Algebra - Study Mode
[#186] If x = $$2 - {2^{frac{1}{3}}} + {2^{frac{2}{3}}},$$ xa0 then find the value of x 3 - 6x 2 + 18x.
Correct Answer
(D) 22
Explanation
Solution: $$eqalign{
& x = 2 - {2^{frac{1}{3}}} + {2^{frac{2}{3}}} cr
& x - 2 = {2^{frac{2}{3}}} - {2^{frac{1}{3}}} cr
& {left( {x - 2}
ight)^3} = {left( {{2^{frac{2}{3}}} - {2^{frac{1}{3}}}}
ight)^3} cr
& {x^3} - 8 - 3 imes 2xleft( {x - 2}
ight) = 4 - 2 - 3 imes {2^{frac{2}{3}}} imes {2^{frac{1}{3}}}left( {{2^{frac{2}{3}}} - {2^{frac{1}{3}}}}
ight) cr
& {x^3} - 8 - 6{x^2} + 12x = 4 - 2 - 6left( {x - 2}
ight) cr
& {x^3} - 8 - 6{x^2} + 12x = 2 - 6x + 12 cr
& {x^3} - 6{x^2} + 12x + 6x = 2 + 12 + 8 cr
& {x^3} - 6{x^2} + 18x = 22 cr} $$
[#187] If a + b + c + d = 2, then the maximum value of (1 + a)(1 + b)(1 + c)(1 + d) is:
Correct Answer
(B) $$frac{{81}}{{16}}$$
Explanation
Solution: $$eqalign{
& a + b + c + d = 2 cr
& { ext{Let }}a = b = c = d = frac{1}{2} cr
& { ext{satisfy the above}} cr
& { ext{so, }}left( {1 + a}
ight)left( {1 + b}
ight)left( {1 + c}
ight)left( {1 + d}
ight) cr
& = left( {1 + frac{1}{2}}
ight)left( {1 + frac{1}{2}}
ight)left( {1 + frac{1}{2}}
ight)left( {1 + frac{1}{2}}
ight) cr
& = frac{3}{2} imes frac{3}{2} imes frac{3}{2} imes frac{3}{2} cr
& = frac{{81}}{{16}} cr} $$
[#188] If 3a = 4b = 6c and a + b + c = $$27sqrt {29} $$ xa0then $$sqrt {{a^2} + {b^2} + {c^2}} $$ xa0 is equal to
Correct Answer
(A) 87
Explanation
Solution: $$eqalign{
& 3a = 4b = 6c cr
& Rightarrow frac{{3a}}{{12}} = frac{{4b}}{{12}} = frac{{6c}}{{12}} Rightarrow frac{a}{4} = frac{b}{3} = frac{c}{2} = k cr
& Rightarrow a = 4k,,b = 3k,,c = 2k cr
& a + b + c = 27sqrt {29} cr
& 9k = 27sqrt {29} cr
& k = 3sqrt {29} cr
& a = 4 imes 3sqrt {29} ,,b = 3 imes 3sqrt {29} ,,c = 2 imes 3sqrt {29} cr
& sqrt {{a^2} + {b^2} + {c^2}} cr
& = sqrt {29left( {144 + 81 + 36}
ight)} cr
& = sqrt {29 imes 261} cr
& = sqrt {29 imes 29 imes 9} cr
& = 29 imes 3 cr
& = 87 cr} $$
[#189] If $$frac{{{x^2} + 1}}{x} = 4frac{1}{4},$$ xa0 then what is the value of $${x^3} + frac{1}{{{x^3}}}?$$
Correct Answer
(D) $$frac{{4097}}{{64}}$$
Explanation
Solution: $$eqalign{
& frac{{{x^2} + 1}}{x} = 4frac{1}{4} cr
& x + frac{1}{x} = frac{{17}}{4} cr
& { ext{On cubing both sides,}} cr
& {left( {x + frac{1}{x}}
ight)^3} = {left( {frac{{17}}{4}}
ight)^3} cr
& {x^3} + frac{1}{{{x^3}}} + 3 imes x imes frac{1}{x}left( {x + frac{1}{x}}
ight) = frac{{4913}}{{64}} cr
& {x^3} + frac{1}{{{x^3}}} + 3left( {frac{{17}}{4}}
ight) = frac{{4913}}{{64}} cr
& {x^3} + frac{1}{{{x^3}}} = frac{{4913}}{{64}} - frac{{51}}{4} cr
& {x^3} + frac{1}{{{x^3}}} = frac{{4097}}{{64}} cr} $$
[#190] If t 2 - 4t + 1 = 0, then the value of $${t^3} + frac{1}{{{t^3}}}$$ xa0is?
Correct Answer
(C) 52
Explanation
Solution: $$eqalign{
& {t^2} - 4t + 1 = 0 cr
& Rightarrow {t^2} + 1 = 4t cr
& Rightarrow frac{{{t^2} + 1}}{t} = frac{{4t}}{t} cr
& Rightarrow t + frac{1}{t} = 4 cr
& left[ {{ ext{Take cube both sides}}}
ight] cr
& Rightarrow {t^3} + frac{1}{{{t^3}}} + 3.t.frac{1}{t}left( {t + frac{1}{t}}
ight) = 64 cr
& Rightarrow {t^3} + frac{1}{{{t^3}}} + 3left( 4
ight) = 64 cr
& Rightarrow {t^3} + frac{1}{{{t^3}}} = 64 - 12 cr
& Rightarrow {t^3} + frac{1}{{{t^3}}} = 52 cr} $$