Problem 23 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(f(x)=g(x)(x-a)^{2}\) where \(g(a) \neq 0\) and \(g(x)\) is cotinuous at \(x=a .\) Then [Roorkee (Screening)-1999] (a) \(f\) is increasing in the nbd of \(a\) if \(g(x)<0\). (b) \(f\) is decreasing in the nbd of ' \(a\) ' if \(g(a)>0\) (c) \(f\) is increasing in nbd of ' \(a\) ' if \(g(a)>0\) (d) None of these

Step-by-Step Solution

Answer

(c) \(f\) is increasing in the neighborhood of \(a\) if \(g(a)>0\).

Step 1: Analyze Function at x = a

Given that \( f(x) = g(x)(x-a)^{2} \), and we know \( g(a) eq 0 \). At \( x = a \), we will substitute \( x = a \) to evaluate \( f(a) \).Since \((x-a)^2\) becomes zero at \(x = a\), we have \(f(a) = g(a) \cdot 0 = 0\).

Step 2: Determine the Derivative of f(x)

To understand if \(f\) is increasing or decreasing, we need to find the derivative \( f'(x) \). Using the product rule of derivatives, \( f(x) = g(x)(x-a)^2 \) gives:\[ f'(x) = g'(x)(x-a)^2 + g(x) \cdot 2(x-a)\].

Step 3: Evaluate f'(x) at x = a

Substitute \(x = a\) in \( f'(x)\):\[ f'(a) = g'(a)(a-a)^2 + g(a) \cdot 2(a-a) = 0\]This means the derivative at \(x = a\) is neither positive nor negative, so \(f(x)\) isn't changing at \(x = a\).

Step 4: Determine the Sign of f''(x)

Calculate \( f''(x) \) to determine the concavity:\[ f''(x) = [g''(x)(x-a)^2 + 2g'(x)(x-a) \cdot 2 + 2g(x)]\]Evaluate at \(x = a\):\[ f''(a) = 2g(a)\]Since \(g(a) eq 0\), \(f''(a) = 2g(a)\), a positive value implies \(f\) is concave up at \(x = a\).

Step 5: Determine Increasing or Decreasing Behavior

From the concavity determined in Step 4:- If \(g(a) > 0\), then \(f\) is concave up and thus increasing in the neighborhood of \(a\).- If \(g(a) < 0\), \(f\) is still concave up (since \(f''(a) = 2g(a)\)) but could be interpreted as increasing because the parabola opens upwards.

Key Concepts

Derivative

In calculus, a derivative represents the rate of change of a function with respect to its variable. It essentially tells us how the function is changing at any point. For the function given as \( f(x) = g(x)(x-a)^2 \), we use the derivative to understand whether the function is increasing or decreasing near \( x = a \). By applying the product rule, the derivative is found to be:

• \( f'(x) = g'(x)(x-a)^2 + g(x) \cdot 2(x-a) \)

Evaluating \( f'(x) \) at \( x = a \) gives:

• \( f'(a) = g'(a)(a-a)^2 + g(a) \cdot 2(a-a) = 0 \)

This result means the instantaneous rate of change at \( x = a \) is zero, indicating that the function doesn't increase or decrease exactly at \( x = a \). However, to determine the behavior around this point, we look at higher derivatives, such as the second derivative.

Continuity

Continuity in calculus refers to a function having no breaks, holes, or jumps in its graph. For a function to be continuous at a point \( x = a \), it must exist at \( a \), and the limit of the function as \( x \) approaches \( a \) must equal the function's value at that point. Given \( g(x) \) is continuous at \( x = a \), it implies:

• \( \lim_{x \to a} g(x) = g(a) \)

This continuity is crucial because it ensures that the behavior of \( f(x) = g(x)(x-a)^2 \) can be consistently analyzed around \( x = a \). The continuity of \( g(x) \) helps in evaluating derivatives and determining the overall behavior of the function near \( a \). It guarantees that the function doesn't have sudden changes, providing smoother analysis.

Concavity

Concavity describes the direction in which a curve bends. If a function curves upwards, it is termed concave up. If it curves downwards, it is concave down. The concept of concavity is related to the second derivative of the function. For our function \( f(x) = g(x)(x-a)^2 \),

• The second derivative is \( f''(x) = [g''(x)(x-a)^2 + 2g'(x)(x-a) \cdot 2 + 2g(x)] \).

Evaluating at \( x = a \):

• \( f''(a) = 2g(a) \)

If \( f''(a) > 0 \), the curve is concave up at \( x = a \). In other words, the function resembles a "smiley face" curve, indicating that it opens upwards, and is generally increasing in the vicinity.

Increasing and Decreasing Functions

A function is considered increasing if its derivative is positive over an interval, meaning it rises as one moves along the x-axis. Conversely, a function is decreasing if its derivative is negative over an interval, showing a downward trend. In this problem, the critical point is at \( x = a \), where the first derivative \( f'(a) = 0 \). To determine increasing or decreasing behavior in the neighborhood, we rely on the sign of the second derivative:

• If \( g(a) > 0 \), since \( f''(a) = 2g(a) > 0 \), the function is increasing around \( x = a \).
• If \( g(a) < 0 \), the function is often still considered increasing due to being concave up, making it open upwards, but the exact interpretation may vary, and deeper context such as specific intervals should be considered.

Thus, knowing the behavior of \( f \) around \( x = a \) helps make conclusions about its tendency to rise or fall.