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

$If x > 0 and the 4^{th} term in the expansion$ $of {(2 + \frac{3}{8} x)}^{10} has maximum value$ $then find the range of x .$
Commented by Tinkutara last updated on 10/Oct/17

$I got the answer but my doubt is that$ $the interval for range will be open or$ $closed ?$
	Answered by mrW1 last updated on 10/Oct/17
	$(2 + (3/8)x)^(10) =Σ_(k=0) ^(10) C_k ^(10) ×2^k ×((3/8))^(10−k) x^(10−k) =((3/8))^(10) ×Σ_(k=0) ^(10) C_k ^(10) ×(((16)/3))^k x^(10−k) let a_k = C_k ^(10) ×(((16)/3))^k x^(10−k) since 4th term is maximum, ⇒a_2 <a_3 ⇒a_4 <a_3 C_2 ^(10) ×(((16)/3))^2 x^8 <C_3 ^(10) ×(((16)/3))^3 x^7 ⇒9x^8 <128x^7 ⇒x^7 (128−9x)>0 ⇒ { ((x>0)),((128−9x>0 ⇒ x<((128)/9))) :} C_4 ^(10) ×(((16)/3))^4 x^6 <C_3 ^(10) ×(((16)/3))^3 x^7 ⇒x^6 (28−3x)<0 ⇒28−3x<0 ⇒x>((28)/3) ⇒ { ((0<x<((128)/9))),((x>((28)/3))) :} ⇒x∈(((28)/3),((128)/9))$
	${(2 + \frac{3}{8} x)}^{10} = \underset{k = 0}{\sum^{10}} C_{k}^{10} \times 2^{k} \times {(\frac{3}{8})}^{10 - k} x^{10 - k}$ $= {(\frac{3}{8})}^{10} \times \underset{k = 0}{\sum^{10}} C_{k}^{10} \times {(\frac{16}{3})}^{k} x^{10 - k}$ $let a_{k} = C_{k}^{10} \times {(\frac{16}{3})}^{k} x^{10 - k}$ $since 4 th term is maximum,$ $\Rightarrow a_{2} < a_{3}$ $\Rightarrow a_{4} < a_{3}$ $C_{2}^{10} \times {(\frac{16}{3})}^{2} x^{8} < C_{3}^{10} \times {(\frac{16}{3})}^{3} x^{7}$ $\Rightarrow {9 x}^{8} < {128 x}^{7}$ $\Rightarrow x^{7} (128 - 9 x) > 0$ $\Rightarrow {\begin{cases} x > 0 \\ 128 - 9 x > 0 \Rightarrow x < \frac{128}{9} \end{cases}$ $C_{4}^{10} \times {(\frac{16}{3})}^{4} x^{6} < C_{3}^{10} \times {(\frac{16}{3})}^{3} x^{7}$ $\Rightarrow x^{6} (28 - 3 x) < 0$ $\Rightarrow 28 - 3 x < 0 \Rightarrow x > \frac{28}{3}$ $\Rightarrow {\begin{cases} 0 < x < \frac{128}{9} \\ x > \frac{28}{3} \end{cases} \Rightarrow x \in (\frac{28}{3}, \frac{128}{9})$
	Commented by mrW1 last updated on 10/Oct/17

	$I made a mistake, see correction .$
	Commented by Tinkutara last updated on 11/Oct/17

	$(2, \frac{64}{21})$
	Commented by mrW1 last updated on 11/Oct/17

	$what is the answer ?$
	Answered by ajfour last updated on 14/Oct/17

	$\Rightarrow {(1 + \frac{3 x}{16})}^{10} h a s i t s 4^{t h} t e r m m a x .$ $\Rightarrow^{10} C_{2} z^{2} <^{10} C_{3} z^{3} >^{10} C_{4} z^{4}$ $w h e r e z = \frac{3 x}{16}$ $\Rightarrow \frac{9 \times 10}{2} < (\frac{8 \times 9 \times 10}{6}) z > (\frac{7 \times 8 \times 9 \times 10}{24}) z^{2}$ $\Rightarrow z > \frac{3}{8} a n d z < \frac{4}{7}$ $o r x > \frac{3}{8} \times \frac{16}{3} a n d x < \frac{4}{7} \times \frac{16}{3}$ $o r x \in (2, \frac{64}{21}) .$
	Commented by Tinkutara last updated on 14/Oct/17

	$Why not closed interval ?$
	Commented by Tinkutara last updated on 14/Oct/17

	$Thank you very much Sir!$ $But my only doubt was that it has to$ $be open or closed interval ?$
	Commented by ajfour last updated on 14/Oct/17

	$f i r s t g u e s s c l o s e d i n t e r v a l . . .$
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