Some Properties Of Definite Integrals

Step 1: Understand the behavior of the function.
The function is an exponential function with base , where . Since the base is greater than 1, this function is decreasing as increases. The rate of decrease becomes slower as increases.

Step 2: Conclusion.
Thus, the function is decreasing, corresponding to option (A).

Step 1: Integrate the expression.
Step 2: Apply the limits.
Step 3: Simplify.
Step 4: Solve for .
Step 5: Conclusion.
The value of is .

Step 1: Set up the constraints.
We are given a set of linear constraints and we need to maximize the objective function . We first graph the constraints and identify the feasible region.

Step 2: Check the corner points.
The maximum value of occurs at one of the corner points of the feasible region. After evaluating the objective function at each corner point, we find that the maximum value of is 36.

Step 3: Conclusion.
Thus, the maximum value of is 36, corresponding to option (C).

Step 1: Simplify the integrand.
The denominator can be factorized as Thus, the integrand becomes Step 2: Use symmetry property.
Let Replace by : Step 3: Add the two integrals.
Step 4: Evaluate the remaining integral.
Let , then . Step 5: Final calculation.

Step 1: Identify the nature of the integrand.
The numerator is an odd function, while the denominator is an even function.
Step 2: Determine the parity of the integrand.
An odd function divided by an even function remains an odd function. Hence, is an odd function.
Step 3: Use the property of definite integrals.
For any odd function , Step 4: Apply the property.
Step 5: Conclusion.
Hence, the value of the given integral is .

Step 1: Substituting and simplifying the integral.
We need to evaluate the integral . Let , so that . The integral becomes: Simplifying this, we find:
Step 2: Conclusion.
Thus, the value of the integral is , which makes option (D) the correct answer.

Step 1: Analyze the integrals.
The integral involves the difference of two functions, and . Both functions are periodic with a period of , and over the interval , their values cancel out.
Step 2: Conclusion.
The integral evaluates to 0. Hence, the correct answer is (B) 0.

Step 1: Use the given conditions.
We are given and , so we can substitute these into the equation to find the values of and .
Step 2: Find .
Now that we know , we calculate : Now, calculate :
Step 3: Conclusion.
The correct answer is 1.

Step 1: Understand the formula for volume of a parallelepiped.
The volume of a parallelepiped is given by the scalar triple product . We are told that the volume of the parallelepiped is 12.

Step 2: Use the relation for the volume of the tetrahedron.
The volume of the tetrahedron is of the volume of the parallelepiped. Given the volume of the parallelepiped is 12, the volume of the tetrahedron is .

Step 3: Conclusion.
Thus, the volume of the tetrahedron is 4 units³, corresponding to option (A).

Step 1: Understanding the integral.
We are asked to evaluate the definite integral: Notice that the two integrand terms are symmetric in nature. Step 2: Symmetry of the integrand.
The terms and are symmetric about the origin. This suggests that the integral will evaluate to 0, as the contributions from and will cancel each other out. Step 3: Conclusion.
Thus, the value of the integral is .

Concept:
Volume of tetrahedron: Scalar projection: Step 1: Use given data.
Step 2: Use projection.
Step 3: Solve.
Step 4: Check scaling. Since projection uses magnitude relation:
Step 5: Conclusion.

Step 1: Identify Curve
and .
.
This is a circle with center and radius 9.
Step 2: Geometric Property
The normal to any point on a circle always passes through the center of that circle.
Step 3: Conclusion
The center is . Therefore, the normal always passes through .
Final Answer: (A)

Concept:
Distance from point to line: Step 1: Find AP.
Step 2: Cross product.
Step 3: Conclusion. Distance =

Concept: For definite integrals of the form we often use the property This symmetry property is very useful for integrals containing trigonometric functions.

Step 1: Let Using the substitution :

Step 2: Add the two integrals.

Step 3: Compute the value of .

Concept: For definite integrals of the form we use the property Also,
Step 1: {Let the given integral be .}
Step 2: {Apply the property .}
Step 3: {Add the two expressions.} Let . Then Thus,
Step 4: {Evaluate the integral.} Hence, the correct option is (B).