Problem 127 - A Complete Resource Book in Mathematics for JEE Main

Question

If the function \(g(x)=\left\\{\begin{array}{ll}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: Ensure the function is continuous at the boundary

First, ensure that the function is continuous at the point where it changes the definition: at \(x = 3\). Evaluate \(g(x)\) as \(x\) approaches 3 from both sides: \(g(3)\) from the left is \(k\sqrt{3+1} = 2k\), and from the right is \(m \cdot 3 + 2\). For continuity, these must be equal: \[2k = 3m + 2\]

Step 2: Differentiate both parts of the function

Differentiate the first part, \(g(x) = k\sqrt{x+1}\), to obtain \(g'(x) = \frac{k}{2\sqrt{x+1}}\). Evaluate at \(x = 3\): \[g'(3) = \frac{k}{2\sqrt{4}} = \frac{k}{4}\]Differentiate the second part, \(g(x) = mx + 2\), to obtain \(g'(x) = m\). The derivative from either side at \(x = 3\) must also be equal for differentiability.

Step 3: Ensure derivatives match at boundary point

Set the derivatives equal at \(x = 3\):\[\frac{k}{4} = m\]Now we have two equations: 1) \(2k = 3m + 2\) 2) \(\frac{k}{4} = m\).

Step 4: Solve the system of equations

Substitute \(m = \frac{k}{4}\) (from the second equation) into the first equation:\[2k = 3\left(\frac{k}{4}\right) + 2\]Simplify and solve for \(k\): \[2k = \frac{3k}{4} + 2\]Multiply through by 4 to clear the fraction:\[8k = 3k + 8\]Thus, \(5k = 8\) so \(k = \frac{8}{5}\).

Step 5: Find m using k value

Substitute \(k = \frac{8}{5}\) back into \(m = \frac{k}{4}\):\[m = \frac{8}{5} \times \frac{1}{4} = \frac{8}{20} = \frac{2}{5}\]

Step 6: Calculate k + m

Calculate the sum \(k + m\):\[k + m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2\]Therefore, \(k + m = 2\).

Key Concepts

Understanding Continuity

Continuity is a fundamental concept in calculus. It means that a function has no jumps, breaks, or holes at a particular point. To check if a function is continuous at a boundary point, we need to ensure that the left-hand limit, the right-hand limit, and the actual value of the function at that point are all the same. For the function in this exercise, the boundary point is where the function changes its definition, which is at \(x = 3\). We check continuity by evaluating the function from both sides of this point. For our function, as \(x\) approaches 3:

• From the left (\(x \leq 3\)), \(g(x) = k\sqrt{x+1}\), which becomes \(2k\) when \(x = 3\).
• From the right (\(x > 3\)), \(g(x) = mx + 2\), becomes \(3m + 2\) at \(x = 3\).

For continuity, the expressions \(2k\) and \(3m + 2\) must be equal. This gives us the first condition we need to work with: \(2k = 3m + 2\).

Exploring Derivatives

Derivatives tell us about how a function is changing at any point and are essential for understanding differentiability. A function is differentiable at a point if its derivative exists there. It also implies that the function is smooth without any sharp edges at that point. To find derivatives, we examine the two pieces of the given piecewise function.

• For the first part, \(g(x) = k\sqrt{x+1}\), the derivative is \(g'(x) = \frac{k}{2\sqrt{x+1}}\).
• For the second part, \(g(x) = mx + 2\), the derivative is simply \(g'(x) = m\).

We need the derivatives from both these pieces to be equal at \(x = 3\) to ensure the function is differentiable. This means that their slopes match up perfectly at the boundary: \(\frac{k}{4} = m\). Solving these derivative conditions together with the continuity equation leads us to the desired values of \(k\) and \(m\).

Analyzing Piecewise Functions

Piecewise functions are defined by different expressions depending on the value of \(x\). These functions can describe a variety of scenarios where different rules apply to different intervals. They are common in real-life situations where a system behaves differently under varying conditions. For the function \(g(x)\) provided in the exercise, we have a piecewise definition split into two parts: one applies when \(0 \leq x \leq 3\), and another when \(3 < x \leq 5\).

• The first piece, \(k\sqrt{x+1}\), usually models scenarios involving roots or variations that depend on the square root function.
• The second piece, \(mx + 2\), represents a linear relationship, indicating a straight-line pattern in another interval.

The challenge with piecewise functions arises at the boundaries, where two segments meet, requiring careful analysis to ensure continuity and differentiability at these points. In our exercise, applying these concepts allows us to solve the problem by enforcing the same limits and slopes, resulting in a function that is smooth and continuous across its entire domain.