Algebra - Study Mode
[#486] If $$x - frac{3}{x} = 6,,x
e 0,$$ xa0 xa0then the value of $$frac{{{x^4} - frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}}$$
Correct Answer
(D) 90
Explanation
Solution: $$eqalign{
& frac{{{x^4} - frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}} cr
& = frac{{xleft( {{x^3} - frac{{27}}{{{x^3}}}}
ight)}}{{xleft( {x - 3 - frac{3}{x}}
ight)}}.....left( { ext{i}}
ight) cr
& {x^3} - frac{{27}}{{{x^3}}} cr
& = {6^3} + 3 imes 3 imes 6 cr
& = 216 + 54 cr
& = 270 cr
& frac{{{x^3} - frac{{27}}{{{x^3}}}}}{{x - frac{3}{x} - 3}} = frac{{270}}{{6 - 3}} = 90 cr} $$
[#487] If $$x = frac{{sqrt 3 }}{2}{ ext{,}}$$ xa0 then the value of $$left( {frac{{sqrt {1 + x} + sqrt {1 - x} }}{{sqrt {1 + x} - sqrt {1 - x} }}}
ight),{ ext{is}} = ?$$
Correct Answer
(D) $$sqrt 3 $$
Explanation
Solution: $$eqalign{
& x = frac{{sqrt 3 }}{2} cr
& = frac{{sqrt {1 + x} + sqrt {1 - x} }}{{sqrt {1 + x} - sqrt {1 - x} }} imes frac{{sqrt {1 + x} + sqrt {1 - x} }}{{sqrt {1 + x} + sqrt {1 - x} }} cr
& = frac{{{{left( {sqrt {1 + x} + sqrt {1 - x} }
ight)}^2}}}{{{{left( {sqrt {1 + x} }
ight)}^2} - {{left( {sqrt {1 - x} }
ight)}^2}}} cr
& = frac{{1 + x + 1 - x + 2sqrt {1 - {x^2}} }}{{1 + x - 1 + x}} cr
& = frac{{2 + 2sqrt {1 - {x^2}} }}{{2x}} cr
& = frac{{1 + sqrt {1 - {x^2}} }}{x} cr
& = frac{{1 + sqrt {1 - frac{3}{4}} }}{{sqrt 3 }} imes 2 cr
& = frac{{left( {1 + frac{1}{2}}
ight)}}{{sqrt 3 }} imes 2 cr
& = frac{{frac{3}{2} imes 2}}{{sqrt 3 }} cr
& = sqrt 3 cr} $$
[#488] If $${4^{4x + 1}} = frac{1}{{64}}{ ext{,}}$$ xa0 then the value of x is?
Correct Answer
(B) -1
Explanation
Solution: $$eqalign{
& {4^{4x + 1}} = frac{1}{{64}} cr
& Rightarrow {4^{4x + 1}} = frac{1}{{{{left( 4
ight)}^3}}} cr
& Rightarrow {4^{4x + 1}} = {left( 4
ight)^{ - 3}} cr
& Rightarrow 4x + 1 = - 3 cr
& Rightarrow 4x = - 4 cr
& Rightarrow x = - 1 cr} $$
[#489] If $${left( {sqrt 5 }
ight)^7} div {left( {sqrt 5 }
ight)^5} = {{ ext{5}}^{ ext{P}}}{ ext{,}}$$ xa0 xa0 then the value of P is?
Correct Answer
(D) 1
Explanation
Solution: $$eqalign{
& {left( {sqrt 5 }
ight)^7} div {left( {sqrt 5 }
ight)^5}{ ext{ = }}{{ ext{5}}^{ ext{P}}} cr
& Rightarrow frac{{{{left( {sqrt 5 }
ight)}^7}}}{{{{left( {sqrt 5 }
ight)}^5}}} = {{ ext{5}}^{ ext{P}}} cr
& Rightarrow {left( {sqrt 5 }
ight)^2} = {{ ext{5}}^{ ext{P}}} cr
& Rightarrow {{ ext{5}}^{ ext{1}}} = { ext{ }}{{ ext{5}}^{ ext{P}}} cr
& Rightarrow x08oxed{{ ext{P}} = 1} cr} $$
[#490] If 1.5a = 0.04b then $$frac{{b - a}}{{b + a}}$$ xa0 is equal to?
Correct Answer
(A) $$frac{{73}}{{77}}$$
Explanation
Solution: $$eqalign{
& 1.5a = 0.04b cr
& frac{a}{b} = frac{{0.04}}{{1.5}} = frac{4}{{100}} imes frac{{10}}{{15}} = frac{2}{{75}} cr
& { ext{Let }}a = 2x,{ ext{ }}b = 75x cr
& herefore frac{{b - a}}{{b + a}} = frac{{75x - 2x}}{{75x + 2x}} = frac{{73}}{{77}} cr
& cr
& {x08f{Alternate:}} cr
& frac{a}{b} = frac{{0.04}}{{1.5}} cr
& herefore frac{{b - a}}{{b + a}} = frac{{1.5 - 0.04}}{{1.5 + 0.04}} cr
& ,,,,,,,,,,,,,,,,,,,,, = frac{{1.46}}{{1.54}} cr
& ,,,,,,,,,,,,,,,,,,,,, = frac{{73}}{{77}} cr} $$