Problem 92 - A Complete Resource Book in Mathematics for JEE Main

Question

\(f(x)\) is a cubic function with \(f(1)=-6, f(-1)=10\) and has maxima at \(x=-1\). If \(f^{\prime}(x)\) has minima at \(x=1\), then (A) \(f(x)=x^{3}+3 x^{2}-9 x+5\) (B) \(f(x)=x^{3}-3 x^{2}-9 x+5\) (C) \(f(x)=x^{3}-3 x^{2}+9 x+5\) (D) \(f(x)=x^{3}-3 x^{2}-9 x+5\)

Step-by-Step Solution

Answer

The correct answer is (D) \(f(x)=x^{3}-3x^{2}-9x+5\).

Step 1: Identify the general form of a cubic function

The general form of a cubic function is given by \( f(x) = ax^3 + bx^2 + cx + d \). We need to find the values of \(a, b, c,\) and \(d\) that satisfy the given conditions.

Step 2: Use condition on maxima

Since \(x=-1\) is a point of local maxima for \(f(x)\), then \(f'(x) = 0\) when \(x=-1\). The derivative of the cubic function is \(f'(x) = 3ax^2 + 2bx + c\). Substitute \(x=-1\) and \(f'(x) = 0\) to get the equation: \(3a(-1)^2 + 2b(-1) + c = 0\). Simplifying, we get \(3a - 2b + c = 0\).

Step 3: Use condition on minima of the derivative

Since \(f'(x)\) has a minima at \(x=1\), then \(f''(x) = 0\) when \(x=1\). The second derivative is \(f''(x) = 6ax + 2b\). Substitute \(x=1\) and \(f''(x) = 0\) to get \(6a(1) + 2b = 0\). Simplifying, we get \(6a + 2b = 0\), or \(3a + b = 0\).

Step 4: Use conditions on values of the function

We are given that \(f(1) = -6\) and \(f(-1) = 10\). These conditions give us two equations: \(a(1)^3 + b(1)^2 + c(1) + d = -6\) simplifies to \(a + b + c + d = -6\), and \(a(-1)^3 + b(-1)^2 + c(-1) + d = 10\) simplifies to \(-a + b - c + d = 10\).

Step 5: Solve the system of equations

We now have the following system of linear equations:1. \(3a - 2b + c = 0\)2. \(3a + b = 0\) 3. \(a + b + c + d = -6\)4. \(-a + b - c + d = 10\)Solving these equations:- From (2), \(b = -3a\).- Substitute \(b = -3a\) into (1): \(3a - 2(-3a) + c = 0\), simplifying yields \(9a + c = 0\), so \(c = -9a\).- Substitute \(b = -3a\) and \(c = -9a\) in (3): \(a - 3a - 9a + d = -6\), simplifying yields \(-11a + d = -6\), so \(d = 11a - 6\).- Substitute into (4): \(-a - 3a + 9a + 11a - 6 = 10\), simplifying yields \(16a - 6 = 10\), solving yields \(a = 1\).Therefore, \(b = -3a = -3\), \(c = -9a = -9\), \(d = 11\times1 - 6 = 5\).Thus, \(f(x) = x^3 - 3x^2 - 9x + 5\).

Step 6: Match the function to the given options

Now compare the function \(f(x) = x^3 - 3x^2 - 9x + 5\) with the provided options. We see that this matches option (D).

Key Concepts

Derivatives

Derivatives offer insight into the behavior of functions. In the context of cubic functions, the first derivative, represented as \( f'(x) \), is particularly important. The first derivative provides rates of change and can help determine the slope of the tangent line at any point on a function. For a cubic function given by \( f(x) = ax^3 + bx^2 + cx + d \), the first derivative is \( f'(x) = 3ax^2 + 2bx + c \). This derivative is a quadratic function that can reveal much about the original cubic function's graph.

Key points to note about the first derivative:

• When \( f'(x) = 0 \), it implies that the function has a critical point where the slope of the tangent is horizontal.
• A critical point can suggest potential maxima, minima, or points of inflection depending on additional analysis using the second derivative or testing the sign changes in \( f'(x) \).
• For this exercise, \( x = -1 \) was identified as a local maximum of \( f(x) \) because \( f'(-1) = 0 \).

Understanding derivatives is crucial when working with functions, especially in identifying and analyzing critical points.

Maxima and Minima

Maxima and minima are the highest and lowest points on a function's graph within a certain range, respectively. These points are crucial in understanding the behavior of the function. When a function achieves its maximum or minimum point, it provides valuable information regarding where the function changes direction.

For a cubic function \( f(x) \), maxima and minima occur where the first derivative \( f'(x) \) is equal to zero \((f'(x) = 0)\). But simply finding where the first derivative equals zero is not enough to confirm if a point is a maximum or minimum. The second derivative \( f''(x) \) can help further confirm this. The second derivative test works as follows:

• If \( f''(x) > 0 \) at a critical point, then \( f(x) \) has a local minimum at that point.
• If \( f''(x) < 0 \) at a critical point, then \( f(x) \) has a local maximum at that point.

For example, in the exercise, \( x = -1 \) was a point of local maxima for \( f(x) \). This is determined by setting \( f'(x) = 0 \) and confirming that this is indeed a maximum because the behavioral change in slope at that point suggests it.

System of Equations

Solving systems of equations is an essential skill in mathematics, particularly in calculus when solving for unknown variables in derivative-related problems. A system of equations consists of two or more equations with the same set of unknowns, which are solved simultaneously. These equations provide conditions that the unknowns must satisfy.

In the given problem, we use conditions derived from the cubic equation, its derivative, and known function values to form a system of linear equations. Specifically, equations arise from:

• The fact that \( f'(x) = 0 \) at a local maximum \( (x = -1) \), resulting in \( equation: 3a - 2b + c = 0 \).
• The minimal point of the derivative \( f''(x) \) equal to zero at \( x = 1, \) so \( equation: 3a + b = 0 \).
• Known values of the function at certain points \( f(1) = -6 \) and \( f(-1) = 10 \), giving equations \( a + b + c + d = -6 \) and \( -a + b - c + d = 10 \).

Each equation provides certain deductions about the coefficients \((a, b, c, d)\). By solving this system, we find the coefficients that define the correct cubic function.