Problem 29 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the function \(f(x)\left\\{\begin{array}{ll}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\\ k_{2} \cos x, & x>\pi\end{array}\right.\) is twice dif- ferentiable, then the ordered pair \(\left(k_{1}, k_{2}\right)\) is equal to: (a) \(\left(\frac{1}{2}, 1\right)\) (b) \((1,0)\) (c) \(\left(\frac{1}{2},-1\right)\) (d) \((1,1)\)

Step-by-Step Solution

Answer

The ordered pair is \( (\frac{1}{2}, 1) \). Choose option (a).

Step 1: Understand the Problem

The function \( f(x) \) is given in a piecewise manner with different expressions on either side of \( x = \pi \). To ensure the function is twice differentiable everywhere, including at \( x = \pi \), the function, its first derivative, and its second derivative must be continuous at \( x = \pi \).

Step 2: Ensure Continuity of \( f(x) \)

To ensure continuity at \( x = \pi \), set the expressions for \( f(x) \) at \( x = \pi \) equal:\[ k_1(\pi - \pi)^2 - 1 = k_2 \cos(\pi) \]\[ -1 = -k_2 \]\[ k_2 = 1 \]

Step 3: Ensure Continuity of \( f'(x) \)

First, find \( f'(x) \) for both parts:- For \( x \leq \pi \): \( f'(x) = 2k_1(x-\pi) \)- For \( x > \pi \): \( f'(x) = -k_2 \sin(x) \)To ensure the first derivative is continuous at \( x = \pi \), set:\[ 2k_1(\pi-\pi) = -k_2 \sin(\pi) \]\[ 0 = 0 \]The condition is automatically met since both sides are zero.

Step 4: Ensure Continuity of \( f''(x) \)

Now, find \( f''(x) \):- For \( x \leq \pi \): \( f''(x) = 2k_1 \)- For \( x > \pi \): \( f''(x) = -k_2 \cos(x) \)Set the second derivatives equal at \( x = \pi \):\[ 2k_1 = -k_2 \cos(\pi) \]\[ 2k_1 = k_2 \]We previously found \( k_2 = 1 \), thus \(2k_1 = 1\) results in \( k_1 = \frac{1}{2} \).

Step 5: Determine the Answer

The values for \( k_1 \) and \( k_2 \) are \( \frac{1}{2} \) and \( 1 \), respectively. Thus, the ordered pair is \( (\frac{1}{2}, 1) \), which corresponds to choice (a).

Key Concepts

Continuity

Continuity is a key property in calculus that ensures a function has no breaks or gaps at any point in its domain, including transition points. For a piecewise function like the one given, ensuring continuity involves checking that the outputs of each piece hover perfectly on either side of the boundary, in this case, at \( x = \pi \). To check this, we ensure

• The left-hand and right-hand limits of the function as \( x \) approaches \( \pi \) from either side must be equal.
• The value of the function at \( x = \pi \) matches these limits.

Continuity guarantees not only that the function has no sudden jumps but also that it's smooth. In mathematical terms, this implies the existence of a finite limit as you approach a given point. For the function \(f(x)\), setting \[ k_1(\pi - \pi)^2 - 1 = k_2 \cos(\pi) \]allows us to establish continuity at \(x = \pi\), leading to solving for \(k_2\).

First Derivative

The first derivative of a function measures its rate of change or slope at any given point, often indicating the function's increasing or decreasing behavior. In the context of piecewise functions, it’s crucial to check for derivatives' continuity at transition points.For the given function, the first derivative is found as follows:

• For \( x \leq \pi \): \( f'(x) = 2k_1(x - \pi) \).
• For \( x > \pi \): \( f'(x) = -k_2 \sin(x) \).

To ensure continuous transition at \( x = \pi \), we equate these derivatives resulting in:\[ 2k_1(\pi - \pi) = -k_2 \sin(\pi) \]This simplifies to zero on both sides, confirming that the first derivative indeed fits the continuity criteria. Derivative continuity is crucial because it implies the absence of abrupt directional changes.

Second Derivative

The second derivative often represents the "curvature" or how the function's slope is changing. It is vital for evaluating whether a function is concave up (like a cup) or concave down (like a frown) at different points. For twice differentiability, at transition points, like \( x = \pi \) here, the continuity of both this derivative and its preceding counterpart must be ensured.For the second derivative:

• When \( x \leq \pi \), \( f''(x) = 2k_1 \).
• As \( x > \pi \), \( f''(x) = -k_2 \cos(x) \).

To maintain a smooth curve, their equality at \( x = \pi \) is required:\[ 2k_1 = -k_2 \cos(\pi) \]Simplifying gives \(2k_1 = k_2\), eventually helping us solve for \(k_1\) and \(k_2\). The presence of such a derivative without discontinuities indicates the function’s shapes merge smoothly between pieces.

Piecewise Function

Piecewise functions are built from multiple sub-functions, each applying to different parts of the function’s domain. They are especially prevalent in modeling scenarios where the nature of the relationship between variables changes beyond certain boundaries.In the given exercise, we handle two distinct expressions defined by:

• \( f(x) = k_1(x-\pi)^2 - 1 \) when \( x \leq \pi \).
• \( f(x) = k_2 \cos(x) \) when \( x > \pi \).

These expressions must be seamlessly aligned at the pivotal boundary \( x = \pi \). Computing their values, first derivatives, and second derivatives at this point ensures the overall function remains well-behaved, attaining the twice-differentiable standard. Understanding piecewise function construction is integral when solving for constants that maintain the concatenated version of the function within calculus frameworks.