Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {a^2} + 4{b^2} + 4b - 4ab - 2a - 8 \cr & = {a^2} - 4ab + 4{b^2} - 2a + 4b - 8 \cr & = {\left( {a - 2b} \right)^2} - 2\left( {a - 2b} \right) - 8 \cr & {\text{Put }} t = a - 2b \cr & = {t^2} - 2t - 8 \cr & = {t^2} - 4t + 2t - 8 \cr & = t\left( {t - 4} \right) + 2\left( {t - 4} \right) \cr & = \left( {t + 2} \right)\left( {t - 4} \right) \cr & = \left( {a - 2b - 4} \right)\left( {a - 2b + 2} \right) \cr & \left( {{\text{Put the value of assume }}t} \right) \cr} $$

$$\frac{1}{{{a^2} + ax + {x^2}}}$$   $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$   $$ + $$ $$\frac{2ax}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$ = \frac{{{a^2} - ax + {x^2} - {a^2} - ax - {x^2}}}{{\left( {{a^2} + {x^2} + ax} \right)\left( {{a^2} + {x^2} - ax} \right)}} + $$       $$\frac{{2ax}}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$\eqalign{ & = \frac{{ - 2ax}}{{{{\left( {{a^2} + {x^2}} \right)}^2} - {{\left( {ax} \right)}^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = \frac{{ - 2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr & = 0 \cr} $$

$$\because $$ x = 11
x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 1
= x⁵ - 11x⁴ - x⁴ + 11x³ + x³ - 11x² - x² + 11x + x - 1
= 11⁵ - 11.11⁴ - 11⁴ + 11.11³ + 11³ - 11.11² - 11² + 11.11 + 11 - 1
= 0 - 0 + 0 + 0 + 11 - 1
= 10

$$\eqalign{ & {x^2} + {y^2} + {z^2} - xy - yz - zx \cr & = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr} $$
  $$ = \frac{1}{2}$$ $$\left[ {{{\left( {997 - 998} \right)}^2} + {{\left( {998 - 999} \right)}^2} + {{\left( {999 - 997} \right)}^2}} \right]$$
$$\eqalign{ & = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr & = 3 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr & = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr & = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr & = \frac{{9 - 4}}{2} \cr & = \frac{5}{2} \cr} $$

$$\eqalign{ & x = p + \frac{1}{p}{\text{ }} \cr & y = p - \frac{1}{p} \cr & \therefore x + y = p + \frac{1}{p} + p - \frac{1}{p} \cr & \Leftrightarrow x + y = 2p \cr & \therefore x - y = p + \frac{1}{p} - p + \frac{1}{p} \cr & \Leftrightarrow x - y = \frac{2}{p} \cr & \therefore {x^4} - 2{x^2}{y^2} + {y^4} \cr & = {x^4} + {y^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left[ {\left( {x + y} \right)\left( {x - y} \right)} \right]^2} \cr & = {\left( {2p \times \frac{2}{p}} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$

$${\left( {a + b - c} \right)^2}{\text{ + }}{\left( {b + c - a} \right)^2}$$     $${\text{ + }}{\left( {c + a - b} \right)^2}$$
$$\eqalign{ & \Rightarrow a + b + c = 0{\text{ }}\left( {{\text{ Given}}} \right) \cr & \Rightarrow a + b = - c \cr & \Rightarrow b + c = - a \cr & \Rightarrow a + c = - b \cr} $$
$$ \Rightarrow {\left( {a + b - c} \right)^2} + {\left( {b + c - a} \right)^2}$$      $$ + {\left( {c + a - b} \right)^2}$$
$$\eqalign{ & \Rightarrow {\left( { - c - c} \right)^2}{\text{ + }}{\left( { - a - a} \right)^2}{\text{ + }}{\left( { - b - b} \right)^2} \cr & \Rightarrow {\left( { - 2c} \right)^2}{\text{ + }}{\left( { - 2a} \right)^2}{\text{ + }}{\left( { - 2b} \right)^2} \cr & \Rightarrow 4{c^2} + 4{a^2} + 4{b^2} \cr & \Rightarrow 4\left( {{a^2} + {b^2} + {c^2}} \right) \cr} $$

$$\eqalign{ & x = 2015 \cr & y = 2014 \cr & z = 2013 \cr & \therefore {x^2} + {y^2} + {z^2} - xy - yz - zx \cr & = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{1}{2}\left[ {{{\left( {2015 - 2014} \right)}^2} + {{\left( {2014 - 2013} \right)}^2} + {{\left( {2013 - 2015} \right)}^2}} \right] \cr & = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr & = 3 \cr} $$

$$\eqalign{ & 3{a^2} = {b^2}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{ }}\frac{{{{\left( {a + b} \right)}^3} - {{\left( {a - b} \right)}^3}}}{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}} \cr} $$
  $$ = \frac{{{a^3} + {b^3} + 3ab\left( {a + b} \right)\, - \,\left( {{a^3} - {b^3} - 3ab\left( {a - b} \right)} \right){\text{ }}}}{{{a^2} + {b^2} + 2ab + {\text{ }}{a^2} + {b^2} - 2ab}}$$
$$\eqalign{ & = \frac{{2{b^3}{\text{ + 6}}{{\text{a}}^2}{\text{b }}}}{{2{a^2} + 2{b^2}{\text{ }}}} \cr & = \frac{{{b^3}{\text{ + 3}}{{\text{a}}^2}{\text{b }}}}{{{a^2} + {b^2}{\text{ }}}} \cr & = \frac{{{b^3} + {b^3}{\text{ }}}}{{\frac{{{b^2}}}{3} + {b^2}{\text{ }}}} \cr & = \frac{{2{b^3}}}{{{b^2}\left( {\frac{1}{3} + 1} \right)}} \cr & = \frac{{2b}}{{\frac{4}{3}}} \cr & = \frac{{3b}}{2} \cr} $$

$$\eqalign{ & x = 9999{\text{ }}\left( {{\text{ Given}}} \right) \cr & \frac{{4{x^3} - x}}{{\left( {2x + 1} \right)\left( {6x - 3} \right)}} \cr & = \frac{{x\left( {4{x^2} - 1} \right)}}{{3\left( {2x + 1} \right)\left( {2x - 1} \right)}} \cr & = \frac{{x\left( {4{x^2} - 1} \right)}}{{3\left( {4{x^2} - 1} \right)}} \cr & = \frac{x}{3} \cr & \therefore \frac{{9999}}{3} = 3333 \cr} $$