Algebra - Study Mode
[#391] If $$x =
oot 3 of {{x^2} + 11} - 2{ ext{,}}$$ xa0xa0 then the value of x 3 + 5x 2 + 12x is?
Correct Answer
(B) 3
Explanation
Solution: $$eqalign{
& x =
oot 3 of {{x^2} + 11} - 2 cr
& Rightarrow x + 2 =
oot 3 of {{x^2} + 11} cr
& Rightarrow { ext{Taking cube on both side}} cr
& Rightarrow {left( {x + 2}
ight)^3} = {x^2} + 11 cr
& Rightarrow {x^3} + 8 + 6xleft( {x + 2}
ight) = {x^2} + 11 cr
& Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 cr
& Rightarrow {x^3} + 5{x^2} + 12x = 3 cr} $$
[#392] If $${p^2} + frac{1}{{{p^2}}} = 47{ ext{,}}$$ xa0xa0 then the value of $$p + frac{1}{p}$$ xa0is?
Correct Answer
(C) 7
Explanation
Solution: $$eqalign{
& {p^2} + frac{1}{{{p^2}}} = 47 cr
& { ext{On adding 2 both side}} cr
& {p^2} + frac{1}{{{p^2}}} + 2 = 47 + 2 cr
& Rightarrow {left( {p + frac{1}{p}}
ight)^2} = 49 cr
& Rightarrow left( {p + frac{1}{p}}
ight) = 7 cr
& Rightarrow p + frac{1}{p} = 7 cr} $$
[#393] The third proportional of the following numbers (x - y) 2 , (x 2 - y 2 ) = ?
Correct Answer
(C) (x - y) (x + y) 2
Explanation
Solution: $$eqalign{
& { ext{Let,}} cr
& a = {left( {x - y}
ight)^2}, cr
& b = left( {{x^2} - {y^2}}
ight){ ext{and}} cr
& c,,{ ext{be}},{ ext{the}},{ ext{third}},{ ext{proportional}} cr
& { ext{Therefore}},,,a:b::b:c cr
& i.e.,,,c = frac{{{b^2}}}{a} cr
& Rightarrow c = frac{{{{left( {{x^2} - {y^2}}
ight)}^2}}}{{left( {x - y}
ight)}} cr
& ,,,,,,,,,,, = frac{{{{left( {x - y}
ight)}^2}{{left( {x + y}
ight)}^2}}}{{left( {x - y}
ight)}} cr
& ,,,,,,,,,,, = left( {x - y}
ight){left( {x + y}
ight)^2} cr} $$
[#394] If $$x + sqrt 5 = 5 + sqrt y $$ xa0 xa0 and x, y are positive integers, then the value of $$frac{{sqrt x + y}}{{x + sqrt y }}$$ xa0 is?
Correct Answer
(A) 1
Explanation
Solution: $$eqalign{
& x + sqrt 5 = 5 + sqrt y cr
& { ext{Put , }}x = 5{ ext{ and }}y = 5 cr
& 5 + sqrt 5 = 5 + sqrt 5 cr
& { ext{L}}{ ext{.H}}{ ext{.S}} = { ext{R}}{ ext{.H}}{ ext{.S}} cr
& frac{{sqrt x + y}}{{x + sqrt y }} cr
& = frac{{sqrt 5 + 5}}{{5 + sqrt 5 }} cr
& = 1 cr} $$
[#395] If x, y and z are real numbers such that (x - 3) 2 + (y - 4) 2 + (z - 5) 2 = 0, then (x + y + z) is equal to?
Correct Answer
(D) 12
Explanation
Solution: This is possible only when $$eqalign{
& {left( {x - 3}
ight)^2} = 0 cr
& x = 3 cr
& {left( {y - 4}
ight)^2} = 0 cr
& y = 4 cr
& {left( {z - 5}
ight)^2} = 0 cr
& z = 5 cr
& { ext{Then, }}left( {x + y + z}
ight) cr
& = 3 + 4 + 5 cr
& = 12 cr} $$