Examine the continuity of the function f(x) = […] at x = 1.

The correct answer is Option (2) → Continuous at $x = 1$ because $\text{LHL} = \text{RHL} = f(1)$. ##

We have,

$f(x) = \begin{cases} \frac{x^2}{2}, & \text{if } 0 \leq x \leq 1 \\ 2x^2 - 3x + \frac{3}{2}, & \text{if } 1 < x \leq 2 \end{cases} \text{ at } x = 1$

At $x = 1$,

$\text{LHL} = \lim\limits_{x \to 1^-} \frac{x^2}{2} = \lim\limits_{h \to 0} \frac{(1 - h)^2}{2}$

Put $x = 1 - h$,

$= \lim\limits_{h \to 0} \frac{1 + h^2 - 2h}{2} = \frac{1}{2} \quad [∵(a - b)^2 = a^2 + b^2 - 2ab]$

$\text{RHL} = \lim\limits_{x \to 1^+} \left( 2x^2 - 3x + \frac{3}{2} \right)$

Put $x = 1 + h$,

$= \lim\limits_{h \to 0} \left[ 2(1 + h)^2 - 3(1 + h) + \frac{3}{2} \right] = \lim\limits_{h \to 0} \left( 2 + 2h^2 + 4h - 3 - 3h + \frac{3}{2} \right)$

$= -1 + \frac{3}{2} = \frac{1}{2} \quad [∵(a + b)^2 = a^2 + b^2 + 2ab]$

and $f(1) = \frac{1^2}{2} = \frac{1}{2}$

$∴\text{LHL} = \text{RHL} = f(1)$

Hence, $f(x)$ is continuous at $x = 1$.