Q.Let $f(x) = \sqrt{x}$ and $g(x) = x$ be two functions defined in the domain $\mathbf{R}^+ \cup \{0\}$. Find

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1. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $f:R \to R$ and $g:R \to R$ are two functions defined by $f(x) = |x|$ and $g(x) = [x]$ then $\{x \in R \mid (gof)(x) = (fog)(x)\} =$ (Here $[x]$ is the greatest integer function) (A) $R$ (B) $[0,\infty)$ (C) $[0,\infty) \cup \{n \mid -n \in N\}$ (D) $Z$

   ›Reveal solutionSolution

   The key idea is to compare the composite functions $g(f(x))$ and $f(g(x))$ for all real $x$, using the definitions of absolute value and greatest integer. The equality holds exactly for $x \ge 0$ and for negative integers, which matches option (C).

   We are given:

   • $f(x) = |x|$ (absolute value)
   • $g(x) = [x]$ (greatest integer less than or equal to $x$)

   We need the set of all $x \in \mathbb{R}$ such that $(g \circ f)(x) = (f \circ g)(x)$, i.e.

   $g(f(x)) = f(g(x))$.

   ---

   Concept and Intuition

   The absolute value makes any input non-negative, while the greatest integer function "rounds down" to the nearest integer.

   When we compose them in different orders, we get different behaviors:

   • $g(f(x))$: First take absolute value (so result is $\ge 0$), then take greatest integer. So $g(f(x)) = [\,|x|\,]$, which is always a non-negative integer.
   • $f(g(x))$: First take greatest integer (could be negative, zero, or positive), then take absolute value. So $f(g(x)) = |\,[x]\,|$, which is also a non-negative integer, but the inner value can be negative.

   The equality will fail when the two processes give different integers. The main source of difference is when $x$ is negative but not an integer: then $|x|$ is positive, but $[x]$ is negative, and their absolute values differ.

   ---

   Step-by-step reasoning

   1. Write the composites explicitly For any real $x$:

   $(g \circ f)(x) = g(f(x)) = [\,|x|\,]$

   $(f \circ g)(x) = f(g(x)) = |\,[x]\,|$

   2. Consider cases based on $x$

      Case 1: $x \ge 0$

      Then $|x| = x$, so $[\,|x|\,] = [x]$.

      Also $[x] \ge 0$, so $|\,[x]\,| = [x]$.

      Hence both composites equal $[x]$, so equality holds for all $x \ge 0$.

      Case 2: $x < 0$

      Here $|x| = -x > 0$, so $[\,|x|\,] = [-x]$, which is a non-negative integer.

      Meanwhile $[x]$ is negative or zero, so $|\,[x]\,| = -[x]$ (since $[x] \le 0$).

      We need:

   $[-x] = -[x]$

   Subcase 2a: $x$ is a negative integer

   Let $x = -n$ where $n \in \mathbb{N}$ (positive integer). Then:

   • $[x] = -n$
   • $|x| = n$, so $[\,|x|\,] = [n] = n$
   • $|\,[x]\,| = |-n| = n$ So equality holds for all negative integers.

   Subcase 2b: $x$ is negative but not an integer …

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2. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $f: R \to R$ and $g: R \to R$ are two functions defined by $f(x) = 2x - 3$ and $g(x) = 5x^2 - 2$, then the least value of the function $(g \circ f)(x)$ is (A) $-2$ (B) $2$ (C) $-4$ (D) $4$

   ›Reveal solutionSolution

   Composing $g$ with $f$ gives $5(2x-3)^2-2$, a perfect-square expression shifted down by 2; its minimum occurs when the square term is zero, giving least value $-2$.

   Concept and Intuition

   When a composite function reduces to (constant)$\times$(something)$^2$ + (constant), the squared term is always $\ge 0$, so the whole expression is minimised exactly when the square vanishes. This is the same idea as completing the square to find a parabola's vertex value — here it's handed to us already in squared form via the composition.

   Step-by-Step Solution

   1. Compute the composite: $(g\circ f)(x) = g(f(x)) = 5\,[f(x)]^2 - 2 = 5(2x-3)^2 - 2$.
   2. Note $(2x-3)^2 \ge 0$ for every real $x$, with equality when $2x-3=0 \Rightarrow x = 3/2$.
   3. Therefore $5(2x-3)^2 \ge 0$, so $(g\circ f)(x) = 5(2x-3)^2 - 2 \ge -2$. …

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3. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If $f(x) = \sqrt{x} - 1$ and $g\{f(x)\} = x + 2\sqrt{x} + 1$ then $g(x) =$ (A) $(x+2)^2$ (B) $(x-2)^2$ (C) $(\sqrt{x}+2)^2$ (D) $(\sqrt{x}-2)^2$

   ›Reveal solutionSolution

   This tests recovering the function $g$ from a composite expression by substituting $t = f(x)$ and rewriting everything in terms of $t$; the algebraic simplification $x+2\sqrt{x}+1 = (\sqrt{x}+1)^2$ is the key step, giving $g(x) = (x+2)^2$.

   Concept and Intuition

   When we are given $g(f(x))$ as an explicit expression in $x$, the standard technique to find $g$ itself is to substitute $t = f(x)$, express $x$ (or, more usefully here, $\sqrt{x}$) in terms of $t$, and rewrite the given expression for $g(f(x))$ purely in terms of $t$. Whatever expression in $t$ results is exactly $g(t)$, and renaming $t \to x$ gives the formula for $g$.

   Step-by-Step Solution

   1. Let $t = f(x) = \sqrt{x} - 1$. Then $\sqrt{x} = t + 1$.
   2. The given composite is $g(f(x)) = x + 2\sqrt{x} + 1$. Notice that $x + 2\sqrt{x} + 1 = (\sqrt{x})^2 + 2\sqrt{x} + 1 = (\sqrt{x} + 1)^2$, a perfect square.
   3. Substitute $\sqrt{x} + 1 = (t+1) + 1 = t + 2$: so $g(f(x)) = (t+2)^2$.
   4. Since $g(f(x)) = g(t)$ by definition (as $t = f(x)$), we get $g(t) = (t+2)^2$.
   5. Renaming the variable, $g(x) = (x+2)^2$.

   Common Mistakes …

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4. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Define the functions f, g and h from R to R such that $f(x)=x^2-1$, $g(x)=\sqrt{x^2+1}$ and $h(x)=\begin{cases}0, & \text{if } x\le 0\\ x, & \text{if } x\ge 0\end{cases}$ consider the following statements (I) fog is invertible (II) h is an identity function (III) fog is not invertible (IV) $(hofog)\,x = x^2$ Then which one of the following is true? (A) II, IV (B) II, III (C) III, IV (D) I, II

   ›Reveal solutionSolution

   Composing $f$ and $g$ collapses to $x^2$, which is not injective (so $fog$ is not invertible, ruling in III and out I); $h$ only matches the identity for $x\ge0$ (ruling out II); and $h(fog(x))=h(x^2)=x^2$ since squares are never negative (ruling in IV).

   Concept and Intuition

   A function is invertible on its whole stated domain/codomain ($\mathbb{R}\to\mathbb{R}$ here) only if it is a bijection there — in particular it must be one-one. Composing two functions doesn't automatically produce something invertible even if the pieces individually look nice; here $g$ squashes distinct inputs to the same output via $f$'s squaring afterward. Also, a piecewise function like $h$ only counts as "the identity function" if it equals $x$ everywhere in its domain — matching on just one branch isn't enough.

   Step-by-Step Solution

   1. Compute $g(x)=\sqrt{x^2+1}$, then $f(g(x))=(g(x))^2-1=(x^2+1)-1=x^2$. So $fog(x)=x^2$ for all real $x$.
   2. Check injectivity of $fog$: $fog(2)=4=fog(-2)$, so it is not one-one on $\mathbb{R}\to\mathbb{R}$, hence not invertible. This makes statement I false and statement III true.
   3. Check $h$: for $x\le 0$, $h(x)=0\ne x$ (except at $x=0$); for $x\ge0$, $h(x)=x$. Since $h$ fails to equal $x$ for all negative $x$, $h$ is not the identity function on $\mathbb{R}$. Statement II is false. …

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5. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $f(x)=x^3-x$ and $g(x)=\sin 2x$, then $f\left(g\left(\dfrac{\pi}{12}\right)\right)=$ (A) 0 (B) 1 (C) $-\dfrac{3}{8}$ (D) 2

   ›Reveal solutionSolution

   Compute the inner function first: $g(\pi/12) = \sin(\pi/6) = 1/2$. Then apply $f$: $f(1/2) = 1/8 - 1/2 = -3/8$.

   Concept and Intuition

   This is a direct function-composition evaluation, $f(g(x))$: evaluate the inner function $g$ at the given point first, then substitute that numerical result into the outer function $f$. No special identities are needed beyond knowing $\sin(\pi/6) = 1/2$, a standard angle value.

   Step-by-Step Solution

   1. Compute the inner function: $g\left(\dfrac{\pi}{12}\right) = \sin\left(2 \cdot \dfrac{\pi}{12}\right) = \sin\left(\dfrac{\pi}{6}\right)$.
   2. Recall the standard value: $\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$.
   3. Substitute this into $f$: $f\left(\dfrac{1}{2}\right) = \left(\dfrac{1}{2}\right)^3 - \dfrac{1}{2}$.
   4. Compute: $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$, so $f\left(\dfrac12\right) = \dfrac{1}{8} - \dfrac{1}{2} = \dfrac{1}{8} - \dfrac{4}{8} = -\dfrac{3}{8}$. …

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6. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$ is defined by $f(x)=x+\dfrac{1}{x}$, then the value of $(f(x))^2=$ (A) $f(x)+f(0)$ (B) $f(x^2)+f(2)$ (C) $f(x^3)+f(0)$ (D) $f(x^2)+f(1)$

   ›Reveal solutionSolution

   Squaring $f(x)=x+1/x$ gives $x^2+1/x^2+2$, which is $f(x^2)$ plus the constant $2$; since $f(1)=2$, this equals $f(x^2)+f(1)$.

   Concept and Intuition

   This is a classic "recognize the pattern" functional-equation question: instead of guessing among the options, just compute $(f(x))^2$ explicitly using algebra, then see which combination of $f$-values reproduces exactly that expression. The key algebraic fact used is $(a+b)^2=a^2+2ab+b^2$ with $a=x,\ b=1/x$, so the cross term $2ab=2$ is a constant, and $a^2+b^2=x^2+1/x^2=f(x^2)$.

   Step-by-Step Solution

   1. $f(x)=x+\dfrac1x$, so $(f(x))^2=\left(x+\dfrac1x\right)^2$.
   2. Expand: $\left(x+\dfrac1x\right)^2=x^2+2\cdot x\cdot\dfrac1x+\dfrac1{x^2}=x^2+2+\dfrac1{x^2}$.
   3. Group: $x^2+\dfrac1{x^2}$ is exactly $f(x^2)$ (substitute $x^2$ in place of $x$ in the definition of $f$): $f(x^2)=x^2+\dfrac1{x^2}$.
   4. So $(f(x))^2=f(x^2)+2$.
   5. Compute $f(1)=1+\dfrac11=1+1=2$. …

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7. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ are defined by $f(x)=x^3-x$ and $g(x)=\sin 2x$, then the value of $x\in(0,2\pi)$ that satisfy $f(g(x))>0$, lie in the interval (A) $\left(\dfrac{\pi}{2},\pi\right)$ (B) $\left(0,\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\pi\right)$ (C) $\left(\dfrac{\pi}{2},\dfrac{3\pi}{4}\right)\cup\left(\dfrac{3\pi}{4},\pi\right)$ (D) $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

   ›Reveal solutionSolution

   f(g(x))>0 requires sin2x strictly between −1 and 0; within (π/2,π) that holds everywhere except the single point x=3π/4 where sin2x=−1 exactly.

   Concept and Intuition

   Since f(t)=t³−t=t(t−1)(t+1) factors into three linear terms, a sign chart on t immediately shows where f(t)>0. Because t=sin2x is confined to [−1,1], only the middle interval of the sign chart (t between −1 and 0) is achievable, and we must also throw out the boundary t=−1 (a strict inequality).

   Step-by-Step Solution

   1. f(t) = t(t−1)(t+1). Test sign in each interval determined by roots −1, 0, 1:
      • t∈(−1,0): (−)(−)(+) = + → f(t)>0.
      • t∈(0,1): (+)(−)(+) = − → f(t)<0.
   2. Since t=sin2x∈[−1,1], f(t)>0 ⟺ sin2x ∈ (−1, 0).
   3. Look for x where sin2x is negative: this happens when 2x lies in (π, 2π) modulo 2π. Restricting to x in (π/2, π) (2x ∈ (π,2π)), sin2x < 0 throughout. …

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8. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.If $f(x) = \sqrt{2 - x^2}$ and $g(x) = \log(1-x)$ are two real valued functions then the domain of the function $(f+g)(x)$ is (A) $[-2, 2]$ (B) $[-2, 1)$ (C) $(-\infty, 1)$ (D) $(1, 2]$

   ›Reveal solutionSolution

   The domain of a sum of two functions is the intersection of their individual domains. Here that intersection is $[-\sqrt2,1)$, since the square-root's domain caps at $\sqrt2$ but the logarithm's domain caps (more tightly, on the right) at $1$.

   Concept and Intuition

   For $(f+g)(x)=f(x)+g(x)$ to be defined, BOTH $f(x)$ and $g(x)$ must individually be defined at $x$ — so the domain of the sum is always $\text{Dom}(f)\cap\text{Dom}(g)$, never a union.

   Step-by-Step Solution

   1. $f(x)=\sqrt{2-x^2}$ needs the radicand non-negative: $2-x^2\ge0\Rightarrow x^2\le2\Rightarrow -\sqrt2\le x\le\sqrt2$. So $\text{Dom}(f)=[-\sqrt2,\sqrt2]$.
   2. $g(x)=\log(1-x)$ needs the argument strictly positive: $1-x>0\Rightarrow x<1$. So $\text{Dom}(g)=(-\infty,1)$.
   3. Domain of $(f+g)$ = $[-\sqrt2,\sqrt2]\cap(-\infty,1)$.
   4. Since $\sqrt2\approx1.414$ is greater than $1$, the upper bound from $f$'s domain ($\sqrt2$) is irrelevant — the tighter constraint on the right comes from $g$, capping at $x<1$ (excluded). …

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9. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Let $f: \mathbb{R} - \left\{\frac{-1}{2}\right\} \to \mathbb{R}$ be defined by $f(x) = \frac{x-2}{2x+1}$. If $\alpha, \beta$ satisfy the equation $f(f(x)) = -x$, then $4(\alpha^2+\beta^2) =$ (A) $17$ (B) $12$ (C) $24$ (D) $34$

   ›Reveal solutionSolution

   Composing $f$ with itself and solving $f(f(x))=-x$ reduces to a quadratic whose roots' sum-of-squares, scaled by 4, equals 17.

   Concept and Intuition

   Composition-of-functions problems like this reduce to algebraic manipulation: compute $f(f(x))$ explicitly as a single rational function, then solve the resulting equation. Once you have a quadratic in $x$, symmetric functions of its roots (sum and product, via Vieta's formulas) let you compute $\alpha^2+\beta^2$ without solving for the roots individually.

   Step-by-Step Solution

   1. Let $y = f(x) = \dfrac{x-2}{2x+1}$. Compute $f(y) = \dfrac{y-2}{2y+1}$.
   2. Numerator: $y - 2 = \dfrac{x-2}{2x+1} - 2 = \dfrac{(x-2)-2(2x+1)}{2x+1} = \dfrac{-3x-4}{2x+1}$.
   3. Denominator: $2y+1 = \dfrac{2(x-2)}{2x+1}+1 = \dfrac{2(x-2)+(2x+1)}{2x+1} = \dfrac{4x-3}{2x+1}$.
   4. So $f(f(x)) = \dfrac{-3x-4}{4x-3}$.
   5. Set $f(f(x)) = -x$: $\dfrac{-3x-4}{4x-3} = -x \Rightarrow -3x-4 = -x(4x-3) = -4x^2+3x$.
   6. Rearranging: $4x^2 - 6x - 4 = 0 \Rightarrow 2x^2 - 3x - 2 = 0$.
   7. By Vieta's formulas for this quadratic, $\alpha+\beta = \dfrac{3}{2}$, $\alpha\beta = \dfrac{-2}{2}=-1$. …

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10. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.Let $f(x) = \sqrt{\dfrac{x+1}{x+3}}$ and $g(x) = \sqrt{\dfrac{2-x}{x+3}}$ be two real valued functions. Then the domain of f/g is (A) $(-\infty, -3) \cup [-1,\infty)$ (B) $[-1,2)$ (C) $(-3,2)$ (D) $(-\infty,-3) \cup [2,\infty)$

    ›Reveal solutionSolution

    Finding where both square roots are defined and then excluding the point where the denominator $g(x)$ vanishes gives the domain of $f/g$ as $[-1, 2)$.

    Concept and Intuition

    For a quotient of functions $f/g$ to be defined at a point, that point must lie in the domain of $f$, in the domain of $g$, AND satisfy $g(x)\ne0$ (since we cannot divide by zero). Since $f$ and $g$ here are square roots, each requires its radicand (the expression under the root) to be non-negative, which itself requires careful sign analysis around the zeros of numerator and denominator.

    Step-by-Step Solution

    1. Domain of $f$: need $\dfrac{x+1}{x+3}\ge 0$. The critical points are $x=-1$ (numerator zero, included since $\sqrt0=0$ is defined) and $x=-3$ (denominator zero, excluded). Sign analysis: for $x<-3$, both $(x+1)$ and $(x+3)$ are negative → ratio positive (valid); for $-3<x<-1$, numerator negative, denominator positive → ratio negative (invalid); for $x\ge-1$, both non-negative → ratio $\ge0$ (valid). So domain$(f) = (-\infty,-3)\cup[-1,\infty)$.
    2. Domain of $g$: need $\dfrac{2-x}{x+3}\ge0$. Critical points $x=2$ (numerator zero, included) and $x=-3$ (denominator zero, excluded). For $x<-3$: numerator positive, denominator negative → ratio negative (invalid). For $-3<x\le2$: numerator $\ge0$ (until $x=2$), denominator positive → ratio $\ge0$ (valid). For $x>2$: numerator negative, denominator positive → invalid. So domain$(g) = (-3,2]$. …

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11. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.Let $f: R \to R$ and $g: R \to R$ be defined by $f(x) = 2x + 1$ & $g(x) = x^2 - 2$ determine $(gof)(x) =$ (A) $2x^2 - 3$ (B) $4x^2 + 4x - 1$ (C) $4x^2 + 4x + 1$ (D) $2x^2 - 4$

    ›Reveal solutionSolution

    Substitute $f(x)$ into $g$ and expand: $(g\circ f)(x) = 4x^2+4x-1$.

    Concept and Intuition

    Function composition $(g\circ f)(x)$ means 'apply $f$ first, then apply $g$ to the result' — i.e., $g(f(x))$.

    Step-by-Step Solution

    1. $f(x) = 2x+1$.
    2. $g(x) = x^2 - 2$, so $g(f(x)) = (f(x))^2 - 2 = (2x+1)^2 - 2$.
    3. Expand $(2x+1)^2 = 4x^2+4x+1$.
    4. So $g(f(x)) = 4x^2+4x+1-2 = 4x^2+4x-1$.

    Common Mistakes …

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12. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $f$ is a function defined on $(0,1)$ by $f(x) = \min\{x - [x], -x - [x]\}$, then $(fofofof)(x)$ is equal to ______ ( $[.]$ greatest integer function ) (A) $x$ (B) $-x$ (C) $4x$ (D) $2x$

    ›Reveal solutionSolution

    Tracking this piecewise-linear map through four successive applications (using the general floor function as each intermediate value moves out of $(0,1)$) shows it returns exactly to $x$.

    Concept and Intuition

    $f(t) = \min(t-[t],\,-t-[t])$ is naturally defined for all real $t$ via the floor function $[t]$ (greatest integer $\le t$). Even though the problem states $f$ on $(0,1)$, applying $f$ repeatedly sends the argument outside $(0,1)$, so we must evaluate the same formula there using the appropriate integer part. Doing this carefully reveals a period-4 orbit that returns to the starting value.

    Step-by-Step Solution

    1. First application, $x\in(0,1)$: $[x]=0$, so $f(x) = \min(x-0,\,-x-0) = \min(x,-x)$. Since $x>0$, $-x<0<x$, so $f(x) = -x$.
    2. Second application, evaluate $f(-x)$ where $-x\in(-1,0)$: $[-x] = -1$. So $f(-x) = \min\big((-x)-(-1),\ -(-x)-(-1)\big) = \min(1-x,\ x+1)$. Since $x\in(0,1)$, $1-x\in(0,1)$ and $x+1\in(1,2)$, so the minimum is $1-x$. Thus $f(f(x)) = 1-x$.
    3. Third application, evaluate $f(1-x)$ where $1-x\in(0,1)$: $[1-x]=0$. So $f(1-x) = \min(1-x,\ -(1-x)) = \min(1-x,\ x-1)$. Since $x-1<0<1-x$, the minimum is $x-1$. Thus $f(f(f(x))) = x-1$. …

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