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Question-26733

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Question Number 26733 by math solver last updated on 28/Dec/17
Commented by prakash jain last updated on 28/Dec/17
((3/2)x^3 −(1/(3x)))^9 =^9 C_i ((3/2))^i x^(2i) (−1)^(9−i) ((1/(3x)))^(9−i) ^9 C_i ((3/2))^i (−1)^(9−i) ((1/3))^(9−i) x^(4i−9) coeffcient of constant term 3i−9=0⇒i=3 ^9 C_2 ((3/2))^3 (−1)^(9−3) ((1/3))^(9−3) =((9!)/(3!6!))×(3^3 /2^3 )×(1/3^6 )=((9×8×7)/(1×2×3))×(1/8)×(1/(3×9))=(7/(18)) 3i−9=−3⇒i=2 2×((9!)/(2!7!))×(3^2 /2^2 )×(−1)^(9−2) ×(1/3^7 )=((9×8)/(4×3^5 ))=−(2/(27)) (7/(18))−(2/(27))=((17)/(54))
$$\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}{x}}\right)^{\mathrm{9}} =\:^{\mathrm{9}} {C}_{{i}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{i}} {x}^{\mathrm{2}{i}} \left(−\mathrm{1}\right)^{\mathrm{9}−{i}} \left(\frac{\mathrm{1}}{\mathrm{3}{x}}\right)^{\mathrm{9}−{i}} \\ $$$$\:^{\mathrm{9}} {C}_{{i}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{i}} \left(−\mathrm{1}\right)^{\mathrm{9}−{i}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}−{i}} {x}^{\mathrm{4}{i}−\mathrm{9}} \\ $$$${coeffcient}\:{of}\:{constant}\:{term} \\ $$$$\mathrm{3}{i}−\mathrm{9}=\mathrm{0}\Rightarrow{i}=\mathrm{3} \\ $$$$\:^{\mathrm{9}} {C}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} \left(−\mathrm{1}\right)^{\mathrm{9}−\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}−\mathrm{3}} \\ $$$$=\frac{\mathrm{9}!}{\mathrm{3}!\mathrm{6}!}×\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }×\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }=\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{3}×\mathrm{9}}=\frac{\mathrm{7}}{\mathrm{18}} \\ $$$$\mathrm{3}{i}−\mathrm{9}=−\mathrm{3}\Rightarrow{i}=\mathrm{2} \\ $$$$\mathrm{2}×\frac{\mathrm{9}!}{\mathrm{2}!\mathrm{7}!}×\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }×\left(−\mathrm{1}\right)^{\mathrm{9}−\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{7}} }=\frac{\mathrm{9}×\mathrm{8}}{\mathrm{4}×\mathrm{3}^{\mathrm{5}} }=−\frac{\mathrm{2}}{\mathrm{27}} \\ $$$$\frac{\mathrm{7}}{\mathrm{18}}−\frac{\mathrm{2}}{\mathrm{27}}=\frac{\mathrm{17}}{\mathrm{54}} \\ $$
Answered by mrW1 last updated on 28/Dec/17
(1+x+2x^3 )((3/2)x^2 −(1/(3x)))^9 =(1/x^9 )(1+x+2x^3 )((3/2)x^3 −(1/3))^9 1×((9×8×7)/(3×2×1))×((3/2))^3 (−(1/3))^6 +2×((9×8)/(2×1))×((3/2))^2 (−(1/3))^7 =((3/2))^2 (−(1/3))^6 [1×((9×8×7)/(3×2×1))×((3/2))+2×((9×8)/(2×1))×(−(1/3))] =((3/2))^2 (−(1/3))^6 (126−24) =(1/2^2 )×(1/3^4 )×102 =((17)/(54)) ⇒Answer (A)
$$\left(\mathrm{1}+{x}+\mathrm{2}{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}{x}}\right)^{\mathrm{9}} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{9}} }\left(\mathrm{1}+{x}+\mathrm{2}{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \\ $$$$ \\ $$$$\mathrm{1}×\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{3}×\mathrm{2}×\mathrm{1}}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{6}} +\mathrm{2}×\frac{\mathrm{9}×\mathrm{8}}{\mathrm{2}×\mathrm{1}}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{7}} \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{6}} \left[\mathrm{1}×\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{3}×\mathrm{2}×\mathrm{1}}×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{2}×\frac{\mathrm{9}×\mathrm{8}}{\mathrm{2}×\mathrm{1}}×\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)\right] \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{6}} \left(\mathrm{126}−\mathrm{24}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }×\mathrm{102} \\ $$$$=\frac{\mathrm{17}}{\mathrm{54}} \\ $$$$\Rightarrow{Answer}\:\left({A}\right) \\ $$
Commented by math solver last updated on 28/Dec/17
plz explain step−3? how you find term independent of x?
$${plz}\:{explain}\:{step}−\mathrm{3}? \\ $$$${how}\:{you}\:{find}\:{term}\:{independent}\:{of}\:{x}? \\ $$$$ \\ $$
Commented by math solver last updated on 28/Dec/17
its something like 9_C_3 and 2. 9_C_7 .
$${its}\:{something}\:{like} \\ $$$$\mathrm{9}_{{C}_{\mathrm{3}} } \:{and}\:\mathrm{2}.\:\mathrm{9}_{{C}_{\mathrm{7}} } \:. \\ $$
Commented by mrW1 last updated on 28/Dec/17
x^0 term from (1/x^9 )(1+x+2x^3 )((3/2)x^3 −(1/3))^9 ≡x^9 term from (1+x+2x^3 )((3/2)x^3 −(1/3))^9 ≡1×x^9 term+x×x^8 term+2x^3 ×x^6 term x^9 term of ((3/2)x^3 −(1/3))^9 is C_3 ^9 ×((3/2)x^3 )^3 ×(−(1/3))^6 there is no x^8 term in ((3/2)x^3 −(1/3))^9 x^6 term of ((3/2)x^3 −(1/3))^9 is C_2 ^9 ×((3/2)x^3 )^2 ×(−(1/3))^7 therefore the coefficient of x^0 term is 1× C_3 ^9 ×((3/2))^3 ×(−(1/3))^6 +2×C_2 ^9 ×((3/2))^2 ×(−(1/3))^7
$${x}^{\mathrm{0}} \:{term}\:{from}\:\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{9}} }\left(\mathrm{1}+{x}+\mathrm{2}{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \\ $$$$\equiv{x}^{\mathrm{9}} \:{term}\:{from}\:\left(\mathrm{1}+{x}+\mathrm{2}{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \\ $$$$\equiv\mathrm{1}×{x}^{\mathrm{9}} \:{term}+{x}×{x}^{\mathrm{8}} \:{term}+\mathrm{2}{x}^{\mathrm{3}} ×{x}^{\mathrm{6}} \:{term} \\ $$$$ \\ $$$${x}^{\mathrm{9}} \:{term}\:{of}\:\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \:{is}\:{C}_{\mathrm{3}} ^{\mathrm{9}} ×\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} \right)^{\mathrm{3}} ×\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{6}} \\ $$$${there}\:{is}\:{no}\:{x}^{\mathrm{8}} \:{term}\:{in}\:\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \\ $$$${x}^{\mathrm{6}} \:{term}\:{of}\:\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \:{is}\:{C}_{\mathrm{2}} ^{\mathrm{9}} ×\left(\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{3}} \right)^{\mathrm{2}} ×\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{7}} \\ $$$$ \\ $$$${therefore}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{0}} \:{term}\:{is} \\ $$$$\mathrm{1}×\:{C}_{\mathrm{3}} ^{\mathrm{9}} ×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} ×\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{6}} +\mathrm{2}×{C}_{\mathrm{2}} ^{\mathrm{9}} ×\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} ×\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{7}} \\ $$

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