First slide

Binomial theorem for positive integral Index

Question

If the 4th term in the expansion of ${(2 + \frac{3}{8} x)}^{10}$ has the maximum numerical value, then the range of values of x is

Moderate

A

$- 2 \leq x \leq 2$
B

$- \frac{64}{21} \leq x \leq - 2$
C

$2 \leq x \leq \frac{64}{21}$
D

none of these

Solution

Since T₄ is the numerically greatest term in the expansion of
$\begin{aligned} 2^{10} {(1 + \frac{3 x}{16})}^{10} \\ ∴ |\frac{T_{4}}{T_{3}}| \geq 1 and |\frac{T_{4}}{T_{5}}| \geq 1 \\ or, |\frac{T_{4}}{T_{3}}| \geq 1 and |\frac{T_{5}}{T_{4}}| \leq 1 \\ Now, \frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} \cdot x \end{aligned}$
Taking r = 3 and r = 4 and replacing x by 3x/16 x and n by 10, in the above two, we get
$|\frac{11 - 3}{3} \cdot \frac{3 x}{16}| \geq 1 and |\frac{11 - 4}{4} \cdot \frac{3 x}{16}| \leq 1$
or, $| x | \geq 2 and | x | \leq \frac{64}{21}$
If x is positive, then | x | = x
$\begin{aligned} ∴ x \geq 2 and x \leq \frac{64}{21} \\ ∴ 2 \leq x \leq \frac{64}{21} - - - - (1) \end{aligned}$
If x is negative, then | x | = – x
$∴ - x \geq 2 and - x \leq \frac{64}{21}$
or, $x \leq - 2 and x \geq - \frac{64}{21}$
$∴ - \frac{64}{21} \leq x \leq - 2$

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