If-x-gt-0-and-the-4-th-term-in-the-expansion-of-2-3-8-x-10-has-maximum-value-then-find-the-range-of-x-

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 + \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 \dots

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