Problem 77 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

For a certain curve \(y=f(x)\) satisfying \(\frac{d^{2} y}{d x^{2}}=6 x-4\), \(f(x)\) has a local minimum value 5 when \(x=1\). (A) Equation of the curve is \(y=x^{3}-2 x^{2}+x+5\) (B) \(f(x)\) has a local maximum at \(x=\frac{1}{3}\) (C) Global maximum value of \(f(x)\) is 7 (D) Global minimum value of \(f(x)\) is 5

Step-by-Step Solution

Answer

A: True, B: True, C: True, D: True

Step 1: Integrate Second Derivative

Given that \(\frac{d^{2} y}{d x^{2}} = 6x - 4\), we need to find the first derivative. Integrate \(6x - 4\) with respect to \(x\) to find the first derivative: \(f'(x) = \int (6x - 4) \, dx = 3x^2 - 4x + C\), where \(C\) is a constant.

Step 2: Use Local Minimum Condition

We know \(f(x)\) has a local minimum at \(x=1\) with \(f(1)=5\). A local minimum implies \(f'(1) = 0\). So, \(3(1)^2 - 4(1) + C = 0\). Simplifying gives \(C = 1\).

Step 3: Determine First Derivative

Substitute \(C=1\) into the first derivative equation: \(f'(x) = 3x^2 - 4x + 1\).

Step 4: Find Original Function

Now that we have \(f'(x) = 3x^2 - 4x + 1\), integrate again to find \(f(x)\): \(f(x) = \int (3x^2 - 4x + 1) \, dx = x^3 - 2x^2 + x + D\). We know \(f(1) = 5\), so substitute \(x = 1\) and \(f(x) = 5\) into the equation: \((1)^3 - 2(1)^2 + 1 + D = 5\). Solving gives \(D = 5\).

Step 5: Write Equation of the Curve

Inserting \(D = 5\) into the function, we get \(f(x) = x^3 - 2x^2 + x + 5\).

Step 6: Analyze Local Maximum

Find critical points by setting \(f'(x)=0\): \(3x^2 - 4x + 1 = 0\). Solving yields \(x = 1\) or \(x = \frac{1}{3}\). Use the second derivative test: \(f''(x) = 6x - 4\); at \(x = \frac{1}{3}\), \(f''\left(\frac{1}{3}\right) = -2\), indicating a local maximum at \(x = \frac{1}{3}\).

Step 7: Determine Global Max/Min Values

The global minimum occurs at \(x = 1\), where \(f(x) = 5\) as already known. Check endpoints or behavior as \(x\) approaches \(\pm\infty\) for the global maximum. As the polynomial is cubic and the limits tend to infinity, check derived points instead. Calculate \(f\left(\frac{1}{3}\right)\): it gives \(f\left(\frac{1}{3}\right) = 7\), thus the global maximum is 7.

Key Concepts

Second Derivative Test

The second derivative test is a valuable tool in calculus for determining the nature of critical points of a function. When a function has a critical point, where its first derivative is zero (\(f'(x) = 0\)), you can use the second derivative to find out whether this point is a local minimum, maximum, or neither. Here's how it generally works:

• If the second derivative at this point is positive (\(f''(x) > 0\)), the function has a local minimum there.
• If the second derivative at this point is negative (\(f''(x) < 0\)), the function has a local maximum.
• If the second derivative is zero (\(f''(x) = 0\)), the test is inconclusive, and higher-derivative tests or alternative methods might be needed.

Understanding the second derivative test aids in analyzing the local behavior of functions, giving insights into the concavity and curvature of the graph of the function.

Local Minimum and Maximum

Local minimums and maximums provide insight into the peaks and troughs of a function over an interval. Understanding these can help you analyze the behavior of functions more deeply. A function has a local minimum at a point if it reaches the lowest value compared to nearby points. Similarly, a local maximum occurs at a point if the function reaches a higher value than any nearby points.

In the context of a problem where \(f(x)\) has a local minimum at \(x = 1\), the derivative \(f'(x)\) equals zero at this point. Checking the second derivative helps confirm this as a minimum since \(f''(1)\) would be positive. Conversely, for a local maximum at \(x = \frac{1}{3}\), you verify it by confirming \(f''\left(\frac{1}{3}\right)\) is negative.

Local extrema are essential in optimization problems and are helpful in sketching graphs, analyzing growth trends, and more.

Cubic Polynomial

Cubic polynomials are algebraic expressions of the form \(ax^3 + bx^2 + cx + d\). Here, the highest degree of the term is three, which means the curve can exhibit interesting behaviors like having up to two turning points, and a point of inflection where the concavity changes.

They are crucial in modeling real-world scenarios and visualizing dynamic processes. Analyzing a cubic polynomial involves finding its roots, determining its end behavior, and understanding the turning points.

• The polynomial \(y = x^3 - 2x^2 + x + 5\) is a specific cubic example.
• It has both linear components affecting the slope and a cubic shape influencing the curve’s asymmetry.

Knowing how to factor and solve cubic equations or to sketch their graphs is a skill that's often applied in calculus problems and real-world applications alike.

Integration Techniques

Integration techniques in calculus are essential for finding antiderivatives and solving differential equations, like deducing a function from its derivative. Integrating involves calculating the integral of a function, which essentially reverses differentiation.

Common techniques include:

• Integration by Substitution: Useful for simplifying functions before integrating.
• Integration by Parts: Specifically beneficial when dealing with products of functions.
• Partial Fractions: Breaking complex fractions into simpler parts that are easier to integrate.

In exercises involving integration, we often start with given derivative forms and work backwards to determine the original function. Here, by integrating the given second derivative \(\frac{d^2 y}{dx^2} = 6x - 4\), we find the first derivative and then continue to integrate to uncover the cubic polynomial representing the curve.