Question Statement

Q16. (a) The function $f(x)$ is defined as: $f(x)=\{\begin{array}{ll}\frac{x^2-4}{x-2}& ,x\ne 2\\ m& ,x=2\end{array}$

Find the value of $m$ for which $f(x)$ is continuous at $x=2$.

(b) A function is defined as: $f(x)=\{\begin{array}{ll}mx-n& ,x<1\\ 2mx+n& ,x>1\\ 5& ,x=1\end{array}$

Find the values of $m$ and $n$ that make $f(x)$ continuous at $x=1$.

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Background and Explanation

For a function to be continuous at a point $x=a$, three conditions must be satisfied: (1) $f(a)$ must exist, (2) ${\lim }_{x\to a}f(x)$ must exist (meaning left-hand and right-hand limits are equal), and (3) ${\lim }_{x\to a}f(x)=f(a)$. For rational functions with removable discontinuities, factorization and cancellation often simplify limit calculations.

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Solution

Part (a): Continuity at $x=2$

To find the limit as $x$ approaches $2$, we evaluate the left-hand and right-hand limits for the rational expression.

Left Hand Limit

$\begin{array}{rl}\underset{x\to 2^{-}}{\lim }f(x)& =\underset{x\to 2^{-}}{\lim }\left(\frac{x^2-4}{x-2}\right)\\ & =\underset{x\to 2^{-}}{\lim }\left(\frac{(x-2)(x+2)}{x-2}\right)\text{(factorizing the numerator)}\\ & =\underset{x\to 2^{-}}{\lim }(x+2)\text{(canceling }x-2\text{ since }x\ne 2)\\ & =2+2\\ \underset{x\to 2^{-}}{\lim }f(x)& =4\end{array}$

Right Hand Limit

By similar calculation (not explicitly shown in raw data but implied by the equality): ${\lim }_{x\to 2^{+}}f(x)=4$

Continuity Condition

From equations (1) and (2), it is clear that: $\begin{array}{rl}\underset{x\to 2^{-}}{\lim }f(x)& =4=\underset{x\to 2^{+}}{\lim }f(x)\\ \Rightarrow \underset{x\to 2}{\lim }f(x)& =4\end{array}$

Also given: $f(2)=m$

Since $f(x)$ is continuous at $x=2$, we require: $\begin{array}{rl}\underset{x\to 2}{\lim }f(x)& =f(2)\\ 4& =m\end{array}$

Therefore, $\boxed{m=4}$.

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Part (b): Continuity at $x=1$

For this piecewise function, we check the left-hand limit, right-hand limit, and function value at $x=1$.

Left Hand Limit

$\begin{array}{rl}\underset{x\to 1^{-}}{\lim }f(x)& =\underset{x\to 1^{-}}{\lim }(mx-n)\\ & =m(1)-n\\ & =m-n\\ \underset{x\to 1^{-}}{\lim }f(x)& =m-n\end{array}$

Right Hand Limit

$\begin{array}{rl}\underset{x\to 1^{+}}{\lim }f(x)& =\underset{x\to 1^{+}}{\lim }(2mx+n)\\ & =2m(1)+n\\ \underset{x\to 1^{+}}{\lim }f(x)& =2m+n\end{array}$

Also given: $f(1)=5$

Continuity Condition

Since $f(x)$ is continuous at $x=1$, all three values must be equal: $\begin{array}{rl}\therefore \underset{x\to 1^{-}}{\lim }f(x)& =\underset{x\to 1^{+}}{\lim }f(x)=f(1)\\ m-n& =2m+n=5\end{array}$

This yields two equations: \begin{align} m-n &=5 \\ 2 m+n &=5 \end{align}

Solving the System

Adding equations (1) and (2): $\begin{array}{rl}m-n& =5\\ 2m+n& =5\\ 3m& =10\\ m& =\frac{10}{3}\end{array}$

Substituting into equation (1): $\frac{10}{3}-n=5$

(The solution continues by solving for $n$, giving $n=\frac{10}{3}-5=-\frac{5}{3}$)

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Key Formulas or Methods Used

• Continuity Condition: ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=f(a)$
• Difference of Squares: $x^2-4=(x-2)(x+2)$
• Limit of Rational Functions: Cancellation of common factors for removable discontinuities
• One-sided Limits: Evaluating piecewise functions by approaching from left and right separately
• System of Linear Equations: Solving simultaneous equations by addition/elimination method

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Summary of Steps

Part (a):

1. Factor the numerator: $x^2-4=(x-2)(x+2)$
2. Cancel the common factor $(x-2)$ from numerator and denominator
3. Evaluate the limit: ${\lim }_{x\to 2}(x+2)=4$
4. Set $f(2)=m$ equal to the limit: $m=4$

Part (b):

1. Calculate left-hand limit: ${\lim }_{x\to 1^{-}}(mx-n)=m-n$
2. Calculate right-hand limit: ${\lim }_{x\to 1^{+}}(2mx+n)=2m+n$
3. Set both equal to $f(1)=5$, creating system: $m-n=5$ and $2m+n=5$
4. Add equations to eliminate $n$: $3m=10\Rightarrow m=\frac{10}{3}$
5. Substitute back to find $n$ (yielding $n=-\frac{5}{3}$)