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

If both the distinct roots of the equation $\displaystyle \left | \sin x \right |^{2}+\left | \sin x \right |+b=0$ in $\displaystyle \left [ 0,\pi \right ]$ are real, then the value of $b$ is
  1. $\displaystyle \left [ -2,0 \right ]$
  2. $\displaystyle \left ( -2,0 \right )$
  3. $\displaystyle \left [ -2,0 \right )$

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

Let t = |sin x|. For x in [0, pi], we have t = sin x in [0, 1]. The equation becomes t^2 + t + b = 0. For each value of t in (0, 1), there are exactly two distinct values of x in [0, pi] satisfying sin x = t. Thus, for the original equation to have exactly two distinct real roots in [0, pi], the quadratic in t must have exactly one root in the interval (0, 1). Since the vertex of the parabola f(t) = t^2 + t + b is at t = -1/2, the function is strictly increasing on [0, 1], meaning we must have f(0) < 0 and f(1) > 0, which simplifies to -2 < b < 0.

Multiple choice

The equation $(cos p -1)x^2 + (cos p) x + sin p = 0$ in the variable x, has real roots. Then p can take any value in the interval
  1. $(0, 2 \pi)$
  2. $(-\pi, 0)$
  3. $(-\pi/2, \pi/2)$
  4. $(0, \pi)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For real roots, the discriminant D = b^2 - 4ac must be >= 0. Here, (cos p)^2 - 4(cos p - 1)(sin p) >= 0. Testing p in (0, pi) shows that for values like pi/2, the expression simplifies to 0, and the inequality holds for the specified range.

Multiple choice

In a triangle PQR, $\angle R$ = $\displaystyle \frac{\pi }{2}$. If tan $\displaystyle \frac{p}{2}$ and tan $\displaystyle \frac{Q }{2}$ are the roots of the equation $ax^{2}+bx+c = 0 (a\neq 0)$ then
  1. $a+b=c$
  2. $b+c=a$
  3. $a+c=b$
  4. $b=c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a triangle, P+Q+R = 180. Since R=90, P+Q=90, so P/2 + Q/2 = 45. tan(P/2 + Q/2) = tan(45) = 1. Using the formula tan(A+B) = (tan A + tan B) / (1 - tan A tan B), we get (sum of roots) / (1 - product of roots) = 1. Sum = -b/a, Product = c/a. (-b/a) / (1 - c/a) = 1. -b / (a-c) = 1. -b = a - c, so a + b = c.

Multiple choice

If $\tan\alpha, \tan\beta$ are the roots of the equation $x^{2}+px+q=0(p\neq 0)$, then
  1. $\displaystyle \tan(\alpha+\beta)=\frac{1-q}{p}$
  2. $\displaystyle \tan(\alpha+\beta)=\frac{-p}{q-1}$
  3. $\displaystyle \tan(\alpha+\beta)=\frac{p}{q-1}$
  4. $\cot(\alpha+\beta)=p$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If tan(alpha) and tan(beta) are roots, tan(alpha) + tan(beta) = -p and tan(alpha) * tan(beta) = q. Using the identity tan(alpha + beta) = (tan(alpha) + tan(beta)) / (1 - tan(alpha) * tan(beta)), we get (-p) / (1 - q) = p / (q - 1).

Multiple choice

The least positive root of the equation $\cos 3x+\sin 5x=0$ is,
  1. $\dfrac{3 \pi}{16}$
  2. $\dfrac{ \pi}{16}$
  3. $\dfrac{7 \pi}{16}$
  4. $\dfrac{9 \pi}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can rewrite the equation as sin(5x) = -cos(3x) = sin(3x - pi/2). This leads to two sets of general solutions, one of which is 5x = (2k + 1)pi - (3x - pi/2), which simplifies to 8x = (2k + 1)pi + pi/2. Setting k = 0 gives the smallest positive root of x = 3pi/16.

Multiple choice

If $\sin x+ \mathrm{cosec} x+ \tan y+ \cot y=4$, where x and y $\epsilon [0, \dfrac {\pi}{2}]$, then $\tan\dfrac {y}{2}$ is a root of the equation-
  1. $\alpha^2+2\alpha+1=0$
  2. $\alpha^2+2\alpha-1=0$
  3. $2\alpha^2-2\alpha-1=0$
  4. $\alpha^2-\alpha -1=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the AM-GM inequality, sin x + cosec x >= 2 and tan y + cot y >= 2. Since the sum is 4, both must equal 2, implying sin x = 1 and tan y = 1. If tan y = 1, then y = pi/4, so tan(y/2) = tan(pi/8) = sqrt(2) - 1. This value satisfies the equation alpha^2 + 2*alpha - 1 = 0.