Area Of The Region Bounded

We are given the function .
First, we check the continuity at the points where and . At , the function is continuous because the left and right limits match.
However, the function is not differentiable at because the slope changes abruptly from negative to positive. At , the function is continuous as both left and right limits match.
However, the function is not differentiable at due to an abrupt change in slope.
Thus, the function is continuous but not differentiable at both points and .

Step 1: Recall the formula for cross product magnitudes
The magnitude of the cross product is:
Step 2: Evaluate each term
For , , and , the contributions along two directions add up, giving:
Step 3: Verify the options
The correct result is , matching option (B).

Step 1: Separating the Variables Rewriting the equation: Step 2: Integrating Both Sides Integrate both sides: The left-hand side becomes: The right-hand side uses the integral of : So the equation becomes: Here, is the constant of integration.
Step 3: Simplify the Expression Combine the logarithms: Exponentiate both sides to remove the logarithm: Step 4: Finding the Particular Solution We are given the condition . Substitute and into the solution: Simplify: Since , we have: Step 5: Final Solution Substitute back into the solution: Final Answer: This is the required particular solution to the given differential equation.

We are asked to find for the given function .
Step 1: Differentiate
Let . To differentiate this, we first take the natural logarithm of both sides: Now, differentiate implicitly with respect to : Using the product rule: We know that , so: Thus, we have:
Step 2: Differentiate
Let . We differentiate this using the chain rule:
Step 3: Combine the results
Since , we use the sum rule for differentiation: Substituting the results from steps 1 and 2: Thus, the derivative is:

Step 1: Simplify the expression inside
The given expression is: Using trigonometric identities, rewrite: Step 2: Transform into a single tangent function
Substituting and : Step 3: Simplify using
Since , we simplify: Conclusion: The simplest form is .

We are given the function .
To show that the function is strictly decreasing in the interval , we first need to compute the derivative of .
Step 1: Find the derivative of
The derivative of is:
Step 2: Solve for critical points by setting
To find the critical points, solve the equation :
This happens when:
Step 3: Determine the sign of in the interval
Now, examine the sign of in the interval .
- For , we know that , so .
Since in this interval, the function is strictly decreasing.
Thus, we have shown that is strictly decreasing in the interval .

Step 1: Definition of a scalar matrix
A scalar matrix is a diagonal matrix where all diagonal elements are equal. This means , and all off-diagonal elements ( , ) are zero.
Step 2: Substitute the values
Using the scalar matrix properties:
Step 3: Verify the options
The correct value is , which corresponds to option (D).

Step 1: Identify the corner points of the feasible region
From the graph, the vertices of the feasible region are: Step 2: Substitute corner points into
Evaluate at each vertex: Step 3: Find the maximum value
The maximum value of occurs at , where .
Step 4: Verify the options
The maximum value is , which corresponds to option (C).