Algebra - Study Mode
[#371] If x and y are positive real numbers and xy = 8, then the minimum value of 2x + y is?
Correct Answer
(D) 8
Explanation
Solution: $$eqalign{
& xy = 8{ ext{ }}left( {{ ext{Given}}}
ight) cr
& { ext{So, }}left( {x,y}
ight) = left( {1,8}
ight) cr} $$ We have to question the options and check them. $$eqalign{
& left( {8,1}
ight) cr
& left( {2,4}
ight) cr
& left( {4,2}
ight) cr
& herefore { ext{ }}2x + y cr
& = 2 imes 1 + 8 cr
& = 10 cr
& 2 imes 8 + 1 = 11 cr
& 2 imes 2 + 4 = 8{ ext{ }}left( {{ ext{Minimum}}}
ight) cr
& 2 imes 4 + 2 = 10 cr} $$ Hence, in this question we have all the options. So, take all the positive factor otherwise we should have to take - ve(negative) values also. $$eqalign{
& left( {x,y}
ight) = left( {1,8}
ight) cr
& { ext{ }}left( {8,1}
ight) cr
& { ext{ }}left( {2,4}
ight) cr
& { ext{ }}left( {4,2}
ight) cr
& { ext{ }}left( { - 1, - 8}
ight) cr
& { ext{ }}left( { - 8, - 1}
ight) cr
& { ext{ }}left( { - 2, - 4}
ight) cr
& { ext{ }}left( { - 4, - 2}
ight) cr} $$
[#372] If a 2 - 4a - 1 = 0 then value of $${a^2} + frac{1}{{{a^2}}} + 3a - frac{3}{a}$$ xa0xa0 is?
Correct Answer
(B) 30
Explanation
Solution: $$eqalign{
& {a^2} - 4a - 1 = 0 cr
& {a^2} - 1 = 4a cr
& a - frac{1}{a} = 4 cr
& { ext{Squring both sides}} cr
& {a^2} + frac{1}{{{a^2}}} - 2 = 16 cr
& Rightarrow {a^2} + frac{1}{{{a^2}}} = 18 cr
& herefore {a^2} + frac{1}{{{a^2}}} + 3a - frac{3}{a} cr
& = {a^2} + frac{1}{{{a^2}}} + 3left( {a - frac{1}{a}}
ight) cr
& = 18 + 3 imes 4 cr
& = 18 + 12 cr
& = 30 cr} $$
[#373] The minimum value of (x - 2)(x - 9) is?
Correct Answer
(D) $$ - frac{{49}}{4}$$
Explanation
Solution: $$eqalign{
& left( {x - 2}
ight)left( {x - 9}
ight) cr
& = {x^2} - 9x - 2x + 18 cr
& = {x^2} - 11x + 18 cr
& = a{x^2} + bx + c = 0 cr
& { ext{For minimum value}} cr
& = frac{{4ac - {b^2}}}{{4a}} cr
& = frac{{4 imes 1 imes 18 - {{left( { - 11}
ight)}^2}}}{{4 imes 1}} cr
& = frac{{72 - 121}}{4} cr
& = frac{{ - 49}}{4} cr
& = - frac{{49}}{4} cr} $$
[#374] If $$sqrt x = sqrt 3 - sqrt 5 { ext{,}}$$ xa0xa0 then the value of x 2 - 16x + 6 is?
Correct Answer
(C) 2
Explanation
Solution: $$eqalign{
& sqrt x = sqrt 3 - sqrt 5 cr
& left( {{ ext{Squaring both sides}}}
ight) cr
& Rightarrow x = 3 + 5 - 2.sqrt {3.} sqrt 5 cr
& Rightarrow x = 8 - 2sqrt {15} cr
& Rightarrow x - 8 = - 2sqrt {15} cr
& left( {{ ext{Squaring both sides}}}
ight) cr
& Rightarrow {x^2} + 64 - 16x = 60 cr
& Rightarrow {x^2} + 4 - 16x = 0 cr
& Rightarrow {x^2} + 6 - 16x = 2 cr
& Rightarrow {x^2} - 16x + 6 = 2 cr} $$
[#375] If x 2 = y + z, y 2 = z + x, z 2 = x + y, then the value of $$frac{1}{{x + 1}}$$ xa0 + $$frac{1}{{y + 1}}$$ xa0 + $$frac{1}{{z + 1}} = ,?$$
Correct Answer
(B) 1
Explanation
Solution: $$eqalign{
& {x^2} = y + z cr
& {y^2} = z + x cr
& {z^2} = x + y cr
& Rightarrow {x^2} + x = x + y + z cr
& { ext{Adding }}x{ ext{ on both sides }} cr
& xleft( {x + 1}
ight) = x + y + z cr
& frac{1}{{left( {x + 1}
ight)}} = frac{x}{{x + y + z}} cr
& { ext{Similarly,}} cr
& frac{1}{{left( {y + 1}
ight)}} = frac{y}{{x + y + z}} cr
& frac{1}{{left( {z + 1}
ight)}} = frac{z}{{x + y + z}} cr
& herefore frac{1}{{left( {x + 1}
ight)}} + frac{1}{{left( {y + 1}
ight)}} + frac{1}{{left( {z + 1}
ight)}} cr
& = frac{x}{{x + y + z}} + frac{y}{{x + y + z}} + frac{z}{{x + y + z}} cr
& = frac{{x + y + z}}{{x + y + z}} cr
& = 1 cr} $$