Step 1: Understand the Domain of a Square Root Function The function \( f(x) = \sqrt{7 - 8x + x^2} \) is defined only when the expression inside the square root is non-negative. That is: \[ 7 - 8x + x^2 \geq 0. \] Step 2: Rewrite the Inequality Rewrite the inequality: \[ x^2 - 8x + 7 \geq 0. \] Step 3: Factor the Quadratic Expression Factor the quadratic expression: \[ x^2 - 8x + 7 = (x - 1)(x - 7). \] So, the inequality becomes: \[ (x - 1)(x - 7) \geq 0. \] Step 4: Solve the Inequality To solve \( (x - 1)(x - 7) \geq 0 \), we determine the intervals where the product is non-negative. The critical points are \( x = 1 \) and \( x = 7 \).
These divide the number line into three intervals:
1. \( x<1 \): Choose \( x = 0 \). Substituting into \( (x - 1)(x - 7) \): \[ (0 - 1)(0 - 7) = (-1)(-7) = 7>0. \] The product is positive in this interval.
2. \( 1<x<7 \): Choose \( x = 4 \). Substituting into \( (x - 1)(x - 7) \): \[ (4 - 1)(4 - 7) = (3)(-3) = -9<0. \] The product is negative in this interval.
3. \( x>7 \): Choose \( x = 8 \). Substituting into \( (x - 1)(x - 7) \): \[ (8 - 1)(8 - 7) = (7)(1) = 7>0. \] The product is positive in this interval.
Step 5: Include the Critical Points At \( x = 1 \) and \( x = 7 \), the expression equals zero, which satisfies the inequality \( \geq 0 \).
Therefore, the critical points are included in the solution.
Step 6: Write the Domain Combining the intervals where the product is non-negative, the domain of \( f(x) \) is: \[ (-\infty, 1] \cup [7, \infty). \]
Step 7: Verify the Answer The domain \( (-\infty, 1] \cup [7, \infty) \) corresponds to option (B).
Final Answer: The domain of \( f(x) \) is: \[ \boxed{(-\infty, 1] \cup [7, \infty)}. \]
Thus, the correct option is (B).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively:
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to: