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Question Number 22047 by Tinkutara last updated on 10/Oct/17

If x > 0 and the 4^(th) term in the expansion of (2 + (3/8)x)^(10) has maximum value then find the range of x.

$$\mathrm{If}\:{x}\:>\:\mathrm{0}\:\mathrm{and}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$ $$\mathrm{of}\:\left(\mathrm{2}\:+\:\frac{\mathrm{3}}{\mathrm{8}}{x}\right)^{\mathrm{10}} \:\mathrm{has}\:\mathrm{maximum}\:\mathrm{value} \\ $$ $$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x}. \\ $$

Commented byTinkutara last updated on 10/Oct/17

I got the answer but my doubt is that the interval for range will be open or closed?

$$\mathrm{I}\:\mathrm{got}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{but}\:\mathrm{my}\:\mathrm{doubt}\:\mathrm{is}\:\mathrm{that} \\ $$ $$\mathrm{the}\:\mathrm{interval}\:\mathrm{for}\:\mathrm{range}\:\mathrm{will}\:\mathrm{be}\:\mathrm{open}\:\mathrm{or} \\ $$ $$\mathrm{closed}? \\ $$

Answered by mrW1 last updated on 10/Oct/17

(2 + (3/8)x)^(10) =Σ_(k=0) ^(10) C_k ^(10) ×2^k ×((3/8))^(10−k) x^(10−k) =((3/8))^(10) ×Σ_(k=0) ^(10) C_k ^(10) ×(((16)/3))^k x^(10−k) let a_k = C_k ^(10) ×(((16)/3))^k x^(10−k) since 4th term is maximum, ⇒a_2 <a_3 ⇒a_4 <a_3 C_2 ^(10) ×(((16)/3))^2 x^8 <C_3 ^(10) ×(((16)/3))^3 x^7 ⇒9x^8 <128x^7 ⇒x^7 (128−9x)>0 ⇒ { ((x>0)),((128−9x>0 ⇒ x<((128)/9))) :} C_4 ^(10) ×(((16)/3))^4 x^6 <C_3 ^(10) ×(((16)/3))^3 x^7 ⇒x^6 (28−3x)<0 ⇒28−3x<0 ⇒x>((28)/3) ⇒ { ((0<x<((128)/9))),((x>((28)/3))) :} ⇒x∈(((28)/3),((128)/9))

$$\:\left(\mathrm{2}\:+\:\frac{\mathrm{3}}{\mathrm{8}}{x}\right)^{\mathrm{10}} \:=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{C}_{\mathrm{k}} ^{\mathrm{10}} ×\mathrm{2}^{\mathrm{k}} ×\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{10}−\mathrm{k}} \:\mathrm{x}^{\mathrm{10}−\mathrm{k}} \\ $$ $$=\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{10}} ×\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{C}_{\mathrm{k}} ^{\mathrm{10}} ×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{k}} \:\mathrm{x}^{\mathrm{10}−\mathrm{k}} \\ $$ $$\mathrm{let}\:\mathrm{a}_{\mathrm{k}} =\:\mathrm{C}_{\mathrm{k}} ^{\mathrm{10}} ×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{k}} \:\mathrm{x}^{\mathrm{10}−\mathrm{k}} \\ $$ $$\mathrm{since}\:\mathrm{4th}\:\mathrm{term}\:\mathrm{is}\:\mathrm{maximum}, \\ $$ $$\Rightarrow\mathrm{a}_{\mathrm{2}} <\mathrm{a}_{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{a}_{\mathrm{4}} <\mathrm{a}_{\mathrm{3}} \\ $$ $$ \\ $$ $$\mathrm{C}_{\mathrm{2}} ^{\mathrm{10}} ×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{2}} \mathrm{x}^{\mathrm{8}} <\mathrm{C}_{\mathrm{3}} ^{\mathrm{10}} ×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{3}} \mathrm{x}^{\mathrm{7}} \\ $$ $$\Rightarrow\mathrm{9x}^{\mathrm{8}} <\mathrm{128x}^{\mathrm{7}} \\ $$ $$\Rightarrow\mathrm{x}^{\mathrm{7}} \left(\mathrm{128}−\mathrm{9x}\right)>\mathrm{0} \\ $$ $$\Rightarrow\begin{cases}{\mathrm{x}>\mathrm{0}}\\{\mathrm{128}−\mathrm{9x}>\mathrm{0}\:\Rightarrow\:\mathrm{x}<\frac{\mathrm{128}}{\mathrm{9}}}\end{cases} \\ $$ $$ \\ $$ $$\mathrm{C}_{\mathrm{4}} ^{\mathrm{10}} ×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{4}} \mathrm{x}^{\mathrm{6}} <\mathrm{C}_{\mathrm{3}} ^{\mathrm{10}} ×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{3}} \mathrm{x}^{\mathrm{7}} \\ $$ $$\Rightarrow\mathrm{x}^{\mathrm{6}} \left(\mathrm{28}−\mathrm{3x}\right)<\mathrm{0} \\ $$ $$\Rightarrow\mathrm{28}−\mathrm{3x}<\mathrm{0}\:\Rightarrow\mathrm{x}>\frac{\mathrm{28}}{\mathrm{3}} \\ $$ $$\Rightarrow\begin{cases}{\mathrm{0}<\mathrm{x}<\frac{\mathrm{128}}{\mathrm{9}}}\\{\mathrm{x}>\frac{\mathrm{28}}{\mathrm{3}}}\end{cases}\:\:\:\:\:\:\Rightarrow\mathrm{x}\in\left(\frac{\mathrm{28}}{\mathrm{3}},\frac{\mathrm{128}}{\mathrm{9}}\right) \\ $$

Commented bymrW1 last updated on 10/Oct/17

I made a mistake, see correction.

$$\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake},\:\mathrm{see}\:\mathrm{correction}. \\ $$

Commented byTinkutara last updated on 11/Oct/17

(2,((64)/(21)))

$$\left(\mathrm{2},\frac{\mathrm{64}}{\mathrm{21}}\right) \\ $$

Commented bymrW1 last updated on 11/Oct/17

what is the answer?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}? \\ $$

Answered by ajfour last updated on 14/Oct/17

⇒ (1+((3x)/(16)))^(10) has its 4^(th ) term max. ⇒^(10) C_2 z^2 < ^(10) C_3 z^3 >^(10) C_4 z^4 where z=((3x)/(16)) ⇒ ((9×10)/2) < ( ((8×9×10)/6))z > (((7×8×9×10)/(24)))z^2 ⇒ z > (3/8) and z < (4/7) or x > (3/8)×((16)/3) and x < (4/7)×((16)/3) or x∈ (2 , ((64)/(21)) ) .

$$\Rightarrow\:\left(\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{16}}\right)^{\mathrm{10}} \:{has}\:{its}\:\mathrm{4}^{{th}\:} {term}\:{max}. \\ $$ $$\Rightarrow\:^{\mathrm{10}} {C}_{\mathrm{2}} {z}^{\mathrm{2}} <\:\:^{\mathrm{10}} {C}_{\mathrm{3}} {z}^{\mathrm{3}} \:>\:^{\mathrm{10}} {C}_{\mathrm{4}} {z}^{\mathrm{4}} \: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{where}\:{z}=\frac{\mathrm{3}{x}}{\mathrm{16}} \\ $$ $$\Rightarrow\:\:\frac{\mathrm{9}×\mathrm{10}}{\mathrm{2}}\:<\:\left(\:\frac{\mathrm{8}×\mathrm{9}×\mathrm{10}}{\mathrm{6}}\right){z}\:\:>\:\left(\frac{\mathrm{7}×\mathrm{8}×\mathrm{9}×\mathrm{10}}{\mathrm{24}}\right){z}^{\mathrm{2}} \\ $$ $$\Rightarrow\:\:{z}\:>\:\frac{\mathrm{3}}{\mathrm{8}}\:\:\:{and}\:\:{z}\:<\:\frac{\mathrm{4}}{\mathrm{7}} \\ $$ $${or}\:\:\:\boldsymbol{{x}}\:>\:\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{16}}{\mathrm{3}}\:\:\:{and}\:\:\boldsymbol{{x}}\:<\:\frac{\mathrm{4}}{\mathrm{7}}×\frac{\mathrm{16}}{\mathrm{3}} \\ $$ $${or}\:\:\boldsymbol{{x}}\in\:\left(\mathrm{2}\:,\:\frac{\mathrm{64}}{\mathrm{21}}\:\right)\:. \\ $$

Commented byTinkutara last updated on 14/Oct/17

Why not closed interval?

$$\mathrm{Why}\:\mathrm{not}\:\mathrm{closed}\:\mathrm{interval}? \\ $$

Commented byTinkutara last updated on 14/Oct/17

Thank you very much Sir! But my only doubt was that it has to be open or closed interval?

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$ $$\mathrm{But}\:\mathrm{my}\:\mathrm{only}\:\mathrm{doubt}\:\mathrm{was}\:\mathrm{that}\:\mathrm{it}\:\mathrm{has}\:\mathrm{to} \\ $$ $$\mathrm{be}\:\mathrm{open}\:\mathrm{or}\:\mathrm{closed}\:\mathrm{interval}? \\ $$

Commented byajfour last updated on 14/Oct/17

first guess closed interval ...

$${first}\:{guess}\:{closed}\:{interval}\:... \\ $$

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