Problem 33 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} &\text { If } \begin{aligned} f(x) &=a x^{2}+b, \quad x \leq 1 \\ &=b x^{2}+a x+c, \quad x>1, \end{aligned}\\\ &\text { where } b \neq 0 \text { . Find } a \text { and } c \text { such that } f(x) \text { is continuous and differentiable at } x=1 \text { . } \end{aligned} $$

Step-by-Step Solution

Answer

The function \(f(x)\) will be continuous and differentiable at \(x=1\) if the coefficient \(a\) is twice the coefficient \(b\) (i.e., \(a = 2b\)), and the coefficient \(c\) is equal to 0 (i.e., \(c = 0\)).

Step 1: Check for continuity at x=1

In order for the function \(f(x)\) to be continuous at \(x=1\), both parts of the function must be equal at \(x=1\). From the given equation, we can write: \(f(1) = a(1)^2 + b\) \(f(1) = b(1)^2 + a(1) + c\) Now, set these two equations equal to each other: \(a + b = b + a + c\)

Step 2: Solve for c

To find the value of \(c\), we can rearrange the equation obtained in Step 1: \(a + b = b + a + c \Rightarrow c = a + b - (b + a) \Rightarrow c = 0\) Now we know that \(c = 0\).

Step 3: Check for differentiability at x=1

In order for the function \(f(x)\) to be differentiable at \(x=1\), the derivatives of both parts of the function must also be equal at \(x=1\). Find the derivatives of both parts of the function: \(f'(x) = 2ax \quad \text{for } x \leq 1\) \(f'(x) = 2bx + a \quad \text{for } x > 1\) Now, set the derivatives equal to each other at \(x=1\): \(2a(1) = 2b(1) + a\)

Step 4: Solve for a

To find the value of \(a\), we can solve the equation obtained in Step 3: \(2a = 2b + a \Rightarrow a = 2b\) So, \(a\) is equal to \(2b\).

Step 5: Final answer

We have found the values of \(a\) and \(c\) that make the function \(f(x)\) continuous and differentiable at \(x=1\): \(a = 2b\) \(c = 0\) Thus, the function will be continuous and differentiable at \(x=1\) if the coefficient \(a\) is twice the coefficient \(b\), and the coefficient \(c\) is equal to 0.

Key Concepts

Piecewise Functions

Piecewise functions are mathematical functions defined by different expressions based on the input values. This means that the function takes on different forms depending on the region of the input. For instance, in our given problem, the function is defined in two parts:

• For values of x less than or equal to 1: the function is defined as \(f(x) = ax^2 + b\).
• For values of x greater than 1: the function transforms to \(f(x) = bx^2 + ax + c\).

This kind of function is useful in real-world applications where a system behaves differently under different conditions.
Understanding how to handle and work with piecewise functions allows us to evaluate and ensure certain properties like continuity and differentiability across different sections of the function. It's crucial to address each "piece" of the function separately while observing the turning point where the function expression changes, often known as the point of interest.

Differentiation

Differentiation is a core concept in calculus and refers to the process of finding the derivative of a function, which represents the rate at which the function values change concerning its inputs. In the context of piecewise functions, differentiating each "piece" separately is essential to analyzing the behavior of the function at specific points.
For our function defined in the exercise:

• The first part is differentiated to give \(f'(x) = 2ax\) when \(x \leq 1\).
• The second part differentiates to \(f'(x) = 2bx + a\) when \(x > 1\).

For the function to be differentiable at a certain point \(x=1\), the derivatives from both sides of the point must be equal. This ensures a smooth transition of the function at the point of interest without any sharp corners or breaks. Solving for equality of derivatives at this point is critical for determining the necessary conditions on the coefficients of the function to achieve differentiability.

Continuity at a Point

When talking about the continuity of functions, particularly at a specific point, it implies that the function does not have any interruptions, jumps, or holes at that point. A function is said to be continuous at a point if the function's value just before and just after that point is consistent and aligns perfectly with the function's value at the point itself.
For our piecewise function to be continuous at \(x = 1\), it requires that:

• The output from the first piece \(\left(ax^2 + b\right)\) at \(x = 1\) should match the output from the second piece \(\left(bx^2 + ax + c\right)\) at the same \(x = 1\).
• This matching condition was solved by ensuring the equation \(a + b = b + a + c\) holds true, thereby using the condition to solve for \(c = 0\).

Thus, enforcing continuity at the point helps in maintaining a smooth function without any disconnection exactly at the point where the two pieces of the function converge.