Problem 2 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

The maximum value of \(5+20 x-4 x^{2}, x \in\) \(R\) is (a) 25 (b) 30 (c) 5 (d) 1

Step-by-Step Solution

Answer

The maximum value is 30.

Step 1: Recognize the Formula

The given expression is a quadratic function of the form \( f(x) = ax^2 + bx + c \). Identify the coefficients: \( a = -4, b = 20, c = 5 \). The quadratic term coefficient \( a \) is negative, indicating the parabola opens downwards, thus the vertex is a maximum point.

Step 2: Find the x-coordinate of the Vertex

Use the formula for the x-coordinate of the vertex of a parabola, \( x = -\frac{b}{2a} \). Substituting the values, we get \[ x = -\frac{20}{2(-4)} = -\frac{20}{-8} = \frac{5}{2}. \] This is the x-coordinate where the maximum value occurs.

Step 3: Calculate the Maximum Value

Substitute \( x = \frac{5}{2} \) into the original quadratic function to find the maximum value. \[ f\left(\frac{5}{2}\right) = 5 + 20\left(\frac{5}{2}\right) - 4\left(\frac{5}{2}\right)^2. \]Simplify this:\[ = 5 + 50 - 4\left(\frac{25}{4}\right) = 5 + 50 - 25 = 30. \] Thus, the maximum value is 30.

Key Concepts

Vertex of Parabola

The vertex of a parabola is a key feature in understanding quadratic functions. It is the point where the parabola reaches its maximum or minimum value. In a quadratic equation of the form \( f(x) = ax^2 + bx + c \), the vertex can be found with the help of the formula for the x-coordinate, \( x = -\frac{b}{2a} \).

• If the coefficient \( a \) is positive, the parabola opens upwards, and the vertex is the minimum point.
• If \( a \) is negative, as in our exercise, the parabola opens downwards, making the vertex the maximum point.

Once you calculate the x-coordinate of the vertex using the formula, you substitute that into the original equation to find the y-coordinate. This y-coordinate gives you the value at the maximum or minimum point. In quadratic equations, this process is crucial because it helps locate the most important point on the curve.

Maximum Value

When analyzing a quadratic function like \( f(x) = ax^2 + bx + c \), knowing how to determine the maximum value is essential when the parabola opens downwards (\( a < 0 \)). The vertex, which we calculated using \( x = -\frac{b}{2a} \), lies at the highest point of the curve in this scenario.
To find the maximum value, you simply substitute the x-coordinate of the vertex back into the original function. This gives you \( f\left( x \right) \), which is the maximum value the quadratic function achieves. For instance, in the exercise, substituting \( x = \frac{5}{2} \) into the function results in a maximum value of 30.
Calculating this value allows us to comprehend the behavior of the function and the bounds within which it operates. This can be particularly useful in numerous applications, such as physics or finance, where understanding limits and extrema is crucial.

Quadratic Formula

The quadratic formula is an essential tool for finding the solutions (roots) of quadratic equations of the form \( ax^2 + bx + c = 0 \). Though not directly needed to find the vertex or maximum value in the exercise, it's worth discussing due to its profound ability to solve any quadratic equation.
The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• The \( \pm \) symbol indicates that there could be two potential solutions: one where you add the square root term and one where you subtract it.
• The part under the square root, \( b^2 - 4ac \), is known as the discriminant and helps determine the nature of the roots—whether they are real or complex.

Using the quadratic formula can confirm the result of factorization or provide solutions when factorization is complicated. It is a versatile and indispensable component of understanding quadratic relationships fully.