Q.Find $gof$ and $fog$, if $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are given by $f(x) = \cos x$ and $g(x) = 3x^2$. Show that $gof \neq fog$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Let $f$ be a function with domain $[0, 7]$ and $g$ be a function defined by $g(x) = |2x + 1|$. Then the domain of $(f \circ g)(x)$ is (A) $[0, 7]$ (B) $[-7, 0]$ (C) $[-4, 3]$ (D) $[-3, 4]$

   ›Reveal solutionSolution

   The domain of the composite function $f \circ g$ consists of all $x$ such that $g(x)$ lies inside the domain of $f$.

   Since $f$ has domain $[0,7]$ and $g(x)=|2x+1|$, we require $0 \le |2x+1| \le 7$.

   Solving gives $x \in [-4, 3]$, so the correct option is (C).

   ---

   Concept and intuition

   When we form $(f \circ g)(x) = f(g(x))$, the “inner” function $g$ must produce outputs that the “outer” function $f$ can accept.

   The domain of $f$ is the set of allowed inputs for $f$. So for $f(g(x))$ to be defined, $g(x)$ must land in that set.

   Here $f$ only accepts numbers from 0 to 7 inclusive. Therefore we need

   $0 \le g(x) \le 7.$

   Since $g(x) = |2x+1|$ is always non‑negative, the lower bound $0$ is automatically satisfied for all $x$ (absolute value is never negative). The real restriction comes from the upper bound:

   $|2x+1| \le 7.$

   ---

   Step‑by‑step solution

   1. Set up the inequality We require

   $|2x+1| \le 7.$

   This is a standard absolute‑value inequality: it means the distance from $2x+1$ to 0 is at most 7.

   2. Rewrite without absolute value For any real number $a$, $|a| \le k$ (with $k \ge 0$) is equivalent to $-k \le a \le k$. Hence

   $-7 \le 2x+1 \le 7.$

   3. Solve the compound inequality Subtract 1 from all three parts:

   $-8 \le 2x \le 6.$

   Then divide by 2 (positive, so inequality signs stay the same):

   $-4 \le x \le 3.$

   4. Check the lower bound …

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2. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.The domain of the real valued function f(x) = $\dfrac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ is (A) $(-3, \infty)$ (B) $(-3,-1) \cup (-1, \infty)$ (C) $(-3,-2) \cup (-2,-1) \cup (-1, \infty)$ (D) $(-3,-2) \cup (-1, \infty)$

   ›Reveal solutionSolution

   Combine the log's domain condition with the square-root-in-denominator condition to get $(-3,-2)\cup(-1,\infty)$.

   Concept and Intuition

   For a quotient with a log in the numerator and a square root in the denominator, every piece imposes its own domain restriction, and the final domain is the intersection of all of them: the log argument must be strictly positive, and the expression under the root must be strictly positive (not just non-negative, since it also sits in the denominator and can't be zero).

   Step-by-Step Solution

   1. Log condition: $x+3>0 \Rightarrow x>-3$.
   2. Denominator condition: $x^2+3x+2>0$. Factor: $(x+1)(x+2)>0$, true when $x<-2$ or $x>-1$.
   3. Intersect $x>-3$ with ($x<-2$ or $x>-1$):
      • $x>-3$ and $x<-2$ gives $(-3,-2)$.
      • $x>-3$ and $x>-1$ gives $(-1,\infty)$. …

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3. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The domain of the real valued function $f(x) = \dfrac{3}{4-x^2} + \log_{10}(x^3 - x)$ is (A) $(1,2) \cup (2,\infty)$ (B) $(-1,0) \cup (1,2)$ (C) $(-1,0) \cup (1,2) \cup (2,\infty)$ (D) $(-\infty,-1) \cup (1,2) \cup (2,\infty)$

   ›Reveal solutionSolution

   The domain needs $x \neq \pm 2$ (denominator non-zero) and $x^3-x>0$ (log argument positive); combining gives $(-1,0)\cup(1,2)\cup(2,\infty)$.

   Concept and Intuition

   For $f(x)=\dfrac{3}{4-x^2}+\log_{10}(x^3-x)$ to be real-valued, every term must individually be defined: the rational term needs a non-zero denominator, and the logarithm needs a strictly positive argument. The domain of $f$ is the intersection of the domains required by each term.

   Step-by-Step Solution

   1. Denominator condition: $4-x^2 \neq 0 \Rightarrow x^2 \neq 4 \Rightarrow x \neq 2$ and $x\neq -2$.
   2. Logarithm condition: need $x^3-x>0$, i.e. $x(x-1)(x+1)>0$. The roots are $x=-1,0,1$, dividing the real line into four intervals; testing a point in each:
      • $x<-1$ (e.g. $x=-2$): $(-2)(-3)(-1) = -6 <0$.
      • $-1<x<0$ (e.g. $x=-0.5$): $(-0.5)(-1.5)(0.5) = 0.375>0$.
      • $0<x<1$ (e.g. $x=0.5$): $(0.5)(-0.5)(1.5)=-0.375<0$.
      • $x>1$ (e.g. $x=2$): $(2)(1)(3)=6>0$. So $x^3-x>0$ exactly on $(-1,0)\cup(1,\infty)$. …

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4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$A \subseteq \mathbb{R}$ and $f: A \to \mathbb{R}$ is a function defined by $f(x) = \dfrac{x-1}{x+1}$. If the domain of the function $f(2x)$ is $B$, then $A \cap B =$ (A) $\mathbb{R} - \{-1, -2\}$ (B) $\mathbb{R} - \left\{-1, -\dfrac{1}{2}\right\}$ (C) $\mathbb{R} - \left\{-1, \dfrac{1}{3}\right\}$ (D) $\mathbb{R} - \{-1, 1\}$

   ›Reveal solutionSolution

   The domain of $f$ excludes $x=-1$; the domain of $f(2x)$ excludes $x=-\frac12$; their intersection excludes both, so $A\cap B = \mathbb{R} - \{-1, -\tfrac12\}$, which is option (B).

   We need to find $A$, the natural domain of $f(x)=\frac{x-1}{x+1}$, then $B$, the natural domain of $f(2x)$, and finally their intersection.

   Why this approach works:

   The domain of a composite function like $f(2x)$ is found by first requiring that the “inside” expression $2x$ belongs to the domain of $f$, and then also requiring that the “inside” expression itself is defined (it always is, since $2x$ is defined for all real $x$). So we simply take the domain of $f$ and replace $x$ by $2x$, solving for the restriction.

   1. Find $A$ — the domain of $f(x)$. $f(x)=\frac{x-1}{x+1}$ is a rational function. The only restriction is that the denominator cannot be zero:

   $x+1 \neq 0 \quad\Rightarrow\quad x \neq -1.$

   Hence

   $A = \mathbb{R} - \{-1\}.$

   2. Find $B$ — the domain of $f(2x)$. The function $f(2x)$ means we substitute $2x$ into $f$:

   $f(2x) = \frac{2x-1}{2x+1}.$

   Again, the denominator cannot be zero:

   $2x+1 \neq 0 \quad\Rightarrow\quad x \neq -\frac12.$

   So

   $B = \mathbb{R} - \left\{-\frac12\right\}.$

   3. Find $A \cap B$. …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The domain of the real valued function $f(x) = \cos^{-1}\left(\dfrac{2-|x|}{4}\right) + [\log(3-x)]^{-1}$ is (A) $(-6,2) \cup (2,3)$ (B) $[-6,2) \cup (2,3)$ (C) $(-\infty,2) \cup (2,3)$ (D) $[-6,2) \cup (2,3]$

   ›Reveal solutionSolution

   The domain is the intersection of the domains of the inverse cosine term and the logarithmic term, excluding points where the log denominator is zero. The final domain is $[-6,2) \cup (2,3)$, which corresponds to option (B).

   We need the set of all real $x$ for which $f(x)$ is defined. The function has two parts:

   • $\cos^{-1}\left(\dfrac{2-|x|}{4}\right)$
   • $[\log(3-x)]^{-1}$, i.e., $\dfrac{1}{\log(3-x)}$

   Each imposes conditions. Let’s find them step by step.

   1. Domain of $\cos^{-1}(u)$ The inverse cosine is defined only when its argument $u$ satisfies $-1 \le u \le 1$. Here $u = \dfrac{2-|x|}{4}$, so we require:

   $-1 \le \frac{2-|x|}{4} \le 1$

   Multiply by 4 (positive, so inequality direction unchanged):

   $-4 \le 2 - |x| \le 4$

   Subtract 2 from all parts:

   $-6 \le -|x| \le 2$

   Multiply by $-1$ (reverses inequalities):

   $6 \ge |x| \ge -2$

   The right inequality $|x| \ge -2$ is always true (absolute value is nonnegative).

   The left inequality $|x| \le 6$ gives:

   $-6 \le x \le 6$

   So the inverse cosine part is defined for $x \in [-6, 6]$.

   2. Domain of $\log(3-x)$ The logarithm (presumably base 10 or natural — domain is the same) requires its argument positive:

   $3 - x > 0 \quad \Rightarrow \quad x < 3$

   So $\log(3-x)$ is defined for $x \in (-\infty, 3)$.

   3. Denominator condition: $[\log(3-x)]^{-1}$ The reciprocal $\dfrac{1}{\log(3-x)}$ is undefined when the denominator is zero:

   $\log(3-x) = 0 \quad \Rightarrow \quad 3 - x = 1 \quad \Rightarrow \quad x = 2$

   So we must exclude $x = 2$ from the domain. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The domain of $f(x) = Cos^{-1}[\log_2(x^2+5x+8)]$ is (A) $[-4,-3]$ (B) $[-3,-2]$ (C) $[-2,-1]$ (D) $[-1,2]$

   ›Reveal solutionSolution

   The domain requires the log expression to lie in $[-1,1]$; one side is automatically satisfied and the other reduces to a simple quadratic inequality, giving $[-3,-2]$.

   Concept and Intuition

   $\cos^{-1}(t)$ is only defined for $t\in[-1,1]$. So finding the domain of $\cos^{-1}[\log_2(\cdots)]$ is really a two-sided inequality problem on the inner expression, which then converts (via the monotonic exponential $2^t$) into a bound on the underlying quadratic.

   Step-by-Step Solution

   1. Require $-1 \le \log_2(x^2+5x+8) \le 1$.
   2. Convert using $2^{(\cdot)}$ (monotonic increasing, preserves inequality direction): $2^{-1} \le x^2+5x+8 \le 2^1$, i.e. $0.5 \le x^2+5x+8 \le 2$.
   3. Lower bound: $x^2+5x+8\ge0.5 \Leftrightarrow x^2+5x+7.5\ge0$. Discriminant $=25-30=-5<0$ and leading coefficient positive, so this quadratic is always positive — the lower bound holds for all real $x$, imposing no restriction.
   4. Upper bound: $x^2+5x+8\le2 \Leftrightarrow x^2+5x+6\le0 \Leftrightarrow (x+2)(x+3)\le0$. …

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7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The domain of the real valued function $f(x) = \dfrac{1}{\sqrt{\log_{0.5}(2x-3)}} + \sqrt{4 - 9x^2}$ is (A) $\left[\dfrac{2}{3}, \dfrac{3}{2}\right)$ (B) Null Set (C) $\left[\dfrac{2}{3}, 2\right)$ (D) $\left[-\dfrac{2}{3}, \dfrac{2}{3}\right]$

   ›Reveal solutionSolution

   The two terms of $f(x)$ impose disjoint requirements on $x$ ($\frac32<x<2$ vs $-\frac23\le x\le\frac23$), so no $x$ satisfies both — the domain is empty, option (B).

   Concept and Intuition

   For a sum of two functions to be defined, both pieces must individually be defined at $x$ — the domain of the sum is the intersection of the individual domains. Here:

   • $\dfrac{1}{\sqrt{\log_{0.5}(2x-3)}}$ requires the expression under the square root, $\log_{0.5}(2x-3)$, to be strictly positive (it can't be zero, since it's in a denominator, and can't be negative, since it's under a real square root).
   • $\sqrt{4-9x^2}$ requires $4-9x^2 \ge 0$.

   For a log with base $b<1$ (here $b=0.5$), $\log_b(t)$ is a decreasing function of $t$, and $\log_b(1)=0$. So $\log_b(t)>0 \iff t<1$ (combined with the log's own domain requirement $t>0$), giving $0<t<1$.

   Step-by-Step Solution

   1. Log argument domain + positivity: need $\log_{0.5}(2x-3) > 0$. Since base $<1$: this holds iff $0 < 2x-3 < 1$. $0 < 2x-3 \Rightarrow x > \dfrac32$; and $2x-3<1 \Rightarrow x<2$. So first term needs $x \in \left(\dfrac32, 2\right)$.
   2. Square-root domain: need $4-9x^2\ge0 \Rightarrow 9x^2\le4 \Rightarrow x^2\le\dfrac49 \Rightarrow x\in\left[-\dfrac23,\dfrac23\right]$. …

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8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.For $x \in \mathbb{R}$ if $f(x)=\sqrt{\log_{10}\left(\dfrac{3-x}{x}\right)}$, then the domain of f is (A) $\left[0,\dfrac{3}{2}\right]$ (B) $\left(0,\dfrac{3}{2}\right]$ (C) $[0,1]$ (D) $(0,1]$

   ›Reveal solutionSolution

   Two conditions — the log's argument must be positive, and (for the outer square root) at least 1 — combine to give the domain $(0,3/2]$.

   Concept and Intuition

   For $f(x)=\sqrt{\log_{10}(g(x))}$ to be real we need (i) $g(x)>0$ so the log is defined, and (ii) $\log_{10}(g(x))\ge0$, i.e. $g(x)\ge1$, so the square root has a non-negative argument.

   Step-by-Step Solution

   1. Let $g(x)=\dfrac{3-x}{x}$.
   2. Condition (i) $g(x)>0$: numerator and denominator must share sign. For $x>0$: need $3-x>0\Rightarrow x<3$, giving $0<x<3$. For $x<0$: need $3-x<0\Rightarrow x>3$, impossible. So condition (i) gives $x\in(0,3)$. …

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9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The domain of the real valued function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$, is (A) $[-15,-12] \cup [12,15]$ (B) $(-\infty,-12] \cup [12,\infty)$ (C) $[-15,15]$ (D) $[-12,12]$

   ›Reveal solutionSolution

   The domain requires the inner square root to exist AND the outer square root's argument to be non-negative. Answer: $[-15,-12]\cup[12,15]$.

   Concept and Intuition

   For $f(x)=\sqrt{9-\sqrt{x^2-144}}$ to be real we need two nested conditions: the inner radical $\sqrt{x^2-144}$ must be defined, and once it is, the quantity $9-\sqrt{x^2-144}$ must be $\ge 0$ for the outer radical to be defined.

   Step-by-Step Solution

   1. Inner radical needs $x^2-144\ge 0 \Rightarrow |x|\ge 12$.
   2. Outer radical needs $9-\sqrt{x^2-144}\ge 0 \Rightarrow \sqrt{x^2-144}\le 9 \Rightarrow x^2-144\le 81 \Rightarrow x^2\le 225 \Rightarrow |x|\le 15$. …

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10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.If $\int\dfrac{\sin x\cos x}{\sqrt{\cos^4x-\sin^4x}}dx=-\dfrac{f(x)}{2}+c$, then domain of $f(x)$ is (A) $[2n\pi,(2n+1)\pi],\ n=0,1,2...$ (B) $\left[(4n-1)\dfrac{\pi}{2},(4n+1)\dfrac{\pi}{2}\right],\ n=0,1,2,...$ (C) $\left[(4n-1)\dfrac{\pi}{4},(4n+1)\dfrac{\pi}{4}\right],\ n=0,1,2,...$ (D) $\left[2n\dfrac{\pi}{4},(2n+1)\dfrac{\pi}{4}\right],\ n=0,1,2,...$

    ›Reveal solutionSolution

    Rewriting $\cos^4x-\sin^4x$ as $\cos2x$ turns the integral into a simple $u=\cos2x$ substitution; the resulting $f(x)=\sqrt{\cos2x}$ is only real where $\cos2x\ge0$.

    Concept and Intuition

    The expression under the root, $\cos^4x-\sin^4x$, is a difference of squares: $(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)$. Since $\cos^2x+\sin^2x=1$, this is simply $\cos2x$. Likewise $\sin x\cos x=\tfrac12\sin2x$ is the double-angle identity. Recognizing both identities turns a scary-looking radical integral into a one-step substitution problem — always hunt for double-angle simplification when you see $\sin x\cos x$ and even powers of sine/cosine together.

    Step-by-Step Solution

    1. Simplify: $\cos^4x-\sin^4x=\cos2x$, and $\sin x\cos x=\tfrac12\sin2x$.
    2. The integral becomes $\displaystyle\int\frac{\tfrac12\sin2x}{\sqrt{\cos2x}}\,dx=\frac12\int\frac{\sin2x}{\sqrt{\cos2x}}\,dx$.
    3. Let $u=\cos2x$, so $du=-2\sin2x\,dx\Rightarrow \sin2x\,dx=-\tfrac12 du$.
    4. The integral becomes $\dfrac12\int\dfrac{-\tfrac12\,du}{\sqrt u}=-\dfrac14\int u^{-1/2}du=-\dfrac14\cdot2u^{1/2}=-\dfrac12\sqrt{\cos2x}+c$.
    5. Matching this to the given form $-\dfrac{f(x)}{2}+c$ gives $f(x)=\sqrt{\cos2x}$. …

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11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.The domain of the real valued function $f(x) = \log_2 \log_3 \log_5 (x^2 - 5x + 11)$ is (A) $(2, \infty)$ (B) $(-\infty, 3)$ (C) $(2, 3)$ (D) $(-\infty, 2) \cup (3, \infty)$

    ›Reveal solutionSolution

    A triple-nested logarithm requires working from the innermost expression outward, each time ensuring the argument of the next-outer log is strictly positive; here this reduces to $x^2-5x+11>5$, giving domain $(-\infty,2)\cup(3,\infty)$.

    Concept and Intuition

    For $\log_b(u)$ to be a real number we need $u>0$ (and $b>0, b\neq1$, which is already satisfied for bases 2, 3, 5). When logs are nested, $\log_2\big(\log_3(\log_5(g))\big)$, we must ensure positivity at every level, working from the inside out:

    • $\log_5(g)$ needs $g>0$.
    • $\log_3(\log_5(g))$ needs $\log_5(g) > 0$, i.e. $g > 5^0 = 1$.
    • $\log_2(\log_3(\log_5(g)))$ needs $\log_3(\log_5(g)) > 0$, i.e. $\log_5(g) > 3^0 = 1$, i.e. $g > 5^1 = 5$.

    So the binding constraint is always the outermost one once you propagate it inward — here it collapses to a single simple quadratic inequality.

    Step-by-Step Solution

    1. Let $g(x) = x^2-5x+11$. Its discriminant is $(-5)^2-4(1)(11)=25-44=-19<0$, and the leading coefficient is positive, so $g(x)>0$ for all real $x$ — this level imposes no restriction.
    2. For $\log_3(\log_5(g))$ to be defined we need $\log_5(g)>0 \Leftrightarrow g>1$.
    3. For $\log_2(\log_3(\log_5(g)))$ to be defined we need $\log_3(\log_5(g))>0 \Leftrightarrow \log_5(g) > 1 \Leftrightarrow g > 5$. …

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12. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The domain of the function $f(x) = \sqrt{\log_e\left(\frac{1}{x^2-4x+4}\right)} + \operatorname{Sin}^{-1}(x^2-2)$ is (A) $[1,3]$ (B) $[1,3)$ (C) $[1,\sqrt{3}]$ (D) $[1,\sqrt{3})$

    ›Reveal solutionSolution

    The domain is the intersection of the constraints from the log-square-root term (giving $x\in[1,3]$) and the arcsine term (giving $x\in[-\sqrt3,-1]\cup[1,\sqrt3]$), which works out to the closed interval $[1,\sqrt3]$.

    Concept and Intuition

    Whenever a function is a sum of two pieces, its domain is the intersection of each piece's individual domain — both must be simultaneously defined. Here one piece needs a square root's argument to be non-negative (which itself requires a logarithm to be non-negative, i.e. its argument $\geq 1$), and the other piece needs an arcsine's argument to lie in $[-1,1]$.

    Step-by-Step Solution

    1. First term: $\sqrt{\log_e\left(\dfrac{1}{x^2-4x+4}\right)}$. Note $x^2-4x+4=(x-2)^2 \geq 0$, equal to $0$ only at $x=2$ (so $x\neq 2$ is required just for the fraction to exist). For the square root to be real, we need $\log_e\!\left(\dfrac{1}{(x-2)^2}\right) \geq 0$, i.e. $\dfrac{1}{(x-2)^2} \geq 1$ (since $\log_e u \geq 0 \Leftrightarrow u\geq 1$ for $u>0$). This gives $(x-2)^2 \leq 1$, i.e. $-1\leq x-2\leq 1$, i.e. $x\in[1,3]$ (and this automatically excludes the problem point $x=2$ from being undefined, since $(x-2)^2\le 1$ still permits $x=2$ where the fraction blows up — but that single point is excluded separately since the fraction itself is undefined there; it doesn't affect the final interval once intersected with the next constraint since $2 \notin [1,\sqrt3]$ anyway).
    2. Second term: $Sin^{-1}(x^2-2)$ requires $-1 \leq x^2-2 \leq 1$, i.e. $1\leq x^2 \leq 3$. Since $x^2\geq 1 \Leftrightarrow x\leq -1$ or $x\geq 1$, and $x^2\leq 3 \Leftrightarrow -\sqrt3\leq x\leq \sqrt3$, combining gives $x\in[-\sqrt3,-1]\cup[1,\sqrt3]$. …

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