Quantitative Aptitude

Algebra

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Algebra Questions

Multiple choice

If $\alpha $ and $\beta $ are the two zeros of the polynomial ${ 25p }^{ 2 }-15p+2,$ find a quadratic polynomial whose zeros are $\dfrac { 1 }{ 2\alpha }$ and $\dfrac { 1 }{ 2\beta } .$
  1. $\dfrac { 1 }{ 8 } \left( { 6p }^{ 2 }+25p-30 \right) $
  2. $\dfrac { 1 }{ 8 } \left( { 6p }^{ 2 }-30p+25 \right) $
  3. $\dfrac { 1 }{ 8 } \left( { 8p }^{ 2 }+30p-25 \right) $
  4. $\dfrac { 1 }{ 8 } \left( { 8p }^{ 2 }-30p+25 \right) $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For 25p^2 - 15p + 2 = 0, the sum of zeros alpha + beta = 15/25 = 3/5 and product alpha*beta = 2/25. The new zeros are 1/(2alpha) and 1/(2beta). Their sum is (alpha+beta)/(2*alpha*beta) = (3/5) / (4/25) = 15/4, and their product is 1/(4*alpha*beta) = 1/(8/25) = 25/8. The polynomial is x^2 - (15/4)x + 25/8, which scales to (1/8)(8x^2 - 30x + 25).

Multiple choice

The equation $2\sin^3 \theta + (2K - 3)\sin^2 \theta - (3K + 2) \sin \theta - 2K = 0$ has exactly three roots in $(0, 2\pi)$ then $K$ can be equal to
  1. $0$
  2. $\dfrac{1}{2}$
  3. $1$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let f(s) = 2s^3 + (2K-3)s^2 - (3K+2)s - 2K = 0, where s = sin(theta). Factoring gives (s-2)(2s^2 + (2K+1)s + K) = 0. Since s=2 is impossible, we solve 2s^2 + (2K+1)s + K = 0, which factors to (2s+1)(s+K) = 0. The roots are s = -1/2 and s = -K. For exactly three roots in (0, 2pi), s = -1/2 provides two roots (7pi/6 and 11pi/6), so s = -K must provide exactly one root in (-1, 1), which happens if -K = 1, -1, or 0 (if 0, it gives two roots, so K=0 is excluded). Testing K=0 gives s=0, which yields two roots in (0, 2pi), totaling four roots.