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Q12:
Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} x^{10} - 1, & \text{if } x \le 1 \\ x^2, & \text{if } x > 1 \end{cases}$

Solution :

Step 1: Identifying the Domain and Potential Points of Discontinuity

The function $f(x)$ is defined in a piecewise manner over the set of all real numbers $\mathbb{R}$. The domain is partitioned at $x = 1$.

  • For $x < 1$, $f(x) = x^{10} - 1$, which is a polynomial function. Polynomial functions are continuous everywhere in their domain [Per the continuity theorem for polynomials].
  • For $x > 1$, $f(x) = x^2$, which is also a polynomial function, hence continuous for all $x > 1$.
  • The only potential point of discontinuity is at the transition point, $x = 1$.

Step 2: Checking Continuity at $x = 1$

A function $f(x)$ is continuous at $x = c$ if and only if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ [Definition of Continuity].

Evaluation of the Left-Hand Limit (LHL):

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1} (x^{10} - 1)$

$= (1)^{10} - 1 = 1 - 1 = 0$

Evaluation of the Right-Hand Limit (RHL):

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1} (x^2)$

$= (1)^2 = 1$

Evaluation of the Function Value at $x = 1$:

$f(1) = 1^{10} - 1 = 0$ [Using the condition $x \le 1$]

Step 3: Analytical Comparison

Limit Type Result
LHL ($\lim_{x \to 1^-} f(x)$) $0$
RHL ($\lim_{x \to 1^+} f(x)$) $1$
Function Value ($f(1)$) $0$

Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, the limit $\lim_{x \to 1} f(x)$ does not exist [By the Limit Existence Theorem]. Consequently, the function fails the criteria for continuity at $x = 1$.

Visual Representation of Discontinuity

x=1 f(1)=0 RHL=1

Final Solution: The function $f$ is discontinuous at $x = 1$. There are no other points of discontinuity in the domain of $f$.


More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.1


CBSE Solutions for Class 12 Mathematics Continuity and Differentiability


Chapters in CBSE - Class 12 Mathematics


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