Question

If \(\mathrm{f}(\mathrm{x})=\mid \begin{array}{ll}\mathrm{m}+3 \mathrm{nx}, & \mathrm{x}>1 \\ 11, & \mathrm{x}=1 \\ 5 \mathrm{nx}-2 \mathrm{~m}, & \mathrm{x}<1\end{array}\) is continuous at \(\mathrm{x}=1\) then \(\mathrm{m}=\ldots .\) and \(\mathrm{n}=\ldots . .\) ? (a) \(\mathrm{m}=2, \mathrm{n}=-3\) (b) \(\mathrm{m}=-2, \mathrm{n}=3\) (c) \(\mathrm{m}=2, \mathrm{n}=3\) (d) \(m=3, n=3\)

Short answer

The solution is \(m = \frac{11}{4}\) and \(n = \frac{11}{4}\). None of the given choices satisfy the condition of continuity at x=1.

Step-by-step solution

Calculate the left-hand limit.

To find the left-hand limit, evaluate the limit of the function as x approaches 1 from the left: \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5nx - 2m).\)

Calculate the right-hand limit.

To find the right-hand limit, evaluate the limit of the function as x approaches 1 from the right: \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (m + 3nx).\)

Set the left-hand limit and the right-hand limit equal.

For the function to be continuous at x=1, the left-hand limit must be equal to the right-hand limit: \(5n - 2m = m + 3n.\)

Solve for m in terms of n.

Solve the equation obtained in step 3 for m, so we can then find n: \(m = 3n - 2m \Rightarrow 3m = 3n \Rightarrow m = n.\)

Check the value of the function at x=1.

The value of the function at x=1 is given as: \(f(1) = 11.\)

Set the right-hand limit equal to the function value at x=1.

For the function to be continuous at x=1, the right-hand limit must be equal to the function value at x=1: \(m(1) + 3n(1) = 11 \Rightarrow m + 3n = 11.\)

Substitute m=n and solve for n.

Use the relation m=n from step 4 and substitute it into the equation from step 6 to solve for n: \(n + 3n = 11 \Rightarrow 4n = 11 \Rightarrow n = \frac{11}{4}.\)

Find the value of m.

Substitute the value of n back into the relation m=n: \(m = n \Rightarrow m = \frac{11}{4}.\) The solution is \(m = \frac{11}{4}\) and \(n = \frac{11}{4}\). None of the choices given in the problem matches this solution, and therefore none of the choices satisfy the condition of continuity at x=1.

Key concepts

Limit of a Function

When we explore the limit of a function, we are seeking to understand what value the function approaches as the input, typically denoted as \(x\), gets nearer to some specific point. To determine continuity at a specific point, we need to evaluate both the left-hand and right-hand limits. For the function \(f(x)\), as \(x\) approaches 1, calculating the limits from both sides is crucial:
- **Left-hand limit:** This is calculated as \(x\) approaches 1 from the left (denoted as \(x \to 1^-\)). For the exercise, this involved the expression \(5nx - 2m\).- **Right-hand limit:** Conversely, this is calculated as \(x\) approaches 1 from the right (denoted as \(x \to 1^+\)), involving the expression \(m + 3nx\).
Safe to say, determining limits precisely helps us ascertain whether a defining point of interest, such as \(x=1\), will register as a point of continuity or not.

Piecewise Functions

Piecewise functions are defined by different expressions over different intervals of the domain. They essentially "piece together" different functions, each valid over a specific interval. In our given scenario, the function \(\mathrm{f}(\mathrm{x})\) is composed as follows:

• \(m+3nx\) when \(x>1\)
• 11 when \(x=1\)
• \(5nx-2m\) when \(x<1\)

This construction allows for flexibility and accommodation of different mathematical behaviors within different segments of the number line. For a piecewise function to be continuous at a certain point, it must not only align across its segments, but must also ensure that computed limits as discussed earlier confirm to the same consistent point value.

Equating Limits

Equating limits focuses on making sure both the left-hand and right-hand limits of a function meet at the same value for continuity to hold at a focused point, here namely \(x=1\).
- **Establishing equal limits:** From the step-by-step solution, we see this involves setting \(5n - 2m = m + 3n\). This equation ensured that both these limits must match at \(x = 1\) to denote the point as continuous.- **Verifying against actual function value:** Further, for absolute continuity verification, the function’s value at \(x=1\) assessed previously should match the confirmed limit. This is what leads to another equality, \(m + 3n = 11\).
In resolving these equations, we find \(m\) and \(n\) must equal \(\frac{11}{4}\) for perfect alignment, verifying none of the given options fit the experimental solution needed to ensure continuity at the specific point of \(x = 1\). By precisely verifying these conditions, we establish a pivotal understanding of how equated limits solidify continuity in piecewise compositions.