Question

Let \(f(x)=\begin{cases}cx^{2}+2x,&\text{if } x<2\\2x+4,&\text{if } x\ge2\end{cases}\). If the function \(f\) is continuous on \((-\infty,\infty)\), then the value of \(c\) is equal to

• A. 4
• B. 2
• C. 3
• D. 1
• E. 5

Hint:

Calculus Tip: "Continuous" geometrically implies the graph has no jumps. Setting the two piecewise equations equal to each other at the exact boundary point guarantees their graphs physically touch at that specific coordinate!

Answer 1

The Correct Option is D

Concept:
For a piecewise function to be continuous everywhere, its individual pieces must smoothly connect at the transition boundary. This means the Left-Hand Limit (LHL) as $x$ approaches the boundary point must exactly equal the Right-Hand Limit (RHL) at that point.

Step 1: Identify the transition point.
The function definition splits at the boundary $x = 2$. This is the only point where a discontinuity could occur, so we must force continuity here.

Step 2: Calculate the Left-Hand Limit (LHL).
Approach $x = 2$ from values less than 2 ($x \to 2^-$). Use the top piece: $$\text{LHL} = \lim_{x\rightarrow2^-} (cx^2 + 2x)$$ Substitute $x = 2$: $$\text{LHL} = c(2)^2 + 2(2) = 4c + 4$$

Step 3: Calculate the Right-Hand Limit (RHL).
Approach $x = 2$ from values greater than 2 ($x \to 2^+$). Use the bottom piece: $$\text{RHL} = \lim_{x\rightarrow2^+} (2x + 4)$$ Substitute $x = 2$: $$\text{RHL} = 2(2) + 4 = 4 + 4 = 8$$

Step 4: Set the LHL and RHL equal to each other.
For the function to be continuous at $x = 2$, the limits from both sides must match perfectly: $$\text{LHL} = \text{RHL}$$ $$4c + 4 = 8$$

Step 5: Solve the resulting equation for c.
Subtract 4 from both sides: $$4c = 4$$ Divide by 4: $$c = 1$$ Hence the correct answer is (D) 1.