Q.The function $f:R\to Z$ defined by $f(x)=[x]$; where $[\,.\,]$ denotes the greatest integer function, is
(A) Continuous at $x=2.5$ but not differentiable at $x=2.5$
(B) Not Continuous at $x=2.5$ but differentiable at $x=2.5$
(C) Not Continuous at $x=2.5$ and not differentiable at $x=2.5$
(D) Continuous as well as differentiable at $x=2.5$

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Let $[y]$ represent the greatest integer less than or equal to $y$. Then set of all $x$ at which $f(x) = \mathrm{Cos}^{-1}[4x + 3]$ is differentiable is (A) $\mathbb{R}$ (B) $[-1, 1]$ (C) $\left[-1, -\dfrac{1}{4}\right) - \left\{-\dfrac{3}{4}, -\dfrac{1}{2}\right\}$ (D) $\left(-\dfrac{3}{4}, \infty\right)$

   ›Reveal solutionSolution

   This tests domain + continuity of a composition with the greatest-integer function.

   The domain of $\cos^{-1}$ forces $[4x+3]\in\{-1,0,1\}$, giving $x\in[-1,-1/4)$; the

   function is differentiable there except at the two points where the floor jumps.

   Concept and Intuition

   $[y]$ denotes the greatest integer $\le y$ (the floor function) — it is an integer-valued

   step function, constant on each interval $[n,n+1)$ and jumping by $1$ at every integer.

   $\cos^{-1}(t)$ needs $t\in[-1,1]$. So for $f(x)=\cos^{-1}[4x+3]$ to even be defined,

   the integer $[4x+3]$ must lie in $\{-1,0,1\}$ (the only integers in $[-1,1]$).

   Differentiability requires first continuity — and a step function composed with a

   continuous one is continuous only where the step itself doesn't jump.

   Step-by-Step Solution

   1. Domain from $[4x+3]=-1$: $-1\le 4x+3<0 \Rightarrow -4\le 4x<-3 \Rightarrow x\in[-1,-3/4)$.
   2. Domain from $[4x+3]=0$: $0\le 4x+3<1 \Rightarrow -3\le 4x<-2 \Rightarrow x\in[-3/4,-1/2)$.
   3. Domain from $[4x+3]=1$: $1\le 4x+3<2 \Rightarrow -2\le 4x<-1 \Rightarrow x\in[-1/2,-1/4)$.
   4. Union of all three: $x\in[-1,-1/4)$ — this is the full domain of $f$.
   5. On $[-1,-3/4)$, $f(x)=\cos^{-1}(-1)=\pi$ (constant); on $[-3/4,-1/2)$, $f(x)=\cos^{-1}(0)=\pi/2$ (constant); on $[-1/2,-1/4)$, $f(x)=\cos^{-1}(1)=0$ (constant).
   6. At $x=-3/4$: $f$ jumps from $\pi$ to $\pi/2$ — discontinuous, so not differentiable. …

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2. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$[\,.\,]$ represents the greatest integer function. At $x = -1$, $\dfrac{d}{dx}\sin\pi[x] =$ (A) 0 (B) 2 (C) -2 (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   $\sin(\pi[x])$ is identically zero for all $x$ (since $[x]$ is always an integer and $\sin$ of any integer multiple of $\pi$ is 0), so its derivative is 0 everywhere, including at $x=-1$.

   Concept and Intuition

   The key insight is to evaluate the outer function first: no matter what real number $x$ is, $[x]$ (the greatest integer function) always returns an integer, and $\sin(n\pi)=0$ for every integer $n$. So the composite function $\sin(\pi[x])$ never depends on the fine structure of $[x]$'s jumps at all — it's simply the zero function, constant everywhere.

   Step-by-Step Solution

   1. For any real $x$, $[x]\in\mathbb Z$.
   2. $\sin(\pi\cdot n)=0$ for every integer $n$.
   3. So $\sin(\pi[x])=0$ for all $x$, not just near $x=-1$ — it's the identically-zero function.
   4. The derivative of a constant (here, the constant $0$) is $0$ everywhere it's differentiable, including at $x=-1$.

   Common Mistakes …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $[.]$ denotes the Greatest integer function, then $f(x)=[x]^2-[x^2]$ is discontinuous at ________ (A) All integers (B) All integers except 0 and 1 (C) All integers except 1 (D) All integers except 0

   ›Reveal solutionSolution

   Testing $f(x)=[x]^2-[x^2]$ around each integer shows the one-sided limits only agree with the function value at $x=1$; every other integer is a genuine discontinuity.

   Concept and Intuition

   The greatest integer function $[x]$ itself is discontinuous at every integer, so a function built from $[x]$ and $[x^2]$ needs care: near an integer $n$, $[x]$ jumps but $[x^2]$ jumps at a different point (since $x^2$ isn't linear), so the two jumps generally don't cancel. The only way $f$ can stay continuous at an integer is if the jumps of $[x]^2$ and $[x^2]$ happen to compensate exactly from both sides.

   Step-by-Step Solution

   1. At an integer $x=n$: $[n]=n$ and $n^2$ is an integer, so $[n^2]=n^2$. Hence $f(n)=n^2-n^2=0$ always.
   2. Right-hand limit ($x=n+h$, $h\to0^+$): $[x]=n$ so $[x]^2=n^2$ (constant). For $n>0$, $x^2=n^2+2nh+h^2$ is slightly above $n^2$, so $[x^2]=n^2$ too (for small $h$), giving $f\to n^2-n^2=0=f(n)$: right-continuous for every positive integer. But for $n\le0$ ($n=0$ or negative), $2nh\le0$ makes $x^2$ approach $n^2$ from below (or stay at $0$ in a way that flips the sign pattern), and one finds $[x^2]=n^2-1$, so $f\to n^2-(n^2-1)=1\ne f(n)$: right-discontinuous at $n\le 0$ (check $n=0$ directly: $x=-h$ from the left gives the real problem — see step 3).
   3. Left-hand limit ($x=n-h$, $h\to0^+$): $[x]=n-1$, so $[x]^2=(n-1)^2$. For $n>0$: $x^2=n^2-2nh+h^2$ is slightly below $n^2$, so $[x^2]=n^2-1$, giving $f\to (n-1)^2-(n^2-1)=2(1-n)$. This equals $f(n)=0$ only when $n=1$.
   4. For $n=0$: left side is $x=-h\to0^-$, so $[x]=-1$, $[x]^2=1$; $x^2=h^2\to0^+$, so $[x^2]=0$; $f\to1-0=1\ne f(0)=0$ — discontinuous at $0$. …

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4. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.Let [t] represents the greatest integer not exceeding t and $C=1-2e^2$. If the function $f(x)=\begin{cases}[e^x], & x<0\\ ae^x+[x-2], & 0\le x<2\\ [e^{-x}]-C, & x\ge2\end{cases}$ is continuous at $x=2$, then $f(x)$ is discontinuous at (A) $x=1$ only (B) $x=0$ and $x=1$ (C) $x=0$ only (D) $x=0,\ x=1$ and $x=\dfrac{1}{2}$

   ›Reveal solutionSolution

   Solving for $a$ from continuity at $x=2$, then explicitly writing out $f$ on each sub-interval (splitting the floor function $[x-2]$ at its own jump inside $[0,2)$) shows the only actual discontinuity is at $x=1$.

   Concept and Intuition

   The greatest-integer function $[t]$ jumps at every integer value of $t$. Wherever a formula involves $[g(x)]$, the function can have discontinuities exactly where $g(x)$ crosses an integer — so the middle piece $ae^x+[x-2]$, valid on $[0,2)$, itself has an internal jump at $x=1$ (where $x-2=-1$, an integer) that has nothing to do with the "continuity at $x=2$" condition used to find $a$.

   Step-by-Step Solution

   1. Find $a$: For $x<0$, $e^x\in(0,1)$ so $[e^x]=0$ always — $f(x)=0$ there.
   2. For $x\ge2$, $e^{-x}\in(0,e^{-2}]\subset(0,1)$ so $[e^{-x}]=0$ — $f(x)=-C=2e^2-1$ (constant) for all $x\ge2$.
   3. Continuity at $x=2$: as $x\to2^-$ (middle piece), $x-2\to0^-$, and for $x\in[1,2)$, $x-2\in[-1,0)$ so $[x-2]=-1$. So $\lim_{x\to2^-}f(x)=ae^2-1$. Setting equal to $f(2)=2e^2-1$: $ae^2-1=2e^2-1\Rightarrow a=2$.
   4. Write $f$ fully on $[0,2)$: for $x\in[0,1)$, $x-2\in[-2,-1)\Rightarrow[x-2]=-2$, so $f(x)=2e^x-2$. For $x\in[1,2)$, $x-2\in[-1,0)\Rightarrow[x-2]=-1$, so $f(x)=2e^x-1$.
   5. Check $x=0$: left limit ($x<0$) is $0$; $f(0)=2e^0-2=0$. Continuous. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The number of points of discontinuity of the function $f(x)=[x]+|x-2|$, $-3<x<3$ is (A) $5$ (B) $3$ (C) $4$ (D) $2$

   ›Reveal solutionSolution

   Only the floor function contributes discontinuities (the absolute-value term is continuous everywhere); counting the integers strictly inside $(-3,3)$ gives $5$ points.

   Concept and Intuition

   A sum of functions is discontinuous exactly where at least one summand is discontinuous, unless the jumps happen to cancel exactly. Here $|x-2|$ is continuous for all real $x$ (it has a corner at $x=2$, but no jump), while $[x]$ (the greatest integer / floor function) jumps by exactly $1$ at every integer. Since $|x-2|$ contributes no jump anywhere, none of the floor function's jumps can be cancelled, so the discontinuities of $f$ are precisely the integers in the domain.

   Step-by-Step Solution

   1. $|x-2|$ is continuous for every real $x$ (piecewise linear, no jump, only a kink at $x=2$).
   2. $[x]$ (floor) is discontinuous at every integer $n$: as $x\to n^-$, $[x]\to n-1$, but $[n]=n$ — a jump of size $1$.
   3. Since $f(x)=[x]+|x-2|$ and the second term never jumps, $f$ is discontinuous at exactly the same points as $[x]$, i.e. at every integer in the domain.
   4. List the integers strictly between $-3$ and $3$: $-2,-1,0,1,2$ — that's $5$ integers. …

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6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Let $[t]$ represents the greatest integer not more than t. Then the number of discontinuous points of $f(x) = \left[x^{\frac{1}{x}}\right]$ in $(0, \infty)$ is (A) $0$ (B) $1$ (C) $2$ (D) $\infty$

   ›Reveal solutionSolution

   Tracking $x^{1/x}$ across $(0,\infty)$ shows the floor function only jumps once, at $x=1$.

   Concept and Intuition

   Analyze $g(x)=x^{1/x}=e^{(\log x)/x}$ across $(0,\infty)$ and see how many times $\lfloor g(x)\rfloor$ jumps.

   Step-by-Step Solution

   1. For $x\in(0,1)$: $\log x<0$, so $(\log x)/x<0$, giving $g(x)<1$; also $g(x)>0$ always. So $0<g(x)<1 \Rightarrow \lfloor g(x)\rfloor=0$.
   2. At $x=1$: $g(1)=1 \Rightarrow \lfloor g(1)\rfloor=1$. This is a jump from the left-hand value $0$.
   3. For $x>1$: $\log x>0$ so $g(x)>1$. The function $(\log x)/x$ has a maximum at $x=e$ (derivative $(1-\log x)/x^2=0$), giving $g(e)=e^{1/e}\approx1.4447$.
   4. As $x\to\infty$, $(\log x)/x\to0^+$, so $g(x)\to1^+$ (approaching but never reaching 1, staying above it). …

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7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \begin{cases} a - \dfrac{\sin[x-1]}{x-1} & \text{if } x > 1 \\ 1 & \text{if } x = 1 \\ b - \left[\dfrac{\sin[x-1] - [x-1]}{([x-1])^3}\right] & \text{if } x < 1 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. If $f$ is continuous at $x=1$, then $a+b=$ (A) 0 (B) 1 (C) 2 (D) 3

   ›Reveal solutionSolution

   Evaluating the two one-sided limits (each becomes a constant, thanks to the floor function freezing $[x-1]$ near $x=1$) and matching them to $f(1)=1$ gives $a=1,\ b=0$, so $a+b=1$.

   Concept and Intuition

   The trick with $[x-1]$ (greatest integer of $x-1$) is that for $x$ in a punctured neighbourhood of $1$ (excluding $1$ itself), $x-1$ never actually reaches an integer boundary except right at $x=1$ — so $[x-1]$ is genuinely constant just to the right of $1$ (equal to $0$, since $0<x-1<1$) and constant just to the left of $1$ (equal to $-1$, since $-1<x-1<0$). This turns each piece of $f$ into an ordinary constant near $x=1$, and continuity just becomes matching these constants (and the outer floor bracket) to $f(1)=1$.

   Step-by-Step Solution

   1. Right-hand limit ($x\to1^+$): for $x$ slightly greater than $1$, $x-1\in(0,1)$, so $[x-1]=0$.

   $f(x) = a-\frac{\sin[x-1]}{x-1} = a-\frac{\sin0}{x-1} = a-0=a$

   So $\displaystyle\lim_{x\to1^+}f(x)=a$.

   2. Left-hand limit ($x\to1^-$): for $x$ slightly less than $1$, $x-1\in(-1,0)$, so $[x-1]=-1$ throughout this range.

   $\sin[x-1]-[x-1] = \sin(-1)-(-1) = 1-\sin1$

   $([x-1])^3 = (-1)^3=-1$

   $\frac{\sin[x-1]-[x-1]}{([x-1])^3} = \frac{1-\sin1}{-1} = \sin1-1$

   Since $\sin1\approx0.8415$, this equals $\approx-0.1585$, a constant strictly between $-1$ and $0$, so its greatest integer (the outer $[\ \cdot\ ]$) is $-1$:

   $f(x) = b-\Big[\sin1-1\Big] = b-(-1) = b+1$ …

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8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.Let [x] represents the greatest integer not more than x. The discontinuous points of the function $f(x) = \dfrac{5+[x]}{\sqrt{11+[x]-6\sqrt{2+[x]}}}$ lies in the interval (A) $[0,\infty)$ (B) $[5,8]$ (C) $[7,8)$ (D) $[7,10)$

   ›Reveal solutionSolution

   A substitution $u=\sqrt{2+[x]}$ turns the messy nested radical into a perfect square $(u-3)^2$; the denominator vanishes (making $f$ undefined) exactly when $[x]=7$, i.e. on $[7,8)$.

   Concept and Intuition

   Functions built from the greatest-integer function are step functions, constant on each interval $[n,n+1)$. The real question here is where the formula itself breaks down (denominator zero), since that is where $f$ fails to be defined/continuous outright, rather than merely jumping between integers.

   Step-by-Step Solution

   1. Let $n=[x]$ (so $n$ is a fixed integer over $x\in[n,n+1)$), and require $n\geq-2$ for $\sqrt{2+n}$ to be real.
   2. Let $u=\sqrt{2+n}\geq0$, so $n=u^2-2$.
   3. The quantity under the outer square root: $11+n-6\sqrt{2+n} = 11+(u^2-2)-6u = u^2-6u+9=(u-3)^2$.
   4. So the denominator is $\sqrt{(u-3)^2}=|u-3|$.
   5. This is zero exactly when $u=3$, i.e. $\sqrt{2+n}=3\Rightarrow n=7$. …

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9. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let $[t]$ represents the greatest integer not exceeding $t$. Then the number of discontinuous points of $[10^x]$ in $(0, 10)$ is (A) $10^{10} - 1$ (B) $10^{10}$ (C) $10^{10} - 2$ (D) $e^{10}$

   ›Reveal solutionSolution

   The floor function jumps at integers; count how many integers $10^x$ passes through as $x$ ranges over $(0,10)$.

   Concept and Intuition

   $g(x)=10^x$ is continuous and strictly increasing, so $[g(x)]$ (the floor of $g$) is discontinuous exactly at those $x$ for which $g(x)$ is an integer — one discontinuity per integer value crossed.

   Step-by-Step Solution

   1. As $x$ ranges over the open interval $(0,10)$, $y=10^x$ ranges over the open interval $(10^0,10^{10})=(1,10^{10})$.
   2. $[y]$ is discontinuous at every integer value of $y$ in this range.
   3. Integers strictly between $1$ and $10^{10}$: $2,3,\dots,10^{10}-1$.
   4. Count $=(10^{10}-1)-2+1=10^{10}-2$. …

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10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.Let $[P]$ denote the greatest integer $\le P$. If $0\le a\le2$, then the number of integral values of 'a' such that $\lim_{x\to a}([x^2]-[x]^2)$ does not exist is (A) 3 (B) 2 (C) 1 (D) 0

    ›Reveal solutionSolution

    The greatest-integer function jumps at integers, so check the one-sided limits of $[x^2]-[x]^2$ at each of the three integral candidates $0,1,2$ in $[0,2]$ directly; the limit fails to exist wherever the two sides disagree.

    Concept and Intuition

    $[x]$ jumps by $1$ at every integer, and $[x^2]$ jumps whenever $x^2$ crosses an integer. Near a non-special point these jumps align smoothly, but exactly at an integer $a$, both $[x]$ and possibly $[x^2]$ can jump asymmetrically from the two sides, which can make the combination $[x^2]-[x]^2$ discontinuous with unequal one-sided limits.

    Step-by-Step Solution

    1. At $a=0$: Left ($x=-\epsilon$): $[x]=-1\Rightarrow[x]^2=1$; $x^2=\epsilon^2\Rightarrow[x^2]=0$. So $g\to 0-1=-1$. Right ($x=\epsilon$): $[x]=0\Rightarrow[x]^2=0$; $[x^2]=0$. So $g\to0-0=0$. Left $\ne$ Right $\Rightarrow$ limit does not exist at $a=0$.
    2. At $a=1$: Left ($x=1-\epsilon$): $[x]=0,[x]^2=0$; $x^2\approx1-2\epsilon<1\Rightarrow[x^2]=0$. $g\to0$. Right ($x=1+\epsilon$): $[x]=1,[x]^2=1$; $x^2\approx1+2\epsilon<2\Rightarrow[x^2]=1$. $g\to1-1=0$. Both sides give $0$ $\Rightarrow$ limit exists at $a=1$.
    3. At $a=2$: Left ($x=2-\epsilon$): $[x]=1,[x]^2=1$; $x^2\approx4-4\epsilon<4$, and $\ge3\Rightarrow[x^2]=3$. $g\to3-1=2$. …

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11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.The domain of the function $f(x) = \dfrac{1}{[x]-1}$, where $[x]$ is greatest integer function of $x$ is (A) $R - (1,2)$ (B) $R - \{1\}$ (C) $R - \{0,1\}$ (D) $R - [1,2)$

    ›Reveal solutionSolution

    The function is undefined whenever $[x]=1$; since $[x]=1$ for the whole interval $x\in[1,2)$ (not just a single point), the excluded set is $[1,2)$, not just $\{1\}$.

    Concept and Intuition

    For a function with $[x]$ (the greatest integer / floor function) in the denominator, you must find every $x$ for which the denominator vanishes — and because $[x]$ is a step function, it stays constant (equal to the same integer) over a whole half-open interval, not just at isolated points. So "$[x]=1$" isn't a single forbidden value of $x$; it's an entire interval of forbidden $x$-values.

    Step-by-Step Solution

    1. $f(x) = \dfrac{1}{[x]-1}$ is undefined whenever $[x] - 1 = 0$, i.e., whenever $[x] = 1$.
    2. Recall $[x] = n$ (an integer) exactly for $x \in [n, n+1)$. So $[x] = 1$ exactly for $x \in [1, 2)$.
    3. Therefore $f$ is undefined for every $x$ in $[1,2)$, and defined everywhere else. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If a real valued function $f(x) = \begin{cases} \log(1+[x]), & x \ge 0 \\ \sin^{-1}[x], & -1 \le x < 0 \\ k([x]+|x|), & x < -1 \end{cases}$ is continuous at $x = -1$, then $k =$ (A) $-\pi/2$ (B) $-\pi$ (C) $\pi$ (D) $\pi/2$

    ›Reveal solutionSolution

    Match the left-hand limit (from the $x<-1$ piece) to the function's actual value at $x=-1$ (from the $-1\le x<0$ piece). Answer: $k=\pi/2$.

    Concept and Intuition

    $x=-1$ belongs to the domain "$-1\le x<0$" of the function's definition, so $f(-1)$ is computed from that piece. Continuity at $x=-1$ then requires the limit from the left (using the "$x<-1$" piece, which is the only piece active for $x$ just below $-1$) to equal this value $f(-1)$.

    Step-by-Step Solution

    1. $f(-1) = \sin^{-1}([-1]) = \sin^{-1}(-1) = -\dfrac{\pi}{2}$ (using the $-1\le x<0$ branch, since $-1$ satisfies $-1\le -1<0$).
    2. For $x\to -1^-$ (i.e. $x$ slightly less than $-1$, so $x<-1$), use the third branch $k([x]+|x|)$.
    3. For such $x$ (e.g. $x=-1-\epsilon$, small $\epsilon>0$): $[x]=\lfloor -1-\epsilon\rfloor = -2$, and $|x| = 1+\epsilon \to 1$.
    4. So $\displaystyle\lim_{x\to-1^-} k([x]+|x|) = k(-2+1) = -k$. …

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