Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Given xepression ,}} \cr & \frac{{\left( {x + y - z} \right)\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)\left( {x + z - y} \right)}} + \frac{{\left( {y + x - z} \right)\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)\left( {x + y - z} \right)}} + \frac{{\left( {z + x - y} \right)\left( {z - x + y} \right)}}{{\left( {y + z + x} \right)\left( {y + z - x} \right)}} \cr & = \frac{{\left( {x + y - z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{\left( {x + y - z} \right) + \left( {y - x + z} \right) + \left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\,\,\,\, \cr & \Rightarrow aqr + bpr + cpq = 0....({\text{i}}) \cr & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\,\, \cr & \Rightarrow {\left( {\frac{p}{a} + \frac{q}{b} + \frac{r}{c}} \right)^2} = 1 \cr & \Rightarrow {\text{ }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\, + 2\left( {\frac{{pq}}{{ab}} + \frac{{pr}}{{ac}} + \frac{{qr}}{{bc}}} \right) = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} + \frac{{2\left( {pqc + prb + qra} \right)}}{{abc}} = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1....\left[ {{\text{using (i)}}} \right] \cr} $$

$$\eqalign{ & x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & {x^3} - \frac{1}{{{x^3}}} \cr & = {\left[ {x - \frac{1}{x}} \right]^3} + 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) \cr} $$
  $$ = {\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)^3} + $$      $$3\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)$$
$$\eqalign{ & = {\left( {2\sqrt 2 } \right)^3} + 3\left( {2\sqrt 2 } \right) \cr & = 16\sqrt 2 + 6\sqrt 2 \cr & = 22\sqrt 2 \cr} $$

(1001)³
=1001 × 1001 × 1001
=1002001 × 1001
=1003003001

($$\sqrt {6} $$  + $$\sqrt {10} $$  - $$\sqrt {21} $$  - $$\sqrt {35} $$) × ($$\sqrt {6} $$  - $$\sqrt {10} $$  + $$\sqrt {21} $$  - $$\sqrt {35} $$)
= {($$\sqrt {6} $$  - $$\sqrt {35} $$) + ($$\sqrt {10} $$  - $$\sqrt {21} $$)} × {($$\sqrt {6} $$  - $$\sqrt {35} $$) - ($$\sqrt {10} $$  - $$\sqrt {21} $$)}
= ($$\sqrt {6} $$  - $$\sqrt {35} $$)² - ($$\sqrt {10} $$  - $$\sqrt {21} $$)²
= 6 + 35 - 2$$\sqrt {210} $$  - 10 - 21 + 2$$\sqrt {210} $$
= 41 - 31
= 10

$$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right] - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left( {\frac{2}{{{x^2} - 1}} - \frac{2}{{{x^2} + 1}}} \right) - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{2\left( {{x^2} + 1} \right) - 2\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}} \right] - \frac{4}{{{x^4} + 1}} \cr & = \frac{4}{{{x^4} - 1}} - \frac{4}{{{x^4} + 1}} \cr & = \frac{{4\left( {{x^4} + 1} \right) - 4\left( {{x^4} - 1} \right)}}{{\left( {{x^4} - 1} \right)\left( {{x^4} + 1} \right)}} \cr & = \frac{8}{{{x^8} - 1}} \cr} $$

Given expression,
$$\left[ {\left( {x + \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} - x.\frac{1}{x}} \right)} \right]$$     $$\left[ {\left( {x - \frac{1}{x}} \right)\left( {{x^2} + \frac{1}{{{x^2}}} + x.\frac{1}{x}} \right)} \right]$$
$$\eqalign{ & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = {x^6} - \frac{1}{{{x^6}}} \cr} $$

$$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$

1² + 3² + 5² + . . . . . + 17²
= (1² + 2² + 3² + 4² + . . . . . + 17²) - (2² + 4² + . . . . . + 16²)
= (1² + 2² + 3² + 4² + . . . . . + 17²) - 2²(1² + 2² + . . . . . + 8²)
Using given formula
$$\eqalign{ & \left[ {\frac{{p\left( {p + 1} \right)\left( {2p + 1} \right)}}{6}} \right] \cr & = \frac{{17 \times 18 \times 35}}{6} - 4\left[ {\frac{{8 \times 9 \times 17}}{6}} \right] \cr & = 1785 - 4 \times 204 \cr & = 1785 - 816 \cr & = 969 \cr} $$

$$\left( {\frac{1}{{\sqrt 9 - \sqrt 8 }} - \frac{1}{{\sqrt 8 - \sqrt 7 }} + \frac{1}{{\sqrt 7 - \sqrt 6 }} - \frac{1}{{\sqrt 6 - \sqrt 5 }}} \right)$$
$$\eqalign{ & = \sqrt 9 + \sqrt 8 - \sqrt 8 - \sqrt 7 + \sqrt 7 + \sqrt 6 - \sqrt 6 - \sqrt 5 \cr & = \sqrt 9 - \sqrt 5 \cr & = 3 - \sqrt 5 \cr} $$