Question

Which one of the following functions is discontinuous at $x=1$?

• A. $f(x) = \sin^2 x + \tan^2 x + \cos^2 x - \sec^2 x$
• B. $f(x) = \dfrac{1}{1 + 2^{\sin x}}$
• C. \( f(x) = \begin{cases} \dfrac{x - 1}{|x - 1| + 2(x - 1)^2}, & \text{if } x \ne 1 \\ 1, & \text{if } x = 1 \end{cases} \)
• D. $f(x) = e^x + 5$

Hint:

Continuity Criteria at a Point:

• Check $\lim_x \to a^-$ and $\lim_x \to a^+$.
• Both must exist and equal $f(a)$ for continuity.
• Pay special attention to piecewise/absolute value definitions.

Answer 1

The Correct Option is C

• (1): Simplifies to constant function $f(x) = 0$, continuous at $x=1$.
• (2): Continuous composition of continuous functions $\Rightarrow$ continuous.
• (3): Left-hand limit $= -1$, right-hand limit $= 1$, $f(1)=1$ $\Rightarrow$ discontinuous.
• (4): Exponential function $+$ constant is continuous.

Only option (3) is discontinuous at $x = 1$.

Answer 2

Step 1: Understand the function and continuity at \(x=1\)
The function is defined as:
\( f(x) = \frac{x - 1}{|x - 1| + 2(x - 1)^2} \) for \( x \neq 1 \), and \( f(1) = 1 \).
To check continuity at \( x = 1 \), the left-hand limit (LHL) and right-hand limit (RHL) as \( x \to 1 \) must be equal and also equal to \( f(1) \).

Step 2: Calculate the right-hand limit (as \( x \to 1^+ \))
For \( x > 1 \), \( |x - 1| = x - 1 \). So,
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x - 1}{(x - 1) + 2(x - 1)^2} = \lim_{h \to 0^+} \frac{h}{h + 2h^2} = \lim_{h \to 0^+} \frac{h}{h(1 + 2h)} = \lim_{h \to 0^+} \frac{1}{1 + 2h} = 1 \]

Step 3: Calculate the left-hand limit (as \( x \to 1^- \))
For \( x < 1 \), \( |x - 1| = -(x - 1) \). So,
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x - 1}{-(x - 1) + 2(x - 1)^2} = \lim_{h \to 0^-} \frac{h}{-h + 2h^2} = \lim_{h \to 0^-} \frac{h}{-h(1 - 2h)} = \lim_{h \to 0^-} \frac{1}{-1 + 2h} = -1 \]

Step 4: Compare limits and function value at \(x=1\)
The left-hand limit is \(-1\), the right-hand limit is \(1\), and the function value at \(x=1\) is \(1\).
Since \(\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)\), the limit does not exist at \(x=1\).
Hence, the function is discontinuous at \(x=1\).