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

The number of real roots of the equation $5+\left| { 2 }^{ x }-1 \right| ={ 2 }^{ x }\left( { 2 }^{ x }-2 \right) $ is:
  1. $2$
  2. $3$
  3. $4$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let y = 2^x. Equation: 5 + |y - 1| = y(y - 2) = y^2 - 2y. Case 1: y >= 1: 5 + y - 1 = y^2 - 2y => y^2 - 3y - 4 = 0 => (y-4)(y+1) = 0. y = 4 (since y > 0). 2^x = 4 => x = 2. Case 2: y < 1: 5 - y + 1 = y^2 - 2y => y^2 - y - 6 = 0 => (y-3)(y+2) = 0. No solution for y < 1. Only 1 real root.

Multiple choice

Let $f(x) = ax^{2} + bx + c, a, b\epsilon R, a\neq 0$ satisfying $f(1) + f(2) = 0$ and and $-1$ is a root of the expression Then the equation $f(x) = 0$ has

  1. No real root
  2. $1$ and $2$ are real roots

  3. Two distinct roots

  4. Two equal roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

f(1)+f(2)=0 => a+b+c + 4a+2b+c = 0 => 5a+3b+2c=0. Root -1 => a-b+c=0 => c=b-a. Substitute: 5a+3b+2(b-a)=0 => 3a+5b=0 => b = -3a/5. c = -3a/5 - a = -8a/5. f(x) = ax^2 - (3/5)ax - (8/5)a = 0. x^2 - 0.6x - 1.6 = 0. Discriminant = (-0.6)^2 - 4(1)(-1.6) = 0.36 + 6.4 = 6.76 > 0. Two distinct real roots.

Multiple choice

The equation: $\sin ^{ 2 }{ \theta } -\cfrac { 4 }{ \sin ^{ 2 }{ \theta } -1 } =1-\cfrac { 4 }{ \sin ^{ 2 }{ \theta } -1 } $ has:

  1. no root

  2. one root

  3. two root

  4. infinite roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is sin^2(theta) - 4/(sin^2(theta) - 1) = 1 - 4/(sin^2(theta) - 1). Subtracting the common term from both sides gives sin^2(theta) = 1. This implies sin^2(theta) - 1 = 0, which makes the denominator zero. Thus, there is no valid root.

Multiple choice

If $x<\theta <\cfrac { \pi }{ 2 } $ & $sin\theta +cos\theta +tan\theta +cot\theta +cec\theta +cosec\theta =7$ then $sin2\theta $ is a root of the equation
  1. ${ x }^{ 2 }-44x+36=0$
  2. ${ x }^{ 2 }-44x-36=0$
  3. ${ x }^{ 2 }+44x+36=0$

  4. None of these.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Interpreting the garbled terms as sec(theta) and 0 < theta < pi/2, let t = sin(theta) + cos(theta). The given sum leads to t^2 - 8t + 9 = 0, and since sin(2theta) = t^2 - 1, eliminating t gives x^2 - 44x + 36 = 0.

Multiple choice

Let $-\frac { \pi }{ 6 } <\theta <\frac { \pi }{ 12 } .$ Suppose $ { \alpha }{ 1 } $ and $ { \beta }{ 1 } $ are the roots of the equation $ { x }^{ 2 }-2xsec\theta +1=0 $ and $ { \alpha }{ 2 }$ and $ { \beta }{ 2 } $ are the roots of the equation $ { x }^{ 2 }+2xtan\theta -1=0. $ If $ { \alpha }{ 1 }>\beta _{ 1 } $ and $ { \alpha }{ 2 }>{ \beta }{ 2 }, $ Then $ { \alpha }{ 1 }+{ \beta }_{ 2 } $ equals
  1. $2\left( sec\theta -tan\theta \right) $
  2. $2sec\theta $
  3. $-2tan\theta $

  4. 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

If $ \alpha $ and $ \beta $ are roots of equation $ \frac { 3 } { 4 } \sin \left( \frac { \theta } { 9 } \right) = \sin ^ { 3 } \theta + 3 \sin ^ { 3 } \left( \frac { \theta } { 3 } \right) + 9 \sin ^ { 3 } \left( \frac { \theta } { 9 } \right) + \frac { 1 } { 4 \sqrt { 2 } } $ for $ 0 < \theta < \frac { \pi } { 2 } $ then $ \tan \alpha + \tan \beta $ is equal to
  1. $

    2 + \sqrt { 3 }

    $
  2. $

    3 + \sqrt { 3 }

    $
  3. $

    3 - \sqrt { 3 }

    $
  4. $

    2 - \sqrt { 3 }

    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

Number of roots of the equation $\displaystyle \sin x+2\sin 2x= 3+\sin 3x$ in $\left [ 0,\pi \right ]$ is/are

  1. 2

  2. 3

  3. 0

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using trigonometric identities, the equation sin(x) + 2sin(2x) = 3 + sin(3x) has a maximum value of 1 + 2 + 1 = 4 for the LHS, but the RHS is 3 + sin(3x). Testing values in [0, pi], the equation simplifies to sin(x) + 4sin(x)cos(x) = 3 + 3sin(x) - 4sin^3(x). There are no real solutions in the interval [0, pi].

Multiple choice

If $\displaystyle \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are roots of the equation $8x^3+4x^2-4x-1=0$. The value of $\displaystyle \cos \frac{\pi}{7}\cos \frac{2\pi}{7}\cos\frac {3\pi}{7} $ is equal to
  1. $\displaystyle \frac{1}{64}$
  2. $\displaystyle \frac{\sqrt{7}}{8}$
  3. $\displaystyle \frac{7}{64}$
  4. $\displaystyle \frac{1}{8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The product of the cosines is given by the identity cos(pi/7)cos(2pi/7)cos(3pi/7) = 1/8. This is a standard result derived from the product of roots of the associated polynomial.

Multiple choice

If $\tan{(\displaystyle\frac{\alpha}{2})}$ and $\tan{(\displaystyle\frac{\beta}{2})}$ are the roots of the equation $8{x}^{2}-26x+15=0$ then $\cos{(\alpha +\beta)}$ is equal to-
  1. $-\displaystyle\frac{627}{725}$
  2. $\displaystyle\frac{627}{725}$
  3. $-\displaystyle\frac{725}{625}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let tan(a/2) = u, tan(b/2) = v. u+v = 26/8 = 13/4. uv = 15/8. cos(a+b) = (1 - tan^2((a+b)/2)) / (1 + tan^2((a+b)/2)). tan((a+b)/2) = (u+v)/(1-uv) = (13/4) / (1 - 15/8) = (13/4) / (-7/8) = -26/7. tan^2((a+b)/2) = 676/49. cos(a+b) = (1 - 676/49) / (1 + 676/49) = (-627/49) / (725/49) = -627/725.