Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - x - 1 = 0$. If $p_k = (\alpha)^k + (\beta)^k , k \ge 1$, then which one of the following statements is not true ?
  1. $p_3 = p_5 - p_4$
  2. $p_5 = p_2 . p_3$
  3. $p_5 = 11$
  4. $(p_1 + p_2 + p_3 + p_4 + p_5) = 26$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots of the equation satisfy the relation p_{k+2} = p_{k+1} + p_k. Calculating the values gives p_1 = 1, p_2 = 3, p_3 = 4, p_4 = 7, and p_5 = 11. Statement B claims p_5 = p_2 * p_3, which is 11 = 12, and is therefore false.

Multiple choice

The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is?
  1. $3$
  2. $4$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let u = e^x. The equation becomes u^4 + u^3 - 4u^2 + u + 1 = 0. Dividing by u^2 (since u > 0): u^2 + u - 4 + 1/u + 1/u^2 = 0. (u^2 + 1/u^2) + (u + 1/u) - 4 = 0. Let t = u + 1/u. Then u^2 + 1/u^2 = t^2 - 2. (t^2 - 2) + t - 4 = 0 => t^2 + t - 6 = 0 => (t+3)(t-2) = 0. Since u > 0, t = u + 1/u >= 2. Thus t = 2. u + 1/u = 2 => u = 1. e^x = 1 => x = 0. Only one real root.

Multiple choice

Let $a,b\epsilon R,a\ne0$ be such that the equation $ax^{2}-2bx+5=0$ has a repeated root $\alpha$, which is also a root of the equation, $x^{2}-2bx-10=0$. If $\beta$ is the other root of this equations then $\alpha^{2}+\beta^{2}$ is equal to:
  1. $24$
  2. $26$
  3. $25$
  4. $28$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For ax^2 - 2bx + 5 = 0, repeated root alpha means discriminant = 0. (-2b)^2 - 4(a)(5) = 0 => 4b^2 = 20a => b^2 = 5a. Root alpha = -(-2b)/(2a) = b/a. Since alpha is a root of x^2 - 2bx - 10 = 0, (b/a)^2 - 2b(b/a) - 10 = 0. Substitute b^2 = 5a: (5a/a^2) - 2b^2/a - 10 = 0 => 5/a - 10 - 10 = 0 => 5/a = 20 => a = 1/4. Then b^2 = 5/4. Alpha = b/a = b/(1/4) = 4b. Alpha^2 = 16b^2 = 16(5/4) = 20. For the second equation, sum of roots alpha + beta = 2b, product alpha * beta = -10. Beta = -10/alpha. Alpha^2 + beta^2 = 20 + (-10/alpha)^2 = 20 + 100/alpha^2 = 20 + 100/20 = 20 + 5 = 25.

Multiple choice

Let $\alpha$ and $\beta$ be two roots of the equation $x^{2} + 2x + 2 = 0$, then $\alpha^{15} + \beta^{15}$ is equal to
  1. $512$
  2. $-512$
  3. $-256$
  4. $256$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots of x^2 + 2x + 2 = 0 are x = (-2 +/- sqrt(4-8))/2 = -1 +/- i. In polar form, -1+i = sqrt(2) * (cos(3pi/4) + i*sin(3pi/4)). By De Moivre's Theorem, alpha^15 + beta^15 = 2 * (sqrt(2))^15 * cos(15 * 3pi/4) = 2 * 2^7.5 * cos(45pi/4) = 2 * 2^7.5 * cos(11pi + pi/4) = 2 * 2^7.5 * (-cos(pi/4)) = 2 * 2^7.5 * (-1/sqrt(2)) = -2 * 2^7 = -256.

Multiple choice

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation, $6x^ 2-11x+ \alpha = 0$ are rational numbers is
  1. $2$
  2. $5$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots are rational if the discriminant D = b^2 - 4ac = (-11)^2 - 4(6)(alpha) = 121 - 24 * alpha is a perfect square. For alpha = 1, D = 97 (no). For alpha = 2, D = 121 - 48 = 73 (no). For alpha = 3, D = 121 - 72 = 49 (yes, 7^2). For alpha = 4, D = 121 - 96 = 25 (yes, 5^2). For alpha = 5, D = 121 - 120 = 1 (yes, 1^2). For alpha > 5, D < 0. There are 3 values.

Multiple choice

The value of $\lambda$ such that sum of the squares of the roots of the quadratic equation, $x^2 + (3 -\lambda) x + 2 = \lambda$ has the least value is:
  1. $2$
  2. $\dfrac{4}{9}$
  3. $\dfrac{15}{8}$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the roots be x1, x2. x1+x2 = -(3-lambda) = lambda-3, x1*x2 = 2-lambda. Sum of squares = (x1+x2)^2 - 2(x1*x2) = (lambda-3)^2 - 2(2-lambda) = lambda^2 - 6lambda + 9 - 4 + 2lambda = lambda^2 - 4lambda + 5. This is a parabola opening upward with minimum at lambda = -(-4)/(2*1) = 2.

Multiple choice

If $\lambda$ be the ratio of the roots of the quadratic equation in $x , 3 \mathrm { m } ^ { 2 } \mathrm { x } ^ { 2 } + \mathrm { m } ( \mathrm { m } - 4 ) \mathrm { x } + 2 = 0 ,$ then the least value of $\mathrm { m }$ for which $\lambda + \dfrac { 1 } { \lambda } = 1 ,$ is :
  1. $2 - \sqrt { 3 }$
  2. $4 - 3 \sqrt { 2 }$
  3. $- 2 + \sqrt { 2 }$
  4. $4 - 2 \sqrt { 3 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given lambda + 1/lambda = 1, we have lambda^2 - lambda + 1 = 0, which implies lambda is a complex cube root of -1. For a quadratic ax^2+bx+c=0, the ratio of roots lambda = alpha/beta satisfies (alpha+beta)^2 / (alpha*beta) = (lambda+1)^2 / lambda = (lambda^2+2lambda+1)/lambda = (lambda+1+2lambda+1)/lambda = (3lambda)/lambda = 3. Using the equation coefficients, (b/a)^2 / (c/a) = b^2/ac = 3. Here, m^2(m-4)^2 / (3m^2 * 2) = 3. Solving (m-4)^2 / 6 = 3 gives (m-4)^2 = 18, so m-4 = +/- 3*sqrt(2). m = 4 +/- 3*sqrt(2). The least value is 4 - 3*sqrt(2).

Multiple choice

If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^2 + x \sin \theta - 2\sin \theta = 0, \theta \in \left(0, \dfrac{\pi}{2}\right)$, then $\dfrac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12}) (\alpha - \beta)^{24}}$ is equal to:
  1. $\dfrac{2^6}{(\sin \theta + 8)^{12}}$
  2. $\dfrac{2^{12}}{(\sin \theta - 8)^6}$
  3. $\dfrac{2^{12}}{(\sin \theta - 4)^{12}}$
  4. $\dfrac{2^{12}}{(\sin \theta + 8)^{12}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The roots satisfy alpha + beta = -sin(theta) and alpha * beta = -2sin(theta). The expression simplifies using the properties of roots and trigonometric identities to the given result.

Multiple choice

Let $\lambda$ be a real number for which the system of linear equations $x+y+z=6$ $4x+\lambda y-\lambda z=\lambda -2$ $3x+2y-4z=-5$ has many infinitely many solutions. Then $\lambda$ is a root of the quadratic equation
  1. ${ \lambda }^{ 2 }-3\lambda -4=0\quad $
  2. ${ \lambda }^{ 2 }-\lambda -6=0$
  3. ${ \lambda }^{ 2 }+3\lambda -4=0$
  4. ${ \lambda }^{ 2 }+\lambda -6=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For infinitely many solutions, the determinant of the coefficient matrix must be 0. Matrix: [[1,1,1],[4,L,-L],[3,2,-4]]. Det = 1(-4L + 2L) - 1(-16 + 3L) + 1(8 - 3L) = -2L + 16 - 3L + 8 - 3L = -8L + 24 = 0, so L=3. Checking the quadratic options, L^2 - L - 6 = 0 has roots 3 and -2. Since L=3 is a root, option B is correct.

Multiple choice

The sum of the real roots of the equation $\begin{vmatrix} x & -6 & -1 \ 2 & -3x & x-3 \ -3 & 2x & x+2 \end{vmatrix}=0$ is equal to
  1. $6$
  2. $1$
  3. $0$
  4. $-4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Expanding the determinant: x(-3x(x+2) - 2x(x-3)) + 6(2(x+2) - (-3)(x-3)) - 1(2(2x) - (-3)(-3x)) = 0. This simplifies to a cubic equation in x. The sum of roots of a polynomial ax^3 + bx^2 + cx + d = 0 is -b/a. Calculating the coefficient of x^2 and x^3 shows the sum is 0.

Multiple choice

If $\alpha$ and $\beta$ are the roots of the equation $375x^2-25x-2=0$, then $\displaystyle\lim_{n\rightarrow \infty}\displaystyle\sum^n_{r=1}\alpha^r+\displaystyle\lim_{n\rightarrow \infty}\displaystyle\sum^n_{r=1}\beta^r$ is equal to?
  1. $\dfrac{21}{346}$
  2. $\dfrac{29}{358}$
  3. $\dfrac{1}{12}$
  4. $\dfrac{7}{116}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sum of the infinite geometric series sum(alpha^r) is alpha / (1 - alpha) and sum(beta^r) is beta / (1 - beta). The sum is (alpha(1-beta) + beta(1-alpha)) / ((1-alpha)(1-beta)) = (alpha + beta - 2*alpha*beta) / (1 - (alpha + beta) + alpha*beta). From 375x^2 - 25x - 2 = 0, alpha + beta = 25/375 = 1/15 and alpha*beta = -2/375. Substituting these: (1/15 - 2(-2/375)) / (1 - 1/15 - 2/375) = (25/375 + 4/375) / (375/375 - 25/375 - 2/375) = (29/375) / (348/375) = 29/348 = 1/12.

Multiple choice

If $\lambda \epsilon R$ is such that the sum of the cubes of the roots of the equation, $x^{2} + (2 - \lambda) x + (10 - \lambda) = 0$ is minimum, then the magnitude of the difference of the roots of this equation is
  1. $20$
  2. $2\sqrt {5}$
  3. $2\sqrt {7}$
  4. $4\sqrt {2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the roots be a and b. a+b = lambda-2 and ab = 10-lambda. Sum of cubes a^3+b^3 = (a+b)^3 - 3ab(a+b). Substituting the expressions, we minimize the function of lambda. The difference of roots is sqrt((a+b)^2 - 4ab).