Quantitative Aptitude

Algebra

3,715 Questions
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Algebra Questions

Multiple choice

Let the function $f(x)$, defined as $f(x)= \begin{cases} 3ax+b\quad & x<1 \ 11\quad \quad \quad \quad & x=1\quad \ 5ax-2b & x>1 \end{cases} $ be continuous at $x=1$. $a$ and $b$ are the roots of a quadratic equation, then the equation is
  1. ${ x }^{ 2 }-5x+6=0$
  2. ${ x }^{ 2 }+5x+6=0$
  3. ${ x }^{ 2 }-5x-6=0$
  4. ${ 3x }^{ 2 }-5x+2=0$
  5. ${ 2x }^{ 2 }-5x+3=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For continuity at x=1, 3a(1)+b = 11 and 5a(1)-2b = 11. Solving: 6a+2b=22 and 5a-2b=11. Adding gives 11a=33, a=3. Then 9+b=11, b=2. The quadratic equation with roots 3 and 2 is (x-3)(x-2) = x^2-5x+6=0.

Multiple choice

$z_1, z_2$ are the roots of the equation $z^2+az+b=0$. Let $z_1$, $z_2$ and the origin be the vertices of an equilateral triangle. Then $a^2-3b =$
  1. $0$
  2. $1$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If roots z1, z2 and origin form an equilateral triangle, then z1^2 + z2^2 + 0^2 = z1*z2 + z1*0 + z2*0. So z1^2 + z2^2 = z1*z2. (z1+z2)^2 - 2z1z2 = z1z2 => (z1+z2)^2 = 3z1z2. From Vieta's, z1+z2 = -a and z1z2 = b. So (-a)^2 = 3b => a^2 = 3b => a^2 - 3b = 0.

Multiple choice

The roots of the equation $x^{3} -14\ {x}^{2}+56x-64=0$ are in
  1. $AGP$
  2. $HP$
  3. $AP$
  4. $GP$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a cubic equation x^3 - ax^2 + bx - c = 0, if roots are in GP (a/r, a, ar), then product of roots = c = a^3. Here 64 = a^3, so a = 4. If 4 is a root, 64 - 14(16) + 56(4) - 64 = 64 - 224 + 224 - 64 = 0. Yes, 4 is a root. The other roots are 4/r and 4r. Sum of roots = 4/r + 4 + 4r = 14 => 4/r + 4r = 10 => 2/r + 2r = 5 => 2r^2 - 5r + 2 = 0 => (2r-1)(r-2)=0. r=2. Roots are 2, 4, 8. They are in GP.

Multiple choice

Condition for the roots of the equation $x^{3}-px^{2}+qx-r=0$ are in $\mathrm{G}.\mathrm{P}$ is
  1. $q=pr$
  2. $q^{2}=p^{2}r$
  3. $q^{3}=pr^{3}$
  4. $q^{3}=p^{3}r$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If roots a/r, a, ar are in G.P., their product is a^3 = r, so a = r^(1/3). The sum of roots is a(1/r + 1 + r) = p. The sum of roots taken two at a time is a^2(1/r + 1 + r) = q. Dividing the second by the first gives a = q/p. Substituting a^3 = r gives (q/p)^3 = r, which simplifies to q^3 = p^3 * r.

Multiple choice

The equation whose roots are $1, 1,-2$ is
  1. $x^{3}+3x-2=0$
  2. $x^{3}+x-2=0$
  3. $x^{3}+2x-3=0$
  4. $x^{3}-3x+2=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If roots are 1, 1, -2, the polynomial is (x - 1)(x - 1)(x + 2) = (x^2 - 2x + 1)(x + 2) = x^3 + 2x^2 - 2x^2 - 4x + x + 2 = x^3 - 3x + 2 = 0.

Multiple choice

The equation whose roots are $-2,1,3$, is
  1. $x^{3}+2x^{2}+3x-6=0$
  2. $x^{3}-2x^{2}-5x+6=0$
  3. $x^{3}+x^{2}-x-1=0$
  4. $x^{3}-x^{2}+x+1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The polynomial with roots -2, 1, 3 is (x+2)(x-1)(x-3) = (x^2+x-2)(x-3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6 = x^3 - 2x^2 - 5x + 6.

Multiple choice

The equation whose roots are $2, 1\pm 3i$ is
  1. $x^{3}-4x^{2}+14x-20=0$
  2. $x^{3}-x^{2}+5x-14=0$
  3. $x^{3}+2x^{2}-3x-10=0$
  4. $x^{3}-3x^{2}-14x+10=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Roots are 2, 1+3i, 1-3i. The quadratic factor for the complex roots is (x - (1+3i))(x - (1-3i)) = (x-1)^2 + 9 = x^2 - 2x + 10. The cubic is (x-2)(x^2 - 2x + 10) = x^3 - 2x^2 + 10x - 2x^2 + 4x - 20 = x^3 - 4x^2 + 14x - 20 = 0.