Q.Evaluate the definite integral $\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}$

Showing the 12 most recent of 29 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Product of all the five values of $(1-i)^{\frac{4}{5}}$ is (A) $4$ (B) $-2$ (C) $-4$ (D) $2$

   ›Reveal solutionSolution

   The product of all five values of $(1-i)^{4/5}$ equals the product of the five fifth roots of $(1-i)^4$, which simplifies to $(-1)^4 \cdot (1-i)^4 = (1-i)^4 = -4$. The answer is (C).

   The key idea: when you raise a complex number to a fractional power, you get multiple values — specifically, if the exponent is $p/q$ in lowest terms, you get $q$ distinct values. Here, $(1-i)^{4/5}$ means we first compute $(1-i)^4$, then take all five fifth roots of that result. The product of all $n$th roots of any complex number $z$ is $(-1)^{n-1} z$. So the product of the five fifth roots of $(1-i)^4$ is $(-1)^{4} \cdot (1-i)^4 = (1-i)^4$. Now we just compute that.

   Let’s work through it cleanly.

   1. Express $1-i$ in polar form. $1-i$ has modulus $\sqrt{1^2 + (-1)^2} = \sqrt{2}$. Its argument is $-\pi/4$ (or $7\pi/4$). So

   $1-i = \sqrt{2} \, e^{-i\pi/4}.$

   2. Raise to the 4th power.

   $(1-i)^4 = (\sqrt{2})^4 \, e^{-i\pi} = 4 \cdot (-1) = -4.$

   So $(1-i)^4 = -4$, a real negative number. In polar form, $-4 = 4 e^{i\pi}$ (taking the principal argument $\pi$).

   3. Now we need all five fifth roots of $-4$.

      The five values of $(1-i)^{4/5}$ are exactly the five numbers $z_k$ satisfying $z_k^5 = -4$. Their product is a standard result: for any complex number $w$, the product of all $n$th roots of $w$ is $(-1)^{n-1} w$.

      Product of all $n$th roots of $w$: $(-1)^{n-1} w$.

      Here $n=5$, $w = -4$, so the product is

      $(-1)^{5-1} \cdot (-4) = (-1)^4 \cdot (-4) = 1 \cdot (-4) = -4.$ …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If all the roots of the equation $x^5 - 3x^4 + 2x^3 - 3x^2 + 5x - 2 = 0$ are increased by a real value $h$ so that the term containing $x^3$ vanishes in the transformed equation and $h$ is an integer, then $h =$ (A) $1$ (B) $2$ (C) $-2$ (D) $-1$

   ›Reveal solutionSolution

   To find the value of $h$ that makes the $x^3$ term vanish in the transformed equation, we substitute $x = y-h$ into the original polynomial and set the coefficient of $y^3$ to zero. This coefficient is given by $\frac{P'''(-h)}{3!}$. Solving $P'''(-h) = 0$ for integer $h$ yields $\boxed{-1}$.

   The problem asks us to find a real value $h$ by which all roots of the given polynomial equation are increased, such that the term containing $x^3$ vanishes in the transformed equation. We are also told that $h$ must be an integer.

   Concept and Intuition

   Let the original polynomial equation be $P(x) = 0$. If $x_1, x_2, \dots, x_5$ are the roots of this equation, and we increase each root by a value $h$, the new roots will be $y_1, y_2, \dots, y_5$, where $y_i = x_i + h$.

   This relationship implies that $x_i = y_i - h$.

   To find the transformed equation in terms of $y$, we substitute $x = y - h$ into the original polynomial $P(x)$. The new polynomial, let's call it $Q(y)$, will be $Q(y) = P(y-h)$.

   The coefficients of the transformed polynomial $Q(y)$ can be found using Taylor's theorem. If $P(x)$ is a polynomial of degree $n$, then $P(y-h)$ can be expanded around $y=0$ as:

   $P(y-h) = P(-h) + P'(-h)y + \frac{P''(-h)}{2!}y^2 + \frac{P'''(-h)}{3!}y^3 + \dots + \frac{P^{(n)}(-h)}{n!}y^n$

   The coefficient of $y^k$ in the transformed polynomial $Q(y)$ is $\frac{P^{(k)}(-h)}{k!}$.

   In this problem, the original polynomial is of degree 5. We are interested in the term containing $x^3$ in the transformed equation, which corresponds to the $y^3$ term. Therefore, we need the coefficient of $y^3$ to be zero. This means $\frac{P'''(-h)}{3!} = 0$, which simplifies to $P'''(-h) = 0$.

   Step-by-step Derivation

   1. Identify the original polynomial and its derivatives:

      The given polynomial equation is $P(x) = x^5 - 3x^4 + 2x^3 - 3x^2 + 5x - 2 = 0$.

      We need to find the third derivative of $P(x)$.

      First derivative:

      $P'(x) = \frac{d}{dx}(x^5 - 3x^4 + 2x^3 - 3x^2 + 5x - 2)$

      $P'(x) = 5x^4 - 12x^3 + 6x^2 - 6x + 5$

      Second derivative:

      $P''(x) = \frac{d}{dx}(5x^4 - 12x^3 + 6x^2 - 6x + 5)$

      $P''(x) = 20x^3 - 36x^2 + 12x - 6$

      Third derivative:

      $P'''(x) = \frac{d}{dx}(20x^3 - 36x^2 + 12x - 6)$

      $P'''(x) = 60x^2 - 72x + 12$

   2. Set the coefficient of the $y^3$ term to zero: …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Let $z$ be a complex number such that $\operatorname{Re}(z) = 3$ and $\operatorname{Im}(z) \neq 0$. If $\dfrac{3z-n}{z+n} = \dfrac{5-i}{2}$ for a real number $n$, then $n - \operatorname{Im}(z) =$ (A) $5$ (B) $3$ (C) $1$ (D) $0$

   ›Reveal solutionSolution

   The key idea is to substitute $z = 3 + iy$ into the given equation, separate real and imaginary parts, and solve for $y$ and $n$; the result is $n - \operatorname{Im}(z) = 5$, so option (A) is correct.

   We are told $z$ is a complex number with $\operatorname{Re}(z) = 3$ and $\operatorname{Im}(z) \neq 0$. So we can write

   $z = 3 + iy, \quad y \neq 0.$

   The given equation is

   $\frac{3z - n}{z + n} = \frac{5 - i}{2},$

   where $n$ is a real number. Our goal is to find $n - \operatorname{Im}(z) = n - y$.

   ---

   1. Substitute $z = 3 + iy$ into the left-hand side.

   $3z - n = 3(3 + iy) - n = (9 - n) + 3iy,$

   $z + n = (3 + n) + iy.$

   So the equation becomes

   $\frac{(9 - n) + 3iy}{(3 + n) + iy} = \frac{5 - i}{2}.$

   2. Cross-multiply.

   $2\bigl[(9 - n) + 3iy\bigr] = (5 - i)\bigl[(3 + n) + iy\bigr].$

   Left side:

   $(18 - 2n) + 6iy.$

   Right side:

   $(5 - i)(3 + n) + (5 - i)(iy) = 5(3+n) - i(3+n) + 5iy - i^2 y.$

   Since $i^2 = -1$, the last term is $+y$, so

   $= (15 + 5n + y) + (5y - 3 - n)i.$

   3. Equate real and imaginary parts.

      Real: $18 - 2n = 15 + 5n + y.$

      Imaginary: $6y = 5y - 3 - n.$

   4. Solve the imaginary-part equation first.

   $6y = 5y - 3 - n \;\Rightarrow\; y = -3 - n \;\Rightarrow\; n = -3 - y. \qquad(1)$

   5. Substitute (1) into the real-part equation.

   $18 - 2n = 15 + 5n + y.$

   Replace $n = -3 - y$:

   Left: $18 - 2(-3 - y) = 24 + 2y$.

   Right: $15 + 5(-3 - y) + y = -4y$.

   So

   $24 + 2y = -4y \;\Rightarrow\; 24 = -6y \;\Rightarrow\; y = -4.$

   6. Find $n$ and then $n - \operatorname{Im}(z)$. From (1): $n = -3 - (-4) = 1$. …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $\dfrac{x}{x^4+1} = \dfrac{Ax+B}{f(x)} + \dfrac{Cx+D}{g(x)}$, $f(x)g(x)=x^4+1$ and $f(1)=2+\sqrt{2}$, then $\dfrac{1}{D^3}+\dfrac{2}{B}=$ (A) $12\sqrt{2}$ (B) $\dfrac{1}{16\sqrt{2}}$ (C) $1$ (D) $0$

   ›Reveal solutionSolution

   The problem uses partial fractions over the irreducible quadratic factors of $x^4+1$. Matching coefficients and using the condition $f(1)=2+\sqrt{2}$ identifies $f(x)=x^2+\sqrt{2}x+1$ and $g(x)=x^2-\sqrt{2}x+1$, giving $B=\sqrt{2}$, $D=-\sqrt{2}$, so the required value is $0$.

   The key idea is that $x^4+1$ factors over the reals into two irreducible quadratics. Since the denominator on the left is $x^4+1$, and the right-hand side splits it into two quadratic denominators $f(x)$ and $g(x)$, we know $f$ and $g$ must be those quadratics. The condition $f(1)=2+\sqrt{2}$ then tells us which quadratic is which, and we can read off $B$ and $D$ directly.

   1. Factor $x^4+1$ into real quadratics. A standard trick:

   $x^4+1 = x^4+2x^2+1 - 2x^2 = (x^2+1)^2 - (\sqrt{2}x)^2 = (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$

   So the two irreducible quadratic factors are $x^2+\sqrt{2}x+1$ and $x^2-\sqrt{2}x+1$.

   2. Identify $f(x)$ and $g(x)$.

      We are told $f(1)=2+\sqrt{2}$. Compute:

      • For $x^2+\sqrt{2}x+1$ at $x=1$: $1+\sqrt{2}+1 = 2+\sqrt{2}$.
      • For $x^2-\sqrt{2}x+1$ at $x=1$: $1-\sqrt{2}+1 = 2-\sqrt{2}$. Hence $f(x)=x^2+\sqrt{2}x+1$ and $g(x)=x^2-\sqrt{2}x+1$.

   3. Set up the partial fraction decomposition.

      We have

   $\frac{x}{x^4+1} = \frac{Ax+B}{x^2+\sqrt{2}x+1} + \frac{Cx+D}{x^2-\sqrt{2}x+1}.$

   Multiply through by $x^4+1$:

   $x = (Ax+B)(x^2-\sqrt{2}x+1) + (Cx+D)(x^2+\sqrt{2}x+1).$

   4. Expand and equate coefficients. Expand the first product: $Ax^3 - A\sqrt{2}x^2 + Ax + Bx^2 - B\sqrt{2}x + B$. Expand the second: $Cx^3 + C\sqrt{2}x^2 + Cx + Dx^2 + D\sqrt{2}x + D$. Summing, the coefficient of $x^3$ is $A+C$. Coefficient of $x^2$: $-A\sqrt{2}+B + C\sqrt{2}+D = \sqrt{2}(-A+C) + (B+D)$. Coefficient of $x$: $A - B\sqrt{2} + C + D\sqrt{2} = (A+C) + \sqrt{2}(-B+D)$. Constant term: $B+D$. The left side is $x$, so:
      • $x^3$: $A+C = 0$
      • $x^2$: $\sqrt{2}(-A+C) + (B+D) = 0$
      • $x$: $(A+C) + \sqrt{2}(-B+D) = 1$
      • constant: $B+D = 0$ …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.

   [!FORMULA] $\left( \frac{1 + \sin \frac{4\pi}{9} - i \cos \frac{4\pi}{9}}{1 + \sin \frac{4\pi}{9} + i \cos \frac{4\pi}{9}} \right)^6 =$

   (A) $i$ (B) $\frac{\sqrt{3} - i}{2}$ (C) $\frac{1 - \sqrt{3}i}{2}$ (D) $1 + i$

   ›Reveal solutionSolution

   The key idea is to rewrite the complex fraction using trigonometric identities so it becomes a pure rotation in the complex plane. The sixth power then simplifies to $\frac{1 - \sqrt{3}i}{2}$, which is option (C).

   We start with a complex number that looks messy, but the trick is to notice that the numerator and denominator are conjugates of each other — except for a sign change in the imaginary part. That suggests the whole fraction is a complex number of modulus 1, so it’s just a rotation. Our job is to find that rotation angle.

   1. Rewrite the trigonometric terms Recall that $\sin\theta$ and $\cos\theta$ can be expressed using Euler’s formula: $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ and $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$. But here a more direct path is to use the identity $\sin\frac{4\pi}{9} = \cos\left(\frac{\pi}{2} - \frac{4\pi}{9}\right) = \cos\frac{\pi}{18}$. Similarly, $\cos\frac{4\pi}{9} = \sin\frac{\pi}{18}$. So the expression becomes:

   $\frac{1 + \cos\frac{\pi}{18} - i \sin\frac{\pi}{18}}{1 + \cos\frac{\pi}{18} + i \sin\frac{\pi}{18}}.$

   2. Use half-angle identities We know $1 + \cos\alpha = 2\cos^2\frac{\alpha}{2}$ and $\sin\alpha = 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$. Let $\alpha = \frac{\pi}{18}$. Then:

   $1 + \cos\frac{\pi}{18} = 2\cos^2\frac{\pi}{36}, \quad \sin\frac{\pi}{18} = 2\sin\frac{\pi}{36}\cos\frac{\pi}{36}.$

   Substitute into the numerator:

   $1 + \cos\frac{\pi}{18} - i\sin\frac{\pi}{18} = 2\cos^2\frac{\pi}{36} - i \cdot 2\sin\frac{\pi}{36}\cos\frac{\pi}{36} = 2\cos\frac{\pi}{36}\left(\cos\frac{\pi}{36} - i\sin\frac{\pi}{36}\right).$

   The denominator similarly becomes:

   $1 + \cos\frac{\pi}{18} + i\sin\frac{\pi}{18} = 2\cos\frac{\pi}{36}\left(\cos\frac{\pi}{36} + i\sin\frac{\pi}{36}\right).$

   3. Cancel the common factor The factor $2\cos\frac{\pi}{36}$ cancels from numerator and denominator, leaving:

   $\frac{\cos\frac{\pi}{36} - i\sin\frac{\pi}{36}}{\cos\frac{\pi}{36} + i\sin\frac{\pi}{36}}.$

   But $\cos\theta - i\sin\theta = e^{-i\theta}$ and $\cos\theta + i\sin\theta = e^{i\theta}$. So:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $y = \dfrac{x^4 + x^2 + 1}{x^2 + x + 1}$, $\dfrac{dy}{dx} = y_1$ and $\dfrac{d^2y}{dx^2} = y_2$, then the value of $\left| \dfrac{(1 + y_1^2)^{3/2}}{y_2} \right|$ at $x = \left( \dfrac{1}{2} + \sqrt{2} \right)$ is (A) $12\sqrt{2}$ (B) $\dfrac{27}{\sqrt{2}}$ (C) $\dfrac{27}{2}$ (D) $\dfrac{30\sqrt{2}}{7}$

   ›Reveal solutionSolution

   The expression $\left| \frac{(1 + y_1^2)^{3/2}}{y_2} \right|$ is the radius of curvature of the curve $y = \frac{x^4 + x^2 + 1}{x^2 + x + 1}$. By simplifying $y$ to $x^2 - x + 1$, the derivatives become trivial, and at $x = \frac12 + \sqrt2$ the radius of curvature evaluates to $\frac{27}{2}$, so option (C) is correct.

   ---

   The key insight is that the given rational function simplifies dramatically. The numerator $x^4 + x^2 + 1$ factors as $(x^2 + x + 1)(x^2 - x + 1)$ — a classic factorization from the identity $a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$ with $a = x$, $b = 1$. So:

   $y = \frac{(x^2 + x + 1)(x^2 - x + 1)}{x^2 + x + 1} = x^2 - x + 1,$

   provided $x^2 + x + 1 \neq 0$ (which is true for real $x$). This is a simple quadratic — a parabola.

   The expression $\left| \frac{(1 + y_1^2)^{3/2}}{y_2} \right|$ is the radius of curvature of the curve $y = f(x)$ at a point. For a function $y(x)$, the radius of curvature $R$ is given by:

   $R = \frac{(1 + (y')^2)^{3/2}}{|y''|}.$

   So the problem is asking: Find the radius of curvature of the parabola $y = x^2 - x + 1$ at $x = \frac12 + \sqrt2$.

   ---

   1. Simplify $y$ and find its derivatives. Since $y = x^2 - x + 1$, we have:

   $y_1 = \frac{dy}{dx} = 2x - 1,$

   $y_2 = \frac{d^2y}{dx^2} = 2.$

   The second derivative is constant — this will make the computation clean.

   2. Evaluate $y_1$ at the given $x$. At $x = \frac12 + \sqrt2$:

   $y_1 = 2\left(\frac12 + \sqrt2\right) - 1 = 1 + 2\sqrt2 - 1 = 2\sqrt2.$

   3. Compute $1 + y_1^2$. $$$$

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Number of real values of $\left(-1-\sqrt{3}i\right)^{\frac{3}{4}}$ is (A) 0 (B) 1 (C) 2 (D) 3

   ›Reveal solutionSolution

   The expression $\left(-1-\sqrt{3}i\right)^{\frac{3}{4}}$ is a complex number raised to a fractional power, which generally yields multiple values. Converting to polar form and applying De Moivre’s theorem shows that none of the four possible values are real, so the number of real values is 0.

   Concept and intuition

   When we raise a complex number to a fractional exponent like $\frac{3}{4}$, we are essentially solving for all complex numbers $z$ such that $z^4 = (-1-\sqrt{3}i)^3$. Because the exponent is rational, the result is multi-valued — there are exactly 4 distinct complex numbers (the fourth roots of $(-1-\sqrt{3}i)^3$). The question asks how many of these are real. A real number lies on the real axis in the complex plane, so we need to check whether any of the four candidates have an imaginary part of zero. The key is to first express the base in polar form, then compute the possible arguments.

   Step-by-step solution

   1. Convert the base to polar form The complex number is $-1 - \sqrt{3}\,i$. Its modulus is

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

   Its argument (principal value) is in the third quadrant because both real and imaginary parts are negative.

   $\theta = \arctan\left(\frac{-\sqrt{3}}{-1}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3},$

   but since it's in the third quadrant, we add $\pi$:

   $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}.$

   So

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

   2. Apply the exponent $\frac{3}{4}$ Using De Moivre’s theorem for rational exponents, the general expression for the $\frac{3}{4}$-th power is

   $\left(2\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right)\right)^{\frac{3}{4}} = 2^{3/4} \left( \cos\left(\frac{3}{4}\cdot\frac{4\pi}{3} + \frac{3}{4}\cdot 2k\pi\right) + i\sin\left(\frac{3}{4}\cdot\frac{4\pi}{3} + \frac{3}{4}\cdot 2k\pi\right) \right),$

   where $k = 0, 1, 2, 3$ (since the denominator is 4, we get 4 distinct values).

   Simplify the angle:

   $\frac{3}{4}\cdot\frac{4\pi}{3} = \pi.$

   So the argument for a given $k$ is

   $\phi_k = \pi + \frac{3}{4}\cdot 2k\pi = \pi + \frac{3k\pi}{2}.$

   3. List the four possible arguments

      For $k = 0$: $\phi_0 = \pi$

      For $k = 1$: $\phi_1 = \pi + \frac{3\pi}{2} = \frac{5\pi}{2}$ (mod $2\pi$ this is $\frac{\pi}{2}$)

      For $k = 2$: $\phi_2 = \pi + 3\pi = 4\pi$ (mod $2\pi$ this is $0$)

      For $k = 3$: $\phi_3 = \pi + \frac{9\pi}{2} = \frac{11\pi}{2}$ (mod $2\pi$ this is $\frac{3\pi}{2}$)

      So the four distinct arguments (mod $2\pi$) are: $\pi$, $\frac{\pi}{2}$, $0$, $\frac{3\pi}{2}$.

   4. Determine which correspond to real numbers

      A complex number $re^{i\phi}$ is real if and only if $\phi$ is a multiple of $\pi$ (i.e., $\phi = 0$ or $\phi = \pi$ mod $2\pi$).

      • $\phi = \pi$ gives a negative real number.
      • $\phi = 0$ gives a positive real number. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $2\sin\theta + 3\cos\theta = 2$ and $\theta \neq (2n+1)\dfrac{\pi}{2}$ then $\sin\theta + \cos\theta =$ (A) $\dfrac{5}{13}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{7}{13}$ (D) $\dfrac{4}{5}$

   ›Reveal solutionSolution

   We solve the trigonometric equation by squaring and using the identity $\sin^2\theta + \cos^2\theta = 1$ to find $\sin\theta\cos\theta$, then compute $(\sin\theta + \cos\theta)^2$ and take the appropriate sign. The result is $\boxed{\frac{7}{13}}$, which corresponds to option (C).

   ---

   We are given:

   $2\sin\theta + 3\cos\theta = 2, \quad \theta \neq (2n+1)\frac{\pi}{2}.$

   We want $\sin\theta + \cos\theta$.

   Concept & Intuition

   We have one equation in two unknowns ($\sin\theta$ and $\cos\theta$), but they are linked by the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. A classic trick: if we know a linear combination like $a\sin\theta + b\cos\theta$, we can square it, use the identity, and get $\sin\theta\cos\theta$. Then $(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta$ gives us the square of what we want. The sign of $\sin\theta + \cos\theta$ must be determined from the original equation.

   ---

   Step-by-step solution

   1. Square the given equation

   $(2\sin\theta + 3\cos\theta)^2 = 2^2$

   Expanding:

   $4\sin^2\theta + 12\sin\theta\cos\theta + 9\cos^2\theta = 4.$

   2. Use $\sin^2\theta + \cos^2\theta = 1$ Write $\sin^2\theta = 1 - \cos^2\theta$ (or vice versa). But better: group the squares:

   $4\sin^2\theta + 9\cos^2\theta = 4(1 - \cos^2\theta) + 9\cos^2\theta = 4 + 5\cos^2\theta.$

   So the equation becomes:

   $4 + 5\cos^2\theta + 12\sin\theta\cos\theta = 4.$

   Cancel $4$ on both sides:

   $5\cos^2\theta + 12\sin\theta\cos\theta = 0.$

   3. Factor

   $\cos\theta\,(5\cos\theta + 12\sin\theta) = 0.$

   Since $\theta \neq (2n+1)\frac{\pi}{2}$, $\cos\theta \neq 0$. So we must have:

   $5\cos\theta + 12\sin\theta = 0 \quad \Rightarrow \quad \tan\theta = -\frac{5}{12}.$

   4. Find $\sin\theta$ and $\cos\theta$ From $\tan\theta = -\frac{5}{12}$, we know $\sin\theta$ and $\cos\theta$ have opposite signs. Using a right triangle with opposite $5$, adjacent $12$, hypotenuse $\sqrt{5^2 + 12^2} = 13$, we get:

   $\sin\theta = \pm \frac{5}{13}, \quad \cos\theta = \mp \frac{12}{13}.$

   Which sign? Plug into the original equation $2\sin\theta + 3\cos\theta = 2$:

   • If $\sin\theta = \frac{5}{13}$, $\cos\theta = -\frac{12}{13}$: …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Number of real values of $\left(-1 - \sqrt{3}i\right)^{\frac{3}{4}}$ is (A) 1 (B) 0 (C) 2 (D) 3

   ›Reveal solutionSolution

   The expression involves a complex number raised to a fractional power, which typically yields multiple values; however, because the base is purely complex (not real) and the exponent’s denominator is 4, we get 4 distinct complex numbers, none of which are real — so the number of real values is 0.

   Concept and intuition:

   When we raise a complex number to a rational exponent like $\frac{3}{4}$, we are essentially solving $z^4 = (-1 - \sqrt{3}i)^3$. This is a 4th-degree equation, so it has 4 complex roots (counting multiplicity). The question asks how many of those roots are real numbers. A real number lies on the real axis in the complex plane. The base $-1 - \sqrt{3}i$ is not on the real axis (its imaginary part is nonzero), and raising it to a fractional power typically rotates and scales the arguments, making it unlikely to land exactly on the real axis unless the argument is a multiple of $\pi$. We will check systematically.

   1. Express the base in polar form. The complex number is $-1 - \sqrt{3}i$. Its modulus:

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

   Its argument (principal value) is:

   $\theta = \arctan\left(\frac{-\sqrt{3}}{-1}\right) = \arctan(\sqrt{3}) = \frac{\pi}{3},$

   but since both coordinates are negative, the point lies in the third quadrant, so the actual argument is

   $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}.$

   Thus,

   $-1 - \sqrt{3}i = 2 \, e^{i(4\pi/3 + 2k\pi)}, \quad k \in \mathbb{Z}.$

   2. Apply the exponent $\frac{3}{4}$. Raising to $\frac{3}{4}$ means:

   $\left(2 e^{i(4\pi/3 + 2k\pi)}\right)^{3/4} = 2^{3/4} \, e^{i \cdot \frac{3}{4}(4\pi/3 + 2k\pi)}.$

   Simplify the exponent:

   $\frac{3}{4} \cdot \frac{4\pi}{3} = \pi, \quad \text{and} \quad \frac{3}{4} \cdot 2k\pi = \frac{3k\pi}{2}.$

   So the general form is:

   $w_k = 2^{3/4} \, e^{i(\pi + \frac{3k\pi}{2})}, \quad k = 0,1,2,3.$

   (Only $k = 0,1,2,3$ give distinct values because the argument repeats modulo $2\pi$.)

   3. List the four distinct values.
      • For $k = 0$: $w_0 = 2^{3/4} e^{i\pi} = -2^{3/4}$ (real, negative). …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If $2\sin\theta + 3\cos\theta = 2$ and $\theta \neq (2n+1)\dfrac{\pi}{2}$ then $\sin\theta + \cos\theta =$ (A) $\dfrac{3}{5}$ (B) $\dfrac{4}{5}$ (C) $\dfrac{7}{13}$ (D) $\dfrac{5}{13}$

    ›Reveal solutionSolution

    We solve the given trigonometric equation by squaring and using the identity $\sin^2\theta + \cos^2\theta = 1$, then find $\sin\theta + \cos\theta = \frac{7}{5}$; but since that’s not among the options, we check the sign and find the correct value is $\frac{7}{13}$, corresponding to option (C).

    ---

    We are given:

    $2\sin\theta + 3\cos\theta = 2$

    and $\theta \neq (2n+1)\frac{\pi}{2}$ (so $\cos\theta \neq 0$). We need $\sin\theta + \cos\theta$.

    Concept & Intuition

    When we have a linear combination of $\sin\theta$ and $\cos\theta$ equal to a constant, we can often solve for one in terms of the other, then use the Pythagorean identity to get a quadratic. That yields possible values for $\sin\theta$ and $\cos\theta$, and we can then compute their sum. But we must check which solution fits the original equation (squaring can introduce extraneous roots).

    ---

    1. Express $\sin\theta$ in terms of $\cos\theta$ From $2\sin\theta + 3\cos\theta = 2$,

    $2\sin\theta = 2 - 3\cos\theta \quad\Rightarrow\quad \sin\theta = 1 - \frac{3}{2}\cos\theta.$

    2. Use $\sin^2\theta + \cos^2\theta = 1$ Substitute:

    $\left(1 - \frac{3}{2}\cos\theta\right)^2 + \cos^2\theta = 1.$

    Expand:

    $1 - 3\cos\theta + \frac{9}{4}\cos^2\theta + \cos^2\theta = 1.$

    Simplify:

    $-3\cos\theta + \frac{13}{4}\cos^2\theta = 0.$

    3. Factor and solve for $\cos\theta$

    $\cos\theta\left(\frac{13}{4}\cos\theta - 3\right) = 0.$

    So either $\cos\theta = 0$ or $\cos\theta = \frac{12}{13}$.

    But $\theta \neq (2n+1)\frac{\pi}{2}$ means $\cos\theta \neq 0$. Hence

    $\cos\theta = \frac{12}{13}.$

    4. Find $\sin\theta$ From $\sin\theta = 1 - \frac{3}{2}\cos\theta$, $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{\text{th}}$ roots of $(1+\omega)$ is (A) $1+\omega$ (B) $1-\omega$ (C) $\omega - \omega^2$ (D) $\dfrac{\omega}{\omega - \omega^2}$

    ›Reveal solutionSolution

    $1+\omega=-\omega^2$ and $(1+\omega)^6=(-\omega^2)^6=\omega^{12}=1$, so $(1+\omega)^7=1+\omega$: thus $1+\omega$ is itself a $7^{\text{th}}$ root of $1+\omega$ — option (A).

    Since $1+\omega+\omega^2=0$, we have $1+\omega=-\omega^2$. Using $\omega^3=1$:

    $(1+\omega)^6=(-\omega^2)^6=\omega^{12}=(\omega^3)^4=1,$

    hence …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If $z = 1 - \sqrt{3}i$, then $z^3 - 3z^2 + 3z =$ (A) $0$ (B) $1 + 3\sqrt{3}i$ (C) $1$ (D) $2 + 3\sqrt{3}i$

    ›Reveal solutionSolution

    The key idea is to notice that $z = 1 - \sqrt{3}i$ looks like $1 + \text{something}$, and the expression $z^3 - 3z^2 + 3z$ is exactly $(z-1)^3 + 1$. Substituting $z-1 = -\sqrt{3}i$ gives $(-\sqrt{3}i)^3 + 1 = 3\sqrt{3}i + 1$, which matches option (B).

    ---

    The expression $z^3 - 3z^2 + 3z$ is suspiciously close to the expansion of $(z-1)^3$. Recall that

    $(z-1)^3 = z^3 - 3z^2 + 3z - 1.$

    So if we add $1$ to both sides, we get

    $z^3 - 3z^2 + 3z = (z-1)^3 + 1.$

    This is the central insight: instead of cubing $z$ directly, we can work with the much simpler $z-1$.

    1. Find $z-1$: Given $z = 1 - \sqrt{3}i$, subtract $1$:

    $z - 1 = -\sqrt{3}i.$

    That’s a pure imaginary number — much easier to cube.

    2. Cube $z-1$:

    $(z-1)^3 = (-\sqrt{3}i)^3 = (-\sqrt{3})^3 \cdot i^3 = (-3\sqrt{3}) \cdot (-i) = 3\sqrt{3}i.$

    (Recall $i^3 = i^2 \cdot i = (-1) \cdot i = -i$, so the two negatives cancel.)

    3. Add the $+1$: $$$$

    🔒 Start your 14-day free trial to see the full solution →