Question

If the function $f(x)=\begin{cases}\frac{x^{2}}{2},& 0\le x<1\\ 2x^{2}-ax+1.5,& 1\le x\end{cases}$ is continuous on $[0, 5]$, then the value of a is

• A. 1
• B. -2
• C. -1
• D. 3
• E. -5

Hint:

Math Tip: When dealing with piecewise functions involving polynomials, you do not need to formally evaluate the limits. Simply plug the transition $x$-value directly into both pieces of the function and set them equal to each other!

Answer 1

The Correct Option is D

Concept:
Calculus - Continuity of a Function.
A function $f(x)$ is continuous at a point $x = c$ if the Left Hand Limit (LHL), Right Hand Limit (RHL), and the function value at $c$ are all equal: $\lim_{x\rightarrow c^-} f(x) = \lim_{x\rightarrow c^+} f(x) = f(c)$.
Step 1: Identify the critical point for continuity.
The function definition changes at $x = 1$. For $f(x)$ to be continuous on the entire interval $[0, 5]$, it must be continuous at the transition point $x = 1$.
Step 2: Calculate the Left Hand Limit (LHL) at $x = 1$.
For $x<1$, the function is $f(x) = \frac{x^2}{2}$. $$ \text{LHL} = \lim_{x\rightarrow 1^-} \frac{x^2}{2} $$ Substitute $x = 1$: $$ \text{LHL} = \frac{(1)^2}{2} = 0.5 $$
Step 3: Calculate the Right Hand Limit (RHL) at $x = 1$.
For $x \ge 1$, the function is $f(x) = 2x^2 - ax + 1.5$. $$ \text{RHL} = \lim_{x\rightarrow 1^+} (2x^2 - ax + 1.5) $$ Substitute $x = 1$: $$ \text{RHL} = 2(1)^2 - a(1) + 1.5 $$ $$ \text{RHL} = 2 - a + 1.5 = 3.5 - a $$
Step 4: Equate LHL and RHL to find 'a'.
For continuity at $x = 1$, LHL must equal RHL: $$ 0.5 = 3.5 - a $$
Step 5: Solve for the constant a.
Rearrange the equation: $$ a = 3.5 - 0.5 $$ $$ a = 3 $$