Q.For the function $f(x) = \begin{cases} x^2+3, & x \neq 0 \\ 1, & x=0 \end{cases}$, which of the following statements is true?
(A) $f(x)$ is continuous and differentiable for all $x \in \mathbb{R}$.
(B) $f(x)$ is continuous for all $x \in \mathbb{R}$.
(C) $f(x)$ is continuous and differentiable for all $x \in \mathbb{R} - \{0\}$.
(D) $f(x)$ is discontinuous at infinite points.

Showing the 12 most recent of 20 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.The value of $k$ for which the function $f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & x \neq 0 \\ k(x+1), & x = 0 \end{cases}$ is a continuous function, is: (A) $\frac{1}{4}$ (B) $2$ (C) $\frac{1}{2}$ (D) $0$

   ›Reveal solutionSolution

   For continuity at $x = 0$, the limit of $x^2 \sin \frac{1}{x}$ as $x \to 0$ must equal the function value $k(0+1) = k$. Since the limit is $0$, we need $k = 0$.

   A function is continuous at a point when three conditions align: the function is defined there, the limit exists as we approach that point, and crucially, the limit equals the function's value at that point. This problem tests whether you can recognize that continuity at $x = 0$ creates a bridge between two different expressions.

   The function behaves as $x^2 \sin \frac{1}{x}$ everywhere except at zero, where it suddenly switches to $k(x+1)$. At $x = 0$, this second piece gives us $f(0) = k(0+1) = k$. For continuity, we need:

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

   Since we approach zero from the region where $x \neq 0$, the relevant limit is:

   $\lim_{x \to 0} x^2 \sin \frac{1}{x} = k$

   Let me find this limit.

   1. Recognize the bounded oscillation

      The sine function satisfies $-1 \leq \sin \frac{1}{x} \leq 1$ for all $x \neq 0$, no matter how wildly $\frac{1}{x}$ oscillates as $x \to 0$.

   2. Apply the squeeze theorem

      Multiplying the inequality by $x^2$ (which is always non-negative):

   $-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$

   3. Evaluate the bounding limits As $x \to 0$:

   $\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$

   4. Conclude via the squeeze theorem Since $x^2 \sin \frac{1}{x}$ is squeezed between two expressions that both approach $0$:

   $\lim_{x \to 0} x^2 \sin \frac{1}{x} = 0$

   5. Match the limit to the function value For continuity at $x = 0$:

   $k = \lim_{x \to 0} x^2 \sin \frac{1}{x} = 0$

   Tip

   Whenever you see $x^n \sin \frac{1}{x}$ or $x^n \cos \frac{1}{x}$ with $n > 0$, the limit as $x \to 0$ is always $0$ because the polynomial term dominates the bounded oscillation.

   Watch out

   Don't try to evaluate $\sin \frac{1}{x}$ as $x \to 0$ directly — it oscillates infinitely and has no limit. The key is that $x^2$ forces the product to zero despite the oscillation.

   ✓Final answer

   The correct option is (D) $0$.

2. CBSE 2026Set ANNUAL1 mark
   Q.Prove that the function f(x) = 5x - 3 is continuous at x = -3.

   ›Reveal solutionSolution

   A function $f$ is continuous at $x=a$ if $\displaystyle\lim_{x\to a} f(x) = f(a)$; check this directly for the linear function $f(x)=5x-3$ at $a=-3$.

   Concept: $f$ is continuous at $x=a$ when: (i) $f(a)$ is defined, (ii) $\displaystyle\lim_{x\to a}f(x)$ exists, and (iii) the two are equal.

   Working:

   $f(-3) = 5(-3) - 3 = -15 - 3 = -18$

   $\lim_{x\to -3} f(x) = \lim_{x\to -3}(5x-3) = 5(-3) - 3 = -18$

   Since $\displaystyle\lim_{x\to -3}f(x) = f(-3) = -18$, all three conditions hold, so $f(x)=5x-3$ is continuous at $x=-3$. (In fact every polynomial is continuous at every real number, by the same reasoning.)

   ✓Final answer

   $f(x)=5x-3$ is continuous at $x=-3$ since $\lim_{x\to-3}f(x) = f(-3) = -18$.

3. CBSE 2025Set 65/4/11 markMCQ
   Q.The function $f$ defined by $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$ is not continuous at : (A) $x = 0$ (B) $x = 1$ (C) $x = 2$ (D) $x = 5$

   ›Reveal solutionSolution

   The function has a jump at $x=1$ because the left-hand limit ($1$) and the right-hand limit ($5$) do not match, so it is discontinuous only at $x=1$. The correct option is (B).

   Continuity at a point means three things must hold: the function is defined there, the limit exists there, and the limit equals the function value. For a piecewise function, the only place where things can go wrong is at the boundary between the pieces — here, at $x=1$. Everywhere else, the function is just a simple rule (either $x$ or the constant $5$), so it's automatically continuous.

   Let’s check each candidate point.

   1. At $x=0$

      For $x \le 1$, the rule is $f(x)=x$. Since $0 \le 1$, we have $f(0)=0$.

      The left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x = 0$.

      The right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$ (because near $0$, $x$ is still $\le 1$).

      So the limit exists and equals $0$, which matches $f(0)$. Continuous here.

   2. At $x=1$ — the critical boundary

      • Left-hand limit: as $x$ approaches $1$ from below, $x \le 1$, so $f(x)=x$. Hence

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

   • Right-hand limit: as $x$ approaches $1$ from above, $x > 1$, so $f(x)=5$. Hence

   $\lim_{x \to 1^+} f(x) = 5.$

   • The left and right limits are different ($1 \neq 5$), so the two-sided limit does not exist.
   • The function value is $f(1)=1$ (since $1 \le 1$). Even though $f(1)$ equals the left-hand limit, the limit itself doesn't exist, so continuity fails.

   Watch out

   A common mistake is to think that because $f(1)=1$ matches the left-hand limit, the function is continuous. But continuity requires the two-sided limit to exist and match — a single side isn't enough.

   3. At $x=2$

      For $x>1$, $f(x)=5$. So $f(2)=5$, and both one-sided limits are $5$. Continuous.

   4. At $x=5$

      Same reasoning: $f(5)=5$, limits are $5$. Continuous.

   Tip

   For a piecewise function with a single break, you only ever need to test the boundary point(s). All other points inherit continuity from the individual pieces.

   ✓Final answer

   The function is not continuous at $x=1$, so the correct option is (B).

4. CBSE 2025Set IX1 mark
   Q.Prove that the function $f(x) = |x|$, is continuous at $x = 0$.

   ›Reveal solutionSolution

   Left limit, right limit and the value all equal $0$, so $|x|$ is continuous at $0$.

   Concept. $f$ is continuous at $x=a$ iff $\displaystyle\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$.

   Here $f(x)=|x|=\begin{cases}-x,&x<0\\ x,&x\ge0\end{cases}$

   • Left-hand limit: $\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(-x)=0.$
   • Right-hand limit: $\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0.$
   • Value: $f(0)=|0|=0.$

   All three coincide, hence $f(x)=|x|$ is continuous at $x=0$.

   ✓Final answer

   $f(x)=|x|$ is continuous at $x=0$ because $\lim_{x\to0^-}f=\lim_{x\to0^+}f=f(0)=0$.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The function $f(x) = |x| - |x + 1|$ is:

   (a) continuous at $x = 0$ as well as at $x = -1$

   (b) continuous at $x = -1$ but not at $x = 0$

   (c) discontinuous at $x = 0$ as well as at $x = -1$

   (d) continuous at $x = 0$ but not at $x = -1$

   ›Reveal solutionSolution

   $|x|$ and $|x+1|$ are each continuous everywhere, and the difference of two continuous functions is continuous.

   $g(x)=|x|$ is continuous on all of $\mathbb{R}$, and $h(x)=|x+1|$ (a shifted absolute value) is also continuous on all of $\mathbb{R}$. Since $f(x)=g(x)-h(x)$ is a difference of two functions continuous everywhere, $f$ is continuous everywhere, including at $x=0$ and $x=-1$.

   ✓Final answer

   $f$ is continuous at both $x=0$ and $x=-1$ — option (a).

6. CBSE 2025Set ANNUAL1 mark
   Q.Check the continuity of the function $f$ given by $f(x) = 2x+3$ at $x=1$. OR Find the value of $k$, so that the function $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ is continuous at $x=2$.

   ›Reveal solutionSolution

   A function is continuous at a point when its limit there equals its value; check both.

   Here $f(x)=2x+3$ (a polynomial), and we test $x=1$.

   Value: $f(1)=2(1)+3=5.$

   Limit: $\displaystyle\lim_{x\to1}(2x+3)=2(1)+3=5.$

   Since $\displaystyle\lim_{x\to1}f(x)=5=f(1)$, the function is continuous at $x=1$.

   ✓Final answer

   $f$ is continuous at $x=1$ (indeed $2x+3$ is continuous everywhere on $\mathbb{R}$).

   Alternative (Or):

   Match the left value $kx^2$ to the right value $3$ at $x=2$.

   We need $f(x)=\begin{cases}kx^2,&x\le2\\3,&x>2\end{cases}$ continuous at $x=2$.

   Left-hand limit and value at $x=2$: $\displaystyle\lim_{x\to2^-}kx^2=k(2)^2=4k.$

   Right-hand limit: $\displaystyle\lim_{x\to2^+}3=3.$

   Continuity requires these equal:

   $4k=3\ \Rightarrow\ k=\frac34.$

   ✓Final answer

   $k=\dfrac34$

7. CBSE 20241 markMCQ
   Q.The value of k, for which $f(x) = \begin{cases} \frac{\sqrt{3} \cos x + \sin x}{3x + \frac{\pi}{2}}, & x \neq -\frac{\pi}{3} \\ k, & x = -\frac{\pi}{3} \end{cases}$ is continuous at $x = -\frac{\pi}{3}$, is : (A) $\frac{2}{3}$ (B) $-\frac{2}{3}$ (C) $\frac{3}{2}$ (D) $6$ $\sim\sim\sim$

   ›Reveal solutionSolution

   Continuity requires $k = \lim_{x\to -\pi/3} f(x)$. The quotient is a $\tfrac{0}{0}$ form at $x=-\tfrac{\pi}{3}$, and the limit evaluates to $\tfrac{2}{3}$ — option (A).

   For $f$ to be continuous at $x = -\tfrac{\pi}{3}$, we need

   $k = \lim_{x \to -\pi/3} \frac{\sqrt{3}\cos x + \sin x}{3\left(x + \tfrac{\pi}{3}\right)}.$

   Simplify the numerator. Writing it as a single sine:

   $\sqrt{3}\cos x + \sin x = 2\left(\tfrac{\sqrt{3}}{2}\cos x + \tfrac{1}{2}\sin x\right) = 2\sin\!\left(x + \tfrac{\pi}{3}\right).$

   At $x = -\tfrac{\pi}{3}$ this is $2\sin 0 = 0$, and the denominator also vanishes — a genuine $\tfrac{0}{0}$ form.

   Evaluate the limit. Let $t = x + \tfrac{\pi}{3}$, so $t \to 0$:

   $k = \lim_{t \to 0} \frac{2\sin t}{3t} = \frac{2}{3}\lim_{t \to 0}\frac{\sin t}{t} = \frac{2}{3}\cdot 1 = \frac{2}{3}.$

   ✓Final answer

   $k = \dfrac{2}{3}$, which is option (A).

8. CBSE 2024Set 65/1/11 markMCQ
   Q.For the function $f(x) = \begin{cases} x^2+3, & x \neq 0 \\ 1, & x=0 \end{cases}$, which of the following statements is true? (A) $f(x)$ is continuous and differentiable for all $x \in \mathbb{R}$. (B) $f(x)$ is continuous for all $x \in \mathbb{R}$. (C) $f(x)$ is continuous and differentiable for all $x \in \mathbb{R} - \{0\}$. (D) $f(x)$ is discontinuous at infinite points.

   ›Reveal solutionSolution

   The function is a parabola with a hole at $x=0$ and a single isolated point at $(0,1)$. Because the limit as $x\to 0$ is $3$, not $1$, the function is discontinuous at $x=0$ — but it is continuous and differentiable everywhere else. The correct option is (C).

   The key to this problem is understanding what continuity and differentiability mean at a point, and then checking the one point where the definition changes.

   Continuity at a point $x = a$ requires three things to match: the function value $f(a)$, the left-hand limit $\lim_{x \to a^-} f(x)$, and the right-hand limit $\lim_{x \to a^+} f(x)$. If any one of these differs, the function is discontinuous there.

   Differentiability at a point requires continuity first — and then the left and right derivatives must also be equal. So if a function is discontinuous at a point, it cannot be differentiable there.

   Here, the function is defined by two pieces: for every $x$ except $0$, it behaves like $x^2 + 3$ (a smooth parabola shifted up by 3). At $x = 0$ alone, it jumps to the value $1$. That single point is the only place where anything unusual can happen.

   Let’s check systematically.

   1. Check continuity at $x = 0$ For $x \neq 0$, $f(x) = x^2 + 3$. As $x$ approaches $0$ from either side, $x^2$ approaches $0$, so

   $\lim_{x \to 0} f(x) = 0^2 + 3 = 3.$

   But $f(0) = 1$. Since $3 \neq 1$, the limit does not equal the function value.

   Watch out

   A common mistake is to think that because the formula $x^2+3$ is continuous everywhere, the whole function is continuous. But the definition at $x=0$ overrides that — the function is piecewise-defined, and the value at the breakpoint must match the limit.

   Hence $f$ is discontinuous at $x = 0$.

   2. Check continuity for $x \neq 0$

      For any $a \neq 0$, near $a$ the function is simply $f(x) = x^2 + 3$, which is a polynomial. Polynomials are continuous everywhere. So $f$ is continuous at every $x \neq 0$.

   3. Check differentiability at $x = 0$

      Since $f$ is not continuous at $0$, it cannot be differentiable there. (Differentiability implies continuity — that’s a theorem you must remember.)

   4. Check differentiability for $x \neq 0$

      For any $a \neq 0$, the function is locally just $x^2 + 3$, whose derivative is $2x$. Polynomials are differentiable everywhere, so $f$ is differentiable at every $x \neq 0$.

   Tip

   You don’t need to compute left and right derivatives at $0$ here — the discontinuity alone kills differentiability. But if the function were continuous at $0$, you’d then check if the slopes from left and right match.

   Now look at the options:

   • (A) says continuous and differentiable for all $x \in \mathbb{R}$. False — fails at $x=0$.
   • (B) says continuous for all $x \in \mathbb{R}$. False — discontinuous at $0$.
   • (C) says continuous and differentiable for all $x \in \mathbb{R} - \{0\}$. True — that’s exactly what we found.
   • (D) says discontinuous at infinite points. False — only one point of discontinuity.

   ✓Final answer

   The correct option is (C).

9. CBSE 2024Set EX1 mark
   Q.Show that the function $f(x) = \begin{cases} x + 2 & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$ is not continuous at $x = 0$.

   ›Reveal solutionSolution

   The limit as $x\to 0$ is $2$ (from $x+2$), but the defined value is $f(0)=1$. Limit $\ne$ value, so $f$ is discontinuous at $0$.

   Concept. $f$ is continuous at $x=0$ iff $\displaystyle\lim_{x\to 0}f(x)=f(0)$.

   Limit. For $x\ne 0$, $f(x)=x+2$, so

   $\lim_{x\to 0}f(x)=\lim_{x\to 0}(x+2)=0+2=2.$

   Value. By definition $f(0)=1$.

   Since $\displaystyle\lim_{x\to 0}f(x)=2\ne 1=f(0)$, the condition for continuity fails.

   ✓Final answer

   $f$ is not continuous at $x=0$ because $\lim_{x\to 0}f(x)=2\ne f(0)=1$.

10. CBSE 2024Set ANNUAL1 mark
    Q.Examine the continuity of the function $f(x) = 5x - 3$ at $x = 5$.

    ›Reveal solutionSolution

    Check that the limit at $x=5$ equals the function value there.

    Given $f(x) = 5x - 3$.

    Function value: $f(5) = 5(5) - 3 = 25 - 3 = 22.$

    Limit: $\displaystyle\lim_{x\to5} f(x) = \lim_{x\to5}(5x-3) = 5(5)-3 = 22.$

    Since

    $\lim_{x\to5} f(x) = 22 = f(5),$

    the function is continuous at $x=5$.

    ✓Final answer

    $f(x)=5x-3$ is continuous at $x=5$, as $\lim_{x\to5}f(x)=f(5)=\boldsymbol{22}$.

11. CBSE 2024Set ANNUAL1 mark
    Q.When is a function $f(x)$ said to be continuous at $x = c$ ?

    ›Reveal solutionSolution

    Standard definition: left-hand limit = right-hand limit = function value at that point.

    A function $f(x)$ is said to be continuous at $x=c$ if:

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

    equivalently, $\displaystyle\lim_{x\to c}f(x)$ exists and is equal to $f(c)$. If any one of these fails — the limit does not exist, or $f(c)$ is undefined, or the limit and $f(c)$ differ — then $f$ is discontinuous at $x=c$.

    ✓Final answer

    $f(x)$ is continuous at $x=c$ if $\displaystyle\lim_{x\to c}f(x)=f(c)$, i.e. LHL $=$ RHL $= f(c)$.

12. CBSE 2023Set 65/1/11 markMCQ
    Q.The value of k for which $f(x) = \begin{cases} 3x+5, & x \ge 2 \\ kx^2, & x < 2 \end{cases}$ is a continuous function, is : (A) $-\frac{11}{4}$ (B) $\frac{4}{11}$ (C) 11 (D) $\frac{11}{4}$

    ›Reveal solutionSolution

    For a piecewise function to be continuous at the join point $x=2$, the left-hand limit and right-hand limit must equal the function value at $x=2$. Equating $k(2)^2$ with $3(2)+5$ gives $4k = 11$, so $k = \frac{11}{4}$. The correct option is (D).

    The Core Idea: Continuity at a Point

    A function is continuous at a point if three things match perfectly — the value from the left, the value from the right, and the actual function value at that point. For a piecewise function like this one, the only place where things could break is at the boundary where the formula changes, which is $x = 2$.

    Think of it like two roads meeting at a junction. For a smooth ride, the elevation of the left road as you approach the junction must exactly match the elevation of the right road as you approach from the other side — and that elevation must also be the height of the junction itself. If they don't match, there's a jump, and the function is discontinuous.

    Here, the left piece ($x < 2$) uses $kx^2$, and the right piece ($x \ge 2$) uses $3x+5$. The function value at $x=2$ is given by the right piece (since $x \ge 2$ includes $2$). So we need the left-hand limit to equal that value.

    Step-by-Step Solution

    1. Find the function value at $x=2$.

    Since $x=2$ falls in the case $x \ge 2$, we use $f(x) = 3x+5$.

    $f(2) = 3(2) + 5 = 6 + 5 = 11$

    2. Find the left-hand limit as $x \to 2^-$.

    For $x < 2$, the function is $f(x) = kx^2$. As $x$ approaches $2$ from the left, we simply substitute $x=2$ into this expression (since $kx^2$ is a polynomial and polynomials are continuous everywhere).

    $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} kx^2 = k(2)^2 = 4k$

    3. Find the right-hand limit as $x \to 2^+$.

    For $x > 2$, the function is $f(x) = 3x+5$. Again, this is a polynomial, so the limit is just the value at $x=2$.

    $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x+5) = 3(2)+5 = 11$

    4. Apply the continuity condition.

    For $f$ to be continuous at $x=2$, we need:

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

    We already have $\lim_{x \to 2^+} f(x) = 11$ and $f(2) = 11$, so the right-hand side is consistent. The only condition left is:

    $4k = 11$

    5. Solve for $k$.

    $k = \frac{11}{4}$

    Watch out

    A common mistake is to forget that $f(2)$ is defined by the $x \ge 2$ piece, not by the $x < 2$ piece. Some students incorrectly set $k(2)^2 = 3(2)+5$ but then forget that $f(2)$ itself is $11$, so they end up solving $4k = 11$ correctly anyway — but the reasoning is incomplete. Always check all three parts: left limit, right limit, and function value.

    Tip

    In piecewise functions where each piece is a polynomial, the only potential trouble is at the boundary. You never need to compute limits using $\epsilon$-$\delta$ here — just substitute the boundary point into each piece, because polynomials are continuous everywhere. The entire problem reduces to solving one simple equation.

    ✓Final answer

    The value of $k$ is $\boxed{\frac{11}{4}}$, which corresponds to option (D).