Algebra - Study Mode

[#216] If $$x = frac{{sqrt 3 }}{2}{ ext{,}}$$ xa0 then $$frac{{sqrt {1 + x} }}{{1 + sqrt {1 + x} }}{ ext{ + }}$$ xa0 $$frac{{sqrt {1 - x} }}{{1 - sqrt {1 - x} }}$$ xa0 is equal to?
Correct Answer

(B) $$frac{2}{{sqrt 3 }}$$

Explanation

Solution: $$eqalign{
& x = frac{{sqrt 3 }}{2} cr
& { ext{or }}1 + x = 1 + frac{{sqrt 3 }}{2} cr
& Rightarrow 1 + x = frac{{2 + sqrt 3 }}{2} cr
& Rightarrow 1 + x = frac{{2left( {2 + sqrt 3 }
ight)}}{{2 imes 2}} cr
& left( {{ ext{Divided and multiply by 2}}}
ight) cr
& Rightarrow 1 + x = frac{{4 + 2sqrt 3 }}{4} cr
& Rightarrow 1 + x = frac{{1 + 3 + 2sqrt 3 }}{4} cr
& Rightarrow 1 + x = frac{{4 + 2sqrt 3 }}{4} cr
& Rightarrow 1 + x = frac{{{{left( 1
ight)}^2} + {{left( {sqrt 3 }
ight)}^2} + 2.1.sqrt 3 }}{4} cr
& Rightarrow 1 + x = frac{{{{left( {1 + sqrt 3 }
ight)}^2}}}{4} cr
& herefore sqrt {1 + x} = frac{{1 + sqrt 3 }}{2} cr
& cr
& {x08f{Similarly:}} cr
& sqrt {1 - x} cr
& = frac{{sqrt 3 - 1}}{2} cr
& herefore frac{{sqrt {1 + x} }}{{1 + sqrt {1 + x} }}{ ext{ + }}frac{{sqrt {1 - x} }}{{1 - sqrt {1 - x} }} cr
& = frac{{frac{{1 + sqrt 3 }}{2}}}{{1 + frac{{1 + sqrt 3 }}{2}}} + frac{{frac{{sqrt 3 - 1}}{2}}}{{1 - frac{{sqrt 3 - 1}}{2}}} cr
& = frac{{1 + sqrt 3 }}{{3 + sqrt 3 }} + frac{{sqrt 3 - 1}}{{3 - sqrt 3 }} cr
& = frac{{1 + sqrt 3 }}{{sqrt 3 left( {sqrt 3 + 1}
ight)}} + frac{{1 - sqrt 3 }}{{sqrt 3 left( {sqrt 3 - 1}
ight)}} cr
& = frac{1}{{sqrt 3 }} + frac{1}{{sqrt 3 }} cr
& = frac{2}{{sqrt 3 }} cr} $$

[#217] If for non-zero, x, x 2 - 4x - 1 = 0, the value of $${x^2} + frac{1}{{{x^2}}}$$ xa0 is?
Correct Answer

(D) 18

Explanation

Solution: $$eqalign{
& Leftrightarrow {x^2} - 4x - 1 = 0 cr
& Leftrightarrow {x^2} - 1 = 4x cr
& left( {{ ext{divide }}x{ ext{ both sides}}}
ight) cr
& Leftrightarrow x - frac{1}{x} = 4 cr
& Leftrightarrow {x^2} + frac{1}{{{x^2}}} - 2 = 16 cr
& Leftrightarrow {x^2} + frac{1}{{{x^2}}} = 18 cr} $$

[#218] $$left( {x + frac{1}{x}}
ight)$$ $$left( {x - frac{1}{x}}
ight)$$ $$left( {{x^2} + frac{1}{{{x^2}}} - 1}
ight)$$xa0 $$left( {{x^2} + frac{1}{{{x^2}}} + 1}
ight)$$ xa0 is equal to?
Correct Answer

(D) $${x^6} - frac{1}{{{x^6}}}$$

Explanation

Solution: $$left( {x + frac{1}{x}}
ight)$$ $$left( {x - frac{1}{x}}
ight)$$ $$left( {{x^2} + frac{1}{{{x^2}}} - 1}
ight)$$xa0 $$left( {{x^2} + frac{1}{{{x^2}}} + 1}
ight)$$ $${ ext{ = }}left( {x + frac{1}{x}}
ight)$$ $$left( {{x^2} + frac{1}{{{x^2}}} - 1}
ight)$$xa0 $$left( {x - frac{1}{x}}
ight)$$ $$left( {{x^2} + frac{1}{{{x^2}}} + 1}
ight)$$ $$eqalign{
& x08ecause left( {{ ext{A}} + { ext{B}}}
ight)left( {{{ ext{A}}^2} - { ext{AB}} + {{ ext{B}}^2}}
ight) = {{ ext{A}}^3} + {{ ext{B}}^3} cr
& x08ecause left( {{ ext{A}} - { ext{B}}}
ight)left( {{{ ext{A}}^2} + { ext{AB}} + {{ ext{B}}^2}}
ight) = {{ ext{A}}^3} - {{ ext{B}}^3} cr
& = left( {{x^3} + frac{1}{{{x^3}}}}
ight)left( {{x^3} - frac{1}{{{x^3}}}}
ight) cr
& = x08oxed{{x^6} - frac{1}{{{x^6}}}} cr} $$

[#219] If a 2x+2 = 1, where is a positive real number other than 1, then x is equal to?
Correct Answer

(B) -1

Explanation

Solution: $$eqalign{
& Rightarrow {a^{2x + 2}} = 1 cr
& Rightarrow {a^{2x + 2}} = {a^0} cr
& Rightarrow 2x + 2 = 0 cr
& Rightarrow x = - frac{2}{2} cr
& Rightarrow x = - 1 cr} $$

[#220] If $${left( {{ ext{ }}a + frac{1}{a}}
ight)^2} = 3{ ext{,}}$$ xa0xa0 then the value of a 18 + a 12 + a 6 + 1 is?
Correct Answer

(C) 0

Explanation

Solution: $$eqalign{
& {left( {{ ext{ }}a + frac{1}{a}}
ight)^2} = 3 cr
& Rightarrow a + frac{1}{a} = sqrt 3 cr
& Rightarrow {a^3} + frac{1}{{{a^3}}} = 0 cr
& Rightarrow {a^6} = - 1 cr
& Rightarrow left( {{a^6} + 1}
ight) = 0 cr
& { ext{So,}}{a^{18}} + { ext{ }}{a^{12}} + { ext{ }}{a^6} + { ext{ 1}} cr
& Rightarrow {a^{12}}left( {{a^6} + 1}
ight) + { ext{ }}{a^6} + { ext{ 1}} cr
& Rightarrow {a^{12}}left( 0
ight) + 0 cr
& Rightarrow 0 cr} $$