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.

GUJCET 2026Set x1 markMCQ
$f$
›Reveal solutionSolution

For continuity at $x=\pi$ , the left value $f(\pi)$ must equal the right-hand limit.

$f(\pi)=k\pi+1$ (from $x\le\pi$ branch).

$\lim_{x\to\pi^+}f(x)=\cos\pi=-1$ .

Setting them equal: …
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GUJCET 2020Set 071 markMCQ
$f(\alpha) = \begin{cases} \frac{1 - \cos 6\alpha}{36\alpha^2} & \text{if } \alpha \ne 0 \\ k & \text{if } \alpha = 0 \end{cases}$
›Reveal solutionSolution

For continuity $k=\lim_{\alpha\to0}\dfrac{1-\cos6\alpha}{36\alpha^2}=\tfrac12$ .

Concept — removable discontinuity. Continuity at $0$ requires $k$ equal the limit. Use $1-\cos\theta = 2\sin^2(\theta/2)$ : …
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GUJCET 2024Set 131 markMCQ
$f$
›Reveal solutionSolution

Evaluating the limit at $x=\tfrac{\pi}{2}$ gives $k$ ; continuity requires $k=2024$ .

Concept. For continuity, $\lim_{x\to\pi/2} f(x) = f(\pi/2) = 2024$ .

Steps. Put $x=\tfrac{\pi}{2}+h$ : $\cos x = -\sin h \approx -h$ and $\pi-2x = -2h$ , so …
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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$f(x) = \dfrac{k\tan 2x}{x-\pi}$
(a) 1
(b) -1
(c) 2
(d) -2
›Reveal solutionSolution

Continuity at $x=\pi$ forces the limit of the expression to equal $f(\pi)=2$ , giving $k=1$ .

Put $x=\pi+h$ , so $h\to0$ as $x\to\pi$ . Then $\tan 2x=\tan(2\pi+2h)=\tan 2h$ , so

$\lim_{x\to\pi}\frac{k\tan 2x}{x-\pi}=\lim_{h\to0}\frac{k\tan 2h}{h}=2k$

…

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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$f(x) = \dfrac{k\cos x}{3\pi - 2x}$
(a) $6$
(b) $3$
(c) $-6$
(d) $-3$
›Reveal solutionSolution

Continuity at $x=\frac{3\pi}{2}$ means the limit of $f(x)$ there must equal $f(3\pi/2)=3$ ; substitute $x=\frac{3\pi}{2}+h$ to resolve the $0/0$ form.

Let $x=\dfrac{3\pi}{2}+h$ . Then $\cos x=\cos\left(\dfrac{3\pi}{2}+h\right)=\sin h$ , and $3\pi-2x=3\pi-3\pi-2h=-2h$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$f(x) = \begin{cases} kx+1, & x \le \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$
(a) $-\frac{2}{\pi}$
(b) $\frac{2}{\pi}$
(c) $1$
(d) $0$
›Reveal solutionSolution

Continuity at a breakpoint requires the two pieces to agree there.

$\lim_{x\to\pi/2^-}f(x)=k\cdot\frac{\pi}{2}+1$ and $f\left(\frac{\pi}{2}\right)=\sin\frac{\pi}{2}=1$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$f$
(a) $-2/3$
(b) $3/2$
(c) $-3/2$
(d) $2/3$
›Reveal solutionSolution

Continuity at $x=3$ requires the two branches of $f$ to agree at $x=3$ ; equate them and solve for $d-c$ .

$f(x)=cx+1$ for $x\le3$ and $f(x)=dx+3$ for $x>3$ . For continuity at $x=3$ , $\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)$ :

…

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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$f(x) = \begin{cases} \frac{\sin 4x}{9x}, & x \ne 0 \\ k^2, & x = 0 \end{cases}$
(a) $-\frac{3}{2}$
(b) $\frac{3}{2}$
(c) $\pm\frac{2}{3}$
(d) $\frac{4}{9}$
›Reveal solutionSolution

Continuity at $x=0$ forces $f(0)=k^2$ to equal the limit of $f(x)$ as $x\to0$ .

$\lim_{x\to0}\frac{\sin4x}{9x} = \lim_{x\to0}\frac{4}{9}\cdot\frac{\sin4x}{4x} = \frac{4}{9}\cdot1 = \frac{4}{9}$ .

…

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

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

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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