Algebra - Study Mode

[#206] If x = -1, then the value of $$frac{1}{{{x^{99}}}}$$xa0 + $$frac{1}{{{x^{98}}}}$$xa0 + $$frac{1}{{{x^{97}}}}$$xa0 + $$frac{1}{{{x^{96}}}}$$xa0 + $$frac{1}{{{x^{95}}}}$$xa0 + $$frac{1}{{{x^{94}}}}$$xa0 + $$frac{1}{x}$$xa0 - 1 is?
Correct Answer

(C) -2

Explanation

Solution: $$frac{1}{{{x^{99}}}}{ ext{ + }}frac{1}{{{x^{98}}}} + frac{1}{{{x^{97}}}} + frac{1}{{{x^{96}}}} + frac{1}{{{x^{95}}}} + frac{1}{{{x^{94}}}} + frac{1}{x} - 1$$ $$ = frac{1}{{{{left( { - 1}
ight)}^{99}}}}{ ext{ + }}frac{1}{{{{left( { - 1}
ight)}^{98}}}} + frac{1}{{{{left( { - 1}
ight)}^{97}}}} + frac{1}{{{{left( { - 1}
ight)}^{96}}}} + $$ xa0 xa0 xa0 xa0 $$frac{1}{{{{left( { - 1}
ight)}^{95}}}} + $$ xa0 $$frac{1}{{{{left( { - 1}
ight)}^{94}}}} + $$ xa0 $$frac{1}{{left( { - 1}
ight)}} - $$ xa0 $$1$$ $$eqalign{
& = - 1 + 1 - 1 + 1 - 1 + 1 + frac{1}{{ - 1}} - 1 cr
& = - 2 cr} $$

[#207] If $$frac{1}{{
oot 3 of 4 +
oot 3 of 2 + 1}}$$ xa0xa0 = $$a
oot 3 of 4 $$xa0 + $$b
oot 3 of 2 $$xa0 + c and a, b, c are rational numbers then a + b + c is equal to?
Correct Answer

(A) 0

Explanation

Solution: $$eqalign{
& frac{1}{{
oot 3 of 4 +
oot 3 of 2 + 1}} = a
oot 3 of 4 + b
oot 3 of 2 + c cr
& Rightarrow frac{1}{{
oot 3 of 4 +
oot 3 of 2 + 1}} = frac{1}{{{{left( {{2^{frac{1}{3}}}}
ight)}^2} + {2^{frac{1}{3}}} + {{left( 1
ight)}^2}}} cr
& Rightarrow herefore {{ ext{A}}^3} - {{ ext{B}}^3}{ ext{ = }}left( {{ ext{A}} - { ext{B}}}
ight)left( {{{ ext{A}}^2} + { ext{AB}} + {{ ext{B}}^2}}
ight) cr
& { ext{Put, A}} = {2^{frac{1}{3}}}{ ext{, B}} = 1 cr
& Rightarrow frac{{left( {{2^{frac{1}{3}}} - 1}
ight)}}{{left( {{2^{frac{1}{3}}} - 1}
ight)left( {{{left( {{2^{frac{1}{3}}}}
ight)}^2} + {2^{frac{1}{3}}} + {{left( 1
ight)}^2}}
ight)}} cr
& Rightarrow frac{{left( {{2^{frac{1}{3}}} - 1}
ight)}}{{{{left( {{2^{frac{1}{3}}}}
ight)}^3} - {{left( 1
ight)}^3}}} cr
& Rightarrow left( {{2^{frac{1}{3}}} - 1}
ight) cr
& herefore {2^{frac{1}{3}}} - 1 cr
& = aleft( {{2^{frac{2}{3}}}}
ight) + b{left( 2
ight)^{frac{1}{3}}} + c cr
& left( {{ ext{Comparing the terms}}}
ight) cr
& a = 0 cr
& b = 1 cr
& c = - 1 cr
& herefore a + b + c cr
& = 0 + 1 - 1 cr
& = 0 cr} $$

[#208] If $$x =
oot 3 of {2 + sqrt 3 } { ext{,}}$$ xa0xa0 then the value of $${x^3}{ ext{ + }}frac{1}{{{x^3}}}$$ xa0 is?
Correct Answer

(D) 4

Explanation

Solution: $$eqalign{
& { ext{ }}x =
oot 3 of {2 + sqrt 3 } cr
& {x^3} = 2 + sqrt 3 cr
& frac{1}{{{x^3}}} = frac{1}{{2 + sqrt 3 }} imes frac{{2 - sqrt 3 }}{{2 - sqrt 3 }} = 2 - sqrt 3 cr
& herefore {x^3}{ ext{ + }}frac{1}{{{x^3}}} cr
& = 2 + sqrt 3 + 2 - sqrt 3 cr
& = 4 cr} $$

[#209] The simplest form of the expression $$frac{{{p^2} - p}}{{2{p^3} + {p^2}}}$$ xa0 + $$frac{{{p^2} - 1}}{{{p^2} + 3p}}$$ xa0 + $$frac{{{p^2}}}{{p + 1}}$$ xa0 = ?
Correct Answer

(B) $$frac{1}{{2{p^2}}}$$

Explanation

Solution: $$frac{{{p^2} - p}}{{2{p^3} + {p^2}}} + frac{{{p^2} - 1}}{{{p^2} + 3p}} + frac{{{p^2}}}{{p + 1}}$$ In such type of question assume values of p $$eqalign{
& herefore { ext{Let }}p = 1 cr
& herefore frac{{1 - 1}}{{2 + 1}} + frac{{1 - 1}}{{1 + 3}} + frac{1}{{1 + 1}} cr
& = 0 + 0 + frac{1}{2} cr
& = frac{1}{2} cr
& { ext{Now check option 'B'}} cr
& frac{1}{{2{p^2}}} = frac{1}{2} cr
& { ext{Hence the answer is option 'B'}} cr} $$

[#210] If $$a = frac{{sqrt 5 + 1}}{{sqrt 5 - 1}}$$ xa0 & $$b = frac{{sqrt 5 - 1}}{{sqrt 5 + 1}}{ ext{,}}$$ xa0xa0 then the value of $$frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}{ ext{ is?}}$$
Correct Answer

(B) $$frac{4}{3}$$

Explanation

Solution: $$eqalign{
& a = frac{{sqrt 5 + 1}}{{sqrt 5 - 1}} cr
& b = frac{{sqrt 5 - 1}}{{sqrt 5 + 1}}{ ext{ }} cr
& herefore a = frac{1}{b} cr
& a + b = a + frac{1}{a} cr
& Rightarrow frac{{sqrt 5 + 1}}{{sqrt 5 - 1}}{ ext{ + }}frac{{sqrt 5 - 1}}{{sqrt 5 + 1}} cr
& Rightarrow frac{{5 + 1 + 2sqrt 5 + 5 + 1 - 2sqrt 5 }}{{{{left( {sqrt 5 }
ight)}^2} - {{left( 1
ight)}^2}}} cr
& Rightarrow frac{{6 + 2sqrt 5 + 6 - 2sqrt 5 }}{{5 - 1}} cr
& Rightarrow frac{{12}}{4} cr
& Rightarrow 3 cr
& herefore frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} cr
& Rightarrow frac{{{a^2} + frac{1}{{{a^2}}} + ab}}{{{a^2} + frac{1}{{{a^2}}} - ab}} cr
& Rightarrow a + frac{1}{a} = 3 cr
& Rightarrow {a^2} + frac{1}{{{a^2}}} cr
& Rightarrow 9 - 2 cr
& Rightarrow 7left( {ab = 1}
ight) cr
& herefore frac{{{a^2} + frac{1}{{{a^2}}} + ab}}{{{a^2} + frac{1}{{{a^2}}} - ab}} cr
& = frac{{7 + 1}}{{7 - 1}} cr
& = frac{8}{6} cr
& = frac{4}{3} cr} $$