2 \cdot \frac{1}{2} = 2\pi \text{ cm}^3/\text{s}\]

Final Answer: The volume is increasing at a rate of \(2\pi\) cm\(^3\)/s when the radius is 1 cm.

Explanation: A balloon, which always remains spherical, has a variable diameter given by
\[d = \dfrac{3}{2}(2x + 1).\]
We are to find the rate of change of its volume with respect to \(x\).

The volume of a sphere is given by:
\[V = \dfrac{4}{3}\pi r^3\]

Since the diameter is \(d = \dfrac{3}{2}(2x + 1)\), the radius is:

\[r = \dfrac{d}{2} = \dfrac{3}{4}(2x + 1)\]

Substitute into the volume formula:

\[V = \dfrac{4}{3}\pi \left( \dfrac{3}{4}(2x + 1) \right)^3 = \dfrac{4}{3}\pi \cdot \dfrac{27}{64} (2x + 1)^3 = \dfrac{9}{16}\pi (2x + 1)^3\]

Now differentiate with respect to \(x\):

\[\dfrac{dV}{dx} = \dfrac{9}{16}\pi \cdot \dfrac{d}{dx}(2x + 1)^3 = \dfrac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot 2 = \dfrac{27}{8}\pi (2x + 1)^2\]

Final Answer:
\[\dfrac{dV}{dx} = \dfrac{27}{8}\pi (2x + 1)^2\]

Explanation: Sand is pouring from a pipe at the rate of \(12,\text{cm}^3/\text{s}\).
The falling sand forms a cone on the ground such that the height is always one-sixth of the radius of the base.
We are to find how fast the height of the cone is increasing when the height is \(4,\text{cm}\).\[\]

The volume of a cone is given by:
\[V = \dfrac{1}{3} \pi r^2 h\]

Since the height is one-sixth of the radius,
\[h = \dfrac{1}{6}r \Rightarrow r = 6h\]

Substitute into the volume formula:

\[V = \dfrac{1}{3} \pi (6h)^2 h = \dfrac{1}{3} \pi \cdot 36h^2 \cdot h = 12\pi h^3\]

Differentiate both sides with respect to time \(t\):

\[\dfrac{dV}{dt} = \dfrac{d}{dt}(12\pi h^3) = 12\pi \cdot 3h^2 \cdot \dfrac{dh}{dt} = 36\pi h^2 \dfrac{dh}{dt}\]

We are given: At \(h = 4,\text{cm}\): \(\dfrac{dV}{dt} = 12,\text{cm}^3/\text{s}\), So


\[12 = 36\pi (4)^2 \dfrac{dh}{dt} = 36\pi \cdot 16 \cdot \dfrac{dh}{dt}\]
\[\Rightarrow \dfrac{dh}{dt} = \dfrac{12}{576\pi} = \dfrac{1}{48\pi},\text{cm/s}\]

Final Answer:
The height of the cone is increasing at a rate of \(\dfrac{1}{48\pi},\text{cm/s}\) when the height is \(4,\text{cm}\).

Explanation: The total cost \(C(x)\) (in rupees) for producing \(x\) units of an item is given by:
\[C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000\]

We are asked to find the marginal cost when \(x = 17\).\[\]
Marginal cost is the rate of change of total cost with respect to the number of units produced, i.e.,

\(MC = \dfrac{dC}{dx}\)
\[\]

Differentiate \(C(x)\):

\[\dfrac{dC}{dx} = 0.007 \cdot 3x^2 - 0.003 \cdot 2x + 15 = 0.021x^2 - 0.006x + 15\]

Now, substitute \(x = 17\):

\[MC = 0.021 \cdot 17^2 - 0.006 \cdot 17 + 15 = 0.021 \cdot 289 - 0.006 \cdot 17 + 15\]

\[= 6.069 - 0.102 + 15 = 20.967\]

Final Answer:
When 17 units are produced, the marginal cost is Rs. 20.967

Explanation: The total revenue (in rupees) from selling \(x\) units of a product is given by:
\[R(x) = 13x^2 + 26x + 15\]

We are asked to find the marginal revenue when \(x = 7\).\[\]

Marginal revenue is the rate of change of total revenue with respect to the number of units sold, i.e.,

\(MR = \dfrac{dR}{dx}\)
\[\]

Differentiate \(R(x)\):

\[\dfrac{dR}{dx} = 13 \cdot 2x + 26 = 26x + 26\]

Now substitute \(x = 7\):

\[MR = 26 \cdot 7 + 26 = 182 + 26 = 208\]

Final Answer:
The marginal revenue when \(x = 7\) is Rs. 208

Correct Answer: 12\(\pi\)

Explanation: The rate of change of the area of a circle with respect to its radius \(r\) at \(r = 6,\text{cm}\) is to be determined.\[\]

We know that the area of a circle is:
\(A = \pi r^2\)

Differentiate with respect to \(r\):

\[\dfrac{dA}{dr} = \dfrac{d}{dr}(\pi r^2) = 2\pi r\]

Now, substitute \(r = 6\):

\[\dfrac{dA}{dr} = 2\pi \cdot 6 = 12\pi,\text{cm}^2/\text{cm}\]

Final Answer:
The rate of change of the area at \(r = 6,\text{cm}\) is \(12\pi,\text{cm}^2/\text{cm}\).

Correct Answer: 126

Explanation: The total revenue (in rupees) from selling \(x\) units of a product is given by:
\[R(x) = 3x^2 + 36x + 5\]

We are asked to find the marginal revenue when \(x = 15\).\[\]

Marginal revenue is the rate of change of total revenue with respect to the number of units sold:

\[\therefore MR = \dfrac{dR}{dx} = 3 \cdot 2x + 36 = 6x + 36\]

Substituting \(x = 15\):

\[MR = 6 \cdot 15 + 36 = 90 + 36 = 126\]

Final Answer:
The marginal revenue when \(x = 15\) is Rs. 126.

Explanation: Let \(x_1\) and \(x_2\) be any two real numbers such that \(x_1 < x_2\).
Then,

\[f(x_1) = 3x_1 + 17 < 3x_2 + 17 = f(x_2)\]

\[\Rightarrow f(x_1) < f(x_2)\]

Hence, \(f\) is strictly increasing on \(\mathbb{R}\).
\[\]

\( \bf Alternate Method:\)

Differentiate the function:

\[f'(x) = \dfrac{d}{dx}(3x + 17) = 3\]

Since \(f'(x) = 3 > 0\) for all \(x \in \mathbb{R}\), the function is strictly increasing on \(\mathbb{R}\).

Explanation: Let \(x_1\) and \(x_2\) be any two real numbers such that \(x_1 < x_2\).

Then:
\[x_1 < x_2 \Rightarrow 2x_1 < 2x_2\]
\[\Rightarrow e^{2x_1} < e^{2x_2}\]
\[\Rightarrow f(x_1) < f(x_2)\]


Hence, \(f(x)\) is strictly increasing on \(\mathbb{R}\).
\[\]

\(\bf Alternate Method:\) Differentiating w.r.x. we get
\[f'(x) =2 e^{2x}\]
It is always positive for any value of x. So \(f(x)\) is strictly increasing on \(\mathbb{R}\).

Explanation: We are given:
\(f(x) = \sin x \Rightarrow f'(x) = \cos x\)
\[\]

In the interval \(x \in \left(0,\dfrac{\pi}{2}\right)\), we have \(\cos x > 0 \Rightarrow f'(x) > 0\)
\(\Rightarrow f(x)\) is strictly increasing in \(\left(0,\dfrac{\pi}{2}\right)\)
\[\]

In the interval \(x \in \left(\dfrac{\pi}{2},\pi\right)\), we have \(\cos x < 0 \Rightarrow f'(x) < 0\)
\(\Rightarrow f(x)\) is strictly decreasing in \(\left(\dfrac{\pi}{2},\pi\right)\)
\[\]

Since \(f\) increases in part of the interval and decreases in another,
\(f\) is neither increasing nor decreasing in the entire interval \((0,\pi)\).

Explanation: We are given:
\(f(x) = 2x^2 - 3x\). \[\]


Differentiate:
\[f'(x) = \dfrac{d}{dx}(2x^2 - 3x) = 4x - 3\]

Set \(f'(x) = 0\) to find critical point:
\[4x - 3 = 0 \Rightarrow x = \dfrac{3}{4}\]

Now analyze the sign of \(f'(x)\):

In the interval \(\left(-\infty,\dfrac{3}{4}\right)\), we have \(f'(x) = 4x - 3 < 0\)
\(\Rightarrow f(x)\) is strictly decreasing on \(\left(-\infty,\dfrac{3}{4}\right)\)
\[\]

In the interval \(\left(\dfrac{3}{4},\infty\right)\), we have \(f'(x) = 4x - 3 > 0\)
\(\Rightarrow f(x)\) is strictly increasing on \(\left(\dfrac{3}{4},\infty\right)\)

\[\]Final Answer:

(a) \(f(x)\) is strictly increasing in \(\left(\dfrac{3}{4}, \infty\right)\)

(b) \(f(x)\) is strictly decreasing in \(\left(-\infty, \dfrac{3}{4}\right)\)

Explanation: Differentiate the function:

\[f'(x) = \dfrac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36\]

Factor the derivative:

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

Set \(f'(x) = 0\) to find critical points:
\[6(x + 2)(x - 3) = 0 \Rightarrow x = -2,,3\]

Now analyze the sign of \(f'(x)\) in the intervals:
\[\]

In \((-\infty, -2)\):
Both \((x + 2) < 0\) and \((x - 3) < 0 \Rightarrow f'(x) > 0\)
\(\Rightarrow f(x)\) is strictly increasing on \((-\infty, -2)\)
\[\]

In \((-2, 3)\):
\((x + 2) > 0\) and \((x - 3) < 0 \Rightarrow f'(x) < 0\)
\(\Rightarrow f(x)\) is strictly decreasing on \((-2, 3)\)
\[\]

In \((3, \infty)\):
Both \((x + 2) > 0\) and \((x - 3) > 0 \Rightarrow f'(x) > 0\)
\(\Rightarrow f(x)\) is strictly increasing on \((3, \infty)\)
\[\]

Final Answer:

\(f(x)\) is strictly increasing on \((-\infty, -2)\) and \((3, \infty)\) and\[\]

\(f(x)\) is strictly decreasing on \((-2, 3)\)

Explanation: We have:
\[f'(x) = 2x + 2\]

Setting \(f'(x) = 0\):
\[2x + 2 = 0 \Rightarrow x = -1\]

In \((-\infty, -1)\), \[f'(x) < 0 ⟹ \text {strictly decreasing}\]

In \((-1, \infty)\), \[f'(x) > 0 ⟹ \text{strictly increasing}\]

Explanation: We have:
\[f'(x) = -6 - 4x\]

Setting \(f'(x) = 0\):
\[-6 - 4x = 0 \Rightarrow x = -\dfrac{3}{2}\]

In \(\left(-\infty, -\dfrac{3}{2}\right)\), \(f'(x) > 0\) ⟹ strictly increasing
\[\]

In \(\left(-\dfrac{3}{2}, \infty\right)\), \(f'(x) < 0\) ⟹ strictly decreasing

Explanation: We have:
\[f'(x) = -6x^2 - 18x - 12 = -6(x + 1)(x + 2)\]

Setting \(f'(x) = 0\):
\[x = -1,\ -2\]

Intervals: \((-\infty, -2),\ (-2, -1),\ (-1, \infty)\)
\[\]

In \((-\infty, -2)\) and \((-1, \infty)\), \(f'(x) < 0\) ⟹ strictly decreasing
\[\]

In \((-2, -1)\), \(f'(x) > 0\) ⟹ strictly increasing

Explanation: We have:
\[f'(x) = -9 - 2x\]

Setting \(f'(x) = 0\):
\[x = -\dfrac{9}{2}\]

In \(\left(-\infty, -\dfrac{9}{2}\right)\), \(f'(x) > 0\) ⟹ strictly increasing
\[\]

In \(\left(-\dfrac{9}{2}, \infty\right)\), \(f'(x) < 0\) ⟹ strictly decreasing

Explanation: We have:
\[
f(x) = (x + 1)^3(x - 3)^3
\]

Differentiate using product rule:
\[
f'(x) = 3(x + 1)^2(x - 3)^3 + 3(x - 3)^2(x + 1)^3
\]

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

Now, set derivative equal to zero:
\[
f'(x) = 0 \Rightarrow (x + 1)^2 = 0 \text{ or } (x - 3)^2 = 0 \text{ or } (x - 1) = 0
\]
\[
\Rightarrow x = -1,\ 1,\ 3
\]

These values divide the number line into intervals:
\[
(-\infty, -1),\ (-1, 1),\ (1, 3),\ (3, \infty)
\]

Now we analyze the sign of \(f'(x)\) in each interval:
\[\]

In \((- \infty, -1)\), choose \(x = -2\):
\(f'(x) = 6(-1)^2(-5)^2(-3) < 0\)
So, \(f(x)\) is \(\textbf{strictly decreasing}\).

\[\]

In \((-1, 1)\), choose \(x = 0\):
\(f'(x) = 6(1)^2(-3)^2(-1) < 0\)
So, \(f(x)\) is \(\textbf{strictly decreasing}\).

\[\]

In \((1, 3)\), choose \(x = 2\):
\(f'(x) = 6(3)^2(-1)^2(1) > 0\)
So, \(f(x)\) is \(\textbf{strictly increasing}\).

\[\]

In \((3, \infty)\), choose \(x = 4\):
\(f'(x) = 6(5)^2(1)^2(3) > 0\)
So, \(f(x)\) is \(\textbf{strictly increasing}\).

\[\]

Conclusion:
\[
f(x) \text{ is strictly decreasing in } (-\infty, -1) \cup (-1, 1)
\]
\[
f(x) \text{ is strictly increasing in } (1, 3) \cup (3, \infty)
\]

Explanation: We are given:
\[
y = \log (1 + x) - \dfrac{2x}{2 + x}
\]

Differentiate with respect to \(x\):
\[
\dfrac{dy}{dx} = \dfrac{1}{1 + x} - \dfrac{(2 + x)(2) - 2x(1)}{(2 + x)^2}
= \dfrac{1}{1 + x} - \dfrac{4}{(2 + x)^2}
\]

Take LCM and simplify:
\[
\dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2}
\]

Now, consider the domain \(x > -1\). Let us examine the sign of the derivative:
\[\] The numerator \(x^2 \geq 0\), and is 0 only when \(x = 0\)
\[\] The denominator \((1 + x)(2 + x)^2 > 0\) for all \(x > -1\), since all terms are positive.

Therefore,
\[
\dfrac{dy}{dx} = \dfrac{x^2}{(1 + x)(2 + x)^2} \geq 0 \text{ for all } x > -1,
\]

\(\textbf{Conclusion:} \)
Since \(\dfrac{dy}{dx} > 0\) for all \(x > -1\), the function is \(\textbf{increasing throughout its domain}\).

Explanation: We start with:
\[
y = [x(x - 2)]^2 = \left(x^2 - 2x\right)^2
\]

Differentiate using chain rule:
\[
\frac{dy}{dx} = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x) = 2(x^2 - 2x)(2x - 2)
\]
\[
\Rightarrow \frac{dy}{dx} = 4x(x - 2)(x - 1)
\]

Now set the derivative equal to zero:
\[
\frac{dy}{dx} = 0 \Rightarrow x = 0,\ 1,\ 2
\]

These critical points divide the real number line into the intervals:
\[
(-\infty, 0),\ (0, 1),\ (1, 2),\ (2, \infty)
\]

Now test the sign of \(\frac{dy}{dx}\) in each interval:
\[\]
In \((- \infty, 0)\), choose \(x = -1\):
\[
\frac{dy}{dx} = 4(-1)(-3)(-2) = -24 < 0
\]
So, \(y\) is decreasing.
\[\]

In \((0, 1)\), choose \(x = 0.5\):
\[
\frac{dy}{dx} = 4(0.5)(-1.5)(-0.5) = 1.5 > 0
\]
So, \(y\) is increasing.
\[\]

In \((1, 2)\), choose \(x = 1.5\):
\[
\frac{dy}{dx} = 4(1.5)(-0.5)(0.5) = -1.5 < 0
\]
So, \(y\) is decreasing.
\[\]

In \((2, \infty)\), choose \(x = 3\):
\[
\frac{dy}{dx} = 4(3)(1)(2) = 24 > 0
\]
So, \(y\) is increasing.
\[\]

\(\textbf{Conclusion:}\)
The function \( y = [x(x - 2)]^2 \) is increasing in the intervals:
\[
(0, 1) \quad \text{and} \quad (2, \infty)
\]

Explanation: We are given:
\[
y = \frac{4\sin \theta}{2 + \cos \theta} - \theta
\]

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

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

Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[
= \frac{8\cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos \theta)^2} - 1 = \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} - 1
\]

Now, set the derivative equal to zero:
\[
\frac{dy}{d\theta} = 0 \Rightarrow \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} = 1
\]

Multiply both sides by \( (2 + \cos \theta)^2 \):
\[
8\cos \theta + 4 = (2 + \cos \theta)^2 = 4 + 4\cos \theta + \cos^2 \theta
\]

Simplify:
\[
8\cos \theta + 4 = 4 + 4\cos \theta + \cos^2 \theta \Rightarrow 4\cos \theta = \cos^2 \theta
\]

\[
\Rightarrow \cos \theta (4 - \cos \theta) = 0
\]

So, \( \cos \theta = 0 \) or \( \cos \theta = 4 \). Since \( \cos \theta \in [0,1] \), the only valid solution is \( \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \).

Now express the derivative in a factored form:
\[
\frac{dy}{d\theta} = \frac{8\cos \theta + 4 - (4 + 4\cos \theta + \cos^2 \theta)}{(2 + \cos \theta)^2} = \frac{4\cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2}
\]

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

Now analyze the sign of \( \frac{dy}{d\theta} \) in \( \left[0, \frac{\pi}{2} \right] \):
\[\]

In this interval, \( \cos \theta > 0 \), and \( 4 - \cos \theta > 0 \) since \( \cos \theta \le 1 \)
\[\] Also, \( (2 + \cos \theta)^2 > 0 \)

So,
\[
\frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2} > 0 \quad \text{for all } \theta \in \left(0, \frac{\pi}{2} \right)
\]

The derivative is positive throughout the open interval, and since the function is continuous on the closed interval \( \left[0, \frac{\pi}{2} \right] \), it is strictly increasing on the entire interval.

Explanation: Let \( f(x) = \log x \).

Then,
\[
f'(x) = \frac{d}{dx}(\log x) = \frac{1}{x}
\]

For \( x > 0 \), we have:
\[
f'(x) = \frac{1}{x} > 0
\]

Since the derivative is positive for all \( x > 0 \),
the function \( f(x) = \log x \) is strictly increasing on \( (0, \infty) \).

Explanation: We have:
\[
f(x) = x^2 - x + 1
\]

Differentiating with respect to \( x \), we get:
\[
f'(x) = 2x - 1
\]

Now,
\[
f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = \frac{1}{2}
\]

So, \( x = \frac{1}{2} \) divides the interval \( (-1, 1) \) into two parts:
\[
(-1, \tfrac{1}{2}) \quad \text{and} \quad (\tfrac{1}{2}, 1)
\]

In \( (-1, \tfrac{1}{2}) \), we have:
\[
f'(x) = 2x - 1 < 0 \Rightarrow f \text{ is strictly decreasing}
\]

In \( (\tfrac{1}{2}, 1) \), we have:
\[
f'(x) = 2x - 1 > 0 \Rightarrow f \text{ is strictly increasing}
\]

\(\textbf{Conclusion:} \)
Since \( f \) is decreasing in one part and increasing in another,
\[\text{the function } f(x) = x^2 - x + 1 \text{ is neither strictly increasing nor strictly decreasing on } (-1, 1).
\]

Explanation: We know a function is strictly increasing on an interval if its derivative is positive on that interval.

\[
f'(x) = \frac{d}{dx}(x^2 + ax + 1) = 2x + a
\]

We require:
\[
f'(x) > 0 \quad \text{for all } x \in (1, 2)
\]

This means:
\[
2x + a > 0 \quad \text{for all } x \in (1, 2)
\]

The minimum value of \( f'(x) = 2x + a \) in the interval \( (1, 2) \) occurs at the left endpoint \( x = 1 \).

\[
\text{So, we need: } 2(1) + a > 0 \Rightarrow 2 + a > 0 \Rightarrow a > -2
\]

Thus, the least value of \( a \) such that \( f(x) \) is strictly increasing on \( (1, 2) \) is:
\[
\boxed{-2}
\]

Explanation: We are given:
\[
f(x) = x + \frac{1}{x}
\]

Differentiate with respect to \( x \):
\[
f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2}
\]

Set \( f'(x) = 0 \):
\[
1 - \frac{1}{x^2} = 0 \Rightarrow \frac{1}{x^2} = 1 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1
\]

So, the critical points are \( x = -1 \) and \( x = 1 \), which divide the real line into intervals:
\[
(-\infty, -1),\quad (-1, 1),\quad (1, \infty)
\]

Now analyze the sign of \( f'(x) \) in each interval:


\[\] In \( (-\infty, -1) \):
\( x < -1 \Rightarrow x^2 > 1 \Rightarrow \frac{1}{x^2} < 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} > 0 \)
So, \( f \) is strictly increasing on \( (-\infty, -1) \).

\[\] In \( (-1, 1) \):
\( -1 < x < 1,\ x \neq 0 \Rightarrow x^2 < 1 \Rightarrow \frac{1}{x^2} > 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} < 0 \)
So, \( f \) is strictly decreasing on \( (-1, 1) \).

\[\] In \( (1, \infty) \):
\( x > 1 \Rightarrow x^2 > 1 \Rightarrow \frac{1}{x^2} < 1 \Rightarrow f'(x) = 1 - \frac{1}{x^2} > 0 \)
So, \( f \) is strictly increasing on \( (1, \infty) \).
\[\]

Hence, \( f(x) = x + \frac{1}{x} \) is strictly increasing on any interval \( \mathbf{I} \) such that \( \mathbf{I} \cap (-1, 1) = \emptyset \).
That is, \( f \) is strictly increasing on any interval disjoint from \( (-1, 1) \).

Explanation: We are given:
\[
f(x) = \log \sin x
\]

Differentiate with respect to \( x \):
\[
f'(x) = \frac{d}{dx} \left( \log \sin x \right) = \frac{1}{\sin x} \cdot \cos x = \cot x
\]

Now analyze the sign of \( f'(x) = \cot x \) in the given intervals:

\[\] In \( \left(0, \dfrac{\pi}{2} \right) \),
\( \cot x > 0 \), since both \( \sin x > 0 \) and \( \cos x > 0 \),
\( \Rightarrow f'(x) > 0 \)
\( \Rightarrow f \) is strictly increasing on \( \left(0, \dfrac{\pi}{2} \right) \).

\[\] In \( \left( \dfrac{\pi}{2}, \pi \right) \),
\( \cot x < 0 \), since \( \sin x > 0 \) and \( \cos x < 0 \),
\( \Rightarrow f'(x) < 0 \)
\( \Rightarrow f \) is strictly decreasing on \( \left( \dfrac{\pi}{2}, \pi \right) \).
\[\]
\(\textbf{Conclusion:}\)
\[
f(x) = \log \sin x \text{ is strictly increasing on } \left(0, \dfrac{\pi}{2} \right) \text{ and strictly decreasing on } \left( \dfrac{\pi}{2}, \pi \right).
\]

Explanation: We are given:
\[
f(x) = \log \cos x
\]

Differentiate with respect to \( x \):
\[
f'(x) = \frac{d}{dx} \left( \log \cos x \right) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x
\]

Now analyze the sign of \( f'(x) = -\tan x \) in the given intervals:

\[\] In \( \left(0, \dfrac{\pi}{2} \right) \),
\( \tan x > 0 \Rightarrow -\tan x < 0 \Rightarrow f'(x) < 0 \)
\( \Rightarrow f \) is strictly decreasing on \( \left(0, \dfrac{\pi}{2} \right) \).

\[\] In \( \left( \dfrac{\pi}{2}, \pi \right) \),
\( \tan x < 0 \Rightarrow -\tan x > 0 \Rightarrow f'(x) > 0 \)
\( \Rightarrow f \) is strictly increasing on \( \left( \dfrac{\pi}{2}, \pi \right) \).
\[\]
\(\textbf{Conclusion:}\)
\[
f(x) = \log \cos x \text{ is strictly decreasing on } \left(0, \dfrac{\pi}{2} \right) \text{ and strictly increasing on } \left( \dfrac{\pi}{2}, \pi \right).
\]

Explanation: Let
\[
f(x) = x^3 - 3x^2 + 3x
\]

Differentiate \( f(x) \) with respect to \( x \):
\[
f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x) = 3x^2 - 6x + 3
\]

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

Now, for all \( x \in \mathbb{R} \), we know:
\[
(x - 1)^2 \geq 0 \Rightarrow f'(x) = 3(x - 1)^2 \geq 0
\]

Also, \( f'(x) = 0 \) only at \( x = 1 \), and \( f'(x) > 0 \) elsewhere.
Thus, \( f'(x) \geq 0 \) for all \( x \in \mathbb{R} \), and \( f'(x) > 0 \) for all \( x \ne 1 \).
\[\]
\(\textbf{Conclusion:}\)
\(
f(x) = x^3 - 3x^2 + 3x \text{ is an increasing function on } \mathbb{R}.
\)

Explanation: We have:
\[
(2x - 1)^2 \geq 0 \text{ for all } x \in \mathbb{R}
\]
\[
f(x) = (2x - 1)^2 + 3 \geq 3 \text{ for all } x \in \mathbb{R}
\]
Minimum value occurs when \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\).
\[
\text{Minimum value of } f\left(\frac{1}{2}\right) = (2 \cdot \frac{1}{2} - 1)^2 + 3 = 3
\]
No maximum value exists since the function is unbounded above.

Explanation: Complete the square:
\[
f(x) = 9x^2 + 12x + 2 = 9\left(x^2 + \frac{4}{3}x\right) + 2
= 9\left(x + \frac{2}{3}\right)^2 - 4 + 2 = 9\left(x + \frac{2}{3}\right)^2 - 2
\]
\[
\text{Minimum value occurs at } x = -\frac{2}{3}
\]
\[
f\left(-\frac{2}{3}\right) = -2
\]
No maximum value exists since the parabola opens upwards.

Explanation: We know:
\[
(x - 1)^2 \geq 0 \Rightarrow - (x - 1)^2 \leq 0
\Rightarrow f(x) = - (x - 1)^2 + 10 \leq 10
\]
Maximum value occurs when \(x = 1\):
\[
f(1) = - (1 - 1)^2 + 10 = 10
\]
No minimum value exists since the function decreases indefinitely.

Explanation: Since \(x^3\) is strictly increasing on \(\mathbb{R}\), the function \(g(x) = x^3 + 1\) is also strictly increasing.

\[
\text{Hence, } g(x) \text{ has neither a maximum nor a minimum value on } \mathbb{R}.
\]

Explanation: Since \(|x + 2| \geq 0\) for all \(x \in \mathbb{R}\),
\[
f(x) = |x + 2| - 1 \geq -1
\]
Equality occurs when \(|x + 2| = 0 \Rightarrow x = -2\).
\[
\text{Minimum value of } f = f(-2) = | -2 + 2 | - 1 = -1
\]
There is no maximum value since \(|x + 2|\) grows unbounded as \(x \to \infty\).

Explanation: Since \(-|x + 1| \leq 0\),
\[
g(x) = -|x + 1| + 3 \leq 3
\]
Equality occurs when \(|x + 1| = 0 \Rightarrow x = -1\).
\[
\text{Maximum value of } g = g(-1) = -| -1 + 1 | + 3 = 3
\]
There is no minimum value since \(-|x + 1|\) decreases without bound.

Explanation: Since \(-1 \leq \sin(2x) \leq 1\),
\[
4 \leq \sin(2x) + 5 \leq 6
\]
\[
\text{Minimum value of } h(x) = 4, \quad \text{Maximum value of } h(x) = 6
\]

Explanation: Since \(-1 \leq \sin(4x) \leq 1\),
\[
2 \leq \sin(4x) + 3 \leq 4
\]
\[
\Rightarrow 2 \leq |\sin(4x) + 3| \leq 4
\]
\[
\text{Minimum value of } f(x) = 2, \quad \text{Maximum value of } f(x) = 4
\]

Explanation: This is a linear function defined on an open interval. \[\]As \(x \to -1\), \(h(x) \to 3\), and as \(x \to 1\), \(h(x) \to 5\). \[\]
However, neither 3 nor 5 is attained in the open interval \((-1, 1)\).
\[
\text{Therefore, } h(x) \text{ has neither a maximum nor a minimum value in } (-1, 1).
\]

Explanation: We have:
\[
f(x) = x^2 \Rightarrow f'(x) = 2x
\]
Setting \( f'(x) = 0 \Rightarrow x = 0 \)

Also,
\[
f''(x) = 2 > 0
\]
So, \( x = 0 \) is a point of local minima.

Local minimum value is:
\[
f(0) = 0
\]

There is no local maximum.

Explanation: \[
g(x) = x^3 - 3x \Rightarrow g'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)
\]
Setting \( g'(x) = 0 \Rightarrow x = \pm 1 \)

\[
g''(x) = 6x
\Rightarrow g''(1) = 6 > 0 \Rightarrow \text{Local minimum at } x = 1
\]
\[
g(1) = 1 - 3 = -2
\]
\[
g''(-1) = -6 < 0 \Rightarrow \text{Local maximum at } x = -1
\]
\[
g(-1) = -1 + 3 = 2
\]

Explanation: \[
h(x) = \sin x + \cos x \Rightarrow h'(x) = \cos x - \sin x
\]
\[
h'(x) = 0 \Rightarrow \cos x = \sin x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}
\]
\[
h''(x) = -\sin x - \cos x < 0 \Rightarrow \text{Local maximum at } x = \frac{\pi}{4}
\]
\[
h\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}
\]

Explanation: \[
f(x) = \sin x - \cos x \Rightarrow f'(x) = \cos x + \sin x
\]
\[
f'(x) = 0 \Rightarrow \cos x = -\sin x \Rightarrow \tan x = -1
\Rightarrow x = \frac{3\pi}{4}, \frac{7\pi}{4}
\]

Second derivative:
\[
f''(x) = -\sin x + \cos x
\]

At \( x = \frac{3\pi}{4} \),
\[
f''\left( \frac{3\pi}{4} \right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0
\Rightarrow \text{Local maximum}
\]
\[
f\left( \frac{3\pi}{4} \right) = \frac{1}{\sqrt{2}} - \left( -\frac{1}{\sqrt{2}} \right) = \sqrt{2}
\]

At \( x = \frac{7\pi}{4} \),
\[
f''\left( \frac{7\pi}{4} \right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} > 0
\Rightarrow \text{Local minimum}
\]
\[
f\left( \frac{7\pi}{4} \right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}
\]

Explanation: \[
f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)
\Rightarrow f'(x) = 0 \Rightarrow x = 1, 3
\]
\[
f''(x) = 6x - 12
\Rightarrow f''(1) = -6 < 0 \Rightarrow \text{Local maximum}
\Rightarrow f(1) = 1 - 6 + 9 + 15 = 19
\]
\[
f''(3) = 6 > 0 \Rightarrow \text{Local minimum}
\Rightarrow f(3) = 27 - 54 + 27 + 15 = 15
\]

Explanation: \[
g'(x) = \frac{1}{2} - \frac{2}{x^2}
\Rightarrow g'(x) = 0 \Rightarrow \frac{1}{2} = \frac{2}{x^2}
\Rightarrow x^2 = 4 \Rightarrow x = 2 \, (\text{since } x > 0)
\]
\[
g''(x) = \frac{4}{x^3}
\Rightarrow g''(2) = \frac{4}{8} = \frac{1}{2} > 0 \Rightarrow \text{Local minimum}
\Rightarrow g(2) = 1 + 1 = 2
\]

Explanation: \[
g'(x) = \frac{-2x}{(x^2 + 2)^2}
\Rightarrow g'(x) = 0 \Rightarrow x = 0
\]

Check sign change:
- For \( x < 0 \), \( g'(x) > 0 \)
- For \( x > 0 \), \( g'(x) < 0 \)

So by first derivative test:
\[
x = 0 \text{ is a local maximum}
\Rightarrow g(0) = \frac{1}{2}
\]

Explanation: Let us differentiate:
\[
f(x) = x \sqrt{1 - x} \Rightarrow f'(x) = \sqrt{1 - x} + x \cdot \left( \frac{-1}{2\sqrt{1 - x}} \right)
= \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}}
\]

Set \( f'(x) = 0 \Rightarrow 2 - 3x = 0 \Rightarrow x = \frac{2}{3} \)

Now second derivative:
\[
f''(x) = \frac{d}{dx} \left( \frac{2 - 3x}{2\sqrt{1 - x}} \right)
= \frac{3x - 4}{4(1 - x)^{3/2}}
\Rightarrow f''\left( \frac{2}{3} \right) = \frac{2 - 4}{2(1/3)^{3/2}} < 0
\Rightarrow \text{Local maximum}
\]
\[
f\left( \frac{2}{3} \right) = \frac{2}{3} \cdot \sqrt{\frac{1}{3}} = \frac{2\sqrt{3}}{9}
\]

Explanation: We have,
\[
f(x) = e^x \Rightarrow f'(x) = e^x
\]
Since \( e^x > 0 \) for all \( x \in \mathbb{R} \), \( f'(x) \neq 0 \) for any real \( x \).
Therefore, there is no critical point \( c \in \mathbb{R} \) such that \( f'(c) = 0 \).
Hence, the function \( f(x) = e^x \) does not have any maximum or minimum.

Explanation: We have,
\[
g(x) = \log x \Rightarrow g'(x) = \frac{1}{x}
\]
The function \( \log x \) is defined for \( x > 0 \), and for all such \( x \),
\[
g'(x) = \frac{1}{x} > 0
\]
So, \( g(x) \) is strictly increasing and \( g'(x) \neq 0 \) for any \( x > 0 \).
Therefore, there does not exist any \( c \in (0, \infty) \) such that \( g'(c) = 0 \).
Hence, the function \( g(x) = \log x \) does not have any maximum or minimum.

Explanation: We have,
\[
h(x) = x^3 + x^2 + x + 1 \Rightarrow h'(x) = 3x^2 + 2x + 1
\]
Now, the discriminant of \( h'(x) = 3x^2 + 2x + 1 \) is:
\[
\Delta = (2)^2 - 4(3)(1) = 4 - 12 = -8 < 0
\]
So, the derivative has no real roots.
This implies \( h'(x) \neq 0 \) for all \( x \in \mathbb{R} \).

Also, since the leading coefficient is positive, \( h'(x) > 0 \) for all \( x \in \mathbb{R} \).
Therefore, the function is strictly increasing and has no maximum or minimum.

Hence, \( h(x) = x^3 + x^2 + x + 1 \) does not have any local maxima or minima.

Explanation: We have \( f(x) = x^3 \)
\[
f'(x) = 3x^2
\]
Setting \( f'(x) = 0 \Rightarrow x = 0 \)

Now evaluate \( f \) at endpoints and critical points:
\[
f(-2) = (-2)^3 = -8 \\
f(0) = 0 \\
f(2) = 2^3 = 8
\]

\(\textbf{Absolute maximum}\) is \( f(2) = 8 \) at \( x = 2 \) and
\(\textbf{Absolute minimum}\) is \( f(-2) = -8 \) at \( x = -2 \)

Explanation: We have \( f(x) = \sin x + \cos x \)

\[
f'(x) = \cos x - \sin x \\
f'(x) = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1 \Rightarrow x = \frac{\pi}{4}
\]

Now evaluate \( f \) at endpoints and critical points:
\[
f\left( \frac{\pi}{4} \right) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \\
f(0) = \sin 0 + \cos 0 = 1 \\
f(\pi) = \sin \pi + \cos \pi = -1
\]

\(\textbf{Absolute maximum}\) is \( \sqrt{2} \) at \( x = \frac{\pi}{4} \) and \(\textbf{Absolute minimum}\) is \( -1 \) at \( x = \pi \)

Explanation: We have \( f(x) = 4x - \frac{1}{2}x^2 \)

\[
f'(x) = 4 - x \\
f'(x) = 0 \Rightarrow x = 4
\]

Now evaluate \( f \) at endpoints and critical point:
\[
f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10 \]
\[f(4) = 4(4) - \frac{1}{2}(16) = 16 - 8 = 8 \]
\[f\left( \frac{9}{2} \right) = 4 \cdot \frac{9}{2} - \frac{1}{2} \cdot \left( \frac{9}{2} \right)^2 = 18 - \frac{81}{8} = 18 - 10.125 = 7.875
\]

\(\textbf{Absolute maximum}\) is \( 8 \) at \( x = 4 \) and
\(\textbf{Absolute minimum}\) is \( -10 \) at \( x = -2 \)

Explanation: We have \( f(x) = (x - 1)^2 + 3 \)

\[
f'(x) = 2(x - 1) \Rightarrow f'(x) = 0 \Rightarrow x = 1
\]

Now evaluate \( f \) at endpoints and critical point:
\[
f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19 \\
f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3
\]

\(\textbf{Absolute maximum}\) is \( 19 \) at \( x = -3 \) and \(\textbf{Absolute minimum}\) is \( 3 \) at \( x = 1 \)

Explanation: We have:
\[
p(x) = 41 - 24x - 18x^2
\Rightarrow p'(x) = -24 - 36x
\Rightarrow p''(x) = -36 < 0
\]

Set derivative to zero:
\[
p'(x) = 0 \Rightarrow -24 - 36x = 0 \Rightarrow x = -\frac{2}{3}
\]

By second derivative test, local maximum occurs at \( x = -\frac{2}{3} \)

Now, compute:
\[
p\left( -\frac{2}{3} \right) = 41 - 24\left( -\frac{2}{3} \right) - 18\left( -\frac{2}{3} \right)^2
= 41 + 16 - 8 = 49
\]
\(\textbf{Maximum profit}\) = 49

Explanation: \[
f'(x) = 3x^2 - \frac{3}{x^4} = \frac{3x^6 - 3}{x^4}
\]

Set derivative to zero:
\[
f'(x) = 0 \Rightarrow x^6 = 1 \Rightarrow x = \pm 1
\]

Sign of derivative:
\[\] \( f'(x) > 0 \) in \( (-\infty, -1) \cup (1, \infty) \) ⟹ Increasing
\[\] \( f'(x) < 0 \) in \( (-1, 1) \) ⟹ Decreasing

Explanation: We are given:
\[
f(x) = \sin 2x
\Rightarrow f'(x) = 2\cos 2x
\]

Set the derivative to zero:
\[
f'(x) = 0 \Rightarrow \cos 2x = 0
\Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}
\Rightarrow x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
\]

Now evaluate \( f(x) \) at these critical points:
\[
f\left( \frac{\pi}{4} \right) = \sin\left( \frac{\pi}{2} \right) = 1 \]
\[f\left( \frac{3\pi}{4} \right) = \sin\left( \frac{3\pi}{2} \right) = -1 \]
\[f\left( \frac{5\pi}{4} \right) = \sin\left( \frac{5\pi}{2} \right) = 1 \]
\[f\left( \frac{7\pi}{4} \right) = \sin\left( \frac{7\pi}{2} \right) = -1
\]

Also check endpoints:
\[
f(0) = \sin 0 = 0, \quad f(2\pi) = \sin(4\pi) = 0
\]

Thus, the absolute maximum value of \( f(x) \) on \([0, 2\pi]\) is
\(
\boxed{1}
\quad \text{at} \quad x = \frac{\pi}{4} \text{ and } x = \frac{5\pi}{4}
\)

Explanation: Let
\[
f(x) = \sin x + \cos x
\]

Differentiate:
\[
f'(x) = \cos x - \sin x
\]

Set \( f'(x) = 0 \):
\[
\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1
\Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4}, \ldots
\]

Now compute the second derivative:
\[
f''(x) = -\sin x - \cos x = -(\sin x + \cos x)
\]

To determine if it's a maximum, evaluate at \( x = \frac{\pi}{4} \):
\[
f''\left( \frac{\pi}{4} \right) = -\left( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \right) = -\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) = -\sqrt{2} < 0
\]

Since \( f''(x) < 0 \), this is a local maximum.

Now compute the value:
\[
f\left( \frac{\pi}{4} \right) = \sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}
\]

\[
\boxed{\text{Maximum value of } \sin x + \cos x \text{ is } \sqrt{2}, \text{ at } x = \frac{\pi}{4}}
\]

Explanation: Let
\[
f(x) = 2x^3 - 24x + 107
\]

Differentiate:
\[
f'(x) = 6x^2 - 24 = 6(x^2 - 4)
\]

Set \( f'(x) = 0 \):
\[
x^2 - 4 = 0 \Rightarrow x = \pm 2
\]



For the Interval \([1, 3]\):

Evaluate the function at:
\(x = 1\):
\[
f(1) = 2(1)^3 - 24(1) + 107 = 2 - 24 + 107 = 85
\]

and at \(x = 2\):
\[
f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75
\]

At \(x = 3\):
\[
f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89
\]

\[
\boxed{\text{Maximum value in } [1,3] \text{ is } 89 \text{ at } x = 3}
\]



For the Interval \([-3, -1]\):

Evaluate the function at:
\(x = -3\):
\[
f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125
\]

and at \(x = -2\):
\[
f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139
\]

and at \(x = -1\):
\[
f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129
\]

\[
\boxed{\text{Maximum value in } [-3, -1] \text{ is } 139 \text{ at } x = -2}
\]

Explanation: We are given:
\[
f(x) = x^4 - 62x^2 + ax + 9
\]

Differentiate:
\[
f'(x) = 4x^3 - 124x + a
\]

Since \(f(x)\) has a maximum at \(x = 1\), we have:
\[
f'(1) = 0
\]

Substitute \(x = 1\):
\[
f'(1) = 4(1)^3 - 124(1) + a = 4 - 124 + a = 0
\]

Solve for \(a\):
\[
a = 120
\]

\[
\boxed{a = 120}
\]

Explanation: Let
\[
f(x) = x + \sin 2x
\]

Differentiate:
\[
f'(x) = 1 + 2\cos 2x
\]

Set derivative to zero for critical points:
\[
f'(x) = 0 \Rightarrow 1 + 2\cos 2x = 0 \Rightarrow \cos 2x = -\frac{1}{2}
\]

Now solve \( \cos 2x = -\frac{1}{2} \):
\[
2x = \frac{2\pi}{3}, \frac{4\pi}{3} \Rightarrow x = \frac{\pi}{3}, \frac{2\pi}{3}
\]

Also in the interval \([0, 2\pi]\), the values of \(x\) satisfying \( \cos 2x = -\frac{1}{2} \) are:
\[
x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
\]

Evaluate \(f(x)\) at critical points and endpoints:
\[
f\left(0\right) = 0 + \sin 0 = 0
\]
\[
f\left(2\pi\right) = 2\pi + \sin 4\pi = 2\pi + 0 = 2\pi
\]
\[
f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \sin\left(\frac{2\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}
\]
\[
f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + \sin\left(\frac{4\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}
\]
\[
f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \sin\left(\frac{8\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}
\]
\[
f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} + \sin\left(\frac{10\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}
\]

Compare all values. The smallest is at \(x = 0\), and the largest is at \(x = 2\pi\).

\[
\boxed{
\text{Maximum value of } f(x) \text{ on } [0, 2\pi] \text{ is } 2\pi \text{ at } x = 2\pi
}
\]
\[
\boxed{
\text{Minimum value of } f(x) \text{ on } [0, 2\pi] \text{ is } 0 \text{ at } x = 0
}
\]

Explanation: Let one of the numbers be \( x \).
Then, the other number is \( 24 - x \).

Let the product of the two numbers be:
\[
P(x) = x(24 - x) = 24x - x^2
\]

Differentiate \( P(x) \):
\[
P'(x) = 24 - 2x, \quad P''(x) = -2
\]

Set the first derivative to zero for critical points:
\[
P'(x) = 0 \Rightarrow 24 - 2x = 0 \Rightarrow x = 12
\]

Check the second derivative:
\[
P''(12) = -2 < 0
\]

Since \( P''(x) < 0 \), the function attains a **maximum** at \( x = 12 \).

Therefore, the two numbers are:
\[
x = 12, \quad 24 - x = 12
\]


The two numbers are 12 and 12. The product is maximum when both are equal.

Explanation: We are given:
\[
x + y = 60 \Rightarrow y = 60 - x
\]

We want to maximize:
\[
f(x) = x y^3 = x(60 - x)^3
\]

Differentiate using the product rule:
\[
f'(x) = \frac{d}{dx} \left[x(60 - x)^3\right] = (60 - x)^3 + x \cdot 3(60 - x)^2 \cdot (-1)
\]
\[
f'(x) = (60 - x)^2 \left[(60 - x) - 3x\right] = (60 - x)^2(60 - 4x)
\]

Set derivative to zero:
\[
f'(x) = 0 \Rightarrow (60 - x)^2(60 - 4x) = 0
\]
\[
\Rightarrow x = 60 \text{ or } x = 15
\]

Check second derivative at \( x = 15 \):

\[
f''(x) = \frac{d}{dx}\left[ (60 - x)^2(60 - 4x) \right]
\]
Using the product rule:
\[
f''(x) = 2(60 - x)(-1)(60 - 4x) + (60 - x)^2(-4)
\]
\[
f''(x) = -2(60 - x)(60 - 4x) - 4(60 - x)^2
\]
\[
f''(x) = -2(60 - x)[(60 - 4x) + 2(60 - x)] = -2(60 - x)(60 - 4x + 2(60 - x))
\]
\[
= -2(60 - x)(180 - 6x) = -12(60 - x)(30 - x)
\]

Now evaluate at \( x = 15 \):
\[
f''(15) = -12(60 - 15)(30 - 15) = -12(45)(15) < 0
\]

Since \( f''(15) < 0 \), \( x = 15 \) is a point of **local maximum**.

Hence,
\[
x = 15, \quad y = 60 - 15 = 45
\]



The two positive numbers are 15 and 45.

Explanation: We are given:
\[
x + y = 35 \Rightarrow y = 35 - x
\]

We want to maximize:
\[
P(x) = x^2 (35 - x)^5
\]

Differentiate using the product rule:
\[
P'(x) = \frac{d}{dx}\left[ x^2 (35 - x)^5 \right] = 2x(35 - x)^5 + x^2 \cdot 5(35 - x)^4 \cdot (-1)
\]
\[
P'(x) = 2x(35 - x)^5 - 5x^2(35 - x)^4
\]
Factor common terms:
\[
P'(x) = x(35 - x)^4 [2(35 - x) - 5x]
\]
\[
= x(35 - x)^4 (70 - 2x - 5x) = x(35 - x)^4 (70 - 7x)
\]
\[
= 7x(35 - x)^4 (10 - x)
\]

Set derivative to zero:
\[
P'(x) = 0 \Rightarrow x = 0,\ 35,\ \text{or}\ 10
\]

Check valid values:
\[ x = 0 \Rightarrow y = 35 \Rightarrow P = 0 \]
\[ x = 35 \Rightarrow y = 0 \Rightarrow P = 0 \]
\[ x = 10 \Rightarrow y = 25 \] Hence valid critical point.\[\]

Now compute second derivative to verify maximum:
\[
P''(x) = \frac{d}{dx} \left[ 7x(35 - x)^4(10 - x) \right]
\]
Use product rule:
Let \( u = x,\ v = (35 - x)^4(10 - x) \), then
\[
P''(x) = 7[u'v + uv']
\]

Compute:
\[
u' = 1,\quad v = (35 - x)^4(10 - x)
\]
\[
v' = \frac{d}{dx}[(35 - x)^4(10 - x)]
= (35 - x)^3[-4(10 - x) + (35 - x)(-1)]
\]
\[
= (35 - x)^3 [-4(10 - x) - (35 - x)] = (35 - x)^3 [-40 + 4x - 35 + x] = (35 - x)^3(-75 + 5x)
\]

So:
\[
P''(x) = 7[(35 - x)^4(10 - x) + x(35 - x)^3(-75 + 5x)]
\]

Evaluate at \( x = 10 \):
\[
(35 - 10) = 25,\quad (10 - 10) = 0\]
\[\Rightarrow P''(10) = 7[25^4 \cdot 0 + 10 \cdot 25^3(-75 + 50)] = 7 \cdot 10 \cdot 25^3(-25)
\]
\[
= -7 \cdot 10 \cdot 25^3 \cdot 25 < 0
\]

Since \( P''(10) < 0 \), the function attains a \({\bf maximum}\) at \( x = 10 \).

Thus, the required numbers are:
\[
\boxed{x = 10,\quad y = 35-x=25}
\]

Explanation: Let one number be \( x \).
Then the other number is \( 16 - x \).
We want to minimize the sum of their cubes:
\[
S(x) = x^3 + (16 - x)^3
\]

Differentiate:
\[
S'(x) = 3x^2 - 3(16 - x)^2
\]

Set \( S'(x) = 0 \) for critical points:
\[
3x^2 = 3(16 - x)^2 \Rightarrow x^2 = (16 - x)^2
\]

Expand:
\[
x^2 = 256 - 32x + x^2 \Rightarrow -32x + 256 = 0
\Rightarrow x = \frac{256}{32} = 8
\]

Now check second derivative:
\[
S''(x) = 6x + 6(16 - x) = 6x + 96 - 6x = 96 > 0
\]

Since \( S''(8) > 0 \), this confirms a local minimum.

\[\] Final Answer:
\[
\boxed{\text{The two numbers are } 8 \text{ and } 16-8=8.}
\]

Explanation: Let the side of the square cut off from each corner be \( x \,\text{cm} \).
\[\]
Then,
the length and breadth of the resulting box will each be \( (18 - 2x)\,\text{cm} \),
\[\] and the height will be \( x \,\text{cm} \).
\[\]
So, the volume \( V(x) \) of the box is:
\[
V(x) = x(18 - 2x)^2
\]

Differentiate:
\[
V'(x) = (18 - 2x)^2 + x \cdot 2(18 - 2x)(-2)
= (18 - 2x)^2 - 4x(18 - 2x)
\]

Factor:
\[
V'(x) = (18 - 2x)[(18 - 2x) - 4x]
= (18 - 2x)(18 - 6x)
\]

Set derivative to zero:
\[
V'(x) = 0 \Rightarrow (18 - 2x)(18 - 6x) = 0
\Rightarrow x = 9 \text{ or } x = 3
\]

If \( x = 9 \), then \( 18 - 2x = 0 \), so length and breadth become zero.
This gives zero volume which is not acceptable.
\[\]

So, \( x = 3 \) is a critical point.
\[\]

Check second derivative:
\[
V''(x) = \frac{d}{dx}\left[(18 - 2x)(18 - 6x)\right]
= -2(18 - 6x) + (18 - 2x)(-6)
= -2(18 - 6x) - 6(18 - 2x)
\]

Evaluate at \( x = 3 \):
\[
V''(3) = -2(18 - 18) - 6(18 - 6) = -6 \cdot 12 = -72 < 0
\]

Since \( V''(3) < 0 \), the volume is \(\bf maximum\) at \( x = 3 \).

\[\]

Final Answer:
\[
\boxed{\text{The side of the square to be cut off is } 3\,\text{cm.}}
\]

Explanation: Let the side of each square cut off be \( x\,\text{cm} \).

- After folding, the height of the box = \( x \)
- Length of the box = \( 45 - 2x \)
- Breadth of the box = \( 24 - 2x \)

So, the volume \( V(x) \) of the box is:
\[
V(x) = x(45 - 2x)(24 - 2x)
\]

Expand:
\[
V(x) = x[(45)(24) - 2x(24) - 2x(45) + 4x^2]
= x[1080 - 48x - 90x + 4x^2]
= x[1080 - 138x + 4x^2]
= 4x^3 - 138x^2 + 1080x
\]

Differentiate:
\[
V'(x) = 12x^2 - 276x + 1080
\]

Factor:
\[
V'(x) = 12(x^2 - 23x + 90)
= 12(x - 18)(x - 5)
\]

Set derivative to zero:
\[
V'(x) = 0 \Rightarrow x = 18 \text{ or } x = 5
\]

Now, \( x = 18 \) is not valid since:
- \( 2x = 36 > 24 \) → length or breadth becomes negative or zero.

So, only possible value is \( x = 5 \).

Check second derivative:
\[
V''(x) = \frac{d}{dx}(12x^2 - 276x + 1080) = 24x - 276
\]

Evaluate at \( x = 5 \):
\[
V''(5) = 24(5) - 276 = 120 - 276 = -156 < 0
\]

So, \( x = 5 \) is a point of **local maximum**.

---

### **Final Answer**:
\[
\boxed{\text{The side of the square to be cut off is } 5\,\text{cm.}}
\]