Question

Let \(f(x)=\begin{cases A+2x,&\text{if }x \\ 1+x^{2},&\text{if }x\ge3 \end{cases}\). If the function \(f(x)\) is continuous at \(x=3\), then the value of \(A\) is}

• A. \(1\)
• B. \(6\)
• C. \(10\)
• D. \(4\)

Hint:

For piecewise functions, \[ \boxed{ \text{Continuity} \Longrightarrow \text{LHL}=\text{RHL}=f(a) } \] Always calculate the left-hand expression and right-hand expression separately before equating them.

Answer 1

The Correct Option is D

Concept: A function is continuous at a point if its left-hand limit (LHL), right-hand limit (RHL), and function value at that point are all equal. Mathematically, \[ \boxed{ \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a) } \] Since the function is piecewise defined, we compute the left-hand value and the function value at \(x=3\).

Step 1: Find the left-hand limit.
For \(x<3\), \[ f(x)=A+2x \] Therefore, \[ \lim_{x\to3^-}f(x)=A+2(3) \] \[ =A+6 \]

Step 2: Find the right-hand limit and function value.
For \(x\ge3\), \[ f(x)=1+x^2 \] Hence, \[ f(3)=1+3^2 \] \[ =1+9 \] \[ =10 \] Also, \[ \lim_{x\to3^+}f(x)=10 \]

Step 3: Apply the continuity condition.
Since the function is continuous, \[ A+6=10 \] Therefore, \[ A=10-6 \] \[ \boxed{A=4} \] Final Answer: \[ \boxed{A=4} \] Hence, \[ \boxed{Option (D) is correct \]