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

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) Need evaluation at bounds, (D) True since it's the minimum at x=1.

Step 1: Integrate the Second Derivative

Given the second derivative \(\frac{d^{2} y}{d x^{2}} = 6x - 4\), we integrate once with respect to \(x\) to find the first derivative \(\frac{dy}{dx}\).\\[\frac{dy}{dx} = \int (6x - 4) \, dx = 3x^2 - 4x + C\]

Step 2: Integrate to Find the Function

Next, integrate the expression for \(\frac{dy}{dx}\) to find the function \(y = f(x)\).\\[y = \int (3x^2 - 4x + C) \, dx = x^3 - 2x^2 + Cx + D\]

Step 3: Use the Local Minimum Condition

We know \(f(x)\) has a local minimum at \(x = 1\) with value 5. Thus, substitute \(x = 1\) into \(y = x^3 - 2x^2 + Cx + D = 5\). This gives:\\[1^3 - 2(1)^2 + C(1) + D = 5\]\\[1 - 2 + C + D = 5\]\\[C + D = 6\]

Step 4: Use the Condition for a Local Minimum in the Derivative

At \(x = 1\), the first derivative \(\frac{dy}{dx}\) equals zero for a local minimum: \(3x^2 - 4x + C = 0\). Substituting \(x = 1\) gives:\\[3(1)^2 - 4(1) + C = 0\]\\[3 - 4 + C = 0\]\\[C = 1\]

Step 5: Find Constant D

Substitute \(C = 1\) back into \(C + D = 6\):\\[1 + D = 6\]\\[D = 5\]

Step 6: Final Equation of the Curve

Substitute \(C = 1\) and \(D = 5\) in \(y = x^3 - 2x^2 + Cx + D\):\\[y = x^3 - 2x^2 + x + 5\]

Step 7: Analyze Critical Points

Set the first derivative \(\frac{dy}{dx} = 3x^2 - 4x + 1\) to zero and solve for \(x\):\\[3x^2 - 4x + 1 = 0\]. Solving gives two critical points: \(x = \frac{1}{3}\) and \(x = 1\).

Step 8: Determine the Nature of Critical Points

Calculate the second derivative \(\frac{d^2 y}{d x^2} = 6x - 4\) at critical points.\\- For \(x = \frac{1}{3}\): \(6\times\frac{1}{3} - 4 = -2\), so it's a local maximum.\\- For \(x = 1\): \(6\times 1 - 4 = 2\), so it's a local minimum.

Step 9: Find Global Maximum and Minimum

Evaluate \(f(x) = x^3 - 2x^2 + x + 5\) at critical points or endpoints to find extremum values. Check values at boundaries or asymptotic behavior for extreme functions, if applicable.

Key Concepts

Second Derivative

In Calculus, the second derivative of a function is a powerful tool that helps us understand the behavior of graphs. The second derivative \(\frac{d^2 y}{d x^2}\) is simply the derivative of the first derivative.
In this exercise, the second derivative is given as \(\frac{d^2 y}{d x^2} = 6x - 4\). By examining this, we can determine the concavity of the original function.

• If \(\frac{d^2 y}{d x^2} > 0\), the function is concave up at that point.
• If \(\frac{d^2 y}{d x^2} < 0\), the function is concave down.

In our case, when \(x = 1\), the second derivative is \(2\) (since \(6 imes 1 - 4 = 2\)), meaning the graph is concave up. This suggests a local minimum at this point.
The second derivative test thus provides key insights into identifying and confirming the nature of critical points, especially in determining local maxima and minima.

Local Minimum

A local minimum is a point on the graph of a function where the function value is lower than at any nearby points.
In mathematical terms, if \(f(x)\) has a local minimum at \(x = a\), then \(f(a) < f(x)\) for all \(x\) in some interval around \(a\).
To find a local minimum, we first look for critical points where \(\frac{dy}{dx} = 0\).
At these points, we use the second derivative test:

• If \(\frac{d^2 y}{d x^2} > 0\), it's a local minimum, as the function is concave up.
• If \(\frac{d^2 y}{d x^2} < 0\), it's a local maximum, as the function is concave down.

In our function, at \(x = 1\), the first derivative is zero, \(3 \times 1^2 - 4 \times 1 + 1 = 0\).
The second derivative \(6 \times 1 - 4 = 2\) is positive, confirming a local minimum.
Thus, the local minimum value of the function is \(5\), aligned with the condition given in the original problem statement.

Critical Points

Critical points are values of \(x\) where the first derivative of a function is zero or undefined. These points are significant as they could indicate local maxima, minima, or points of inflection.
To identify them, we set the first derivative to zero: \(\frac{dy}{dx} = 3x^2 - 4x + 1\).
Solving \(3x^2 - 4x + 1 = 0\), we find two critical points: \(x = \frac{1}{3}\) and \(x = 1\).
Once found, the second derivative test can determine their nature.

• For \(x = \frac{1}{3}\), \(\frac{d^2 y}{d x^2} = 6\times\frac{1}{3} - 4 = -2\), indicating a local maximum.
• For \(x = 1\), \(\frac{d^2 y}{d x^2} = 2\), confirming it as a local minimum.

These critical points help us sketch the curve and understand the function's behavior, providing all points where changes in direction and slope occur.