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

If $\alpha,\beta,\gamma$ are the roots of the equation $\mathrm{x}^{3}+4\mathrm{x}+1=0$, then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}=$
  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given alpha + beta + gamma = 0, alpha*beta + beta*gamma + gamma*alpha = 4, and alpha*beta*gamma = -1. We need 1/(alpha+beta) + 1/(beta+gamma) + 1/(gamma+alpha). Since alpha+beta = -gamma, the expression is -1/gamma - 1/alpha - 1/beta = -(alpha*beta + beta*gamma + gamma*alpha) / (alpha*beta*gamma) = -4 / -1 = 4.

Multiple choice

If $\alpha,\beta,\gamma$ are the roots of the equation $x^{3}-10x^{2}+7x+8=0$ Match the elements of list I with elements of list II: List I List II P) $\alpha+\beta+\gamma$ a) $\dfrac{43}{4}$ Q) $\alpha^{2}+\beta^{2}+\gamma^{2}$ b) $-\dfrac{7}{8}$ R) $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}$ c) $86$ S) $\displaystyle \dfrac{\alpha}{\beta\gamma}+\dfrac{\beta}{\gamma\alpha}+\dfrac{\gamma}{\alpha\beta}$ d) $0$ e) $10$ Then correct arrangement of P, Q, R, S is respectively
  1. $e,c,a,b$
  2. $d,c,a,b$
  3. $e,c,b,a$
  4. $e,b,c,a$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

Match elements of List I with elements in List II. List 1 List 2 A. If the roots of $x^{3}-9x^{2}+26x-k=0$ are positive and in A.P., then $k$ is 1. $56$ B. If the roots of $x^{3}-14x^{2}+kx-64=0$ are positive and in G.P., then $k$ is 2. $25$ C. If the roots of $6x^{3}-kx^{2}+6x-1=0$ are positive and in H.P., then $3k$ is 3. $-3$ D. In the equation $x^{3}-kx+6=0,$ the sum of two roots is $3$, then the value of the third root is 4. $33$ 5. $24$
  1. $A- 5; B- 1; C- 4; D- 3$
  2. $A- 5; B- 1; C- 3; D- 4$
  3. $A- 1; B- 5; C- 4; D- 3$
  4. $A- 3; B- 1; C- 4; D- 5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

If ${x}{1},{x}{2},{x}{3},{x}{4}$ are the roots of the equation ${x}^{4}-{x}^{3}\sin{2\beta}+{x}^{2}\cos{2\beta}-x\cos{\beta}-\sin{\beta}=0$, then $\tan ^{ -1 }{ { x }{ 1 } } +\tan ^{ -1 }{ { x }{ 2 } } +\tan ^{ -1 }{ { x }{ 3 } } +\tan ^{ -1 }{ { x }{ 4 } } =n\pi +\cfrac { \pi }{ 2 } -\beta $.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a known identity related to the roots of the given quartic equation involving trigonometric terms.

Multiple choice

If both $\displaystyle :\alpha, \beta $ and $\displaystyle :6x^{2}+11x+3= 0$ are real roots of the equation then
  1. both $\displaystyle \:\cos ^{-1}\alpha $ and $\displaystyle \:\cos ^{-1}\beta $ are real
  2. both $\displaystyle cosec ^{-1}\alpha $ and $\displaystyle \cos ^{-1}\beta $ are real
  3. both $\displaystyle \:\cot ^{-1}\alpha $ and $\displaystyle \:\cot ^{-1}\beta $ are real

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For cos^-1(x) or cot^-1(x) to be real, the input must be in the domain [-1, 1] for cos^-1 and any real number for cot^-1. The roots of 6x^2 + 11x + 3 = 0 are found via quadratic formula: (-11 +/- sqrt(121 - 72))/12 = (-11 +/- 7)/12. Roots are -4/12 = -1/3 and -18/12 = -3/2. Since -3/2 is outside [-1, 1], cos^-1(-3/2) is not real. However, cot^-1(x) is defined for all real x.

Multiple choice

Let $a,\ b$ and $c$ be three real numbers satisfying $\left[ \begin{matrix} a & b & c \end{matrix} \right] \begin{bmatrix} 1 & 9 & 7 \ 8 & 2 & 7 \ 7 & 3 & 7 \end{bmatrix}=\left[ \begin{matrix} 0 & 0 & 0 \end{matrix} \right]$ ........ $(E)$ Let $b=6$, with $a$ and $c$ satisfying (E). lf $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$, then $\displaystyle \sum_{n=0}^{\infty} \left (\displaystyle \frac{1}{\alpha}+\displaystyle \frac{1}{\beta}\right )^{n} $ is:
  1. $6$
  2. $7$
  3. $\dfrac{6}{7}$
  4. $\infty $
Reveal answer Fill a bubble to check yourself
B Correct answer