Quantitative Aptitude

Algebra

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Algebra Questions

Multiple choice

Consider the equation $|x^{2}-4|x|+3|=p$
  1. For $p=2$ the equation has four solutions
  2. For $p=2$ the equation has eight solutions
  3. There exists only one real value of $p$ for which the equation has odd number of solutions
  4. Sum of roots of the equation is zero irrespective of value of $p$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

The determinant $\left| {\begin{array}{*{20}{c}}a&b&{a\alpha + b}\b&c&{b\alpha + c}\{a\alpha + b}&{b\alpha + c}&0\end{array}} \right|$ is equal to zero, if-
  1. $a, b, c$ are in AP
  2. $a, b, c$ are in GP
  3. $\alpha $ is a root of the equation $a{x^2} + bx + c = 0$
  4. $\left( {x - \alpha } \right)$ is a factor of $a{x^2} + 2bx + c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The determinant represents the condition for the quadratic ax^2 + 2bx + c = 0 to have a root alpha. Expanding the determinant or using properties of determinants shows that it equals -(a*alpha^2 + 2b*alpha + c)^2. For this to be zero, a*alpha^2 + 2b*alpha + c must be zero, meaning (x-alpha) is a factor of ax^2 + 2bx + c.

Multiple choice

Let $a,b,c \in R$ such that no two of them are equal and satisfy $\left| {\begin{array}{*{20}{c}}{2a} & b & c\b & c & {2a}\c & {2a} & b\end{array}} \right| = 0$, then equation $24a{x^2} + 4bx + c = 0$ has
  1. at least one root $[0,1]$
  2. at least one root in $\left[ { - \frac{1}{2},\frac{1}{2}} \right]$
  3. at least one root $[ - 1.0]$
  4. at least two root $[ 0.2]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The determinant condition imposes a relationship among a, b, and c that forces the quadratic 24ax^2 + 4bx + c to change sign, or have a zero, within the interval [-1/2, 1/2]. Hence it must have at least one root in that interval. The other listed intervals are malformed or do not follow from the determinant condition.

Multiple choice

If ${ b }{ 1 }{ b }{ 2 }=2\left( { c }{ 1 }+{ c }{ 2 } \right) $ and ${ b }{ 1, }{ b }{ 2 },{ c }{ 1 },{ c }{ 2 }\quad $ are all real numbers, then atleast one of the equations ${ x }^{ 2 }+{ b }{ 1 }x+c_1=0$ and ${ x }^{ 2 }+{ b }{ 2 }x+c_2=0$ has

  1. real roots

  2. purely imaginary roots
  3. roots of the form $a+ib(a,b\in R,ab\ne 0$)

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

The sum of the discriminants of the two equations is D1 + D2 = (b1^2 - 4c1) + (b2^2 - 4c2) = b1^2 + b2^2 - 4(c1 + c2). Given b1*b2 = 2(c1 + c2), we substitute 2(c1 + c2) = b1*b2 into the expression: D1 + D2 = b1^2 + b2^2 - 2(b1*b2) = (b1 - b2)^2. Since (b1 - b2)^2 >= 0, at least one discriminant must be non-negative, ensuring at least one equation has real roots.