If \[ f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x}, & x \ne 0, \\ b, & x = 0, \end{cases} \] is continuous at \( x = 0 \), then \( a + b \) is equal to.

Question

Let \[ f(x)=\lim_{\theta\to 0}\frac{\cos(\pi x)-x^{2/\theta}\sin(x-1)}{1-x^{2/\theta}(x-1)}. \] Statement 1: \(f(x)\) is discontinuous at \(x=1\).
Statement 2: \(f(x)\) is continuous at \(x=-1\).

Answer 1

To determine the value of \(a + b\) such that the function \(f(x)\) is continuous at \(x=0\), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at \(x=0\) are all equal.

The function is given by:

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

To check continuity at \(x=0\), we calculate the limit of \(f(x)\) as \(x\) approaches 0:

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

As \(x \to 0\), \(|x| = x\) because the limit is approached from both sides:

\lim_{x \to 0} (a|x| + x^2) = ax + x^2 \to 0

Now, we simplify the trigonometric expression. Using the identities \(\sin|x| \approx |x|\) and \(\cos|x| \approx 1\) as \(x \to 0\):

\sin|x|\cos|x| \approx |x| \cdot 1 = x

Thus,

\lim_{x \to 0} 2(\sin|x|)(\cos|x|) = 2x

Substitute back:

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

Factor and simplify the expression:

= \lim_{x \to 0} \frac{x(a + x - 2)}{x} = \lim_{x \to 0} (a + x - 2) = a - 2

For continuity at \(x=0\), the limit should equal the function value at \(x=0\), which is \(b\):

a - 2 = b

From \(a - 2 = b\), we find \(b = a - 2\).

To find \(a+b\):

a + b = a + (a - 2) = 2a - 2

Given that the expression should satisfy the continuity, typically \(a = 2\), so:

a + b = 2 \times 1 = 2

Hence, the value of \(a+b\) is 2.

Answer 2