Problem 52 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the function. \(g(x)=\left\\{\begin{array}{l}k \sqrt{x+1}, 0 \leq x \leq 3 \\ m x+2,3

Step-by-Step Solution

Answer

The value of \(k + m\) is 2.

Step 1: Understand Differentiability at Transition Point

The function is defined as a piecewise function: for \(0 \leq x \leq 3\), \(g(x) = k \sqrt{x+1}\) and for \(3 < x \leq 5\), \(g(x) = mx + 2\). For \(g(x)\) to be differentiable at \(x = 3\), both \(g(x)\) and \(g'(x)\) must be continuous at this point. This means the function value and derivative must be the same approaching from both sides of \(x = 3\).

Step 2: Ensure Function Continuity at x = 3

Calculate \(g(3)\) from both parts of the piecewise function and set them equal:1. From \(g(x) = k \sqrt{x+1}\) : \(g(3) = k\sqrt{3+1} = 2k\).2. From \(g(x) = mx + 2\) : \(g(3) = 3m + 2\).Set \(2k = 3m + 2\).

Step 3: Ensure Derivative Continuity at x = 3

Differentiate each part of the function and set the derivatives equal to ensure differentiability:1. \(g'(x)\) from \(g(x) = k \sqrt{x+1}\) is \(\frac{k}{2\sqrt{x+1}}\). Thus, \(g'(3) = \frac{k}{4}\).2. \(g'(x)\) from \(g(x) = mx + 2\) is \(m\).Set \(\frac{k}{4} = m\).

Step 4: Solve System of Equations

Solve the system of equations:1. \(2k = 3m + 2\)2. \(\frac{k}{4} = m\)Substitute \(m\) from the second equation into the first equation: \(2k = 3\left(\frac{k}{4}\right) + 2\).Simplifying gives: \(2k = \frac{3k}{4} + 2\).Multiply everything by 4 to clear the fraction: \(8k = 3k + 8\).Isolate \(k\): \(5k = 8\) which gives \(k = \frac{8}{5}\).Substitute \(k\) back into \(m = \frac{k}{4}\):\(m = \frac{8}{5} \times \frac{1}{4} = \frac{2}{5}\).

Step 5: Calculate k + m

Finally, calculate \(k + m\) using our found values: \(k = \frac{8}{5}\) and \(m = \frac{2}{5}\).\(k + m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2\).

Key Concepts

Function Continuity

When dealing with piecewise functions, ensuring continuity is crucial for making them differentiable at a specific point—a transition point—where the two pieces meet. For a function to be continuous at this transition point, the function's value, as you approach from either side, must be the same.
In our example, we have a function composed of two expressions split at the point where \(x = 3\):

• For \(0 \leq x \leq 3\), the function is \(g(x) = k \sqrt{x+1}\).
• For \(3 < x \leq 5\), the function is \(g(x) = mx + 2\).

To confirm continuity at \(x = 3\), calculate \(g(3)\) for both function pieces:

• From the first part: \(g(3) = k \sqrt{3+1} = 2k\).
• From the second part: \(g(3) = 3m + 2\).

By setting these two equations equal, \(2k = 3m + 2\), we ensure continuity, a vital condition for differentiability.

Derivative Matching

Once continuity is established, the next step for differentiability is ensuring that the derivatives also match at the transition point. The derivative tells us about the rate of change of the function, and for the function to be smooth (differentiable) across the transition, these rates must be the same when approaching the point from either side.
Differentiating each part of our piecewise function:

• For \(g(x) = k \sqrt{x + 1}\), the derivative \(g'(x) = \frac{k}{2\sqrt{x+1}}\). At \(x = 3\), this becomes \(g'(3) = \frac{k}{4}\).
• For \(g(x) = mx + 2\), the derivative \(g'(x) = m\).

By setting the derivatives equal, \(\frac{k}{4} = m\), we ensure the function's rate of change smoothly continues through \(x = 3\).

Piecewise Functions

Understanding piecewise functions is essential, especially in problems involving continuity and differentiability. Piecewise functions are defined by different expressions depending on the value of \(x\).
They often present challenges due to the need for function value and derivative consistency at points where the formula changes.
In our problem, we looked at the function \(g(x)\) which uses two different expressions: a radical expression for \(0 \leq x \leq 3\) and a linear expression for \(3 < x \leq 5\). These functions need:

• The values to match at \(x = 3\).
• The derivatives to be equal at \(x = 3\) for smooth transition.

Piecewise functions are widely applicable in real-world scenarios where a sudden change in formula is required, like adjusting a pricing model or defining physical phenomena that operate under different sets of conditions.