Q.A rational function is a function of the form $f(x) = \dfrac{p(x)}{q(x)}$, where $p(x)$ and $q(x)$ are polynomial functions of $x$ and $q(x) \neq 0$. Prove that every such rational function is continuous (i.e. continuous at every point of its domain, the set of all real $x$ for which $q(x) \neq 0$).

A function is continuous everywhere if it is continuous at every point in its domain. For a piecewise function, the individual pieces are typically continuous within their defined intervals (polynomials and trigonometric functions are continuous). The critical points to check for continuity are the boundary points where the function's definition changes.

In this problem, the function $f(x)$ is defined differently for $x \leq 0$, $0 < x \leq 2$, and $x > 2$. The potential points of discontinuity are $x=0$ and $x=2$. For $f(x)$ to be continuous everywhere, it must be continuous at these two points.

We will apply this condition at $x=0$ and $x=2$ to find the values of $a$ and $b$.

1. Check continuity at $x=0$: For $f(x)$ to be continuous at $x=0$, the left-hand limit, right-hand limit, and the function value at $x=0$ must all be equal.
   • Left-hand limit at $x=0$: For $x < 0$, $f(x) = 1 + \cos x$.

*   **Right-hand limit at $x=0$**: For $x > 0$, $f(x) = a - x$.

*   **Function value at $x=0$**: For $x \leq 0$, $f(x) = 1 + \cos x$.

For continuity at $x=0$, we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
Therefore, $2 = a = 2$. This implies that $a=2$.

2. Check continuity at $x=2$:

Similarly, for $f(x)$ to be continuous at $x=2$, the left-hand limit, right-hand limit, and the function value at $x=2$ must all be equal.

* Left-hand limit at $x=2$: For $x < 2$, $f(x) = a - x$.

*   **Right-hand limit at $x=2$**: For $x > 2$, $f(x) = x^2 - b^2$.

*   **Function value at $x=2$**: For $0 < x \leq 2$, $f(x) = a - x$.

For continuity at $x=2$, we must have $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$.
Therefore, $a - 2 = 4 - b^2 = a - 2$. This gives us the equation $a - 2 = 4 - b^2$.

3. Solve for $a$ and $b$ and calculate $a^2 + b^2$:

From step 1, we found $a=2$.

Substitute $a=2$ into the equation from step 2:

Now we need to find $a^2 + b^2$.

The key idea is that a piecewise function is continuous everywhere if and only if it is continuous at each boundary where the definition changes. At those points, the left-hand limit, right-hand limit, and the function’s value must all be equal. Here we have three boundaries: $x=0$, $x=1$, and $x=2$. Each gives one equation, and three unknowns ($a,b,c$) means we can solve uniquely.

Let’s work through each boundary carefully.

1. Continuity at $x=0$ For $x \leq 0$, $f(x) = \cos(2x + \pi)$. At $x=0$, this gives $f(0) = \cos(\pi) = -1$. For $0 < x < 1$, $f(x) = ax^2 + b$. As $x \to 0^+$, this approaches $a(0)^2 + b = b$. Continuity at $x=0$ requires

so $-1 = b$. Hence

2. Continuity at $x=1$ For $0 < x < 1$, $f(x) = ax^2 + b$. As $x \to 1^-$, this approaches $a(1)^2 + b = a + b$. For $1 \leq x \leq 2$, $f(x) = cx + 4$. At $x=1$, this gives $f(1) = c(1) + 4 = c + 4$. Continuity at $x=1$ requires

so $a + b = c + 4$. Substituting $b = -1$ gives

3. Continuity at $x=2$ For $1 \leq x \leq 2$, $f(x) = cx + 4$. As $x \to 2^-$, this approaches $c(2) + 4 = 2c + 4$. For $x \geq 2$, $f(x) = 3a + 1$. At $x=2$, this gives $f(2) = 3a + 1$. Continuity at $x=2$ requires

so $2c + 4 = 3a + 1$. Rearranging:

Now solve the system:

From the first equation, $a = c + 5$. Substitute into the second:

Then $a = -12 + 5 = -7$. But wait — check: $a = -7$, $c = -12$? Let’s verify with equation (2): $3(-7) - 2(-12) = -21 + 24 = 3$, correct. So $a = -7$, $b = -1$, $c = -12$.

Now compute $b^2 - bc + c^2$:

Sum: $1 - 12 + 144 = 133$.

Concept & Intuition

A piecewise function is continuous on its domain if it doesn’t “jump” at the points where the formula changes. At each such boundary, the left-hand limit (from the piece before) must equal the right-hand limit (from the piece after), and both must equal the function’s value there (if defined). Here the boundaries are $x=1$ and $x=4$. We compute the limits using standard calculus facts: $\lim_{u\to0} \frac{\tan u}{u}=1$ and $\lim_{u\to0} \frac{b^u-1}{u} = \log b$ (but careful — the third piece is $\frac{b^x-1}{x}$ as $x\to4^+$, not near 0, so we just plug in directly). The second piece is a rational function that simplifies.

Step-by-step solution

1. Continuity at $x=1$ For $0<x<1$, $f(x) = \frac{\tan a(x-1)}{x-1}$. Let $u = x-1$; as $x\to1^-$, $u\to0^-$. Then

For $1\le x\le4$, at $x=1$ we have

Continuity requires $\lim_{x\to1^-} f(x) = f(1)$, so

2. Continuity at $x=4$ For $1\le x\le4$, at $x=4$:

For $x>4$, $f(x) = \frac{b^x - 1}{x}$. The limit as $x\to4^+$ is simply

Continuity demands

Multiply both sides by 36: $9(b^4 - 1) = 244 \Rightarrow 9b^4 - 9 = 244 \Rightarrow 9b^4 = 253 \Rightarrow b^4 = \frac{253}{9}$.

3. Compute $6a + 9b^4$ Substitute $a = \frac{31}{6}$ and $b^4 = \frac{253}{9}$:

Hence

When evaluating limits, especially those involving differentiable functions, the first step is always to substitute the limiting value to check for indeterminate forms. If we encounter $\frac{0}{0}$ or $\frac{\infty}{\infty}$, it indicates that further analysis is required, often involving L'Hôpital's Rule or algebraic manipulation to use the definition of the derivative.

The definition of the derivative is a fundamental concept:

Our strategy will be to transform the given limit expression into a form that allows us to directly apply this definition. This usually involves adding and subtracting a specific term in the numerator to create the necessary $(f(x) - f(a))$ and $(x-a)$ components.

Here's how we can solve the problem:

1. Check for indeterminate form:

   Let's substitute $x=a$ into the given expression:

   Numerator: $af(a) - af(a) = 0$

   Denominator: $a-a = 0$

   Since we have the indeterminate form $\frac{0}{0}$, we can proceed with further evaluation.

2. Manipulate the numerator:

   The numerator is $xf(a) - af(x)$. To use the definition of the derivative, we need terms like $f(x) - f(a)$. Notice that we have $af(x)$. If we could pair it with $af(a)$, we would get $a(f(x) - f(a))$.

   Let's add and subtract $af(a)$ in the numerator:

Now, we can factor terms:

3. Substitute back into the limit and split: Substitute this manipulated numerator back into the limit expression:

Now, split the fraction into two separate terms:

As $x \to a$, $x-a \neq 0$, so we can cancel the $(x-a)$ term in the first part:

4. Evaluate the limits: Using the properties of limits, we can evaluate each part separately:

Since $f(a)$ is a constant with respect to $x$:

By the definition of the derivative:

So, the expression becomes:

5. Apply the given condition: The problem states that $f'(a) = af(a)$. Substitute this into our result:

Factor out $f(a)$:

We evaluate $\displaystyle \lim_{x\to 3} \frac{11 - [2-x]}{[x+10]}$, where $[t]$ is the greatest integer $\le t$.

Left-hand limit ($x \to 3^-$): take $x$ slightly less than $3$ (e.g. $x = 2.9$).

So the expression is $\dfrac{11 - (-1)}{12} = \dfrac{12}{12} = 1.$

Right-hand limit ($x \to 3^+$): take $x$ slightly more than $3$ (e.g. $x = 3.1$).

So the expression is $\dfrac{11 - (-2)}{13} = \dfrac{13}{13} = 1.$

Both one-sided limits equal $1$, so the limit exists:

Concept and intuition:

The greatest integer function $[t]$ (also called the floor function) jumps at every integer. When we take a limit as $x$ approaches a point, if the argument of the floor function passes through an integer, the floor value changes abruptly. That often makes the two-sided limit fail to exist. Here, as $x \to 3$, both $[2 - x]$ and $[x + 10]$ cross integer boundaries, so we must check the left and right separately.

1. Identify the critical integer boundaries.

   For $x \to 3$, consider the expressions inside the floor functions:

   • $2 - x$: when $x = 3$, $2 - 3 = -1$. For $x$ just less than 3, $2 - x > -1$; for $x$ just greater than 3, $2 - x < -1$. So the floor $[2 - x]$ will change at $x = 3$.
   • $x + 10$: when $x = 3$, $x + 10 = 13$. For $x$ near 3, $x+10$ is near 13, but does it cross 13? Since 13 is an integer, we need to see if $x+10$ passes through 13 from below or above as $x$ passes through 3. Actually, $x+10 = 13$ exactly at $x=3$. For $x < 3$, $x+10 < 13$; for $x > 3$, $x+10 > 13$. So $[x+10]$ also changes at $x=3$.

2. Compute the left-hand limit ($x \to 3^-$).

   Let $x = 3 - h$ with $h \to 0^+$.

   • $2 - x = 2 - (3 - h) = -1 + h$. Since $h > 0$ small, $-1 + h$ is slightly greater than $-1$ but less than 0. Hence $[2 - x] = -1$.
   • $x + 10 = (3 - h) + 10 = 13 - h$. Since $h > 0$, $13 - h$ is slightly less than 13, so $[x + 10] = 12$.
   • The expression becomes $\frac{11 - (-1)}{12} = \frac{12}{12} = 1$.

3. Compute the right-hand limit ($x \to 3^+$).

   Let $x = 3 + h$ with $h \to 0^+$.

   • $2 - x = 2 - (3 + h) = -1 - h$. This is slightly less than $-1$, so $[2 - x] = -2$.
   • $x + 10 = (3 + h) + 10 = 13 + h$. This is slightly greater than 13, so $[x + 10] = 13$.
   • The expression becomes $\frac{11 - (-2)}{13} = \frac{13}{13} = 1$.

4. Compare the two one-sided limits.

   Both left-hand and right-hand limits equal 1. That suggests the two-sided limit might exist and equal 1. But wait — we must check carefully: is there any subtlety? The floor function is discontinuous, but if both sides give the same value, the limit exists. However, we must ensure that the denominator is never zero near $x=3$. Here $[x+10]$ is 12 or 13, both nonzero, so no division by zero.

   So the limit is 1.

Concept & Intuition

Rolle’s theorem says: if a function is continuous on $[p,q]$, differentiable on $(p,q)$, and $f(p)=f(q)$, then there is at least one $c$ in $(p,q)$ with $f'(c)=0$. Here we are told $-5$ and $4$ are roots, so $f(-5)=0$ and $f(4)=0$. That gives $f(-5)=f(4)=0$, satisfying the equal‑value condition. The theorem then guarantees some $c$ in $(-5,4)$ where the derivative is zero. Our job: find the cubic’s coefficients from the root information, then solve $f'(c)=0$ and pick the $c$ that lies in the interval.