Algebra - Study Mode
[#126] If $$x = a + frac{1}{a}$$ xa0 and $$y = a - frac{1}{a},$$ xa0 then the value of x 4 + y 4 - 2x 2 y 2 is?
Correct Answer
(C) 16
(G) 16
Explanation
Solution: $$eqalign{
& x = a + frac{1}{a} cr
& {x^2} = {a^2} + frac{1}{{{a^2}}} + 2 cr
& {y^2} = {a^2} + frac{1}{{{a^2}}} - 2 cr
& { ext{Now, }} cr
& {x^4} + {y^4} - 2{x^2}{y^2} cr
& = {left( {{x^2} - {y^2}}
ight)^2} cr
& = {left( {{a^2} + frac{1}{{{a^2}}} + 2 - {a^2} - frac{1}{{{a^2}}} + 2}
ight)^2} cr
& = {left( 4
ight)^2} cr
& = 16 cr} $$
[#127] If $$x + frac{1}{x} = sqrt 3 { ext{,}}$$ xa0 then find the value of $${x^3} + frac{1}{{{x^3}}}$$ xa0 = ?
Correct Answer
(B) 0
Explanation
Solution: $$eqalign{
& x + frac{1}{x} = sqrt 3 cr
& { ext{Cubing both side}} cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} + 3.x.frac{1}{x}left( {x + frac{1}{x}}
ight) = 3sqrt 3 cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} + 3left( {sqrt 3 }
ight) = 3sqrt 3 cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} = 3sqrt 3 - 3sqrt 3 cr
& Rightarrow {x^3} + frac{1}{{{x^3}}} = 0 cr} $$
[#128] If $${ ext{2}}x - frac{1}{{2x}} = 5{ ext{,}}$$ xa0xa0 $${ ext{x}}
e { ext{0,}}$$ xa0 then find the value of $${x^2} + frac{1}{{16{x^2}}} - 2$$ xa0xa0 = ?
Correct Answer
(A) $$frac{{19}}{4}$$
Explanation
Solution: $$eqalign{
& 2x - frac{1}{{2x}} = 5 cr
& { ext{Divide by 2 both side}} cr
& x - frac{1}{{4x}} = frac{5}{2} cr
& { ext{Squaring both side}} cr
& Rightarrow {x^2} + frac{1}{{16{x^2}}} - 2 imes x imes frac{1}{{4x}} = frac{{25}}{4} cr
& Rightarrow {x^2} + frac{1}{{16{x^2}}} - frac{1}{2} = frac{{25}}{4} cr
& Rightarrow {x^2} + frac{1}{{16{x^2}}} = frac{{25}}{4} + frac{1}{2} cr
& Rightarrow {x^2} + frac{1}{{16{x^2}}} = frac{{27}}{4} cr
& { ext{So, }} cr
& {x^2} + frac{1}{{16{x^2}}} - 2 cr
& = frac{{27}}{4} - 2 cr
& = frac{{19}}{4} cr} $$
[#129] If $$x = frac{1}{{left( {sqrt 2 + 1}
ight)}}{ ext{,}}$$ xa0xa0 the value of x 2 + 2x - 1 is?
Correct Answer
(C) 0
Explanation
Solution: $$eqalign{
& { ext{Given that,}} cr
& { ext{ }}x = frac{1}{{left( {sqrt 2 + 1}
ight)}}{ ext{ }} cr
& { ext{Then, }}{x^2} + 2x - 1 cr
& = {x^2} + 2x - 1 + 1 - 1 cr
& = {x^2} + 2x + 1 - 2 cr
& = {left( {x + 1}
ight)^2} - 2 cr
& { ext{Now put the value of }}x cr
& = {left( {frac{1}{{left( {sqrt 2 + 1}
ight)}} + 1}
ight)^2} - 2 cr
& = {left( {frac{{1 + sqrt 2 + 1}}{{sqrt 2 + 1}}}
ight)^2} - 2 cr
& = {left( {frac{{sqrt 2 + 2}}{{sqrt 2 + 1}}}
ight)^2} - 2 cr
& = {left( {frac{{left( {sqrt 2 + 1}
ight) imes sqrt2}}{{sqrt 2 + 1}}}
ight)^2} - 2 cr
& = {left( {sqrt 2 }
ight)^2} - 2 cr
& = 2 - 2 cr
& = 0 cr} $$
[#130] If $$x + frac{1}{x} = sqrt {13} { ext{,}}$$ xa0 xa0then $$frac{{3x}}{{left( {{x^2} - 1}
ight)}}$$ xa0 equal to?
Correct Answer
(C) 1
Explanation
Solution: $$eqalign{
& { ext{Given, }}x + frac{1}{x} = sqrt {13} { ext{ }} cr
& { ext{then ,}}frac{{3x}}{{left( {{x^2} - 1}
ight)}} cr
& = frac{3}{{x - frac{1}{x}}},.............(i) cr
& { ext{Now, }}x + frac{1}{x} = sqrt {13} cr
& { ext{On squaring both side}} cr
& = {x^2} + frac{1}{{{x^2}}} cr
& = 13 - 2 cr
& = 11 cr
& = {x^2} + frac{1}{{{x^2}}} - 2 cr
& = 11 - 2 cr
& = 9 cr
& Rightarrow {left( {x - frac{1}{x}}
ight)^2} = 9 cr
& Rightarrow {left( {x - frac{1}{x}}
ight)^2} = {3^2} cr
& Rightarrow x - frac{1}{x} = 3 cr
& { ext{Put this value in equation (i)}} cr
& Rightarrow frac{3}{{x - frac{1}{x}}} cr
& Rightarrow frac{3}{3} cr
& Rightarrow 1 cr} $$