Algebra - Study Mode
[#291] If $${a^3} + frac{1}{{{a^3}}} = 2{ ext{,}}$$ xa0 then the value of $$frac{{{a^2} + 1}}{a}$$ xa0is (a positive number) ?
Correct Answer
(B) 2
Explanation
Solution: $$eqalign{
& { ext{But }}a = 1 cr
& {a^3} + frac{1}{{{a^3}}} = 2 cr
& Rightarrow {1^3} + frac{1}{{{1^3}}} = 2 cr
& Rightarrow 2 = 2{ ext{ }}left( {{ ext{Satisfy}}}
ight) cr
& { ext{So, }}frac{{{a^2} + 1}}{a} cr
& = frac{{{1^2} + 1}}{1} cr
& = 2 cr} $$
[#292] If $$frac{a}{{q - r}}$$xa0 = $$frac{b}{{r - p}}$$xa0 = $$frac{c}{{p - q}}{ ext{,}}$$ xa0 find the value of pa + qb + rc is?
Correct Answer
(A) 0
Explanation
Solution: $$eqalign{
& { ext{Let }}frac{a}{{q - r}} = frac{b}{{r - p}} = frac{c}{{p - q}} = k cr
& frac{a}{{q - r}} = k{ ext{ }}left( {{ ext{On multiplying by }}p}
ight) cr
& pa = kleft( {pq - pr}
ight),......(i) cr
& { ext{In the same way we can write}} cr
& { ext{qb = k}}left( {qr - qp}
ight),........(ii) cr
& { ext{And }}rc = kleft( {rp - rp}
ight),.....iii) cr
& { ext{On adding equation (i), (ii) and (iii) }} cr
& pa + qb + rc cr
& = kleft( {pq - pr + qr - qp + rp - rq}
ight) cr
& = 0 cr} $$
[#293] If $${left( {a + frac{1}{a}}
ight)^2} = 3{ ext{,}}$$ xa0xa0 the value of $${a^3} + frac{1}{{{a^3}}}$$ xa0 = ?
Correct Answer
(A) 0
Explanation
Solution: $$eqalign{
& {left( {a + frac{1}{a}}
ight)^2} = 3 cr
& Rightarrow a + frac{1}{a} = sqrt 3 cr
& { ext{Taking cube on both sides}} cr
& Rightarrow { ext{ }}{a^3} + frac{1}{{{a^3}}} + 3.a.frac{1}{a}left( {a + frac{1}{a}}
ight) = 3sqrt 3 cr
& Rightarrow { ext{ }}{a^3} + frac{1}{{{a^3}}} + 3left( {sqrt 3 }
ight) = 3sqrt 3 cr
& Rightarrow { ext{ }}{a^3} + frac{1}{{{a^3}}} = 3sqrt 3 - 3sqrt 3 cr
& Rightarrow {a^3} + frac{1}{{{a^3}}} = 0 cr} $$
[#294] If $$frac{{{a^2} + {b^2}}}{{{c^2}}}$$xa0 = $$frac{{{b^2} + {c^2}}}{{{a^2}}}$$xa0 = $$frac{{{c^2} + {a^2}}}{{{b^2}}}$$xa0 = $$frac{1}{k}{ ext{,}}$$ $$left( {k
e 0}
ight)$$ xa0 then k = ?
Correct Answer
(D) $$frac{1}{2}$$
Explanation
Solution: $$eqalign{
& frac{{{a^2} + {b^2}}}{{{c^2}}} = frac{{{b^2} + {c^2}}}{{{a^2}}} = frac{{{c^2} + {a^2}}}{{{b^2}}} = frac{1}{k} cr
& { ext{Put }}a = b = c = 1 cr
& Rightarrow frac{{1 + 1}}{1} + frac{{1 + 1}}{1} + frac{{1 + 1}}{1} = frac{1}{k} cr
& Rightarrow 2 = 2 = 2 = frac{1}{k} cr
& Rightarrow k = frac{1}{2} cr} $$
[#295] If $$2x + frac{2}{{9x}} = 4{ ext{,}}$$ xa0 then the value of $$27{x^3} + frac{1}{{27{x^3}}}$$ xa0 is?
Correct Answer
(B) 198
Explanation
Solution: $$eqalign{
& 2x + frac{2}{{9x}} = 4 cr
& { ext{Multiply by }}frac{3}{2}{ ext{ on both sides}} cr
& Rightarrow 3x + frac{1}{{3x}} = 6 cr
& { ext{Taking cube on both sides}} cr
& Rightarrow {left( {3x + frac{1}{{3x}}}
ight)^3} = {6^3} cr
& Rightarrow 27{x^3} + frac{1}{{27{x^3}}} + 3 imes 3x imes frac{1}{{3x}}left( {3x + frac{1}{{3x}}}
ight) = 216 cr
& Rightarrow 27{x^3} + frac{1}{{27{x^3}}} + 3 imes 6 = 216 cr
& Rightarrow 27{x^3} + frac{1}{{27{x^3}}} = 216 - 18 cr
& Rightarrow 27{x^3} + frac{1}{{27{x^3}}} = 198 cr} $$