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Let \(u = \frac{1}{x} = x^{-1}\)

Differentiating \(u\) with respect to \(x\) : \( du = -\frac{1}{x^2} dx\)

Substitute \(u\) and \(du\) into the integral:

Using the standard integral formula \(\displaystyle \int a^u du = \frac{a^u}{\ln a} + C\) (with \(a=2\)):

Hence

Determining the Value of \(k\)

By comparing our result with the given form \(\displaystyle k \cdot 2^{\frac{1}{x}} + C\):

The value of \(k\) is:

Since \(\log 2\) in the options typically denotes the natural logarithm \(\ln 2\) in calculus, the answer is:

\(\displaystyle \frac{-1}{\log 2}\)

$$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1)$$$$\cos 2x - \cos 2\alpha = 2\cos^2 x - 2\cos^2 \alpha$$$$\cos 2x - \cos 2\alpha = 2(\cos^2 x - \cos^2 \alpha)$$ $$=2(\cos x - \cos \alpha)(\cos x + \cos \alpha)$$ Hence $$\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha} = \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x-\cos \alpha}$$ $$ = 2(\cos x + \cos \alpha)$$ Hence integral becomes:$$I = \int 2(\cos x + \cos \alpha) dx$$ $$I = \mathbf{2(\sin x + x \cos \alpha) + C}$$

\[f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3\] Set the derivative equal to zero to find the critical points:\[3x^2 - 3 = 0\]\[3(x^2 - 1) = 0\]\[x^2 = 1\]\[x = \pm 1\]The critical points are \(x = -1\) and \(x = 1\). Since the interval is \([0, 2]\), we only consider the critical point \(x = 1\), as \(x = -1\) is outside the interval.2. Evaluate Function at Critical Points and Endpoints

📍The absolute maximum value of a continuous function on a closed interval must occur at a critical point within the interval or at the endpoints of the interval.

We evaluate \(f(x)\) at \(x=1\), \(x=0\), and \(x=2\).At the critical point \(x=1\):\[f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = \mathbf{0}\]At the left endpoint \(x=0\):\[f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = \mathbf{2}\]At the right endpoint \(x=2\):\[f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = \mathbf{4}\]

The largest value is 4.Therefore, the absolute maximum value of the function \(f(x) = x^3 - 3x + 2\) in the interval \([0, 2]\) is \(\mathbf{4}\).

\[f(x) = x^2 - 4x + 6\] \[f'(x) = 2x - 4\] For Critical Point set \(f'(x) = 0\) implies \[2x - 4 = 0 \implies 2x = 4 \implies x = 2\] The function is increasing when \(f'(x) > 0\) implies \[2x - 4 > 0\]\[2x > 4\]\[\mathbf{x > 2}\]

The rate at which the height of sugar inside the cylindrical tank increases can be determined using the formula for the volume of a cylinder:

\(V = \pi r^2 h\)

Given:

• Radius, \(r = 10 \text{ cm}\)
• Rate of change of volume, \(\dfrac{dV}{dt} = 100\pi \text{ cm}^3/\text{s}\)

Since the radius remains constant, differentiate both sides of the volume equation with respect to time \(t\):

\(\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}\)

Substitute the known values:

\(100\pi = \pi (10)^2 \dfrac{dh}{dt}\)

\(100\pi = 100\pi \dfrac{dh}{dt}\)

Dividing both sides by \(100\pi\):

\(\dfrac{dh}{dt} = 1 \text{ cm/s}\)

Therefore, the height of the sugar in the tank is increasing at a rate of \(1 \text{ cm/s}\).

To determine the values of \(\lambda \) for which the function \(f(x)=\sin x-\cos x-\lambda x+C\) is always decreasing, we need to find the values of \(\lambda \) for which the derivative of \(f(x)\) is less than or equal to zero for all real \(x\).

The derivative of \(f(x)\) with respect to \(x\) is:

\(f^{\prime }(x)=\cos x+\sin x-\lambda \)

For \(f(x)\) to be a decreasing function for all real values of \(x\), we must have:\(f^{\prime }(x)\le 0\)\(\cos x+\sin x-\lambda \le 0\) that is

\(\cos x+\sin x\le \lambda \)

Note that for an expression of the form \(a\cos x+b\sin x\), the maximum value is \(\sqrt{a^{2}+b^{2}}\).

Here, \(a=1\) and \(b=1\). So the maximum value of \(\cos x+\sin x\) is:\(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)

Now the condition \(\cos x+\sin x\le \lambda \) for all real \(x\).implies that \(\lambda \) must be greater than or equal to the maximum possible value of \(\cos x+\sin x\).

Therefore, we must have:\(\lambda \ge \sqrt{2}\)

We differentiate $f(x)$ with respect to $x$:$$f'(x) = 2 - \sin x$$ We know that the range of the sine function is $-1 \le \sin x \le 1$. This means that $f'(x)$ is always positive ($f'(x) > 0$) for all real values of $x$.

Since the first derivative $f'(x)$ is strictly positive for all $x$, the function $f(x) = 2x + \cos x$ is an increasing function throughout its domain. This also rules out (A) and (B) because an increasing function has no local maxima or minima.

**Correct Option:** A **Reasoning:** * First derivative: \( \frac{dy}{dx} = -3x^2 + 6x + 8 \) * Set second derivative to zero: \( \frac{d^2y}{dx^2} = -6x + 6 = 0 \implies x = 1 \) * Substitute \( x=1 \) into original equation: \( y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -10 \). The point is (1, -10).

**Correct Option:** D **Reasoning:** * \(f(x)=|x|\) has \(f(x) \ge 0\) and \(f(0)=0\), so minimum at \(x=0\). * \( \lim_{x \to \pm\infty} |x| = \infty \), so no maximum value. * \( \lim_{x \to 0} |x| = 0 = f(0) \), so continuous at \(x=0\). * LHD at \(x=0\) is -1, RHD at \(x=0\) is 1. Not differentiable at \(x=0\).

The diameter of the spherical ball is given by $D = \frac{5}{2}(3x+1)$. The radius of the ball is $r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Substitute the expression for $r$: $V = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)\right)^3$ $V = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3\right)$ $V = \frac{125\pi}{48}(3x+1)^3$ To find the rate of change of its volume with respect to $x$, we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}\left(\frac{125\pi}{48}(3x+1)^3\right)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^{3-1} \cdot \frac{d}{dx}(3x+1)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3$ $\frac{dV}{dx} = \frac{125\pi \cdot 9}{48}(3x+1)^2$ $\frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2$ $\frac{dV}{dx} = \frac{375\pi}{16}(3x+1)^2$ Now, we evaluate this rate of change when $x=1$: $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(4)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16} \cdot 16$ $\left.\frac{dV}{dx}\right|_{x=1} = 375\pi$ The final answer is $\boxed{\text{375\pi}}$.

The solution is as follows: **Step 1:** Recall the relationship between the derivative of a function and its increasing/decreasing behavior. A function is strictly increasing on an interval if its derivative is strictly positive on that interval. **Step 2:** Analyze the given options in light of this relationship. Option A states \(f^{\prime}(x) \lt 0\), which implies the function is strictly decreasing. Option B states \(f^{\prime}(x) \gt 0\), which implies the function is strictly increasing. Option C states \(f^{\prime}(x) = 0\), which implies the function is constant. Option D is about the function's value, not its rate of change. **Step 3:** Conclude based on the definition of a strictly increasing function. For \(f(x)\) to be strictly increasing in \((a, b)\), its derivative \(f^{\prime}(x)\) must be greater than 0 for all \(x\) in \((a, b)\). The final answer is $\boxed{B}$.

Here's the step-by-step solution: 1. **Find the first derivative of the function:** $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 12x - 18) = 3x^2 - 6x + 12$. 2. **Analyze the sign of the first derivative:** The derivative is a quadratic function. To determine its sign, we can find its discriminant. The discriminant of $ax^2 + bx + c$ is $b^2 - 4ac$. For $3x^2 - 6x + 12$, the discriminant is $(-6)^2 - 4(3)(12) = 36 - 144 = -108$. 3. **Interpret the discriminant:** Since the discriminant is negative and the leading coefficient (3) is positive, the quadratic $f'(x) = 3x^2 - 6x + 12$ is always positive for all real values of $x$. 4. **Conclude the function's behavior:** Because $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing on $\mathbb{R}$. The final answer is $\boxed{B}$.