Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & \frac{{{{\left( {{a^2} - 4{b^2}} \right)}^3} + 64{{\left( {{b^2} - 4{c^2}} \right)}^3} + {{\left( {16{c^2} - {a^2}} \right)}^3}}}{{{{\left( {a - 2b} \right)}^3} + {{\left( {2b - 4c} \right)}^3} + {{\left( {4c - a} \right)}^3}}} \cr & {\text{Put }}a = b = c \cr & = \frac{{{{\left( { - 3{a^2}} \right)}^3} + 64{{\left( { - 3{a^2}} \right)}^3} + {{\left( {15{a^2}} \right)}^3}}}{{{{\left( { - a} \right)}^3} + {{\left( { - 2a} \right)}^3} + {{\left( {3a} \right)}^3}}} \cr & = \frac{{{a^6}\left[ { - 27 - 27 \times 64 + {{\left( {15} \right)}^3}} \right]}}{{{a^3}\left[ { - 1 - 8 + 27} \right]}} \cr & = \frac{{3{a^3}\left[ { - 9 - 576 + 1125} \right]}}{{18}} \cr & = \frac{{{a^3} \times 540}}{6} \cr & = 90{a^3} \cr & {\text{From option put }}a = b = c \cr & \left( {\text{A}} \right)\, - 45{a^3} \cr & \left( {\text{B}} \right)\,90{a^3} \cr & {\text{Hence option B is right answer}}{\text{.}} \cr} $$

Solution:

 $$\eqalign{ & 3{x^2} - 9x + 3 = 0 \cr & 3x\left( {x - 3 + \frac{1}{x}} \right) = 0,\,x + \frac{1}{x} = 3 \cr & \therefore \,{\left( {x + \frac{1}{x}} \right)^3} = {\left( 3 \right)^3} = 27 \cr} $$

Solution:

 a + b = 5, ab = 3 (a3 + b3) = (a + b)[(a + b)2 - 3ab] = 5[52 - 3 × 3] = 5(25 - 9) = 80

Solution:

 $$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}{\text{ }} \cr & = \frac{{{x^6}}}{{{x^3}}} + \frac{{{x^4}}}{{{x^3}}} + \frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}} \cr & = {x^3} + \frac{1}{{{x^3}}} + \frac{1}{x} + x \cr & = 18 + 3 \cr & = 21 \cr} $$

Solution:

 $$\eqalign{ & 2x + 3y = k\left( {2 = x,{\text{ }}0 = y} \right) \cr & \therefore 2 \times 2 + 3 \times 0 = k \cr & \Leftrightarrow k = 4 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let }}\sqrt {\frac{a}{b}} = x \cr & \therefore \,x - 13 = \frac{{ - 1}}{x} - 11 \cr & x + \frac{1}{x} = 2 \cr & \therefore \,x = 1 \cr & \sqrt {\frac{a}{b}} = 1{\text{ and }}a + b = 10 \cr & \therefore \,a = b = 5 \cr & 3ab + 4{a^2} + 5{b^2} \cr & = 3{a^2} + 4{a^2} + 5{a^2} \cr & = 12{a^2} \cr & = 12 \times 25 \cr & = 300 \cr} $$

Solution:

 $$\eqalign{ & \root 3 \of a + \root 3 \of b = \root 3 \of c \cr & {\text{Take cube both sides}} \cr & {\left( {\root 3 \of a + \root 3 \of b } \right)^3} = {\left( {\root 3 \of c } \right)^3} \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {\root 3 \of a + \root 3 \of b } \right) = c \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} = c \cr & \Rightarrow a + b - c = - 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again take cube both sides}} \cr & \Rightarrow {\left( {a + b - c} \right)^3} = - 27abc \cr & \Rightarrow {\left( {a + b - c} \right)^3} + 27abc = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }} \cr & a = 0 \cr & b = 1 \cr & c = 1 \cr & {\text{Value of }}{\left( {a + b - c} \right)^3} + 27abc \cr & = {\left( {0 + 1 - 1} \right)^3} + 27 \times 0 \times 1 \times 1 \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & {x^2} = y + z \cr & {y^2} = z + x \cr & {z^2} = x + y \cr & \Rightarrow {x^2} + x = x + y + z \cr & {\text{Adding }}x{\text{ on both sides }} \cr & x\left( {x + 1} \right) = x + y + z \cr & \frac{1}{{\left( {x + 1} \right)}} = \frac{x}{{x + y + z}} \cr & {\text{Similarly,}} \cr & \frac{1}{{\left( {y + 1} \right)}} = \frac{y}{{x + y + z}} \cr & \frac{1}{{\left( {z + 1} \right)}} = \frac{z}{{x + y + z}} \cr & \therefore \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {y + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}} \cr & = \frac{x}{{x + y + z}} + \frac{y}{{x + y + z}} + \frac{z}{{x + y + z}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & {\text{Find, }}{x^3} + {y^3} + {z^3} = ? \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 2 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 1 - 1 \cr & \Rightarrow \left( {{x^2} + 1 - 2x} \right) + {y^2} + \left( {{z^2} + 1 - 2z} \right) = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} + {y^2} + {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow {\left( {x - 1} \right)^2} = 0 \cr & \Rightarrow x = 1 \cr & \Rightarrow {y^2} = 0 \cr & \Rightarrow y = 0 \cr & \Rightarrow {\left( {z - 1} \right)^2} = 0 \cr & \Rightarrow z = 1 \cr & {\text{Value substituted in question,}} \cr & \Rightarrow {x^3} + {y^3} + {z^3} \cr & \Rightarrow {1^3} + 0 + {1^3} \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & {a^2} + {b^2} + {c^2} = 2\left( {a - b - c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 2a - 2b - 2c - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} - 2a + 2b + 2c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} + 2b + 1 + {c^2} + 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c + 1} \right)^2} = 0 \cr & {\left( {a - 1} \right)^2} = 0{\text{ }}{\left( {b + 1} \right)^2} = 0{\text{ }}{\left( {c + 1} \right)^2} = 0 \cr & \Rightarrow a - 1 = 0{\text{ }} \Rightarrow b + 1 = 0{\text{ }} \Rightarrow c + 1 = 0 \cr & \Rightarrow a = 1{\text{ }} \Rightarrow b = - 1{\text{ }} \Rightarrow c = - 1 \cr & \therefore a + b + c \cr & = 1 - 1 - 1 \cr & = - 1 \cr} $$