Question
Let be defined as . If and , then the value of equals
Options
Solution
Key Concepts and Formulas
- Differentiation of Exponential Functions: The derivative of is .
- Properties of Logarithms and Exponentials: and . Also, .
- Definite Integration: The definite integral of a function from to is found by evaluating its antiderivative at the limits: .
- Solving Systems of Linear Equations: Using algebraic manipulation to find the values of unknown variables.
Step-by-Step Solution:
We are given the function and three conditions that will allow us to determine the values of the constants , , and .
Step 1: Use the condition to form the first equation.
We are given . Substitute into the function: Using the property : We are given that . Therefore, our first equation is: Explanation: This step utilizes the definition of a function and the property of the exponential function at zero to establish a linear relationship between the constants and .
Step 2: Use the condition to form the second equation.
First, we need to find the derivative of , denoted by . Differentiating term by term with respect to :
- The derivative of is .
- The derivative of is .
- The derivative of is . So, the derivative function is: Now, we substitute into : Using the properties and :
- .
- . Substituting these values: We are given that . Therefore, our second equation is: Explanation: This step involves differentiating the function and then applying the given condition using properties of logarithms and exponentials to establish another linear relationship between , , and .
Step 3: Use the condition to form the third equation.
First, let's find the expression for : Now, we need to evaluate the definite integral of this expression from to : Let's find the antiderivative of :
- The antiderivative of is .
- The antiderivative of is . So, the antiderivative of is . Now, we apply the Fundamental Theorem of Calculus: Evaluate the expression at the upper limit () and subtract the evaluation at the lower limit (): At : At : Now, subtract the lower limit evaluation from the upper limit evaluation: Combine like terms: To eliminate the fractions, multiply the entire equation by 2: We can simplify this equation by dividing by 3: Explanation: This step involves simplifying the integrand, finding its antiderivative, and then applying the definite integral formula by evaluating at the limits. Properties of logarithms and exponentials are crucial here, as is algebraic simplification to obtain a linear equation in terms of and .
Step 4: Solve the system of linear equations to find the values of , , and .
We have the following system of equations:
Let's first solve for and using Equations 1 and 3. From Equation 1, we can express in terms of : Substitute this expression for into Equation 3: Now, substitute the value of back into the expression for : Now that we have the values of and , we can substitute them into Equation 2 to find : So, the values of the constants are , , and . Explanation: This step involves a standard procedure for solving a system of linear equations. We used substitution to solve for two variables first and then used those values to find the third variable.
Step 5: Calculate .
We have found , , and . Now, we calculate the sum : Finally, we need to find the absolute value: Explanation: This final step involves substituting the determined values of the constants into the expression and computing its absolute value.
Common Mistakes & Tips
- Logarithm and Exponential Properties: Ensure correct application of rules like and . Mistakes here can lead to incorrect values in Steps 2 and 3.
- Algebraic Errors: Be meticulous with algebraic manipulations, especially when solving systems of equations or simplifying expressions with fractions. Double-checking calculations can prevent errors.
- Differentiation/Integration Errors: Carefully apply the rules of differentiation and integration for exponential functions. The constants of integration are not needed for definite integrals.
Summary
The problem was solved by systematically using the three given conditions to form a system of linear equations for the unknown constants , , and . The first condition, , provided a direct relation between and . The second condition, involving the derivative , yielded another linear equation in , , and . The third condition, a definite integral, was evaluated to produce a third linear equation involving and . Solving this system of equations yielded , , and . Finally, the absolute value of the sum of these constants, , was calculated to be 8.
The final answer is .