Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 5 \cr & \therefore \frac{{2x}}{{3{x^2} - 5x + 3}}\,\left( {{\text{Divide by }}x} \right) \cr & = \frac{{\frac{{2x}}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{5x}}{x} + \frac{3}{x}}} \cr & = \frac{2}{{3x + \frac{3}{x} - 5}} \cr & = \frac{2}{{3\left( {x + \frac{1}{x}} \right) - 5}} \cr & = \frac{2}{{3 \times 5 - 5}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 3x - 1 = 0 \cr & x\left( {x - 3 - \frac{1}{x}} \right) = 0 \cr & x - \frac{1}{x} = 3 \cr & {x^2} + \frac{1}{{{x^2}}} = 11 \cr & \frac{{\left( {{x^2} + 8x - 1} \right)}}{{{x^3} + \frac{1}{x}}} \cr & = \frac{{x\left( {\frac{{x - 1}}{{x + 8}}} \right)}}{{x\left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)}} \cr & = \frac{{3 + 8}}{{11}} \cr & = 1 \cr} $$

Solution:

 ab(a - b) + bc(b - c) + ca(c - a) Let c = 0 ab(a - b) Now from option C (b - a)(b - c)(c - a) = -(a-b)b(-a) = ab(a - b)

Solution:

 x3 + 27y3 + 64z3 = 36xyz ......(i) since ⇒ a + b + c = 0 If a3 + b3 + c3 = 3abc similarly from equation (i) x + 3y + 4z = 0

Solution:

 $$\eqalign{ & \because \,{x^2} + \frac{1}{{25{x^2}}} = \frac{8}{5} \cr & \Rightarrow {x^2} + \frac{1}{{25{x^2}}} + 2x \times \frac{1}{{5x}} = \frac{8}{5} + \frac{2}{5} \cr & \Rightarrow {\left( {x + \frac{1}{{5x}}} \right)^2} = 2 \cr & \Rightarrow x + \frac{1}{{5x}} = \sqrt 2 \cr & {\text{On cubing both sides}} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} + 3 \times \frac{1}{5}\left( {\sqrt 2 } \right) = {\left( {\sqrt 2 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} = 2\sqrt 2 - \frac{{3\sqrt 2 }}{5} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} = \frac{{7\sqrt 2 }}{5} \cr} $$

Solution:

  $${\text{Given}}\,x = \root 3 \of {a + \sqrt {{a^2} + {b^3}} } \,\,+ $$      $$\root 3 \of {a - \sqrt {{a^2} + {b^3}} } $$ $$\eqalign{ & {\text{Let}}\,z = \sqrt {{a^2} + {b^3}} \cr & \therefore x = \root 3 \of {a + z} + \root 3 \of {a - z} \cr & {\text{Cubing both side}} \cr} $$   $$\therefore {x^3} = {\left( {\root 3 \of {a + z} } \right)^3} + {\left( {\root 3 \of {a - z} } \right)^3} \, + $$      $$\,3{\left( {a + z} \right)^{\frac{1}{3}}}$$  $${\left( {a - z} \right)^{\frac{1}{3}}} \times\, $$   $$\left( {\root 3 \of {a + z} + \root 3 \of {a - z} } \right)$$  $$ \,\Rightarrow {x^3} = a + z + a - z \,+\, $$    $$3{\left( {{a^2} + az - az - {z^2}} \right)^{\frac{1}{3}}}$$    $$ \times \left( x \right)$$ $$\eqalign{ & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {z^2}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & {\text{Put the value of }}z \cr & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {a^2} - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a + 3{\left( { - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a - 3bx \cr & \therefore {x^3} + 3bx = 2a \cr} $$

Solution:

 $$\eqalign{ & x - \frac{1}{x} = 1{\text{ }} \cr & \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr & {\text{Divide and multiply by }}x \cr & \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr & \Rightarrow \frac{4}{8} \cr & \Rightarrow \frac{1}{2} \cr} $$

Solution:

 $$\eqalign{ & {\text{ }}n + \frac{2}{3}n + \frac{1}{2}n + \frac{1}{7}n = 97 \cr & \Rightarrow \frac{{42n + 28n + 21n + 6n}}{{42}} = 97 \cr & \Rightarrow \frac{{97n}}{{42}} = 97 \cr & \Rightarrow n = 42 \cr} $$

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:

 $$\eqalign{ & {\text{For minimum value of}} \cr & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \cr & a = b = c \cr & a + b + c = 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore a = b = c = \frac{1}{3} \cr & \frac{1}{a} = \frac{1}{b} = \frac{1}{c} = 3 \cr & \therefore {\text{Minimum value of,}} \cr & = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \cr & = 3 + 3 + 3 \cr & = 9 \cr} $$