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

> [!TLDR] > Each piece is a polynomial, so $f$ is continuous everywhere except possibly at $x=1$. There the left limit is $0$ but the right limit is $1$, so $f$ is discontinuous only at $x=1$. The function is $$f(x) = \begin{cases} x^{10} - 1, & x \leq 1 \\ x^2, & x > 1. \end{cases}$$ **1. Away from $x = 1$.** For $x 1$, $f(x) = x^2$ is a polynomial and hence continuous. So $f$ is continuous at every point except possibly $x = 1$. **2. At $x = 1$.** From the definition ($x \leq 1$ branch), $f(1) = 1^{10} - 1 = 0$. - Left-hand limit: $\displaystyle\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1 - 1 = 0.$ …

Find all points of discontinuity of f, where f is – Class 12 Mathematics

Exercise 5.1 · Q12

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Concept understanding — Continuity At A Point

Continuity at a Point

Imagine drawing the graph of a function and putting your pen down at $x = a$ . If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.

The Three-Condition Test

For $f(x)$ to be continuous at $x = a$ , all three must hold. If even one fails, $f$ is discontinuous there.

Important

Continuity at $x = a$ requires:

$f(a)$ is defined,
$\displaystyle \lim_{x \to a} f(x)$ exists (left- and right-hand limits are equal),
$\displaystyle \lim_{x \to a} f(x) = f(a)$ .

Condition 1 says $a$ is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at $a$ — no misplaced point.

Why All Three Are Needed

$f(x) = \dfrac{x^2 - 1}{x - 1}$ has $\lim_{x \to 1} f(x) = 2$ , yet $f(1)$ is undefined (zero denominator). Condition 1 fails, leaving a hole at $(1, 2)$ .

A piecewise function shows the opposite can be fine:

f(x) = \begin{cases} x + 1 & x < 2 \\ 3 & x = 2 \\ x + 1 & x > 2 \end{cases}

Here $f(2) = 3$ , both one-sided limits equal $3$ , and they match $f(2)$ — so all three hold and $f$ is continuous at $x = 2$ .

Common Pitfalls

Watch out

"Limit exists" does not mean "continuous." The hole example has a limit but no continuity — the limit must equal the function value.

Watch out

"Defined everywhere" does not mean "continuous." A piecewise function can have a value at every point and still jump. Always check the one-sided limits.

A Quick Check …

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Each piece is a polynomial, so $f$ is continuous everywhere except possibly at $x=1$ . There the left limit is $0$ but the right limit is $1$ , so $f$ is discontinuous only at $x=1$ .

The function is

f(x) = \begin{cases} x^{10} - 1, & x \leq 1 \\ x^2, & x > 1. \end{cases}

1. Away from $x = 1$ . For $x < 1$ , $f(x) = x^{10} - 1$ is a polynomial and hence continuous. For $x > 1$ , $f(x) = x^2$ is a polynomial and hence continuous. So $f$ is continuous at every point except possibly $x = 1$ .

2. At $x = 1$ . From the definition ( $x \leq 1$ branch), $f(1) = 1^{10} - 1 = 0$ .

Left-hand limit: $\displaystyle\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} - 1) = 1 - 1 = 0.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2022Set eng-2022-P2-B14 marksMCQ
$f(x)=\begin{cases}x+2, & \text{for } x<1 \\ 4x-1, & \text{for } 1\le x\le 3 \\ x^2+5, & \text{for } x>3\end{cases}$
›Reveal solutionSolution
$f$ is continuous at $x=1$ .

Concept and Intuition

Check continuity only at the junction points $x=1$ and $x=3$ ; each piece is polynomial (continuous) elsewhere, so the interior points named in wrong options are trivially continuous.

Step-by-Step Solution

At $x=1$ : left piece $x+2\to3$ , right piece $4x-1=3$ , value $=3$ . Matched $\Rightarrow$ continuous.

At $x=3$ : piece $4x-1=11$ , right piece $x^2+5=14$ . Mismatch $\Rightarrow$ discontinuous (so (C) is false). …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$f : \mathbb{R} \to \mathbb{R}$
›Reveal solutionSolution
f is continuous at x = 0 (both sides give 3) but jumps at x = 1 (left 7, right -5), so f is continuous everywhere except x = 1, i.e. on R \ {1}.

Concept and Intuition

Check the two join points of the piecewise definition; continuity holds where the one-sided limits and the value agree.

Step-by-Step Solution

At x = 0: left limit 3e^0 = 3; right value 0 + 0 + 3 = 3. Equal, so continuous at 0.

At x = 1: left limit 1 + 3 + 3 = 7; right value 1 - 3 - 3 = -5. Not equal, so discontinuous at 1. …
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KEAM 2026Set eng-2026-04204 marksMCQ
$f(x) = xe^{-\frac{2}{x}}$
›Reveal solutionSolution

The function is continuous everywhere it is defined; it fails only at $x=0$ , so the false claim is (E).

$f(x)=xe^{-2/x}$ is a product/composition of continuous functions for all $x\neq 0$ , so it is continuous at $x=1,-1,2,-2$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$f(x) = \begin{cases} ax+3 & x<1 \\ \frac{4x}{a} & x \geq 1 \end{cases}$
›Reveal solutionSolution

Match the left- and right-hand values at $x=1$ .

Left limit ( $x<1$ ): $a(1)+3=a+3$ . Right limit ( $x\ge1$ ): $\frac{4(1)}{a}=\frac{4}{a}$ . …
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KEAM 2026Set eng-2026-04174 marksMCQ
$f(x) = \begin{cases} \dfrac{1-\sec^2(\alpha x)}{\alpha x^2}, & \text{for } x \neq 0 \\ -3, & \text{for } x = 0 \end{cases}$
›Reveal solutionSolution

Continuity requires $\lim_{x\to0}f(x)=f(0)=-3$ .

$1-\sec^2(\alpha x)=-\tan^2(\alpha x)$ . As $x\to0$ , $\tan(\alpha x)\approx\alpha x$ , so …
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KEAM 2025Set eng-2025-04234 marksMCQ
$f(x) = \begin{cases} \dfrac{\tan \alpha x + (\beta + 1)\tan x}{x}, & \text{for } x \neq 0 \\ 5, & \text{for } x = 0 \end{cases}$
›Reveal solutionSolution
$\alpha+\beta=4$ .

Concept and Intuition

Continuity at $0$ means the limit of $f$ equals $f(0)=5$ . Use $\tan(kx)/x\to k$ .

Step-by-Step Solution

$\lim_{x\to0}\dfrac{\tan(\alpha x)+(\beta+1)\tan x}{x}=\alpha+(\beta+1)$ .

Set equal to $5$ : $\alpha+\beta+1=5$ .

$\alpha+\beta=4$ .

Common Mistakes …
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KEAM 2025Set eng-2025-04284 marksMCQ
$f(x) = \frac{x}{x^2 - 4}$
›Reveal solutionSolution

A rational function is discontinuous only where its denominator vanishes: $x^2-4=0\Rightarrow x=\pm2$ .

The function
$f(x)=\frac{x}{x^2-4}$
is a ratio of polynomials, hence continuous everywhere its denominator is non-zero.

The denominator vanishes when

$x^2-4=0\;\Rightarrow\;x=2\ \text{or}\ x=-2.$ …
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←Previous question · 5.2.1Q11 — Find all points of discontinuity of f, where f is defined by f(x) = x³…→Next question · 5.2.1Q13 — Is the function defined by f(x) = x+5, if x ≤ 1 x-5, if x > 1 a contin…

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

Concept understanding — Continuity At A Point

Continuity at a Point

The Three-Condition Test

Why All Three Are Needed

Common Pitfalls

A Quick Check …

Method: Numerically Verifying Whether Two Polynomial Pieces Agree at Their Boundary

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.Find all points of discontinuity of fff, where fff is defined by f(x)={x10−1,if x≤1x2,if x>1f(x) = \begin{cases} x^{10}-1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases}f(x)={x10−1,x2,​if x≤1if x>1​

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

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