Problem 179 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f: R \rightarrow R\) be defined by \(f(x)= \begin{cases}k-2 x, & \text { if } x \leq-1 \\ 2 x+3 & \text { if } x>-1\end{cases}\) If \(f\) has a local minimum at \(x=-1\), then apossible value of \(k\) is (A) 0 (B) \(-\frac{1}{2}\) (C) \(-1\) (D) 1

Step-by-Step Solution

Answer

A possible value of \(k\) is \(-1\).

Step 1: Define the condition for a local minimum

A function has a local minimum at a point if it changes from decreasing to increasing at that point. For a piecewise function, we must ensure continuity at the point of interest, here at \(x = -1\).

Step 2: Check function continuity at \(x = -1\)

For the function \(f(x)\) to be continuous at \(x = -1\), the left-hand limit, right-hand limit, and the function value must all be equal at \(x=-1\). Calculate as follows: - Left-hand limit as \(x \to -1^-\) is \(f(-1) = k - 2(-1) = k + 2\).- Right-hand limit as \(x \to -1^+\) is \(2(-1) + 3 = 1\).Set these equal for continuity: \[ k + 2 = 1 \]

Step 3: Solve for the value of \(k\)

From the continuity equation, \(k + 2 = 1\), solve for \(k\): Move 2 to the other side: \[ k = 1 - 2 = -1 \]

Step 4: Confirm if \(k = -1\) ensures a local minimum

For \(f(x)\) to have a local minimum at \(x = -1\), the function must decrease before \(x = -1\) and increase after. - Before \(x = -1\), \(f(x) = k - 2x\) is decreasing if \(-2 < 0\).- After \(x = -1\), \(f(x) = 2x + 3\) is increasing since \(2 > 0\).The conditions are satisfied.

Key Concepts

Local Minimum

In calculus, understanding local minimum points of a function is crucial as they help us identify where a function reaches its lowest value within a small, specific range. The concept of a local minimum relates to how the function behaves around a particular point. To identify a local minimum at a point, say at \(x = c\), the function must satisfy the following conditions:

• The function decreases on one side of \(c\) and increases on the other side.
• The derivative changes from negative to positive at \(x = c\).

For piecewise functions, to determine if \(x = c\) is a local minimum, we must verify that:

• The function decreases before \(c\).
• The function increases after \(c\).

In our example, the behavior of the function before and after \(x = -1\) indicates that there's a local minimum at this point when \(k = -1\). Ensuring both continuity and the change in direction from decreasing to increasing is fundamental for confirming a local minimum.

Piecewise Function

Functions can sometimes be defined by different expressions over different parts of their domain; these are known as piecewise functions. This means the rule for calculating the function's value differs based on the input's range.
In the context of calculated limits, piecewise functions require special attention to ensure correct transitions at the boundaries between different pieces. The steps involve:

• Identifying the domain for each piece of the function.
• Ensuring the outputs transition smoothly at the boundaries, especially where the function’s defining expressions change.
• Analyzing limits from both sides of transition points to confirm continuity or identify discontinuities if present.

In our case, the function \(f(x)\) is given by two different expressions based on the condition \(x \leq -1\) and \(x > -1\). We had to check the function at \(x = -1\) to solve for \(k\) ensuring continuity and the switch in behavior required for a local minimum.

Continuity at a Point

Continuity is a fundamental concept in calculus and is essential for ensuring that functions behave predictably at every point within a given interval. When we say a function is continuous at a certain point, it means that there are no "jumps" or breaks in the graph of the function at that point. More formally, a function \(f(x)\) is continuous at \(x = c\) if:

• The left-hand limit as \(x\) approaches \(c\) is equal to the right-hand limit.
• Both the left-hand and right-hand limits are equal to the function value at \(x = c\).

For piecewise functions, checking continuity at the boundary between different pieces is vital. These checks confirm that different pieces connect smoothly. In our exercise, for \(f(x)\) to be continuous at \(x = -1\), both the left side expression, \(k - 2x\), and the right side expression, \(2x + 3\), had to meet at this point.
This requirement led us to solve the equation \(k + 2 = 1\) to ensure there is no jump, resulting in \(k = -1\). Continuity links smoothly back to the concept of local minima by ensuring the transition from decreasing to increasing happens cleanly at the point of interest.