Algebra - Study Mode
[#196] x 2 + y 2 + z 2 = 2(x + z - 1), then the value of x 3 + y 3 + z 3 = ?
Correct Answer
(B) 2
Explanation
Solution: $$eqalign{
& { ext{Given,}} cr
& {x^2} + {y^2} + {z^2} = 2left( {x + z - 1}
ight) cr
& { ext{Find, }}{x^3} + {y^3} + {z^3} = ? cr
& Rightarrow {x^2} + {y^2} + {z^2} = 2left( {x + z - 1}
ight) cr
& Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 2 cr
& Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 1 - 1 cr
& Rightarrow left( {{x^2} + 1 - 2x}
ight) + {y^2} + left( {{z^2} + 1 - 2z}
ight) = 0 cr
& Rightarrow {left( {x - 1}
ight)^2} + {y^2} + {left( {z - 1}
ight)^2} = 0 cr
& Rightarrow {left( {x - 1}
ight)^2} = 0 cr
& Rightarrow x = 1 cr
& Rightarrow {y^2} = 0 cr
& Rightarrow y = 0 cr
& Rightarrow {left( {z - 1}
ight)^2} = 0 cr
& Rightarrow z = 1 cr
& { ext{Value substituted in question,}} cr
& Rightarrow {x^3} + {y^3} + {z^3} cr
& Rightarrow {1^3} + 0 + {1^3} cr
& Rightarrow 2 cr} $$
[#197] If $$x + frac{1}{x} = 1,$$ xa0 then the value of $$frac{2}{{{x^2} - x + 2}} = ,?$$
Correct Answer
(B) 2
Explanation
Solution: $$eqalign{
& { ext{Given, }}x + frac{1}{x} = 1 cr
& { ext{Find }}frac{2}{{{x^2} - x + 2}} = ? cr
& x + frac{1}{x} = 1 cr
& {x^2} + 1 = x cr
& left( {{x^2} - x}
ight) = - 1 cr
& { ext{Putting value in,}} cr
& = frac{2}{{{x^2} - x + 2}} cr
& = frac{2}{{ - 1 + 2}} cr
& = 2 cr} $$
[#198] If $$x = frac{{sqrt 5 - sqrt 3 }}{{sqrt 5 + sqrt 3 }}$$ xa0 and $$y = frac{{sqrt 5 + sqrt 3 }}{{sqrt 5 - sqrt 3 }}$$ xa0 then the value of $$frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ?$$
Correct Answer
(D) $$frac{{63}}{{61}}$$
Explanation
Solution: $$eqalign{
& { ext{Given,}} cr
& x = frac{{sqrt 5 - sqrt 3 }}{{sqrt 5 + sqrt 3 }}{ ext{ , }}y = frac{{sqrt 5 + sqrt 3 }}{{sqrt 5 - sqrt 3 }} cr
& { ext{Find, }}frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? cr
& Rightarrow { ext{ }}frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} cr
& Rightarrow { ext{ }}frac{{{{left( {x + y}
ight)}^2} - xy}}{{{{left( {x - y}
ight)}^2} + xy}} = ? cr
& { ext{Now,}} cr
& { ext{ }}x + y = frac{{sqrt 5 - sqrt 3 }}{{sqrt 5 + sqrt 3 }} + frac{{sqrt 5 + sqrt 3 }}{{sqrt 5 - sqrt 3 }} cr
& Rightarrow x + y = frac{{{{left( {sqrt 5 - sqrt 3 }
ight)}^2} + {{left( {sqrt 5 + sqrt 3 }
ight)}^2}}}{{{{sqrt 5 }^2} - {{sqrt 3 }^2}}} cr
& Rightarrow { ext{ }}x + y = frac{{2left( {{{sqrt 5 }^2} + {{sqrt 3 }^2}}
ight)}}{{5 - 3}} cr
& Rightarrow x + y = 8,.......(i) cr
& Again, cr
& x - y = frac{{sqrt 5 - sqrt 3 }}{{sqrt 5 + sqrt 3 }} - frac{{sqrt 5 + sqrt 3 }}{{sqrt 5 - sqrt 3 }} cr
& Rightarrow { ext{ }}x - y = frac{{4 imes sqrt 5 imes sqrt 3 }}{2} cr
& Rightarrow x - y = 2sqrt {15} ..............(ii) cr
& { ext{And, }}xy = frac{{sqrt 5 - sqrt 3 }}{{sqrt 5 + sqrt 3 }} imes frac{{sqrt 5 + sqrt 3 }}{{sqrt 5 - sqrt 3 }} cr
& Rightarrow { ext{ }}xy = 1 cr
& { ext{Substitutes values in the question}}{ ext{.}} cr
& Rightarrow frac{{{{left( {x + y}
ight)}^2} - xy}}{{{{left( {x - y}
ight)}^2} + xy}} cr
& Rightarrow frac{{{8^2} - 1}}{{{{left( {2sqrt {15} }
ight)}^2} + 1}} cr
& Rightarrow frac{{63}}{{61}} cr} $$
[#199] Simplified value of $$left[ {,left( {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight),left( {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight),, - ,,left( {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight),left( {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight),}
ight]$$ xa0 xa0 xa0 xa0 xa0 xa0xa0 $$ ÷ $$xa0 $$left[ {,left( {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight),, + ,,left( {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight),}
ight]$$ xa0 xa0 xa0 = ?
Correct Answer
(A) $$frac{{20}}{{101}}$$
Explanation
Solution: $$left[ {,left( {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight),left( {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight),, - ,,left( {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight),left( {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight),}
ight]$$ xa0 xa0 xa0 xa0 xa0 xa0xa0 $$ ÷ $$xa0 $$left[ {,left( {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight),, + ,,left( {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight),}
ight]$$ $$ { ext{Let }}left[ {1 + frac{1}{{10 + frac{1}{{10}}}}}
ight] = a,$$ xa0 xa0$${ ext{ }}left[ {1 - frac{1}{{10 + frac{1}{{10}}}}}
ight] = b $$ $$eqalign{
& Rightarrow left( {{a^2} - {b^2}}
ight) div left( {a + b}
ight) = a - b = ? cr
& Rightarrow a = 1 + frac{{10}}{{101}} cr
& Rightarrow a = frac{{111}}{{101}} cr
& Rightarrow b = 1 - frac{{10}}{{101}} cr
& Rightarrow b = frac{{91}}{{101}} cr
& Rightarrow a - b = frac{{111}}{{101}} - frac{{91}}{{101}} cr
& Rightarrow a - b = frac{{20}}{{101}} cr} $$
[#200] If $$2x + frac{2}{x} = 3{ ext{,}}$$ xa0 then the value of $${x^3} + frac{1}{{{x^3}}} + 2$$ xa0 is?
Correct Answer
(C) $$frac{7}{8}$$
Explanation
Solution: $$eqalign{
& { ext{ }}2x + frac{2}{x} = 3 cr
& Rightarrow { ext{ }}x + frac{1}{x} = frac{3}{2} cr
& { ext{Taking cube on both sides}} cr
& Rightarrow {left( {x + frac{1}{x}}
ight)^3} = {left( {frac{3}{2}}
ight)^3}{ ext{ }} cr
& x + frac{1}{x} = a cr
& {x^3} + frac{1}{{{x^3}}} = {a^3} - 3a cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} + 3left( {x + frac{1}{x}}
ight) = frac{{27}}{8} cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} + 3 imes frac{3}{2} = frac{{27}}{8} cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} = frac{{27}}{8} - frac{9}{2} cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} = frac{{ - 9}}{8} cr
& herefore {x^3} + frac{1}{{{x^3}}} + 2 cr
& = frac{{ - 9}}{8} + 2 cr
& = frac{{ - 9 + 16}}{8} cr
& = frac{7}{8} cr} $$