Q.Show that the function defined by $g(x) = x - [x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

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1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.If $[.]$ here denotes the greatest integer function, $\lim_{x \to 0} x^7 \left[ \frac{1}{x^3} \right] =$ (A) $1$ (B) $0$ (C) $-1$ (D) Does not exist

   ›Reveal solutionSolution

   Splitting the greatest-integer function into its exact value minus a bounded fractional part shows the limit is $0$. Answer: $0$.

   Concept and Intuition

   For the greatest integer (floor) function, $[y] = y - \{y\}$ where $\{y\}\in[0,1)$ always, whatever the sign of $y$. This lets us replace an unwieldy floor expression with an exact term plus a bounded "error" term that is easy to control.

   Step-by-Step Solution

   1. Let $y=1/x^3$. Then $[y] = y - \{y\}$, with $0\le\{y\}<1$.
   2. So $x^7[y] = x^7 y - x^7\{y\} = x^7\cdot\dfrac{1}{x^3} - x^7\{y\} = x^4 - x^7\{y\}$.
   3. As $x\to0$ (from either side), $x^4\to0$. …

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9. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$[x]$ denotes integral part of $x$. For $n \in N$, if $f(x) = \begin{cases} \left[|x|\left[\dfrac{1}{|x|}\right]\right], & \text{for } |x| \neq \dfrac{1}{n} \\ 0, & \text{for } |x| = \dfrac{1}{n} \end{cases}$, then for $|x| \neq \dfrac{1}{n}$, $f(x) =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{n}$ (D) $n$

   ›Reveal solutionSolution

   For any $x$ with $|x|\ne 1/n$, the quantity $\big[|x|[1/|x|]\big]$ always simplifies to $0$ — a classic floor-function identity. Answer: (A) 0.

   Concept and Intuition

   This is testing the identity $[t\,[1/t]] = 0$ for all $t>0$ not of the form $1/n$. The intuition: $[1/t]$ is the largest integer $n$ with $n \le 1/t$ (loosely), so $nt \le 1$, and as long as $t$ isn't exactly $1/n$, the product $nt$ stays strictly below 1 — never reaching the next integer — so its floor is always 0.

   Step-by-Step Solution

   1. Let $t = |x| > 0$, and suppose $t \ne 1/n$ for every positive integer $n$.
   2. Case $t \ge 1$: Then $0 < 1/t \le 1$. Since $t\ne 1$ (which would be $1/n$ with $n=1$), we actually have $1/t < 1$, so $[1/t] = 0$. Hence $t\cdot[1/t] = 0$ and $[\,t[1/t]\,] = 0$.
   3. Case $0 < t < 1$: Then $1/t > 1$. Let $n = [1/t]$, an integer $\ge 1$. By definition of floor, $n \le 1/t < n+1$. Since $t \ne 1/n$, the equality $n = 1/t$ cannot hold, so strictly $n < 1/t < n+1$.
   4. Multiplying through by $t>0$: $nt < 1 < (n+1)t$. In particular $0 < nt < 1$ (since $nt>0$ trivially and $nt<1$ from the left inequality). …

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10. 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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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.Let [.] denote the greatest integer function. Assertion (A): $\lim_{x\to\infty}\frac{[x]}{x}=1$ Reason (R): $f(x)=x-1$, $g(x)=[x]$, $h(x)=x$ and $\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}\frac{h(x)}{x}=1$ (A) A is true, R is true; R is correct explanation of A (B) A, R are true; R is not the correct explanation of A (C) A is true, R is false (D) A is false, R is true

    ›Reveal solutionSolution

    Both the assertion and reason are true, and the reason is exactly the squeeze/sandwich-theorem setup that proves the assertion.

    Concept and Intuition

    The greatest integer function satisfies $x-1<[x]\le x$ for all real $x$. Dividing through by $x>0$ and letting $x\to\infty$, both bounding expressions tend to $1$, so by the Sandwich (Squeeze) theorem the middle expression $[x]/x$ is forced to tend to $1$ as well.

    Step-by-Step Solution

    1. Assertion A: for $x-1<[x]\le x$, dividing by $x$ (positive, $x\to\infty$): $1-\tfrac1x<\tfrac{[x]}{x}\le1$.
    2. As $x\to\infty$, $1-\tfrac1x\to1$, so by squeeze $\tfrac{[x]}{x}\to1$ — A is true.
    3. Reason R defines $f(x)=x-1$, $g(x)=[x]$, $h(x)=x$, satisfying $f(x)\le g(x)\le h(x)$ for all $x$ (the standard inequality above), and states $\lim f(x)/x=\lim h(x)/x=1$.
    4. This is exactly the hypothesis of the Sandwich theorem applied to $g(x)/x=[x]/x$, so R is true and directly explains why A holds. …

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12. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.If [ . ] denotes greatest integer function, then $\displaystyle\lim_{x \to \frac{\pi}{2}^{+}} \dfrac{[\sin x] - [\cos x] + 1}{2} =$ (A) $0$ (B) $\dfrac{-1}{2}$ (C) $\dfrac{1}{2}$ (D) $1$

    ›Reveal solutionSolution

    Evaluate the floor of $\sin x$ and $\cos x$ just to the right of $\pi/2$ using the sign/magnitude behaviour there.

    Concept and Intuition

    $[\cdot]$ (greatest integer function) needs the actual numeric range of $\sin x,\cos x$ near $\pi/2^+$, not just their limiting values — floors are sensitive to which side of an integer a quantity approaches from.

    Step-by-Step Solution

    1. For $x=\dfrac\pi2+\varepsilon$ ($\varepsilon\to0^+$): $\sin x=\sin\left(\dfrac\pi2+\varepsilon\right)=\cos\varepsilon$, which is slightly less than $1$ (in $(0,1)$), so $[\sin x]=0$.
    2. $\cos x=\cos\left(\dfrac\pi2+\varepsilon\right)=-\sin\varepsilon$, a small negative number (in $(-1,0)$), so $[\cos x]=-1$.
    3. Substitute: $\dfrac{[\sin x]-[\cos x]+1}{2}=\dfrac{0-(-1)+1}{2}=\dfrac{2}{2}=1$.

    Common Mistakes …

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