Q.Differentiate w.r.t. $x$: $\cos^{-1}\left(\dfrac{\sin x + \cos x}{\sqrt{2}}\right),\ -\dfrac{\pi}{4} < x < \dfrac{\pi}{4}$.

Showing the 12 most recent of 17 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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. $3<x\le4$. …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Domain of the real valued function $f(x)=\log(x^2-1)+x\,\mathrm{Coth}^{-1}x$ is (A) $\mathbb{R}$ (B) $(-1,1)$ (C) $\mathbb{R}-[-1,1]$ (D) $\mathbb{R}-[0,1]$

   ›Reveal solutionSolution

   Both pieces of $f$ demand $|x|>1$, so the domain is all reals except the closed interval $[-1,1]$.

   Concept and Intuition

   A sum/product function's domain is the intersection of the domains of its pieces. Here we have a logarithm term and an inverse hyperbolic cotangent term multiplying $x$; both must individually be defined.

   Step-by-Step Solution

   1. $\log(x^2-1)$ requires its argument to be strictly positive: $x^2-1>0 \Rightarrow x^2>1 \Rightarrow x>1 \text{ or } x<-1$.
   2. $\mathrm{Coth}^{-1}(x)$, the inverse of $\coth$, has range restricted since $\coth(y)$ for real $y\ne 0$ takes all values with $|\coth(y)|>1$. So $\mathrm{Coth}^{-1}(x)$ is only real-valued for $|x|>1$, i.e. $x>1$ or $x<-1$.
   3. Multiplying by $x$ (which is defined for all real $x$) doesn't add any further restriction.
   4. Intersecting both conditions: $x>1$ or $x<-1$ in both cases — the domains coincide exactly. …

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9. 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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10. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.The domain of the function $y=f(x)$, where $x$ and $y$ are related by $2^x+2^y=2$ is (A) $(-\infty,\infty)$ (B) $(-\infty,1)$ (C) $(0,\infty)$ (D) $(1,\infty)$

    ›Reveal solutionSolution

    Since $2^y$ must be strictly positive, $2-2^x>0$ forces $x<1$, giving domain $(-\infty,1)$.

    Concept and Intuition

    An exponential $2^y$ is always positive for real $y$, and conversely, for any positive real number there is a real $y$ with $2^y$ equal to it. So the equation $2^y=2-2^x$ has a real solution $y$ exactly when the right side is positive.

    Step-by-Step Solution

    1. Rearranged: $2^y = 2 - 2^x$.
    2. For $y \in \mathbb{R}$ to exist, need $2-2^x>0$. …

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11. 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).

    ---

    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.

    ---

    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: $x \neq 2$ …

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12. 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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