Question

Let \( f: \mathbb{R} \to \mathbb{R} \) be defined by \[ f(x) = \begin{cases} a - \frac{\sin[x-1]}{x-1} & , \text{if } x>1 \\ 1 & , \text{if } x = 1 \\ b - \frac{\sin([x-1] - [x-1]^3)}{([x-1]^2)} & , \text{if } x<1 \end{cases} \] where \([t]\) denotes the greatest integer less than or equal to t. If f is continuous at \(x=1\), then \(a+b=\)

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For a function to be continuous at \(x=c\), \( \lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = f(c) \). Evaluate LHL and RHL carefully, especially with greatest integer functions \([t]\) and absolute values \(|t|\). For \(x \to c^+\), let \(x=c+h\) with \(h \to 0^+\). For \(x \to c^-\), let \(x=c-h\) with \(h \to 0^+\). For \([t]\): if \(t \to N^+\) (N integer), \([t]=N\). If \(t \to N^-\), \([t]=N-1\). Trigonometric values: know the approximate range of \(\sin(x)\) for \(x\) in radians. \(\sin(2 \text{ rad}) \approx 0.909\).

Answer 1

The Correct Option is B

Let \( f: \mathbb{R} \to \mathbb{R} \) be defined by:
\[ f(x) = \begin{cases} a - \frac{\sin(x-1)}{x-1} & , \text{if } x > 1 \\ 1 & , \text{if } x = 1 \\ b - \frac{\sin([x-1] - [x-1]^3)}{([x-1]^2)} & , \text{if } x < 1 \end{cases} \] where \( [t] \) denotes the greatest integer less than or equal to \( t \).
We need to find \( a + b \) if \( f \) is continuous at \( x = 1 \).
**Continuity Condition:**
For \( f(x) \) to be continuous at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 1 \] **Right-Hand Limit (RHL):**
Let \( x = 1 + h \), where \( h \to 0^+ \).
Then for \( x > 1 \), the expression becomes: \[ f(x) = a - \frac{\sin(x - 1)}{x - 1} \] As \( h \to 0^+ \), we get: \[ \lim_{x \to 1^+} f(x) = a - \frac{\sin(0)}{0} = a \] For continuity, \( a = 1 \).
**Left-Hand Limit (LHL):**
Let \( x = 1 - h \), where \( h \to 0^+ \).
Then for \( x < 1 \), the expression becomes: \[ f(x) = b - \frac{\sin([x - 1] - [x - 1]^3)}{([x - 1]^2)} \] Substituting \( [x - 1] = -1 \), we get: \[ \lim_{x \to 1^-} f(x) = b - \frac{\sin(-2)}{1} = b - [-\sin(2)] = b + 1 \] For continuity, \( b + 1 = 1 \), so \( b = 0 \).
Therefore, \( a = 1 \) and \( b = 0 \). Hence, \( a + b = 1 + 0 = 1 \).
The answer is \( \boxed{1} \).