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 roots of the equation $z^n = (z + 1)^n$ on the complex plane lie on the line
  1. $2x + 1 = 0$
  2. $2x - 1 = 0$
  3. $x + 1 = 0$
  4. $x - 1 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For z^n = (z + 1)^n, the magnitudes must be equal, so |z| = |z + 1|. This implies z is equidistant from 0 and -1 on the complex plane, which describes the perpendicular bisector of the segment connecting 0 and -1, specifically the line x = -1/2, or 2x + 1 = 0.

Multiple choice

If ${z}{1}$ and ${z}{2}$ are roots of quadratic equation $a{z}^{2}+bz+c=0$ such that $Im({z}{1}{z}{2})\ne 0$ then
  1. $a,b,c$ are all real
  2. atleast one of $a,b,c$ is real
  3. atleast one of $a,b,c$ is imaginary
  4. $a,b,c$ are all imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the coefficients a, b, c are all real, the roots must be either real or complex conjugates. If they are complex conjugates, their product z1*z2 = c/a is real, so Im(z1*z2) = 0. Since Im(z1*z2) is not 0, the coefficients cannot all be real.

Multiple choice

Let ${ z }{ k }\left( k=0,1,2,...,6 \right) $ be the roots of the equation ${ \left( z+1 \right) }^{ 7 }+{ z }^{ 7 }=0$, then $\displaystyle \sum _{ k=0 }^{ 6 }{ Re\left( { z }{ k } \right) } $ is equal to
  1. $0$
  2. $\displaystyle \frac { 3 }{ 2 } $
  3. $\displaystyle -\frac { 7 }{ 2 } $
  4. $\displaystyle \frac { 7 }{ 2 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of (z+1)^7 + z^7 = 0 satisfy |z+1| = |z|. This implies Re(z) = -1/2. Since there are 7 roots, the sum of the real parts is 7 * (-1/2) = -7/2.

Multiple choice

If the sum of the roots of the equation $ax^{2} + bx + c = 0$ is equal to sum of the squares of their reciprocals, then $bc^{2}, ca^{2}, ab^{2}$ are in
  1. $A.P$
  2. $G.P$
  3. $H.P$
  4. $A.G.P$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of roots = -b/a. Sum of squares of reciprocals = (1/r1)^2 + (1/r2)^2 = (r1^2 + r2^2) / (r1*r2)^2 = ((r1+r2)^2 - 2r1r2) / (r1r2)^2. Substituting r1+r2 = -b/a and r1r2 = c/a, we get (b^2/a^2 - 2c/a) / (c^2/a^2) = (b^2 - 2ac) / c^2. Equating -b/a = (b^2 - 2ac) / c^2 leads to the condition for AP.

Multiple choice

The abcissae of two points A and B are the roots of the equation $\displaystyle x^{2}+2ax-b^{2}=0 $ and their ordinates are the roots of the equation $\displaystyle x^{2}+2px-q^{2}=0 $ The radius of the circle with AB as diameter is
  1. $\displaystyle \sqrt{a^{2}+b^{2}+p^{2}+q^{2}} $
  2. $\displaystyle \sqrt{a^{2}+p^{2}} $
  3. $\displaystyle \sqrt{b^{2}+q^{2}} $
  4. $\displaystyle \sqrt{a^{2}-b^{2}+p^{2}-q^{2}} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The difference between the two abscissae is 2sqrt(a^2 + b^2), and the difference between the ordinates is 2sqrt(p^2 + q^2). The diameter squared is therefore 4(a^2 + b^2 + p^2 + q^2), so the radius is sqrt(a^2 + b^2 + p^2 + q^2).

Multiple choice

let$\alpha \left( a \right) and\beta \left( a \right) $ be the roots of the equation $\left( \sqrt [ 3 ]{ 1+a } -1 \right) { x }^{ 2 }+\left( \sqrt { 1+a } -1 \right) x+\left( \sqrt [ 6 ]{ 1+a } \right) $ where a>-1. Then ${ lim }{ a\rightarrow o^{ + } }\alpha \left( a \right) and\ { lim }{ a\rightarrow 0^{ + } }\beta \left( a \right) $ are
  1. $-\dfrac { 5 }{ 2 } and1$
  2. $-\dfrac { 1 }{ 2 } and-1$
  3. $-\dfrac { 7 }{ 2 } and2$
  4. $-\frac { 9 }{ 2 } and3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Let $f\left ( x \right )=\begin{cases}x^{2}-1 ;0< x< 2 \ 2x+3; 2\leq x< 3\end{cases}$ The quadratic equation whose roots are $\displaystyle \lim_{x\to 2^-}f\left ( x \right )$ and $\displaystyle \lim_{x\to 2^+}f\left ( x \right )$ is
  1. $x^{2}-6x+9=0$
  2. $x^{2}-10x+21=0$
  3. $x^{2}-14x+49=0$

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

Limit x->2- of f(x) = 2^2 - 1 = 3. Limit x->2+ of f(x) = 2(2) + 3 = 7. Roots are 3 and 7. Equation: (x-3)(x-7) = x^2 - 10x + 21 = 0.

Multiple choice

If $\displaystyle \alpha $ and $\displaystyle \beta$ are the roots of the equation $\displaystyle ax^{2}+bx+c= 0$, then the value of $\displaystyle \underset{x\rightarrow a}{lim}\left [ ax^{2}+bx+c+1 \right ]^{2/\left ( x-a \right )}$ is:
  1. $\displaystyle 2a\left ( \alpha -\beta \right )$
  2. $\displaystyle 2\log \left | a\left ( \alpha -\beta \right ) \right |$
  3. $\displaystyle e^{2a\left ( \alpha -\beta \right )}$
  4. $\displaystyle e^{e^{2}}\left | \alpha -\beta \right |$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

If $f(x)=\displaystyle \frac { 4-7x }{ 7x+4 } $, $\underset{ x\rightarrow 0 }{Lim}{ f(x)=l } $ and $\underset{ x\rightarrow \infty }{Lim}{ f(x)=m } $ the quadratic equation having roots as $\displaystyle \dfrac{1}{l}$ and $\displaystyle \dfrac{1}{m}$ is
  1. $ x^{2}-1=0 $
  2. $ x^{2}-x+1=0 $
  3. $x^2-\dfrac{1}{2}=0$
  4. $ x^{3}-1=0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

l = lim(x->0) (4-7x)/(7x+4) = 4/4 = 1. m = lim(x->inf) (4-7x)/(7x+4) = -7/7 = -1. Roots are 1/l = 1 and 1/m = -1. Equation: (x-1)(x+1) = x^2 - 1 = 0.