Q.Prove that the greatest integer function defined by $f(x) = [x]$, $0 < x < 3$, is not differentiable at $x = 1$ and $x = 2$.

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1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$[x]$ represent greatest integer function. The difference of the maximum and minimum values of the function $f(x) = \sin\dfrac{[x]\pi}{2} + \cos 2[x]\pi$ is (A) $0$ (B) $1$ (C) $2$ (D) $4$

   ›Reveal solutionSolution

   Writing $n=[x]$ reduces the function to $\sin(n\pi/2)+1$ for integer $n$; since $\sin(n\pi/2)$ only ever takes the values $-1,0,1$, the function's range is $\{0,1,2\}$, giving a max$-$min difference of $2$.

   Concept and Intuition

   Since $[x]$ (the greatest integer function) only ever outputs integers, we can replace $[x]$ by an integer variable $n$ and analyze the resulting discrete-valued function directly, rather than treating $x$ as continuous.

   Step-by-Step Solution

   1. Let $n=[x]\in\mathbb Z$. Then $f(x) = \sin\left(\dfrac{n\pi}{2}\right) + \cos(2n\pi)$.
   2. Since $n$ is an integer, $2n\pi$ is an integer multiple of $2\pi$, so $\cos(2n\pi) = 1$ always.
   3. So $f = \sin\left(\dfrac{n\pi}{2}\right) + 1$.
   4. Evaluate $\sin(n\pi/2)$ for consecutive integers (period 4): $n=0\Rightarrow 0$; $n=1\Rightarrow1$; $n=2\Rightarrow0$; $n=3\Rightarrow-1$; then it repeats.
   5. So $\sin(n\pi/2) \in \{-1,0,1\}$ for all integers $n$, hence $f = \sin(n\pi/2)+1 \in \{0,1,2\}$. …

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2. 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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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\displaystyle\lim_{n \to 5^-} \left(\dfrac{[n]}{2}\right)^3 - \left(\dfrac{[n]^3}{2^4}\right) = k$, then $\displaystyle\lim_{n \to k^+} \left(\dfrac{[n]}{2}\right)^3 - \left(\dfrac{[n]^3}{2^4}\right) =$ (here $[n]$ denotes the greatest integer function) (A) $k + 1$ (B) $k - 1$ (C) $k + 2$ (D) $k$

   ›Reveal solutionSolution

   Both one-sided limits land on integer values where $[n]=4$ throughout the punctured neighbourhood, so both expressions evaluate to the same number $4$; the second limit equals $k$.

   Concept and Intuition

   The greatest-integer function $[n]$ is constant on any half-open interval that avoids an integer boundary. As $n\to5^-$, for $n$ slightly less than $5$ (e.g. $4.9$), $[n]=4$ throughout that neighbourhood, so the limit is just the function evaluated at $[n]=4$. Similarly for $n\to4^+$ (e.g. $4.1$), $[n]=4$ throughout.

   Step-by-Step Solution

   1. As $n\to5^-$: for $n$ in $(4,5)$, $[n]=4$. So

   $\lim_{n\to5^-}\left(\frac{[n]}{2}\right)^3-\frac{[n]^3}{2^4}=\left(\frac42\right)^3-\frac{4^3}{16}=8-4=4.$

   So $k=4$.

   2. Now find $\displaystyle\lim_{n\to k^+}\left(\frac{[n]}{2}\right)^3-\frac{[n]^3}{2^4}=\lim_{n\to4^+}(\cdots)$. …

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4. 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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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 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$[x]$ denotes the greatest integer less than or equal to x. If $\{x\}=x - [x]$ and $\displaystyle\lim_{x\to 0^-} \dfrac{\mathrm{Sin}^{-1}(x+[x])}{2-\{x\}} = \theta$, then $\sin \theta + \cos \theta =$ (A) -1 (B) 0 (C) 1 (D) $\sqrt{2}$

   ›Reveal solutionSolution

   This tests reading the floor and fractional-part functions correctly on the left of $0$. Evaluating the limit gives $\theta=-\dfrac{\pi}{2}$, so $\sin\theta+\cos\theta=-1$.

   Concept and Intuition

   For $x$ approaching $0$ from the left, $x$ stays in $(-1,0)$, so the greatest integer function $[x]$ is constant at $-1$ throughout this approach (it does not vary continuously — it only "jumps" at integers). Similarly $\{x\}=x-[x]=x+1$ tends to $1$ as $x\to 0^-$. Substituting these constants turns the seemingly complicated limit into a simple evaluation.

   Step-by-Step Solution

   1. For $x\in(-1,0)$ (which covers $x\to0^-$): $[x]=-1$.
   2. So $x+[x] = x-1$. As $x\to0^-$, $x-1\to -1$.
   3. Also $\{x\}=x-[x]=x+1$. As $x\to0^-$, $\{x\}\to 1$.
   4. Substitute into the limit: $\theta=\lim_{x\to0^-}\frac{\sin^{-1}(x+[x])}{2-\{x\}} = \frac{\sin^{-1}(-1)}{2-1} = \frac{-\pi/2}{1}=-\frac{\pi}{2}$ …

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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 2025Set eng-2025-05-23-AN1 markMCQ
   Q.[ t ] denotes the greatest integer function and $[t-m]=[t]-m$ when $m \in Z$. If $k = 2[2x-1]-1$ and $3[2x-2]+1=2[2x-1]-1$ then the range of $f(x)=[k+5x]$ is (A) $\{7, 8, 9\}$ (B) $\{4, 5, 6\}$ (C) $\{5, 6, 7\}$ (D) $\{6, 7, 8\}$

   ›Reveal solutionSolution

   This tests the property $[t-m]=[t]-m$ for integer $m$, used to pin down $x$ from a floor-function equation, then reading off the range of another floor function over that interval. The answer is $\{6,7,8\}$.

   Concept and Intuition

   The greatest integer function $[t]$ only changes as $t$ crosses an integer. When you subtract an integer $m$ from $t$, the floor just shifts by that integer: $[t-m]=[t]-m$. This lets every floor expression in the problem be rewritten in terms of a single unknown integer, $n=[2x]$, turning a floor equation into an ordinary integer equation.

   Step-by-Step Solution

   1. Let $n=[2x]$. Using $[t-m]=[t]-m$ with $t=2x$: $[2x-1]=[2x]-1=n-1$, and $[2x-2]=[2x]-2=n-2$.
   2. Substitute into $3[2x-2]+1=2[2x-1]-1$:

   $3(n-2)+1=2(n-1)-1 \implies 3n-5=2n-3 \implies n=2.$

   3. So $[2x]=2$, i.e. $2\le 2x<3$, i.e. $x\in[1,1.5)$.
   4. Compute $k=2[2x-1]-1=2(n-1)-1=2(1)-1=1$.
   5. Now $f(x)=[k+5x]=[1+5x]$ for $x\in[1,1.5)$. As $x$ increases continuously from $1$ to just under $1.5$, $1+5x$ increases continuously from $6$ to just under $8.5$. …

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9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.[x] represents the greatest integer function. If $\displaystyle\lim_{x \to 0^+} \dfrac{\cos[x] - \cos(kx - [x])}{x^2} = 5$ then $k =$ (A) $\sqrt{10}$ (B) $\sqrt{11}$ (C) $3$ (D) $9$

   ›Reveal solutionSolution

   Tests the greatest-integer function combined with a standard $1-\cos$ small-angle limit; the answer is $k=\sqrt{10}$.

   Concept and Intuition

   The greatest integer function $[x]$ is locally constant near any non-integer point. As $x\to0^+$ (approaching 0 from the right, through small positive values less than 1), $[x]=0$ throughout — this is the key simplification that turns an odd-looking limit into a familiar one. Once $[x]$ is replaced by $0$, the limit is just the classical $\lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta^2}=\dfrac12$ in disguise.

   Step-by-Step Solution

   1. For $0<x<1$, $[x]=0$. So $\cos[x]=\cos 0=1$, and $\cos(kx-[x])=\cos(kx-0)=\cos(kx)$.
   2. The limit becomes $\displaystyle\lim_{x\to0^+}\frac{1-\cos(kx)}{x^2}$.
   3. Use the standard expansion $1-\cos\theta = 2\sin^2(\theta/2)\approx \dfrac{\theta^2}{2}$ for small $\theta$, with $\theta=kx$: $\frac{1-\cos(kx)}{x^2}\to \frac{k^2x^2/2}{x^2}=\frac{k^2}{2}.$ …

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10. 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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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If a real valued function $f(x) = \begin{cases} e^{\frac{\sin a(x-[x])}{x-[x]}}, & \text{if } x < 1 \\ b+1, & \text{if } x = 1 \\ \dfrac{|x^2+x-2|}{x-1}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then $b \sin a =$ ($[x]$ denotes the greatest integer function) (A) $6$ (B) $4$ (C) $\log_e 9$ (D) $\log_e 2$

    ›Reveal solutionSolution

    Tests continuity of a piecewise function using the greatest-integer function; matching the three one-sided values at $x=1$ gives $b\sin a=\log_e 9$.

    Concept and Intuition

    Continuity at a point where a function is defined piecewise means the left-hand limit, the right-hand limit, and the value at the point must all agree. Since $[x]$ (the greatest integer function) only changes value at integers, for $x$ just below $1$ (i.e. $0<x<1$) we have $[x]=0$, so the messy exponent simplifies a lot.

    Step-by-Step Solution

    1. Left-hand limit ($x\to 1^-$): for $x$ slightly less than $1$, $[x]=0$, so $x-[x]=x$. Thus

    $\lim_{x\to1^-} e^{\frac{\sin(a(x-[x]))}{x-[x]}} = \lim_{x\to1^-} e^{\frac{\sin(ax)}{x}} = e^{\sin a}.$

    2. Value at $x=1$: $f(1)=b+1$.
    3. Right-hand limit ($x\to1^+$): $x^2+x-2=(x-1)(x+2)$. For $x>1$, $x-1>0$ and $x+2>0$, so $|x^2+x-2|=(x-1)(x+2)$, giving

    $f(x)=\frac{(x-1)(x+2)}{x-1}=x+2 \;\Rightarrow\; \lim_{x\to1^+} f(x) = 3.$

    4. Match all three: $e^{\sin a} = b+1 = 3$. …

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12. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.Let $[x]$ denote the greatest integer less than or equal to x. Then $\displaystyle\lim_{x \to 2^+} \left( \dfrac{[x]^3}{3} - \left[\dfrac{x}{3}\right]^3 \right) =$ (A) $0$ (B) $\dfrac{8}{3}$ (C) $\dfrac{64}{27}$ (D) $\dfrac{1}{3}$

    ›Reveal solutionSolution

    Evaluate the two floor functions just to the right of $x=2$: $[x]=2$ and $[x/3]=0$. Answer: $\dfrac83$.

    Concept and Intuition

    The greatest-integer function $[x]$ is constant on each interval $[n,n+1)$, so a one-sided limit "at" an integer just means evaluating $[x]$ on the corresponding half-open interval immediately next to that integer, not at the integer's own possibly-different value on the other side.

    Step-by-Step Solution

    1. As $x\to2^+$, $x$ ranges just above $2$, i.e. $x\in(2,2+\varepsilon)\subset[2,3)$, so $[x]=2$.
    2. For this same range, $\dfrac x3\in\left(\dfrac23,\dfrac{2+\varepsilon}{3}\right)$, which lies inside $[0,1)$ since $\dfrac{2+\varepsilon}3<1$ for small $\varepsilon$. So $\left[\dfrac x3\right]=0$.
    3. Substitute: $\dfrac{[x]^3}{3}-\left[\dfrac x3\right]^3=\dfrac{2^3}{3}-0^3=\dfrac83-0=\dfrac83$.
    4. Since this value is constant for all $x$ in a right-neighbourhood of $2$, the limit equals $\dfrac83$.

    Common Mistakes …

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