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

Match the following List I List II A. If $\alpha,\beta$ are the two roots of the equation$\displaystyle \dfrac{1-8(\log_{10}x)^2}{\log_{10}x-2(\log_{10}x^2)} =1$ then the value of $\displaystyle \frac{(\beta^3+1)}{\alpha^2\beta^3}$ is equal to 1. 5 B. Number of solution (s) of the equation $\log_2(x^2+3)=\dfrac{1}{2}\log_{1/3}\left(x+\dfrac{1}{x}\right),x >0$ is 2. 11 C. Integers satisfying the equation $|x|+\left|\dfrac{4-x^2}{x}\right|=\left|\dfrac{4}{x}\right|$ is\are 3. 0 D. If $\log_c2 \log_b 125=\log_{c}8$ where $c>0,c \neq1,b >1, b\neq$ 1, then $b$ is 4. 1
  1. $A- 2, B- 3, C- 4, D- 1$
  2. $A- 3, B- 2, C- 4, D- 1$
  3. $A- 1, B- 3, C- 4, D- 2$
  4. $A- 2, B- 3, C- 1, D- 4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Let $\displaystyle f\left ( x \right )=\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )\left ( x-6 \right )$ then
  1. $\displaystyle f'\left ( x \right )=0$ has five real roots
  2. four roots of $\displaystyle f'\left ( x \right )=0$ lie in $\displaystyle \left ( 2, 3 \right )\cup \left ( 3, 4 \right )\cup \left ( 4, 5 \right )\cup \left ( 5, 6 \right )$
  3. the equation $\displaystyle f\left ( x \right )$ has only three roots
  4. four roots of $\displaystyle f\left ( x \right )= 0$ lie in $\displaystyle \left ( 1, 2 \right )\cup \left ( 2, 3 \right )\cup \left ( 3, 4 \right )\cup \left ( 4, 5 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Rolle's Theorem, if f(x) has roots at 2, 3, 4, 5, 6, then f'(x) must have a root in each interval (2,3), (3,4), (4,5), and (5,6). This accounts for all 4 roots of the derivative of a degree 5 polynomial.

Multiple choice

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

x^2+2x+5=0. alpha+beta = -2, alpha*beta = 5. alpha^2+beta^2 = (alpha+beta)^2 - 2*alpha*beta = 4 - 10 = -6. a = -2i + 5j. b = 5i - 2j - 6k. a x b = (-2i + 5j + 0k) x (5i - 2j - 6k). Determinant: i(5*-6 - 0) - j(-2*-6 - 0) + k(-2*-2 - 5*5) = -30i - 12j - 21k.

Multiple choice

If the quadratic polynomial, $\displaystyle y=(\cot \alpha )x^{2}+2\left ( \sqrt{\sin \alpha } \right )x +\frac{1}{2}\tan \alpha, \ \ \alpha \, \varepsilon \, [0,2\pi ],$ can take negative values for all $\displaystyle x \, \varepsilon \, R,$ then the value of $\displaystyle \alpha $ must in the interval
  1. $\displaystyle  \left ( \frac{5 \pi}{6},\pi \right )$
  2. $\displaystyle \left ( \frac{5 \pi}{6}, \pi \right )\cup \left ( \frac{11 \pi}{6},2 \pi\ \right )$
  3. $\displaystyle \left ( \frac{11\pi}{6},2 \pi \right )$
  4. $\displaystyle \left ( 0,\frac{\pi}{6} \right )\cup \left ( \frac{5 \pi}{6}, \pi \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadratic to be negative for all x, the coefficient of x^2 (cot alpha) must be negative and the discriminant must be negative. cot alpha < 0 implies alpha is in (pi/2, pi) or (3pi/2, 2pi). Solving the discriminant condition leads to the interval (5pi/6, pi).

Multiple choice

If $(\alpha + \sqrt {\beta})$ and $(\alpha - \sqrt {\beta})$ are the roots of the equation $x^{2} + px + q = 0$, where $\alpha, \beta, p$ and $q$ are real, then the roots of the equation $(p^{2} - 4q) (p^{2} x^{2} + 4px) - 16q = 0$ are
  1. $\left (\dfrac {1}{\alpha} + \dfrac {1}{\sqrt {\beta}}\right )$ and $\left (\dfrac {1}{\alpha} - \dfrac {1}{\sqrt {\beta}}\right )$
  2. $\left (\dfrac {1}{\sqrt {\alpha}} + \dfrac {1}{\beta}\right )$ and $\left (\dfrac {1}{\sqrt {\alpha}} - \dfrac {1}{\beta}\right )$
  3. $\left (\dfrac {1}{\sqrt {\alpha}} + \dfrac {1}{\sqrt {\beta}}\right )$ and $\left (\dfrac {1}{\sqrt {\alpha}} - \dfrac {1}{\sqrt {\beta}}\right )$
  4. $(\sqrt {\alpha} + \sqrt {\beta})$ and $(\sqrt {\alpha} - \sqrt {\beta})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of x^2 + px + q = 0 are alpha +/- sqrt(beta). Thus, sum of roots = -p = 2*alpha and product = q = alpha^2 - beta. Substituting these into the second equation and solving for x reveals the roots are 1/(alpha +/- sqrt(beta)).

Multiple choice

Let common root condition of equations $2a{x^3} + b{x^2} + cx + d = 0$ and $2a{x^2} + 3bx + 4c = 0$ be ${(\lambda bc + ad)^2} = {9 \over 2}(\mu bd + 4{c^2})(m{b^2} - nac),\lambda ,\mu ,m,n \in N$. If the equation ${\left( {{{mx} \over {1 + {x^2}}}} \right)^2} + k\left( {{{nx} \over {1 + {x^2}}}} \right) + \lambda - \mu = 0$ has exactly two real roots which are distinct, then the set of possible real value of $k$ is
  1. $\left( { - \infty ,{{ - 13} \over 2}} \right)$
  2. $\left( {{{ - 13} \over 2},0} \right)$
  3. $\left( {{{ - 13} \over 2},{{13} \over 2}} \right)$
  4. $\left( {{{13} \over 2},\infty } \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

if $\alpha$, $\beta$ are the roots of the equation $ { x }^{ 2 }-px+q=0$ , then find the quadratic equation with the roots $ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) \left( { \alpha }^{ 3 }-{ \beta }^{ 3 } \right) $ and $ { \alpha }^{ 3 }{ \beta }^{ 2 }+{ \alpha }^{ 2 }{ \beta }^{ 3 }$
  1. $ { y }^{ 2 }+p\left\{ { q }^{ 2 }+\left( { p }^{ 2 }-4q \right) \left( { p }^{ 2 }-q \right) \right\} y+{ p }^{ 2 }{ q }^{ 2 }\left( { p }^{ 2 }-4q \right) \left( { p }^{ 2 }-q \right) =0$
  2. $ { y }^{ 2 }+p\left\{ { q }^{ 2 }+\left( { p }^{ 2 }-4pq \right) \left( { p }^{ 2 }-q \right) \right\} y+{ p }^{ 2 }{ q }^{ 2 }\left( { p }^{ 2 }-4q \right) \left( { p }^{ 2 }-q \right) =0$
  3. $ { y }^{ 2 }+p\left\{ { q }^{ 2 }+\left( { p }^{ 2 }-4q \right) \left( { p }^{ 2 }-q \right) \right\} y+4{ p }^{ 2 }{ q }^{ 2 }\left( { p }^{ 2 }-4q \right) \left( { p }^{ 2 }-q \right) =0$
  4. $ { y }^{ 2 }+p\left\{ { q }^{ 2 }+\left( { p }^{ 2 }-4q^2 \right) \left( { p }^{ 2 }-q \right) \right\} y+{ p }^{ 2 }{ q }^{ 2 }\left( { p }^{ 2 }-4q \right) \left( { p }^{ 2 }-q \right) =0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Let $a, b, c, p, q$ be real numbers. Suppose $\displaystyle \alpha , \beta$ are the roots of the equation $\displaystyle x^{2}+2px+q= 0$ and $\displaystyle \alpha , \frac{1}{\beta }$ are the roots of the equation $\displaystyle ax^{2}+2bx+c= 0$, where $\displaystyle \beta ^{2}\notin\left { -1, 0, 1 \right }.$ $\displaystyle \left ( p^{2}-q \right )\left ( b^{2}-ac \right )\geq 0$

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Let $\mathrm{a},\ \mathrm{b},\ \mathrm{c},\ \mathrm{p},\ \mathrm{q}$ be real numbers. Suppose $\alpha,\ \beta$ are the roots of the equation $\mathrm{x}^{2}+2\mathrm{p}\mathrm{x}+\mathrm{q}=0$ and $\alpha,\ \displaystyle \dfrac{1}{\beta}$ are the roots of the equation $\mathrm{a}\mathrm{x}^{2}+2\mathrm{b}\mathrm{x}+\mathrm{c}=0$, where $\beta^{2}\not\in{-1,0,1}$ . STATEMENT 1 : $(\mathrm{p}^{2}-\mathrm{q})(\mathrm{b}^{2}- ac )\geq 0$ STATEMENT 2: $\mathrm{b}\notin pa$ or $\mathrm{c}\notin qa$.

  1. STATEMENT 1 is True, STATEMENT2 is True; STATEMENT 2 is a correct explanation for STATEMENT 1.

  2. STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is NOT a correct explanation for STATEMENT 1.

  3. STATEMENT 1 is True, STATEMENT 2 is False.

  4. STATEMENT 1 is False, STATEMENT 2 is True.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The roots of the first equation are alpha and beta, so alpha*beta = q and alpha+beta = -2p. The roots of the second are alpha and 1/beta, so their product is alpha/beta = c/a and sum is alpha + 1/beta = -2b/a. Both statements are mathematically true, but Statement 2 does not logically derive or explain the inequality in Statement 1.

Multiple choice

If $\alpha, \beta$ are roots of $ax^{2} + bx + c = 0$ and $ \alpha^{1} ,\beta^{1}$ are roots if $a^{1}x^{2}+b^{1}x+c^{1}=0$. Then the equation of circle having $A(\alpha ,\alpha ^{1})$ and $B(\beta ,\beta ^{1})$ as diameter is
  1. $\mathrm{b}\mathrm{b}^{1}(\mathrm{x}^{2}+\mathrm{y}^{2})+\mathrm{a}^{1}\mathrm{b}\mathrm{x} +\mathrm{a}\mathrm{b}^{1}\mathrm{y} +\mathrm{a}^{1}\mathrm{c}+\mathrm{a}\mathrm{c}^{1}=0$
  2. $\mathrm{a}\mathrm{a}^{1}(\mathrm{x}^{2}+\mathrm{y}^{2})+\mathrm{a}^{1}\mathrm{b}\mathrm{x} +\mathrm{a}\mathrm{b}^{1}\mathrm{y} +\mathrm{a}^{1}\mathrm{c}+\mathrm{a}\mathrm{c}^{1}=0$
  3. $\mathrm{c}\mathrm{c}^{1}(\mathrm{x}^{2}+\mathrm{y}^{2})+\mathrm{a}^{1}\mathrm{c}\mathrm{x} +\mathrm{a}\mathrm{c}^{1}\mathrm{x} +\mathrm{a}^{1}\mathrm{b}+\mathrm{a}\mathrm{b}^{1}=0$
  4. $\mathrm{c}\mathrm{c}^{1}(\mathrm{x}^{2}+\mathrm{y}^{2})-\mathrm{a}^{1}\mathrm{c}\mathrm{x} +\mathrm{a}\mathrm{c}^{1}\mathrm{x} -\mathrm{a}^{1}\mathrm{b}+\mathrm{a}\mathrm{b}^{1}=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The circle with diameter endpoints (x1, y1) and (x2, y2) is (x-x1)(x-x2) + (y-y1)(y-y2) = 0. Expanding this with roots alpha, beta and alpha', beta' leads to the equation involving coefficients a, b, c and a', b', c'.

Multiple choice

Statement-1: If the equation ${ ax }^{ 2 }+bx+c=0\left( a,b,c\epsilon Randa\neq 0 \right) $ and ${ 2x }^{ 2 }+7x+10=0$ have a common root, then $\dfrac { 2a+c }{ b } =2.$ Statement-2: If both roots of ${ a }{ 1 }{ x }^{ 2 }+{ b }{ 1 }x+{ c }{ 1 }=0$ and ${ a }{ 2 }{ x }^{ 2 }+{ b }{ 2 }x+{ c }{ 2 }=0$ are same,then $\dfrac { { a }{ 1 } }{ { a }{ 2 } } =\dfrac { { b }{ 1 } }{ { b }{ 2 } } =\dfrac { { c }{ 1 } }{ { c }{ 2 } } .$ Given ${ a }{ 1 },{ b }{ 1 },{ c }{ 1, }{ a }{ 2 },{ b }{ 2, }{ c }{ 2 }\epsilon R$ and ${ a }{ 1 }{ a }{ 2 }\neq 0.$

  1. Statement-1 is true, statement-2 is true and statement -2 is correct explanation for statement-1.

  2. Statement-1 is true, statement-2 is true and statement -2 is NOT the correct explanation for statement-1

  3. Statement-1 is true, statement-2 is false.

  4. Statement-1 is false, statement-2 is true.
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Find the roots of the equation $\displaystyle z^{10}-z^{5}-992= 0,$ whose real part is negative.
  1. $\displaystyle \sqrt[5]{31}\left ( \cos 108^{\circ}+i\sin 108^{\circ} \right ), -\sqrt[5]{31}, \sqrt[5]{31}\left ( \cos 252^{\circ}+i\sin 252^{\circ} \right ).$
  2. $\displaystyle \sqrt[5]{21}\left ( \cos 108^{\circ}+i\sin 108^{\circ} \right ), -\sqrt[5]{21}, \sqrt[5]{21}\left ( \cos 252^{\circ}+i\sin 252^{\circ} \right ).$
  3. $\displaystyle \sqrt[5]{31}\left ( \cos 108^{\circ}-i\sin 108^{\circ} \right ), -\sqrt[5]{31}, \sqrt[5]{31}\left ( \cos 252^{\circ}-i\sin 252^{\circ} \right ).$
  4. $\displaystyle \sqrt[5]{31}\left ( \cos 108^{\circ}+i\sin 108^{\circ} \right ), \sqrt[5]{31}, \sqrt[5]{31}\left ( \cos 252^{\circ}+i\sin 252^{\circ} \right ).$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let u = z^5. u^2 - u - 992 = 0. Roots are u = (1 +/- sqrt(1 + 3968))/2 = (1 +/- 63)/2. u = 32 or u = -31. z^5 = 32 gives z = 2, 2w, 2w^2, 2w^3, 2w^4. z^5 = -31 gives z = -31^(1/5) * (cos(180+360k)/5 + i sin(180+360k)/5). The roots with negative real parts correspond to the second case.

Multiple choice

The quadratic polynomials defined on real coefficients $\displaystyle P\left( x \right) ={ a }{ 1 }{ x }^{ 2 }+2{ b }{ 1 }x+{ c }{ 1 }$ $\displaystyle Q\left( x \right) ={ a }{ 12 }{ x }^{ 2 }+2{ b }{ 2 }x+{ c }{ 2 }$ where $\displaystyle { a }{ 1 }\neq 0,\quad { a }{ 2 }\neq 0$ and $\displaystyle P\left( x \right) $ and $\displaystyle Q\left( x \right) $ both take positive values $\displaystyle \forall \times \in Rg\left( x \right) ={ a }{ 1 }{ a }{ 2 }{ x }^{ 2 }+{ b }{ 1 }{ b }{ 2 }x+{ c }{ 1 }{ c }{ 2 }$ then
  1. $g(x)$ takes positive value only
  2. $g(x)$ takes negative value only
  3. $g(x)$ takes both positive and negative value
  4. nothing can be said about $g(x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For each polynomial to be positive for every real x, we have a1 > 0, a2 > 0, c1 > 0, c2 > 0, and b1^2 < a1c1, b2^2 < a2c2. Therefore, (b1b2)^2 < (a1a2)(c1c2), so the quadratic g(x) has a positive minimum and is positive for every real x.

Multiple choice

If a, b, c are the sides of the triangle ABC and $a^2, b^2, c^2$ are the roots of the equation $x^3-px^2+qx - \lambda =0$, then match the entries of col. I with those of col. II. Column-I Column-II (a) $\dfrac{cos A}{a} + \dfrac{cos B}{b} + \dfrac{cos C}{c} = \dfrac{1}{2}$ (p) $p=2, q=1, \lambda=4$ (b) $a cos A + b cos B + c cos C =0$ (q) $p=1, q=1/4, \lambda=1$ (c) $sin 2A + sin 2B + sin 2C =0$ (r) $p=3, q=9/4, \lambda=9$ (d) $a sin A + b sin B + c sin C = 2 \Delta$ (s) $p=4, q=4, \lambda=2$

  1. (a), (d)-(p, q, r); (b), (c)-(p, q, r, s)

  2. (a), (d)-(p, r, s); (b), (c)-(p, q, r, s)

  3. (a), (d)-(p, q, r); (b), (c)-(p, q, , s)

  4. (a), (d)-(p, q, r); (b), (c)-(p, s)
Reveal answer Fill a bubble to check yourself
A Correct answer