Q.Let $f: \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}$ and $g: \{3, 4, 5, 9\} \to \{7, 11, 15\}$ be functions defined as $f(2) = 3$, $f(3) = 4$, $f(4) = f(5) = 5$ and $g(3) = g(4) = 7$ and $g(5) = g(9) = 11$. Find $gof$.

Showing the 12 most recent of 17 on this concept.

1. 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 $\log_3(\log_5(g))>0 \Leftrightarrow \log_5(g) > 1 \Leftrightarrow g > 5$.
   4. This last condition ($g>5$) is stronger than $g>1$, so it's the actual binding domain condition: $x^2-5x+11>5$.
   5. Simplify: $x^2-5x+6>0 \Rightarrow (x-2)(x-3)>0$.
   6. This product is positive when both factors are positive ($x>3$) or both negative ($x<2$): domain $= (-\infty,2)\cup(3,\infty)$.

   Common Mistakes

   • Stopping at $g>0$ or $g>1$ instead of propagating all the way through to the outermost log's requirement ($g>5$).
   • Forgetting to check that the innermost quadratic is always positive (a quick discriminant check saves needing to intersect with an extra condition).

   ✓Final answer

   The correct option is (D) — $(-\infty, 2) \cup (3, \infty)$.

   ANSWER: D

2. 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

      Because $g(x) = |2x+1|$ is always $\ge 0$, the condition $g(x) \ge 0$ is automatically true for every real $x$. So no extra restriction comes from the lower end of $f$’s domain.

   5. Conclusion

      The set of $x$ for which $g(x)$ lies in $[0,7]$ is exactly the interval $[-4, 3]$.

      Therefore the domain of $(f \circ g)(x)$ is $[-4, 3]$.

   ---

   Watch out

   A common mistake is to think the domain of $f \circ g$ is the same as the domain of $f$ or the domain of $g$.

   It is neither: you must force the output of $g$ to fit inside the input set of $f$.

   Tip

   Since absolute value is always non‑negative, the lower bound $0 \le g(x)$ never eliminates any $x$.

   The only work is solving $|2x+1| \le 7$, which is a quick linear inequality.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. 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 $9-\sqrt{x^2-144}\ge 0 \Rightarrow \sqrt{x^2-144}\le 9 \Rightarrow x^2-144\le 81 \Rightarrow x^2\le 225 \Rightarrow |x|\le 15$.
   3. Combining: $12\le |x|\le 15$, i.e. $x\in[-15,-12]\cup[12,15]$.

   Common Mistakes

   • Forgetting the inner domain restriction ($|x|\ge 12$) and only imposing the outer one.
   • Squaring the inequality $\sqrt{x^2-144}\le 9$ incorrectly (sign errors).

   ✓Final answer

   The correct option is (A) — $[-15,-12]\cup[12,15]$.

   ANSWER: A

4. 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 $4-9x^2\ge0 \Rightarrow 9x^2\le4 \Rightarrow x^2\le\dfrac49 \Rightarrow x\in\left[-\dfrac23,\dfrac23\right]$.
   3. Intersect the two conditions: $\left(\dfrac32,2\right) \cap \left[-\dfrac23,\dfrac23\right]$. Since $\dfrac32 = 1.5$ and $\dfrac23\approx0.667$, the first interval starts well above where the second interval ends — there is no overlap.
   4. Hence the overall domain is $\varnothing$ — the Null Set.

   Common Mistakes

   • Allowing $\log_{0.5}(2x-3)\ge 0$ (including zero) — but zero would make the whole first term undefined (division by $\sqrt0=0$), so the inequality must be strict.
   • Forgetting that base $<1$ flips the usual direction of the log inequality (many students reflexively write $t>1$ as for base $>1$).

   ✓Final answer

   The correct option is (B) — Null Set.

   ANSWER: B

5. 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 $A \cap B$. $A$ excludes $-1$; $B$ excludes $-\frac12$. Their intersection excludes both numbers:

   $A \cap B = \mathbb{R} - \{-1, -\tfrac12\}.$

   Watch out

   A common mistake is to think $f(2x)$ also excludes $x=-1$ because $f(x)$ does. But the restriction shifts: $f(2x)$ is undefined when $2x=-1$, i.e. $x=-\frac12$, not when $x=-1$.

   Tip

   Always replace the variable inside the restriction: if $f$ is undefined at $x=a$, then $f(g(x))$ is undefined when $g(x)=a$. Here $g(x)=2x$, so $2x=-1$ gives $x=-\frac12$.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

6. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.The domain of the real valued function $f(x) = \dfrac{\sqrt{\log_{0.5}(x-3)}}{\sqrt{x-1}}$ is (A) $(3,4]$ (B) $[4,\infty)$ (C) $(1,\infty)$ (D) $(1,3)$

   ›Reveal solutionSolution

   Combining the log's domain, the non-negativity needed under the numerator's square root (using base $<1$ log behaviour), and the denominator's need to be strictly positive gives the domain $(3,4]$.

   Concept and Intuition

   For $f(x)=\dfrac{\sqrt{\log_{0.5}(x-3)}}{\sqrt{x-1}}$ to be real and defined: (i) the argument of the log must be positive: $x-3>0$; (ii) since it sits under a square root, $\log_{0.5}(x-3)\ge0$; (iii) the denominator's radicand must be strictly positive (can't divide by zero): $x-1>0$. Because the log base $0.5$ is less than 1, $\log_{0.5}$ is a decreasing function, so $\log_{0.5}(t)\ge 0$ happens for $0<t\le1$ (opposite of the base $>1$ case).

   Step-by-Step Solution

   1. Log domain: $x-3>0 \Rightarrow x>3$.
   2. Non-negativity under the square root: $\log_{0.5}(x-3)\ge0$. Since base $0.5<1$: this holds iff $0<x-3\le1$, i.e. $3<x\le4$.
   3. Denominator: $x-1>0 \Rightarrow x>1$ (automatically satisfied since $x>3$).
   4. Intersect all conditions: $3<x\le4$, i.e. $(3,4]$.

   Common Mistakes

   • Using the base-$>1$ rule ($\log_a(t)\ge0 \iff t\ge1$) for this base-$0.5$ log, which would wrongly give $x\ge4$ and lose the upper bound.
   • Forgetting the denominator's strict positivity, or forgetting to intersect with the log's positivity requirement.

   ✓Final answer

   The correct option is (A) — $(3,4]$.

   ANSWER: A

7. 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 $(-1,0)\cup(1,\infty)$.
   3. Now impose $x\neq2$ (from step 1) on $(-1,0)\cup(1,\infty)$: the point $x=2$ lies inside $(1,\infty)$, so it must be excised, splitting that interval into $(1,2)\cup(2,\infty)$. ($x=-2$ was already outside the log-domain so it removes nothing further.)
   4. Final domain: $(-1,0)\cup(1,2)\cup(2,\infty)$.

   Common Mistakes

   • Forgetting to also exclude $x=2$ from the log-derived interval $(1,\infty)$ — many stop after finding the cubic's sign pattern and skip re-checking the denominator restriction against it.
   • Mis-sequencing the cubic's sign pattern (it isn't simply "positive for $x>0$"); a careful test-point check on all four sub-intervals is needed since the sign alternates.

   ✓Final answer

   The correct option is (C) — $(-1,0) \cup (1,2) \cup (2,\infty)$.

   ANSWER: C

8. 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: $x^2+5x+8\le2 \Leftrightarrow x^2+5x+6\le0 \Leftrightarrow (x+2)(x+3)\le0$.
   5. This holds when $x$ is between the roots $-3$ and $-2$: $x\in[-3,-2]$.
   6. So the full domain (intersection of both conditions) is $[-3,-2]$.

   Common Mistakes

   • Forgetting to check the lower bound at all (it happens to always hold here, but that must be verified, not assumed).
   • Sign error factoring $x^2+5x+6=(x+2)(x+3)$, or misreading which interval satisfies $\le 0$ between the roots vs. outside them.

   ✓Final answer

   The correct option is (B) — $[-3,-2]$.

   ANSWER: B

9. 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 $(-1,\infty)$.
   4. Union: $(-3,-2)\cup(-1,\infty)$.

   Common Mistakes

   • Allowing $x^2+3x+2\ge 0$ (including zero), which would wrongly include $x=-1,-2$ where the denominator vanishes.
   • Forgetting to intersect with the log's domain, keeping the interval $(-\infty,-2)\cup(-1,\infty)$ instead of trimming to $x>-3$.

   ✓Final answer

   The correct option is (D) — $(-3,-2)\cup(-1,\infty)$.

   ANSWER: D

10. 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.

    4. Intersection of all conditions

       • From step 1: $x \in [-6, 6]$
       • From step 2: $x \in (-\infty, 3)$
       • From step 3: $x \neq 2$

       Intersection: $[-6, 6] \cap (-\infty, 3) = [-6, 3)$.

       Then remove $x = 2$: we get $[-6, 2) \cup (2, 3)$.

    Watch out

    A common mistake is forgetting to exclude $x=2$ where $\log(3-x)=0$, or incorrectly including $x=3$ (where $\log(0)$ is undefined). Also note that $x=2$ is already inside $[-6,3)$, so we must split the interval.

    Tip

    The inverse cosine part gave $|x| \le 6$, which is symmetric. The log part cuts off everything $\ge 3$, so the final domain is just the left part of $[-6,6]$ up to (but not including) 3, with a hole at 2.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. 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 $x\in(0,3)$.
    3. Condition (ii) $g(x)\ge1$: $\dfrac{3-x}{x}-1\ge0\Rightarrow\dfrac{3-2x}{x}\ge0$. Since we already have $x>0$ from step 2, this reduces to $3-2x\ge0\Rightarrow x\le\dfrac32$.
    4. Combining: $0<x\le\dfrac32$, i.e. $\left(0,\dfrac32\right]$.

    Common Mistakes

    • Forgetting condition (ii) entirely and stopping at just "log defined" (which would wrongly give $(0,3)$).
    • Sign errors when combining the two fractions in $(3-x)/x - 1$.

    ✓Final answer

    The correct option is (B) — $\left(0,\dfrac{3}{2}\right]$.

    ANSWER: B

12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.The domain of the real valued function $f(x) = \log_{x-1}(3x+1)$ is (A) $(1, \infty)$ (B) $\mathbb{R}$ (C) $(1,2) \cup (2, \infty)$ (D) $\mathbb{R} - \{2\}$

    ›Reveal solutionSolution

    The domain of a logarithmic function requires the base to be positive and not equal to 1, and the argument to be positive. For $f(x)=\log_{x-1}(3x+1)$, this gives $x>1$, $x\neq2$, and $x>-\frac13$, so the domain is $(1,2)\cup(2,\infty)$, which is option (C).

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    Concept and Intuition

    When we see a logarithm like $\log_{\text{base}}(\text{argument})$, we must remember two fundamental restrictions:

    1. The base must be positive and not equal to 1.

       Why? A logarithm is the inverse of an exponential function. If the base were 0 or negative, the exponential would behave badly (e.g., not be defined for all real exponents). If the base were 1, the exponential is constant, so its inverse wouldn't be a function.

    2. The argument (the number inside the log) must be positive.

       Why? Because you can only take the log of a positive number in real-valued functions — logs of zero or negative numbers are not real.

    So for $f(x) = \log_{x-1}(3x+1)$, we need to enforce both conditions simultaneously.

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    Step-by-step reasoning

    1. Base condition: $x-1 > 0$ The base of the logarithm is $x-1$. It must be positive:

    $x-1 > 0 \quad \Rightarrow \quad x > 1.$

    2. Base cannot be 1: $x-1 \neq 1$ The base also cannot equal 1, because $\log_1(\cdot)$ is undefined.

    $x-1 \neq 1 \quad \Rightarrow \quad x \neq 2.$

    3. Argument condition: $3x+1 > 0$ The expression inside the log must be positive:

    $3x+1 > 0 \quad \Rightarrow \quad x > -\frac13.$

    4. Combine all conditions

       • From step 1: $x > 1$
       • From step 2: $x \neq 2$
       • From step 3: $x > -\frac13$

       The condition $x > -\frac13$ is automatically satisfied if $x > 1$, so it doesn't further restrict the domain.

       Thus the domain is all real numbers greater than 1, except 2:

    $(1, \infty) \setminus \{2\} = (1,2) \cup (2,\infty).$

    Watch out

    A common mistake is to forget that the base cannot be 1. Many students only check $x-1>0$ and $3x+1>0$, getting $(1,\infty)$, which is option (A). But $x=2$ makes the base 1, which is invalid — so (A) is a trap.

    Tip

    Always write the base condition as two separate inequalities: $>0$ and $\neq 1$. This makes it harder to miss the exclusion.

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    ✓Final answer

    The correct option is (C).

    ANSWER: C