A function $f(x) = |\,1 - x + |x|\,|$ is: (A) discontinuous at $x = 1$ only (B) discontinuous at $x = 0$ only (C) discontinuous at $x = 0,\ 1$ (D) continuous everywhere

> [!TLDR] > The function $f(x) = |\,1 - x + |x|\,|$ simplifies to a piecewise constant function that is continuous at all points, including $x=0$ and $x=1$. The correct option is **(D)** continuous everywhere. The key to this problem is to first remove the absolute value signs by breaking the domain into intervals where the expressions inside them are non-negative or negative. The function involves two absolute values: one around $|x|$ and an outer absolute value around the whole expression. We handle them from the inside out. The inner absolute value $|x|$ changes behaviour at $x=0$. So we first consider $x 1$ (since $-2x > 0$), so it is always positive. Hence the outer absolute value does nothing: $$f(x) = |1 - 2x| = 1 - 2x \quad \text{for } x [!TIP] > The outer absolute value turned out to be redundant for $x \ge 0$ because the inner expression simplified to $1$. For $x < 0$, the inner expression $1-2x$ is always positive, so the outer absolute value also does nothing. So $f(x)$ is actually a simple piecewise linear function with no absolute value complications. 3. **Check continuity at $x = 0$:** - Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1 - 2x) = 1$. - Right-hand limit: $\lim_{x \to 0^+} f(x) = 1$. - Function value: $f(0) = 1$. All three match,

A function f(x) = |1 - x + |x|| is: (A) discontinuous a — CBSE Class 12 Mathematics 2024 PYQ

Question

f(x) = |\,1 - x + |x|\,|

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Concept understanding — Continuity Condition

The Continuity Condition: When a Function Has No "Breaks"

If you can trace a curve without ever lifting your pen — no jumps, gaps, or leaps — that curve is continuous. That's the core intuition: the graph passes through a point without interruption, and the value there matches what the surrounding values predict.

The Intuition: Three Things Must Align

For $f(x)$ to be continuous at $x = a$ , three things must hold:

$f$ is defined at $a$ — there is a point $(a, f(a))$ .
$f$ approaches a single value as $x \to a$ — the left and right sides agree.
That value equals $f(a)$ — no "hole" with a different value plugged in.

If any of these fails, $f$ is discontinuous at $a$ .

Note

Continuity is a local property — we check it point by point, so a function can be continuous at some points and discontinuous at others.

The Precise Statement

$f$ is continuous at $x = a$ if and only if:

\lim_{x \to a} f(x) = f(a)

That one equation packs all three conditions: the limit exists (left and right limits equal and finite), $f(a)$ is defined, and they are equal. If $f$ is continuous at every point of $(a, b)$ , it is continuous on that interval.

\text{Continuity at } x = a: \quad \lim_{x \to a} f(x) = f(a)

Common Pitfalls

The "hole" mistake: $f(x) = \frac{x^2 - 1}{x - 1}$ is undefined at $x = 1$ . Even though $\lim_{x \to 1} f(x) = 2$ exists, $f(1)$ doesn't — discontinuous.

The "jump" mistake: piecewise functions often cause this. For

f(x) = \begin{cases} x + 1 & \text{if } x < 2 \\ x^2 & \text{if } x \geq 2 \end{cases}

at $x = 2$ the left limit is $3$ , the right limit is $4$ — they don't match, so the limit doesn't exist.

The "blow-up" mistake: $f(x) = \frac{1}{x}$ at $x = 0$ is undefined and the limit goes to $\pm\infty$ — discontinuous.

Why It Matters

Continuity is the foundation for calculus. Without it, derivatives don't exist (a corner or jump breaks differentiability), the Intermediate Value Theorem fails, and integrals become tricky. …

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The function $f(x) = |\,1 - x + |x|\,|$ simplifies to a piecewise constant function that is continuous at all points, including $x=0$ and $x=1$ . The correct option is (D) continuous everywhere.

The key to this problem is to first remove the absolute value signs by breaking the domain into intervals where the expressions inside them are non-negative or negative. The function involves two absolute values: one around $|x|$ and an outer absolute value around the whole expression. We handle them from the inside out.

The inner absolute value $|x|$ changes behaviour at $x=0$ . So we first consider $x < 0$ and $x \ge 0$ . Then, after simplifying the expression $1 - x + |x|$ , we apply the outer absolute value, which may introduce another break point where the inner expression changes sign. Let's find that point.

Case $x < 0$ : Here $|x| = -x$ . Then

1 - x + |x| = 1 - x - x = 1 - 2x.

For $x < 0$ , $1 - 2x > 1$ (since $-2x > 0$ ), so it is always positive. Hence the outer absolute value does nothing:

f(x) = |1 - 2x| = 1 - 2x \quad \text{for } x < 0.

Case $x \ge 0$ : Here $|x| = x$ . Then

1 - x + |x| = 1 - x + x = 1.

So for all $x \ge 0$ , the inner expression is the constant $1$ , which is positive. Thus

f(x) = |1| = 1 \quad \text{for } x \ge 0.

Now we have a clean piecewise definition:

f(x) = \begin{cases} 1 - 2x, & x < 0, \\ 1, & x \ge 0. \end{cases}

Tip

The outer absolute value turned out to be redundant for $x \ge 0$ because the inner expression simplified to $1$ . For $x < 0$ , the inner expression $1-2x$ is always positive, so the outer absolute value also does nothing. So $f(x)$ is actually a simple piecewise linear function with no absolute value complications.

Check continuity at $x = 0$ :
- Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1 - 2x) = 1$ .
- Right-hand limit: $\lim_{x \to 0^+} f(x) = 1$ .
- Function value: $f(0) = 1$ . All three match, so $f$ is continuous at $x=0$ .
Check continuity at $x = 1$ : …

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

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set ANNUAL1 mark
$k$
›Reveal solutionSolution

For continuity at $x=1$ , $k$ must equal $\lim_{x\to1}\dfrac{x^2-1}{x-1}$ .

For $x\neq1$ : $\dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}=x+1$ .

So $\lim_{x\to1}f(x)=\lim_{x\to1}(x+1)=2$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If f(x) = {tan 5x / 4x, x \neq 0; m² - 1, x = 0} is continuous at x = 0 (m > 0) then value of m is:$
(a) 9/4
(b) 5/4
(c) 25/16
(d) 3/2
›Reveal solutionSolution

Continuity at $x=0$ requires $\lim_{x\to0} f(x) = f(0)$ ; the standard limit $\lim_{x\to0}\frac{\tan kx}{x} = k$ gives the left/right value, which is then matched to $f(0)=m^2-1$ .

For continuity at $x=0$ :

$\lim_{x\to0} f(x) = f(0)$

Compute the limit using $\lim_{\theta\to0}\dfrac{\tan\theta}{\theta}=1$ : …

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CBSE 2026Set ANNUAL1 mark
$k$
›Reveal solutionSolution

A piecewise function is continuous at the junction point when the left-hand limit, right-hand limit, and the function value there all agree; equate them to solve for $k$ .

Given $f(x) = \begin{cases} kx^2, & x \le 1 \\ 5, & x > 1 \end{cases}$

Left-hand limit at $x=1$ :

$\lim_{x\to1^-} f(x) = \lim_{x\to1^-} kx^2 = k(1)^2 = k$

Right-hand limit at $x=1$ :

$\lim_{x\to1^+} f(x) = \lim_{x\to1^+} 5 = 5$

…

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CBSE 2025Set 65/1/11 markMCQ
$f(x) = \begin{cases} 3x-8, & x \le 5 \\ 2k, & x > 5 \end{cases}$
›Reveal solutionSolution
For continuity at $x=5$ , the left-hand limit and right-hand limit must equal $f(5)$ . Computing these gives $3(5)-8 = 7$ on the left and $2k$ on the right; setting them equal yields $k = \frac{7}{2}$ , not $\frac{5}{2}$ . So Assertion is false, Reason is true.

The core idea here is the Continuity Condition — a function is continuous at a point if the function value and both one-sided limits agree. This isn't just a formula to plug into; it's a logical check that the graph has no jump, hole, or break at that point.

For piecewise functions, the potential trouble spot is always the boundary where the definition changes. At $x=5$ , the function switches from $3x-8$ (for $x \le 5$ ) to $2k$ (for $x > 5$ ). The value $f(5)$ is given by the first piece because of the " $\le$ " sign: $f(5) = 3(5)-8 = 7$ .

Now, continuity demands that as we approach $5$ from either side, the outputs must both land on $7$ .

Left-hand limit ( $x \to 5^-$ ): For $x$ just less than $5$ , the function is $3x-8$ . Since this is a polynomial, it's continuous everywhere, so the limit is simply the value at $x=5$ :

$\lim_{x \to 5^-} f(x) = 3(5)-8 = 7$ .

Right-hand limit ( $x \to 5^+$ ): For $x$ just greater than $5$ , the function is the constant $2k$ . The limit of a constant is the constant itself:

$\lim_{x \to 5^+} f(x) = 2k$ .

Continuity condition: We need

$\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$ .

That gives $7 = 2k = 7$ . The equation $2k = 7$ solves to $k = \frac{7}{2}$ . …
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CBSE 2025Set 65/4/11 markMCQ
$f(x) = \begin{cases} 3x - 2, & 0 < x \le 1 \\ 2x^2 + ax, & 1 < x < 2 \end{cases}$
›Reveal solutionSolution
For a piecewise function to be continuous across the boundary at $x = 1$ , the left and right pieces must meet at the same value; equating $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$ gives $a = -3$ , but since that's not an option, we verify by checking $f(1) = \lim_{x \to 1^+} f(x)$ , yielding $a = -1$ .

The heart of this problem is understanding what continuity means at the point where the definition of $f$ changes. A function is continuous at a point if there's no "jump" — the value you approach from the left, the value at the point itself, and the value you approach from the right must all agree.

For $x \in (0, 2)$ , the only potential trouble spot is $x = 1$ , where the formula switches. Everywhere else within each piece, polynomial functions are automatically continuous.

At $x = 1$ , we need to check three things:

The left-hand limit as $x \to 1^-$ (using the first piece)

The value $f(1)$ itself

The right-hand limit as $x \to 1^+$ (using the second piece)

Let me work through each carefully.

1. Find the value at $x = 1$

The domain specification says $0 < x \le 1$ for the first piece, so $x = 1$ belongs to the first formula:
$f(1) = 3(1) - 2 = 1$
2. Find the left-hand limit

As $x$ approaches $1$ from the left (values slightly less than $1$ ), we use the first piece:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 2) = 3(1) - 2 = 1$
3. Find the right-hand limit …
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CBSE 2025Set ANNUAL1 markMCQ
$a$
(i) 0
(ii) -3
(iii) 1
(iv) 3
›Reveal solutionSolution

Match $f(0)$ to $\lim_{x\to 0} f(x) = e^3$ to get $a = 0$ .

For $f$ to be continuous at $x=0$ , we need $\displaystyle\lim_{x\to 0} f(x) = f(0)$ .

Finding the limit: Using the standard limit $\displaystyle\lim_{x\to 0}(1+kx)^{1/x}=e^{k}$ with $k=3$ :

$\lim_{x\to 0}(1+3x)^{1/x} = e^{3}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$f(x)=\begin{cases}2x-5 & \text{if } x\le 3 \\ 2k & \text{if } x>3\end{cases}$
(a) $1$
(b) $\dfrac{1}{6}$
(c) $6$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

For continuity at $x=3$ , the left-hand value must equal the right-hand limit.

At $x=3$ (using the first piece, since it applies for $x\le3$ ): $f(3)=2(3)-5=1$ .

Right-hand limit as $x\to3^+$ : $\lim_{x\to3^+}2k=2k$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If f(x) = { sin 8x / 5x, x \neq 0 ; m + 1, x = 0 } is continuous at x = 0 then value of m is:$
(a) 5/8
(b) 8/5
(c) 3/5
(d) 5/3
›Reveal solutionSolution

Continuity at $x=0$ requires $\displaystyle\lim_{x\to0} f(x) = f(0)$ ; evaluate the limit using $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ .

$\lim_{x\to0} \frac{\sin 8x}{5x} = \lim_{x\to0} \frac{8}{5}\cdot\frac{\sin 8x}{8x} = \frac85 \times 1 = \frac85.$

…

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CBSE 2024Set 65/2/11 markMCQ
$f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$
›Reveal solutionSolution
The function is defined piecewise with three linear/absolute-value pieces. The only potential breakpoints are at $x = -3$ and $x = 3$ . Checking one-sided limits shows $x = -3$ is continuous but $x = 3$ is not, so the number of discontinuities is 1.

The key idea: a piecewise function can only be discontinuous at the boundaries where the definition changes — here at $x = -3$ and $x = 3$ . Everywhere else, each piece is a polynomial (or absolute value, which is continuous), so continuity is automatic. We just need to check whether the left-hand limit, right-hand limit, and function value match at those two points.

Let's examine each boundary carefully.

At $x = -3$

The function is defined by:

For $x \le -3$ : $f(x) = |x| + 3$ . Since $x$ is negative here, $|x| = -x$ , so $f(x) = -x + 3$ .

For $-3 < x < 3$ : $f(x) = -2x$ .

Compute the left-hand limit (as $x \to -3^-$ ):

$f(x) = -x + 3$ , so $\lim_{x \to -3^-} f(x) = -(-3) + 3 = 3 + 3 = 6$ .

Compute the right-hand limit (as $x \to -3^+$ ):

$f(x) = -2x$ , so $\lim_{x \to -3^+} f(x) = -2(-3) = 6$ .

The function value at $x = -3$ comes from the first piece: $f(-3) = |-3| + 3 = 3 + 3 = 6$ .

All three are equal to 6. So $f$ is continuous at $x = -3$ .

At $x = 3$

The function is defined by:

For $-3 < x < 3$ : $f(x) = -2x$ .

For $x \ge 3$ : $f(x) = 6x + 2$ .

Left-hand limit: $\lim_{x \to 3^-} f(x) = -2(3) = -6$ .

Right-hand limit: $\lim_{x \to 3^+} f(x) = 6(3) + 2 = 18 + 2 = 20$ .

The function value at $x = 3$ comes from the third piece: $f(3) = 6(3) + 2 = 20$ . …
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CBSE 2024Set 65/3/11 markMCQ
$f(x) = |\,1 - x + |x|\,|$
›Reveal solutionSolution
The function $f(x) = |\,1 - x + |x|\,|$ simplifies to a piecewise constant function that is continuous at all points, including $x=0$ and $x=1$ . The correct option is (D) continuous everywhere.

The key to this problem is to first remove the absolute value signs by breaking the domain into intervals where the expressions inside them are non-negative or negative. The function involves two absolute values: one around $|x|$ and an outer absolute value around the whole expression. We handle them from the inside out.

The inner absolute value $|x|$ changes behaviour at $x=0$ . So we first consider $x < 0$ and $x \ge 0$ . Then, after simplifying the expression $1 - x + |x|$ , we apply the outer absolute value, which may introduce another break point where the inner expression changes sign. Let's find that point.

Case $x < 0$ : Here $|x| = -x$ . Then

$1 - x + |x| = 1 - x - x = 1 - 2x.$
For $x < 0$ , $1 - 2x > 1$ (since $-2x > 0$ ), so it is always positive. Hence the outer absolute value does nothing:
$f(x) = |1 - 2x| = 1 - 2x \quad \text{for } x < 0.$

Case $x \ge 0$ : Here $|x| = x$ . Then

$1 - x + |x| = 1 - x + x = 1.$
So for all $x \ge 0$ , the inner expression is the constant $1$ , which is positive. Thus
$f(x) = |1| = 1 \quad \text{for } x \ge 0.$
Now we have a clean piecewise definition:
$f(x) = \begin{cases} 1 - 2x, & x < 0, \\ 1, & x \ge 0. \end{cases}$
Tip

The outer absolute value turned out to be redundant for $x \ge 0$ because the inner expression simplified to $1$ . For $x < 0$ , the inner expression $1-2x$ is always positive, so the outer absolute value also does nothing. So $f(x)$ is actually a simple piecewise linear function with no absolute value complications.

Check continuity at $x = 0$ :

Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (1 - 2x) = 1$ .

Right-hand limit: $\lim_{x \to 0^+} f(x) = 1$ .

Function value: $f(0) = 1$ . All three match, so $f$ is continuous at $x=0$ .

Check continuity at $x = 1$ : …
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CBSE 2024Set ANNUAL1 mark
$f(x) = \begin{cases} \dfrac{1-\cos x}{x^2}, & x \neq 0 \\ \lambda, & x = 0 \end{cases}$
›Reveal solutionSolution

Continuity at $x=0$ requires $\lambda = \lim_{x\to 0}\dfrac{1-\cos x}{x^2}$ , which evaluates to $\dfrac{1}{2}$ .

For $f$ to be continuous at $x=0$ , we need $\lambda = \displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x^2}$ .

Using $1-\cos x = 2\sin^2(x/2)$ : …

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CBSE 2024Set ANNUAL1 mark
$k$
›Reveal solutionSolution

Match the left-hand limit, right-hand limit, and function value at $x=1$ .

For continuity at $x=1$ we need

$\lim_{x\to1^-} f(x) = \lim_{x\to1^+} f(x) = f(1).$

Left-hand limit ( $x<1$ , so $f(x)=4$ ):

$\lim_{x\to1^-} f(x) = 4.$

…

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Q.A function f(x)=∣ 1−x+∣x∣ ∣f(x) = |\,1 - x + |x|\,|f(x)=∣1−x+∣x∣∣ is: (A) discontinuous at x=1x = 1x=1 only (B) discontinuous at x=0x = 0x=0 only (C) discontinuous at x=0, 1x = 0,\ 1x=0, 1 (D) continuous everywhere

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Q.A function $f(x) = |\,1 - x + |x|\,|$ is: (A) discontinuous at $x = 1$ only (B) discontinuous at $x = 0$ only (C) discontinuous at $x = 0,\ 1$ (D) continuous everywhere

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