Q.Which of the following is the principal value branch of $\csc^{-1}x$?
(A) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(B) $[0,\pi]-\left\{\frac{\pi}{2}\right\}$
(C) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(D) $\left[\frac{-\pi}{2},\frac{\pi}{2}\right]-\{0\}$

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The domain of the real valued function $f(x) = \sqrt{\cos(\sin x)} + \cos^{-1} \left( \frac{1 + x^2}{2x} \right)$ is (A) $(-1, 1)$ (B) $[-1, 1]$ (C) $\mathbb{R} - (-1, 1)$ (D) $\{-1, 1\}$

   ›Reveal solutionSolution

   The domain is the set of real numbers where both terms are defined: the square root requires $\cos(\sin x) \ge 0$ (true for all real $x$), and the inverse cosine requires $\left|\frac{1+x^2}{2x}\right| \le 1$ and $x \neq 0$, which forces $x = \pm 1$. So the domain is $\{-1, 1\}$, option (D).

   ---

   The key idea: a function defined by a sum is only valid where every piece is defined. Here we have a square root and an inverse cosine. The square root’s domain is where its inside is non‑negative; the inverse cosine’s domain is where its argument lies in $[-1, 1]$. We find the intersection of these conditions.

   Why this approach works:

   Instead of guessing, we treat each term separately. The square root term looks scary but simplifies because $\sin x$ is bounded and cosine is always non‑negative on the range of $\sin x$. The inverse cosine term is the restrictive one — it forces a very narrow set of $x$ values.

   ---

   1. First term: $\sqrt{\cos(\sin x)}$

      For the square root to be real, we need $\cos(\sin x) \ge 0$.

      Since $\sin x \in [-1, 1]$, we check: for any $y \in [-1, 1]$, is $\cos y \ge 0$?

      The cosine function is non‑negative on $[-\pi/2, \pi/2]$, and $[-1, 1]$ is a subset of that interval (because $1 < \pi/2 \approx 1.57$).

      Therefore $\cos(\sin x) > 0$ for all real $x$.

      So the first term imposes no restriction — its domain is $\mathbb{R}$.

   2. Second term: $\cos^{-1}\left( \frac{1 + x^2}{2x} \right)$

      The inverse cosine function $\cos^{-1}(t)$ is defined only for $t \in [-1, 1]$.

      So we require

   $-1 \le \frac{1 + x^2}{2x} \le 1.$

   Also, the denominator $2x$ cannot be zero, so $x \neq 0$.

   3. Solve the inequality Consider the right inequality first:

   $\frac{1 + x^2}{2x} \le 1.$

   Bring to one side:

   $\frac{1 + x^2}{2x} - 1 \le 0 \quad\Rightarrow\quad \frac{1 + x^2 - 2x}{2x} \le 0 \quad\Rightarrow\quad \frac{(x-1)^2}{2x} \le 0.$

   The numerator $(x-1)^2$ is always $\ge 0$ and equals $0$ only at $x=1$.

   So the fraction is $\le 0$ only when the denominator is negative (since numerator is non‑negative).

   That means $2x < 0$, i.e. $x < 0$, or numerator = 0 (i.e. $x=1$).

   So from this inequality: $x < 0$ or $x = 1$.

   Now the left inequality:

   $-1 \le \frac{1 + x^2}{2x} \quad\Rightarrow\quad \frac{1 + x^2}{2x} + 1 \ge 0 \quad\Rightarrow\quad \frac{1 + x^2 + 2x}{2x} \ge 0 \quad\Rightarrow\quad \frac{(x+1)^2}{2x} \ge 0.$

   The numerator $(x+1)^2 \ge 0$, zero only at $x=-1$. …

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2. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $f(x) = \tan\left(\dfrac{\pi}{\sqrt{x+1}+4}\right)$ is a real valued function then the range of $f$ is (A) $[-1,1]$ (B) $(0,1]$ (C) $[-1,\infty)$ (D) $\mathbb{R}$

   ›Reveal solutionSolution

   The function is a composition of a decreasing inner square-root expression and the tangent function over a restricted domain; the range is $(0,1]$, so option (B) is correct.

   We need the range of $f(x) = \tan\left(\dfrac{\pi}{\sqrt{x+1}+4}\right)$ for real $x$ where the function is defined.

   The key is to find what values the argument of $\tan$ can take, then see what outputs $\tan$ gives on that interval.

   ---

   Concept & Intuition

   The tangent function is periodic and unbounded, but here its argument is squeezed into a small interval because the denominator $\sqrt{x+1}+4$ is always at least $4$. That means the fraction $\frac{\pi}{\text{big number}}$ stays small — specifically between $0$ and $\frac{\pi}{4}$. On $(0, \frac{\pi}{4})$, $\tan$ is positive, increasing, and bounded between $0$ and $1$. So the range is $(0,1]$.

   ---

   Step-by-step reasoning

   1. Domain of $f$

      We need $\sqrt{x+1}$ defined, so $x+1 \ge 0 \Rightarrow x \ge -1$.

      Also, the denominator $\sqrt{x+1}+4$ is never zero (it’s always $\ge 4$), so no further restriction.

      Domain: $[-1, \infty)$.

   2. Behavior of the inner expression

      Let $t = \sqrt{x+1}$. For $x \ge -1$, $t \ge 0$.

      Then the argument of $\tan$ is

   $\theta = \frac{\pi}{t+4}.$

   As $x$ increases from $-1$ to $\infty$, $t$ increases from $0$ to $\infty$.

   3. Range of $\theta$

      • When $x = -1$: $t = 0$, so $\theta = \frac{\pi}{4}$.
      • As $x \to \infty$: $t \to \infty$, so $\theta \to 0^+$. Since $t+4$ increases, $\theta$ decreases continuously. Hence $\theta \in (0, \frac{\pi}{4}]$.

   4. Apply $\tan$

      On the interval $(0, \frac{\pi}{4}]$, $\tan$ is:

      • Continuous and strictly increasing.
      • $\tan(0^+) = 0^+$ (approaches 0 from above). …

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3. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The domain of the function $f(x) = \dfrac{1}{\sqrt{|x| - x}}$ is (A) $\mathbb{R}$ (B) $(-\infty, 0)$ (C) $(0, \infty)$ (D) $(-\infty, 1)$

   ›Reveal solutionSolution

   The function $f(x) = \frac{1}{\sqrt{|x| - x}}$ is defined only when the denominator is positive and real, which happens when $|x| - x > 0$. This inequality holds only for $x < 0$, so the domain is $(-\infty, 0)$.

   The key here is understanding what the square root in the denominator demands. A square root is defined only when its argument is non-negative, and since it's in the denominator, it must be strictly positive — zero would make the function undefined. So we need $|x| - x > 0$.

   Now, the absolute value function $|x|$ behaves differently depending on whether $x$ is negative or non-negative. That's the natural place to split the analysis.

   1. Case 1: $x \ge 0$

      For $x \ge 0$, we have $|x| = x$. Then $|x| - x = x - x = 0$.

      The denominator becomes $\sqrt{0} = 0$, which is not allowed. So no $x \ge 0$ works.

   2. Case 2: $x < 0$

      For $x < 0$, we have $|x| = -x$. Then $|x| - x = (-x) - x = -2x$.

      Since $x < 0$, $-2x$ is positive. So $|x| - x > 0$ holds for every negative $x$.

      The square root is defined and the denominator is non-zero.

   Thus the function is defined for all $x < 0$, and for no $x \ge 0$. …

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4. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.The set of all real values of $x$ for which the expansion of $\left(125x^2 - \dfrac{27}{x}\right)^{-\dfrac{2}{3}}$ is valid, is (A) $\left(-\dfrac{3}{5}, \dfrac{3}{5}\right)$ (B) $\left(-\infty, -\dfrac{3}{5}\right) \cup \left(\dfrac{3}{5}, \infty\right)$ (C) $\left(-\dfrac{5}{3}, \dfrac{5}{3}\right)$ (D) $\left(-\infty, -\dfrac{1}{3}\right) \cup \left(\dfrac{1}{3}, \infty\right)$

   ›Reveal solutionSolution

   The binomial expansion of $(a+b)^n$ is valid only when $|a| > |b|$ for negative fractional exponents. Here we require $\left|125x^2\right| > \left|-\frac{27}{x}\right|$, which simplifies to $|x| > \frac{3}{5}$, so the valid set is $(-\infty, -3/5) \cup (3/5, \infty)$.

   Concept & Intuition

   The expression $\left(125x^2 - \frac{27}{x}\right)^{-2/3}$ is a binomial with a negative fractional exponent. For such an expansion to converge (be valid as an infinite series), the term with the larger absolute value must be written first, and the ratio of the smaller term to the larger term must have magnitude less than 1. In other words, if we write $(A + B)^n$ with $n$ not a nonnegative integer, the expansion is valid only when $|A| > |B|$. Here we choose $A = 125x^2$ and $B = -\frac{27}{x}$ (or vice versa), and enforce $|A| > |B|$.

   Step-by-step reasoning

   1. Identify the form

      We have $(125x^2 - 27/x)^{-2/3}$. Let $A = 125x^2$ and $B = -27/x$. Then the expression is $(A + B)^{-2/3}$.

   2. Condition for binomial expansion with negative fractional exponent

      For $(A + B)^n$ with $n$ not a nonnegative integer, the binomial series $\sum_{k=0}^\infty \binom{n}{k} A^{n-k} B^k$ converges absolutely if $|B/A| < 1$ (i.e., $|A| > |B|$). This is because we factor out $A^n$ and expand $(1 + B/A)^n$, which converges when $|B/A| < 1$.

   3. Set up the inequality

      We require $|125x^2| > \left| -\frac{27}{x} \right|$.

      Since $125 > 0$ and $27 > 0$, this becomes $125|x|^2 > \frac{27}{|x|}$.

   4. Solve the inequality

      Multiply both sides by $|x|$ (positive for $x \neq 0$; note $x=0$ is not allowed anyway because of the $1/x$ term):

      $125|x|^3 > 27 \quad \Rightarrow \quad |x|^3 > \frac{27}{125} \quad \Rightarrow \quad |x| > \sqrt[3]{\frac{27}{125}} = \frac{3}{5}.$ …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The amplitude of the complex number $\dfrac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(-1+i)(-1-i)}$ is (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $-\dfrac{5\pi}{12}$ (D) $-\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The amplitude (argument) is $-\dfrac{\pi}{6}$, so the answer is (D).

   Write each factor in polar form and combine arguments (add for products, subtract for the quotient).

   Numerator

   • $\sqrt{3}+i$: modulus $2$, argument $\dfrac{\pi}{6}$.
   • $1-\sqrt{3}\,i$: modulus $2$, argument $-\dfrac{\pi}{3}$.

   Numerator argument $=\dfrac{\pi}{6}-\dfrac{\pi}{3}=-\dfrac{\pi}{6}$.

   Denominator …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The amplitude of the complex number $\dfrac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(-1+i)(-1-i)}$ is (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $-\dfrac{5\pi}{12}$ (D) $-\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The expression reduces to $\sqrt{3}-i$, whose amplitude is $-\dfrac{\pi}{6}$. Option (D).

   Solution

   Simplify numerator and denominator directly.

   Numerator:

   $(\sqrt{3}+i)(1-\sqrt{3}\,i)=\sqrt{3}-3i+i-\sqrt{3}\,i^{2}=\sqrt{3}-2i+\sqrt{3}=2\sqrt{3}-2i.$

   Denominator: the factors are conjugates, so

   $(-1+i)(-1-i)=(-1)^{2}-(i)^{2}=1+1=2.$

   Therefore

   $z=\frac{2\sqrt{3}-2i}{2}=\sqrt{3}-i.$ …

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7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $z = \dfrac{(2-i)(1+i)^3}{(1-i)^2}$, then Arg$(z)=$ (A) $\tan^{-1}\left(\dfrac{1}{3}\right) - \pi$ (B) $\tan^{-1}\left(\dfrac{3}{4}\right) - \pi$ (C) $\pi - \tan^{-1}\left(\dfrac{3}{4}\right)$ (D) $\tan^{-1}\left(\dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   The key idea is to simplify the complex number $z$ into standard form $a+bi$ by expanding powers and rationalising, then compute its argument using $\tan^{-1}(b/a)$ and adjust for the correct quadrant. The final argument is $\tan^{-1}(1/3) - \pi$, so the correct option is (A).

   We are given

   $z = \frac{(2-i)(1+i)^3}{(1-i)^2}.$

   We need $\operatorname{Arg}(z)$, the principal argument (usually in $(-\pi, \pi]$). The trick is to simplify $z$ to a form $a+bi$ and then find the angle.

   ---

   Concept & Intuition

   The argument of a complex number is the angle it makes with the positive real axis. When a complex number is expressed as a fraction of products, we can either simplify algebraically or use properties of arguments (sum/difference of angles). Here, direct algebraic simplification is straightforward because the powers are small. The pitfall: forgetting to check which quadrant the simplified number lies in — the arctan formula alone gives only the reference angle.

   ---

   Step-by-step solution

   1. Simplify $(1+i)^3$

      First, $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$.

      Then $(1+i)^3 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$.

   2. Simplify $(1-i)^2$

      $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.

   3. Write $z$ with these simplifications

   $z = \frac{(2-i)(-2 + 2i)}{-2i}.$

   4. Multiply numerator

      $(2-i)(-2+2i) = 2(-2+2i) - i(-2+2i) = -4 + 4i + 2i - 2i^2$.

      Since $i^2 = -1$, $-2i^2 = -2(-1) = 2$.

      So numerator = $-4 + 4i + 2i + 2 = -2 + 6i$.

   5. Now $z = \dfrac{-2 + 6i}{-2i}$

      Multiply numerator and denominator by $i$ to rationalise:

   $z = \frac{(-2+6i) \cdot i}{-2i \cdot i} = \frac{-2i + 6i^2}{-2i^2}.$

   Since $i^2 = -1$, we get:

   $z = \frac{-2i + 6(-1)}{-2(-1)} = \frac{-2i - 6}{2} = \frac{-6 - 2i}{2} = -3 - i.$

   6. Identify real and imaginary parts

      So $z = -3 - i$, i.e. $a = -3$, $b = -1$. Both are negative, so $z$ lies in the third quadrant.

   7. Find the argument

      The reference angle $\theta_0 = \tan^{-1}\left|\frac{b}{a}\right| = \tan^{-1}\left(\frac{1}{3}\right)$. …

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8. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $X_n = \cos \dfrac{\pi}{2^n} + i \sin \dfrac{\pi}{2^n}$, then $\prod_{n=1}^{\infty} X_n =$ (A) $0$ (B) $1$ (C) $-1$ (D) $i$

   ›Reveal solutionSolution

   The infinite product simplifies by recognizing each term as a complex exponential; the product telescopes into a single complex number, which evaluates to $-1$, so the answer is (C).

   The key insight is that each factor $X_n = \cos(\pi/2^n) + i \sin(\pi/2^n)$ is exactly $e^{i\pi/2^n}$ by Euler’s formula. Multiplying exponentials adds their exponents, so the infinite product becomes $e^{i\pi(1/2 + 1/4 + 1/8 + \cdots)}$. The geometric series sums to $1$, giving $e^{i\pi} = -1$.

   1. Rewrite each term as a complex exponential Euler’s formula tells us $e^{i\theta} = \cos\theta + i\sin\theta$. Hence

   $X_n = e^{i\pi/2^n}.$

   2. Express the product as a single exponential The product from $n=1$ to $\infty$ is

   $\prod_{n=1}^{\infty} X_n = \prod_{n=1}^{\infty} e^{i\pi/2^n} = e^{i\pi \sum_{n=1}^{\infty} \frac{1}{2^n}}.$

   3. Sum the geometric series The series $\sum_{n=1}^{\infty} 1/2^n$ is a standard geometric series with first term $1/2$ and ratio $1/2$:

   $\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1/2}{1 - 1/2} = 1.$

   4. Evaluate the exponential Thus the product equals $e^{i\pi \cdot 1} = e^{i\pi} = -1.$ …

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9. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.Let $k > 0$ and $t = \text{Sech}^{-1}\left(\frac{1}{2}\right) - \text{Cosech}^{-1}\left(\frac{3}{k}\right)$. If $3e^t = 2 + \sqrt{3}$, then $k =$ (A) $2$ (B) $4$ (C) $3\sqrt{3}$ (D) $3\sqrt{2}$

   ›Reveal solutionSolution

   This problem requires converting inverse hyperbolic functions into their logarithmic forms, then simplifying the expression for $t$ and solving the given exponential equation. We find that $k=4$.

   The core idea here is to express the inverse hyperbolic functions in their equivalent logarithmic forms. This transforms the problem from one involving less familiar functions into a standard algebraic manipulation of logarithms and exponentials.

   Here are the relevant logarithmic identities for the inverse hyperbolic functions involved:

   For $0 < x \le 1$, $\text{Sech}^{-1}(x) = \log\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)$.

   For $x > 0$, $\text{Cosech}^{-1}(x) = \log\left(\frac{1 + \sqrt{1 + x^2}}{x}\right)$.

   Let's apply these formulas step-by-step.

   1. Evaluate $\text{Sech}^{-1}\left(\frac{1}{2}\right)$: We use the formula for $\text{Sech}^{-1}(x)$ with $x = \frac{1}{2}$.

   $\text{Sech}^{-1}\left(\frac{1}{2}\right) = \log\left(\frac{1 + \sqrt{1 - \left(\frac{1}{2}\right)^2}}{\frac{1}{2}}\right)$

   First, calculate the term under the square root:
   

   $1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$

   So, $\sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
   Substitute this back into the logarithm:
   

   $\text{Sech}^{-1}\left(\frac{1}{2}\right) = \log\left(\frac{1 + \frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \log\left(\frac{\frac{2 + \sqrt{3}}{2}}{\frac{1}{2}}\right) = \log(2 + \sqrt{3})$

   2. Evaluate $\text{Cosech}^{-1}\left(\frac{3}{k}\right)$: We use the formula for $\text{Cosech}^{-1}(x)$ with $x = \frac{3}{k}$. Since $k > 0$, we know $\frac{3}{k} > 0$.

   $\text{Cosech}^{-1}\left(\frac{3}{k}\right) = \log\left(\frac{1 + \sqrt{1 + \left(\frac{3}{k}\right)^2}}{\frac{3}{k}}\right)$

   Simplify the term under the square root:
   

   $1 + \left(\frac{3}{k}\right)^2 = 1 + \frac{9}{k^2} = \frac{k^2 + 9}{k^2}$

   So, $\sqrt{1 + \left(\frac{3}{k}\right)^2} = \sqrt{\frac{k^2 + 9}{k^2}} = \frac{\sqrt{k^2 + 9}}{k}$ (since $k > 0$).
   Substitute this back into the logarithm:
   

   $\text{Cosech}^{-1}\left(\frac{3}{k}\right) = \log\left(\frac{1 + \frac{\sqrt{k^2 + 9}}{k}}{\frac{3}{k}}\right) = \log\left(\frac{\frac{k + \sqrt{k^2 + 9}}{k}}{\frac{3}{k}}\right) = \log\left(\frac{k + \sqrt{k^2 + 9}}{3}\right)$

   3. Substitute these into the expression for $t$: We are given $t = \text{Sech}^{-1}\left(\frac{1}{2}\right) - \text{Cosech}^{-1}\left(\frac{3}{k}\right)$.

   $t = \log(2 + \sqrt{3}) - \log\left(\frac{k + \sqrt{k^2 + 9}}{3}\right)$

   Using the logarithm property $\log a - \log b = \log\left(\frac{a}{b}\right)$:
   

   $t = \log\left(\frac{2 + \sqrt{3}}{\frac{k + \sqrt{k^2 + 9}}{3}}\right) = \log\left(\frac{3(2 + \sqrt{3})}{k + \sqrt{k^2 + 9}}\right)$

   4. Use the given equation $3e^t = 2 + \sqrt{3}$ to solve for $k$: First, isolate $e^t$:

   $e^t = \frac{2 + \sqrt{3}}{3}$

   Now, substitute the expression for $t$ we found:
   $$ e^{\log\left(\frac{3(2 + \sqrt{3})}{k + \sqrt{k^2 + 9}}\right)} = \frac{2 + \sqrt{3}}{3} $$ …
   

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10. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.If $\sqrt[3]{i} = \text{cis} \alpha$, $\alpha$ belongs to second quadrant and $\sqrt[3]{-i} = \text{cis} \beta$, $\beta$ belongs to third quadrant then $\text{cis} \alpha + \text{cis} \beta =$ (A) $\sqrt{3}$ (B) $i$ (C) $-i$ (D) $-\sqrt{3}$

    ›Reveal solutionSolution

    The cube roots of $i$ and $-i$ are found by De Moivre’s theorem; using the quadrant conditions picks the correct angles, and their sum simplifies to $-\sqrt{3}$.

    The problem asks for the sum of two complex numbers, each given in cis form (cis $\theta = \cos\theta + i\sin\theta$). The cube root of a complex number is not unique — there are three distinct cube roots. The quadrant condition for $\alpha$ and $\beta$ picks exactly one of those three for each. Once we have the correct angles, adding the two cis values is straightforward trigonometry.

    ---

    1. Find the cube roots of $i$. Write $i$ in polar form: $i = \text{cis}\, \frac{\pi}{2}$. By De Moivre’s theorem, the cube roots are

    $\text{cis}\left(\frac{\pi/2 + 2k\pi}{3}\right),\quad k = 0,1,2.$

    That gives the three angles:

    $\frac{\pi}{6},\quad \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6},\quad \frac{\pi}{6} + \frac{4\pi}{3} = \frac{3\pi}{2}.$

    The second quadrant contains angles between $\frac{\pi}{2}$ and $\pi$. Among these, only $\frac{5\pi}{6}$ lies there.

    Hence $\alpha = \frac{5\pi}{6}$.

    2. Find the cube roots of $-i$. Write $-i = \text{cis}\left(-\frac{\pi}{2}\right)$ or equivalently $\text{cis}\,\frac{3\pi}{2}$. The cube roots are

    $\text{cis}\left(\frac{-\pi/2 + 2k\pi}{3}\right),\quad k = 0,1,2.$

    The three angles are:

    $-\frac{\pi}{6},\quad -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2},\quad -\frac{\pi}{6} + \frac{4\pi}{3} = \frac{7\pi}{6}.$

    The third quadrant covers angles from $\pi$ to $\frac{3\pi}{2}$. Only $\frac{7\pi}{6}$ falls there.

    Hence $\beta = \frac{7\pi}{6}$. …

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11. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.If one of the values of $\sqrt{-1-\sqrt{3}i}$ is $a+i\beta$, $\alpha < 0$ and $\beta > 0$, then $\alpha =$ (A) $-\dfrac{1}{\sqrt{2}}$ (B) $-\dfrac{\sqrt{3}}{\sqrt{2}}$ (C) $-\sqrt{3}$ (D) $-\sqrt{2}$

    ›Reveal solutionSolution

    $-1-\sqrt3\,i = 2\,\text{cis}\tfrac{4\pi}{3}$; its square roots are $\sqrt2\,\text{cis}\tfrac{2\pi}{3}$ and $\sqrt2\,\text{cis}\tfrac{5\pi}{3}$. The root with $\alpha<0,\beta>0$ gives $\alpha = -\tfrac{1}{\sqrt2}$.

    Write the radicand in polar form. For $-1 - \sqrt3\,i$: modulus $= \sqrt{1+3} = 2$, and the point $(-1,-\sqrt3)$ is in the third quadrant, so its argument is $\tfrac{4\pi}{3}$ (equivalently $-\tfrac{2\pi}{3}$):

    $-1 - \sqrt3\,i = 2\left(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\right).$

    The square roots have modulus $\sqrt2$ and arguments $\tfrac{1}{2}\cdot\tfrac{4\pi}{3} = \tfrac{2\pi}{3}$ and $\tfrac{2\pi}{3} + \pi = \tfrac{5\pi}{3}$: …

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12. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.The set of all real values of $x$ for which $f(x) = \log_2(2^x - 2) + \sqrt{1-x}$ is also real is (A) $\mathbb{R}$ (B) $(1, \infty)$ (C) $(-\infty, 1]$ (D) $\phi$

    ›Reveal solutionSolution

    For $f(x)$ to be real, the argument of the logarithm must be positive and the radicand must be non-negative. Solving these gives $x > 1$ from the log and $x \le 1$ from the square root — no $x$ satisfies both, so the domain is empty.

    The key idea is that a real-valued function imposes two separate conditions: the expression inside a logarithm must be strictly positive, and the expression inside a square root must be non-negative. Both conditions must hold simultaneously for $f(x)$ to output a real number. When they contradict each other, the domain is empty.

    1. Logarithm condition.

       For $\log_2(2^x - 2)$ to be real, we need $2^x - 2 > 0$.

       That gives $2^x > 2$, so $x > 1$.

    2. Square root condition.

       For $\sqrt{1-x}$ to be real, we need $1 - x \ge 0$.

       That gives $x \le 1$.

    3. Intersection of conditions.

       The first condition demands $x > 1$. The second demands $x \le 1$.

       No real number can be simultaneously greater than 1 and less than or equal to 1. …

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