Q.If a function defined by 𝑓(𝑥) = { 𝑘𝑥 + 1, 𝑥 ≤ 𝜋 cos 𝑥 , 𝑥 > 𝜋 is continuous at 𝑥 = 𝜋, then the value of 𝑘 is
(A) 𝜋
(B) −1 𝜋
(C) 0
(D) −2 𝜋

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 …

   🔒 Start your 14-day free trial to see the full solution →
2. 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:

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

   🔒 Start your 14-day free trial to see the full solution →
3. 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:

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

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $1 \le 1$). …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $x = 0$ …

   🔒 Start your 14-day free trial to see the full solution →
6. 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:

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

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

   …

   🔒 Start your 14-day free trial to see the full solution →
7. 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: $\displaystyle\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0.$ …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 eve …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $x=1$.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $f(0)=1$.

    …

    🔒 Start your 14-day free trial to see the full solution →
11. 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

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

    🔒 Start your 14-day free trial to see the full solution →
12. 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:

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

    🔒 Start your 14-day free trial to see the full solution →