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

Let a, b, c be non-zero real numbers such the : $\displaystyle \int_{0}^{1}\left ( 1+\cos ^{8}x \right )\left ( ax^{2}+bx+c \right )dx=\int_{0}^{2}\left ( 1+\cos ^{8}x \right )\left ( ax^{2}+bx+c \right )dx,$ then the quadratic equation $\displaystyle ax^{2}+bx+c=0$ has
  1. no root in $\displaystyle \left ( 0,2 \right )$
  2. atleast one root in $\displaystyle \left ( 0,2 \right )$
  3. a double root in $\displaystyle \left ( 0,2 \right )$

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

Let f(x) = ax^2 + bx + c and g(x) = 1 + cos^8(x). The equation is integral from 0 to 1 of f(x)g(x) dx = integral from 0 to 2 of f(x)g(x) dx. This implies the integral from 1 to 2 of f(x)g(x) dx = 0. Since g(x) is always positive, f(x) must change sign in the interval (1, 2), which is within (0, 2). By the Intermediate Value Theorem, f(x) must have at least one root in (0, 2).

Multiple choice

Let $f(x)$ is a quadratic function such that $f(0) = 1 ,f(1)=7,f(-1)=1$ and $\displaystyle {\int \frac {f(x) dx}{x^2 (x\, +\, 1)^3}}$ is a rational function. Find the value of $f'(0)$.
  1. $3$
  2. $1$
  3. $2$
  4. $6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given f(0)=1, f(1)=7, f(-1)=1, let f(x) = ax^2 + bx + c. From f(0)=1, c=1. From f(1)=a+b+1=7, a+b=6. From f(-1)=a-b+1=1, a-b=0. Solving gives a=3, b=3. Thus f(x) = 3x^2 + 3x + 1. Then f'(x) = 6x + 3, so f'(0) = 3.

Multiple choice

Let $\displaystyle \alpha ,\beta $ be the distincet positive roots of the equation $\displaystyle \tan x=2x$ then evaluate $\displaystyle \int_{0}^{1}\left ( \sin \alpha x.\sin \beta x \right )dx$ independent of $\displaystyle \alpha ,\beta $.

  1. 0

  2. -2

  3. 1

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For each root, tan x = 2x implies sin x = 2x cos x. Thus sin(alpha - beta)/(alpha - beta) and sin(alpha + beta)/(alpha + beta) are both equal to 2 cos(alpha)cos(beta). Substitution into the product-to-sum integral makes the two terms cancel, giving 0.

Multiple choice

Let $g(x) = \cos x^{2}, f(x) = \sqrt {x}$, and $\alpha, \beta (\alpha < \beta)$ be the roots of the quadratic equation $18x^{2} - 9\pi x + \pi^{2} = 0$. Then the area (in $sq.\ units$) bounded by the curve $y = (gof)(x)$ and the lines $x = \alpha, x = \beta$ and $y = 0$, is
  1. $\dfrac {1}{2}(\sqrt {3} - \sqrt {2})$
  2. $\dfrac {1}{2}(\sqrt {2} - 1)$
  3. $\dfrac {1}{2}(\sqrt {3} - 1)$
  4. $\dfrac {1}{2}(\sqrt {3} + 1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots of 18x^2 - 9pi*x + pi^2 = 0 are (9pi +/- sqrt(81pi^2 - 72pi^2))/36 = (9pi +/- 3pi)/36. Alpha = 6pi/36 = pi/6, Beta = 12pi/36 = pi/3. Area = integral from pi/6 to pi/3 of cos(sqrt(x)^2) dx = integral of cos(x) dx = sin(x) from pi/6 to pi/3 = sin(pi/3) - sin(pi/6) = sqrt(3)/2 - 1/2 = (sqrt(3)-1)/2.

Multiple choice

If $\alpha,\ \beta$ are roots of the equation $x^{2}+2x+5=0$ and $\vec{a}=(\alpha+\beta)\hat{i}+\alpha\beta\hat{j} \vec{b}=\alpha\beta\hat{i}+(\alpha+\beta)\hat{j}+(\alpha^{2}+\beta^{2})\hat{k}$ then $\vec{a}\times\vec{b}=$
  1. $\hat{i}+12\hat{j}+12\hat{k}$
  2. $-30\hat{i}+12\hat{j}-4\hat{k}$
  3. $-30\hat{i}-12\hat{j}-21\hat{k}$
  4. $\hat{i}-12\hat{j}+29\hat{k}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For x^2 + 2x + 5 = 0, alpha + beta = -2 and alpha * beta = 5. Then alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha * beta) = 4 - 10 = -6. Thus, a = -2i + 5j and b = 5i - 2j - 6k. The cross product a x b is (-2i + 5j + 0k) x (5i - 2j - 6k) = (-30i - 12j - 21k).

Multiple choice

If $(0, b, 0)$ is the centroid of the triangle formed by the points $(4, 2, -3)$ , $({a}, -5, 1)$ and $(2, -6, 2)$ . If $a ,b$ are the roots of the quadratic equation $ x^2+px+q = 0 $, then $p,q$ are
  1. $9,18$
  2. $-9,-18$
  3. $3,-18$
  4. $-3,18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Centroid = ((4+a+2)/3, (2-5-6)/3, (-3+1+2)/3) = ((6+a)/3, -3, 0). Given centroid is (0, b, 0), so (6+a)/3 = 0 -> a = -6. And b = -3. The quadratic equation is x^2 + px + q = 0 with roots -6 and -3. Sum of roots = -p = -9 -> p = 9. Product of roots = q = 18.

Multiple choice

If $\displaystyle p_1$ and $\displaystyle p_2$ are the lengths of the perpendiculars from the points $\displaystyle \left ( 2,3,4 \right )$ and $\displaystyle \left ( 1,1,4 \right )$ respectively from the plane $\displaystyle 3x-6y+2z+11=0$, then $\displaystyle p_1$, $\displaystyle p_2$ are the roots of the equation
  1. $\displaystyle p^2-23p+7=0$
  2. $\displaystyle 7p^2-23p+16=0$
  3. $\displaystyle p^2-17p+16=0$
  4. $\displaystyle p^2-16p+7=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Perpendicular distance p = |ax+by+cz+d| / sqrt(a^2+b^2+c^2). Plane: 3x-6y+2z+11=0. Denominator = sqrt(9+36+4) = 7. p1 = |3(2)-6(3)+2(4)+11| / 7 = |6-18+8+11| / 7 = 7/7 = 1. p2 = |3(1)-6(1)+2(4)+11| / 7 = |3-6+8+11| / 7 = 16/7. Roots are 1 and 16/7. Equation: (p-1)(p-16/7) = 0 => p^2 - (23/7)p + 16/7 = 0 => 7p^2 - 23p + 16 = 0.

Multiple choice

If $p$ is chosen at random in the closed interval $[0,5]$. The probability of the equation $\displaystyle { x }^{ 2 }+px+\frac { 1 }{ 4 } \left( p+2 \right) =0$ to have the real roots is
  1. $\displaystyle \frac { 3 }{ 4 } $
  2. $\displaystyle \frac { 3 }{ 5 } $
  3. $\displaystyle \frac { 2 }{ 5 } $

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

For the quadratic equation to have real roots, the discriminant D = p^2 - 4(1/4)(p+2) must be >= 0. This simplifies to p^2 - p - 2 >= 0, or (p-2)(p+1) >= 0. Given p is in [0, 5], the condition holds for p in [2, 5]. The probability is the length of this interval (5-2=3) divided by the total interval length (5-0=5), which is 3/5.

Multiple choice

If coefficients $a,b,c$ of quadratic equation $a{ x }^{ 2 }+bx+c=0$ are chosen at random with replacement from the set $S={1,2,3,4,5,6}$, find out the probability that roots of quadratic are real and distinct.
  1. $\cfrac{20}{108}$
  2. $\cfrac{19}{108}$
  3. $\cfrac{21}{108}$
  4. $\cfrac{22}{108}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the roots to be real and distinct, the discriminant D = b^2 - 4ac must be greater than 0. By testing all 216 possible combinations (6*6*6), there are 19 cases where b^2 > 4ac.

Multiple choice

Let $f(x)=x^{3}-12x$ be function such that the equation $|f(|x|)|=n(n\ \in\ N)$ has exactly $6$ distinct real roots then number of possible values of $n$ are :
  1. $15$
  2. $16$
  3. $17$
  4. $14$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

f(x) = x^3 - 12x. |f(|x|)| = |x^3 - 12x|. The graph of |x^3 - 12x| for x >= 0 has local max at x=2 (value 16) and local min at x=sqrt(12) (value 0). For 6 roots, n must be between 0 and 16. There are 15 such natural numbers (1 to 15).

Multiple choice

The number of roots of the equation $1+3^{x/2}=2^{ x}$ is
  1. $0$
  2. $1$
  3. $2$

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

Let f(x) = 1 + 3^(x/2) - 2^x. By testing values, f(0) = 1 + 1 - 1 = 1, f(2) = 1 + 3 - 4 = 0, and f(4) = 1 + 9 - 16 = -6. Since the function is continuous and changes sign, there is at least one root; checking the derivatives shows it is monotonic for x > 2, confirming exactly one root.