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$$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1)$$$$\cos 2x - \cos 2\alpha = 2\cos^2 x - 2\cos^2 \alpha$$$$\cos 2x - \cos 2\alpha = 2(\cos^2 x - \cos^2 \alpha)$$ $$=2(\cos x - \cos \alpha)(\cos x + \cos \alpha)$$ 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 \[2x - 4 > 0\]\[2x > 4\]\[\mathbf{x > 2}\]

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:** 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 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 $\boxed{B}$.

The solution proceeds as follows: **Step 1: Find the derivative of the function.** The derivative of \(f(x) = kx - \sin~x\) is \(f'(x) = k - \cos~x\). **Step 2: Determine the condition for a strictly increasing function.** For a function to be strictly increasing, its derivative must be strictly positive, i.e., \(f'(x) > 0\). **Step 3: Apply the condition to the derivative.** We need \(k - \cos~x > 0\) for all \(x\). This means \(k > \cos~x\) for all \(x\). **Step 4: Find the maximum value of $\cos~x$.** The maximum value of $\cos~x$ is 1. **Step 5: Determine the condition for $k$.** For \(k > \cos~x\) to hold for all \(x\), \(k\) must be greater than the maximum value of $\cos~x$. Therefore, \(k > 1\). **Final Answer:** The function \(f(x)=kx-\sin~x\) is strictly increasing for \(k\gt1\). The final answer is $\boxed{k\gt1}$.

To determine the correct statement about the function $f(x) = -2x^8$, we can analyze its first derivative to find critical points and intervals of increase/decrease. 1. **Calculate the first derivative:** $$f'(x) = \frac{d}{dx}(-2x^8) = -16x^7$$ 2. **Find critical points:** Set $f'(x) = 0$: $$-16x^7 = 0 \implies x^7 = 0 \implies x = 0$$ The only critical point is $x=0$. 3. **Analyze the sign of $f'(x)$:** * For $x < 0$, $x^7 < 0$, so $f'(x) = -16(\text{negative number}) > 0$. This means $f(x)$ is increasing for $x < 0$. * For $x > 0$, $x^7 > 0$, so $f'(x) = -16(\text{positive number}) < 0$. This means $f(x)$ is decreasing for $x > 0$. 4. **Identify the extremum:** Since $f'(x)$ changes from positive to negative at $x=0$, there is a local maximum at $x=0$. The value of the function at this point is $f(0) = -2(0)^8 = 0$. Furthermore, for any $x \neq 0$, $x^8 > 0$, so $-2x^8 < 0$. Since $f(0) = 0$ and $f(x) < 0$ for all other $x$, $(0,0)$ is a global maximum. The correct statement is: The function $f(x)$ has a global maximum at $x=0$.

Every differentiable function is continuous

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