Question

Find the values of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) for which the function of the form \(\mathrm{f}(\mathrm{x})=\mathrm{Ax}^{2}+\mathrm{Bx}+\mathrm{C}\) satisfies the conditions \(f^{\prime}(1)=8, f(2)+f^{\prime}(2)=33, \int_{0}^{1} f(x) d x=\frac{7}{3}\).

Short answer

Based on the solution above, the values of A, B, and C that satisfy the given conditions are \(A = \frac{9}{2}, B = -1, C = \frac{1}{3}\). Therefore, the quadratic function \(f(x)\) is given by: \[f(x) = \frac{9}{2}x^2 - x + \frac{1}{3}\]

Step-by-step solution

Find the derivative of f(x)

To find the derivative of f(x), we should differentiate the function with respect to x. So, the derivative of \(f(x)=Ax^2+Bx+C\) is: \(f'(x) = \frac{d}{dx}(Ax^2+Bx+C) = 2Ax+B\)

Use the first condition: \(f'(1)=8\)

We have the condition that \(f'(1)=8\). Substituting x=1 in the derivative we found earlier, we get: \(8 = 2A(1)+B\) \(8 = 2A+B\)

Use the second condition: \(f(2)+f'(2)=33\)

We have the condition that \(f(2)+f'(2)=33\). Substituting x=2 in f(x) and f'(x), we get: \(33 = (4A+2B+C) + (4A + B)\). Simplifying this equation, we have: \(33 = 8A + 3B + C\)

Use the third condition: \(\int_{0}^{1} f(x) dx = \frac{7}{3}\)

We have the condition that \(\int_{0}^{1} f(x) dx = \frac{7}{3}\). Integrating f(x) with respect to x, we get: \(\int f(x)dx = \int (Ax^2+Bx+C)dx = \frac{1}{3}Ax^3 + \frac{1}{2}Bx^2 + Cx\) Now, we need to evaluate this integral from 0 to 1: \(\int_{0}^{1} f(x) dx = \left[ \frac{1}{3}A(1)^3 + \frac{1}{2}B(1)^2 + C(1) \right] - \left[ \frac{1}{3}A(0)^3 + \frac{1}{2}B(0)^2 + C(0) \right]\) Simplifying this equation, we get: \(\frac{7}{3}=\frac{1}{3}A+\frac{1}{2}B+C\)

Solve the system of equations

You now have a system of 3 linear equations with 3 unknowns: 1. \(8 = 2A+B\) 2. \(33 = 8A + 3B + C\) 3. \(\frac{7}{3} = \frac{1}{3}A+\frac{1}{2}B+C\) From equation 1, we can express B in terms of A: \(B = 8 - 2A\) Substitute this expression for B into equations 2 and 3: 2. \(33 = 8A + 3(8 - 2A) + C\) 3. \(\frac{7}{3} = \frac{1}{3}A + \frac{1}{2}(8 - 2A) + C\) Solving equation 2 gives us an expression for C in terms of A: \(33 = 8A + 24 - 6A\) \(9 = 2A\) \(A = \frac{9}{2}\) Now substitute this value for A back into the expression for B: \(B = 8 - 2\left(\frac{9}{2}\right)\) \(B = 8 - 9\) \(B = -1\) Finally, substitute the values of A and B into equation 3 to find C: \(\frac{7}{3} = \frac{1}{3}\left(\frac{9}{2}\right) - \frac{1}{2} + C\) \(C = \frac{7}{3} - \frac{3}{2} + \frac{1}{2}\) \(C = \frac{1}{3}\)

Final solution

The values of A, B, and C that satisfy the conditions are: \(A = \frac{9}{2}, B = -1, C = \frac{1}{3}\) So, the given function is: \[f(x) = \frac{9}{2}x^2 - x + \frac{1}{3}\]

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus, especially in the field of integral calculus IIT JEE. It deals with finding the derivative of a function, which represents the rate of change of that function with respect to a variable. Derivatives are used to calculate slopes of tangent lines to curves and can be applied in various branches of physics and engineering for understanding motion and changes in systems.

For the given problem, we have a quadratic function of the form
\( f(x) = Ax^2 + Bx + C \). To find the derivative \( f'(x) \), we apply the power rule, which states that \( d/dx(x^n) = nx^{n-1} \). Thus, the derivative of our quadratic function is \( f'(x) = 2Ax + B \). This step is crucial as it leads to the first condition given in the exercise \( f'(1) = 8 \), which is used to find the first relationship between the coefficients A and B.

Definite Integration

Definite integration calculates the accumulation of quantities, which can represent areas under curves within a certain interval. In integral calculus, particularly for the IIT JEE syllabus, understanding definite integration is crucial to solving a myriad of problems.

In the context of the given exercise, the third condition \( \int_{0}^{1} f(x) dx = \frac{7}{3} \) requires evaluating a definite integral. The process involves finding the antiderivative of the function \( f(x) \) and then applying the Fundamental Theorem of Calculus, which states that evaluating the antiderivative at the upper and lower limits of the integral and finding the difference gives us the value of the definite integral. By integrating \( Ax^2 + Bx + C \) with respect to x over the interval from 0 to 1 and equating it to \( \frac{7}{3} \), we obtain a third equation involving A, B, and C.

Solving System of Linear Equations

Solving systems of linear equations is an essential skill in algebra and underpins many problems in calculus, including those found in the IIT JEE curriculum. A system of linear equations consists of two or more equations with multiple variables. The solution to the system is the set of variable values that satisfies all equations simultaneously.

In our problem, after applying differentiation and using the conditions given, we arrived at three equations:

• \( 8 = 2A + B \)
• \( 33 = 8A + 3B + C \)
• \( \frac{7}{3} = \frac{1}{3}A + \frac{1}{2}B + C \)

Here, there are various ways to solve such a system, including substitution, elimination, and matrix methods like using determinants or row reduction. In the step-by-step solution provided, the substitution method is primarily used, starting with solving one equation for one variable and substituting that result into the other equations. This approach leads us to the values for A, B, and C that meet the initial conditions of the problem, completing the exercise.