Question

Let \[ f(x)= \begin{cases} \dfrac{|a|x + 2x^2 - 2\sin|x|\cos|x|}{x}, & x \neq 0 \\ b, & x = 0 \end{cases} \] If \( f(x) \) is continuous, then the value of \( a + b \) is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For continuity at a point, the left-hand limit, right-hand limit, and function value must be equal.

Answer 1

The Correct Option is A

To determine the value of \(a + b\) for the given function to be continuous at \(x = 0\), we need to ensure the limit of \(f(x)\) as \(x\) approaches 0 matches \(f(0) = b\). We start by evaluating the limit: 

The function \(f(x)\) is given by:

\(f(x) = \begin{cases} \dfrac{|a|x + 2x^2 - 2\sin|x|\cos|x|}{x}, & x \neq 0 \\ b, & x = 0 \end{cases}\)

The limit as \(x\) approaches 0 is:

\(\lim_{{x \to 0}} \dfrac{|a|x + 2x^2 - 2\sin|x|\cos|x|}{x}\)

We simplify this expression. The term \(2\sin|x|\cos|x|\) can be rewritten using the trigonometric identity \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) as:

\(2\sin|x|\cos|x| = \sin(2|x|)\)

This gives:

\(|a|x + 2x^2 - \sin(2|x|)\)

Dividing by \(x\), we get:

\(\dfrac{|a|x + 2x^2 - \sin(2|x|)}{x} = |a| + 2x - \dfrac{\sin(2|x|)}{x}\)

As \(x \to 0\), using the small angle approximation \(\sin(2x) \approx 2x\), we find that \(\dfrac{\sin(2|x|)}{x} = 2\). Hence:

\(\lim_{{x \to 0}} \left( |a| + 2x - 2 \right) = |a| - 2\)

For continuity at \(x = 0\), it must be that:

\(|a| - 2 = b\)

We know \(a + b = 2\), so:

\(a + (|a| - 2) = 2\)

This simplifies to:

\(a + |a| = 4\)

There are two cases:

• If \(a \geq 0\), then \(a = |a|\) and \(a = 2\).
• If \(a < 0\), then \(a = -|a|\) which is not possible for this equality.

Thus, \(a = 2\) and \(b = 0\) yielding \(a + b = 2\).

Therefore, the answer is 2.