SATHEE: Binomial Theorem 1 Question 5

Binomial Theorem 1 Question 5

6.

If the fourth term in the binomial expansion of $(\frac{2}{x} + x^{\log_{8} x)^{6}} (x > 0)$ is $20 \times 8^{7}$ , then the value of $x$ is

(a) $8^{- 2}$

(b) $8^{3}$

(d) $8^{2}$

(2019 Main, 9 April I)

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Answer:

Correct Answer: 6. (d)

Solution:

Given binomial is $(\frac{2}{x} + x^{\log_{8} x)^{6}}$

Since, general term in the expansion of $(x + a)^{n}$ is $T_{r + 1} =^{n} C_{r} x^{n - r} a^{r}$

$∴ T_{4} = T_{3 + 1} =^{6} C_{3} (\frac{2}{x})^{6 - 3} {(x^{\log_{8} x})}^{3} = 20 \times 8^{7} (given)$

$\Rightarrow 20 (\frac{2^{3}}{x}) x^{3 \log_{8} x} = 20 \times 8^{7}$

$∵ [^{6} C_{3} = 20]$

$\Rightarrow 2^{3} x^{[3 (\log_{8} x) - 3]}$ $= (2^{3})^{7} \Rightarrow x^{(} \frac{3}{3} \log_{2} x - 3)$ = $(2^{3})^{6}$

$∵ [\log_{a^{n}} (x) = \frac{1}{n} \log_{a} x for x > 0; a > 0, \neq 1]$

$\Rightarrow x^{\log_{2} x - 3} = 2^{18}$

On taking $\log_{2}$ both sides, we get

$\begin{aligned} (\log_{2} x - 3) \log_{2} x = 18 \\ \Rightarrow {(\log_{2} x)}^{2} - 3 \log_{2} x - 18 = 0 \\ \Rightarrow {(\log_{2} x)}^{2} - 6 \log_{2} x + 3 \log_{2} x - 18 = 0 \\ \Rightarrow \log_{2} x (\log_{2} x - 6) + 3 (\log_{2} x - 6) = 0 \\ \Rightarrow (\log_{2} x - 6) (\log_{2} x + 3) = 0 \\ \Rightarrow \log_{2} x = - 3, 6 \\ \Rightarrow x = 2^{- 3}, 2^{6} \Rightarrow x = \frac{1}{8}, 8^{2} \end{aligned}$

Binomial Theorem 1 Question 5

6.

Answer:

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Binomial Theorem 1 Question 5

6.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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