SATHEE: Binomial Theorem 1 Question 12

Binomial Theorem 1 Question 12

13.

If the third term in the binomial expansion of ${(1 + x^{\log_{2} x})}^{5}$ equals 2560 , then a possible value of $x$ is

(a) $4 \sqrt{2}$

(b) $\frac{1}{4}$

(d) $2 \sqrt{2}$

(2019 Main, 10 Jan I)

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Answer:

Correct Answer: 13. (b)

Solution:

The $(r + 1)$ th term in the expansion of $(a + x)^{n}$ is given by $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$

$∴ 3^{rd}$ term in the expansion of ${(1 + x^{\log_{2} x})}^{5}$ is

$^{5} C_{2} (1)^{5 - 2} {(x^{\log_{2} x})}^{2}$

$\Rightarrow^{5} C_{2} (1)^{5 - 2} {(x^{\log_{2} x})}^{2} = 2560$ (given)

$\Rightarrow 10 {(x^{\log_{2} x})}^{2} = 2560$

$\Rightarrow x^{(2 \log_{2} x)} = 256$

$\Rightarrow \log_{2} x^{2 \log_{2} x} = \log_{2} 256$

taking $\log_{2} on both sides$ $(∵ \log_{2} 256 = \log_{2} 2^{8} = 8)$

$\Rightarrow 2 (\log_{2} x) (\log_{2} x) = 8$

${(\log_{2} x)}^{2} = 4$

$\Rightarrow \log_{2} x = \pm 2$

$\Rightarrow \log_{2} x = 2 or \log_{2} x = - 2$

$\Rightarrow x = 4 or x = 2^{- 2} = \frac{1}{4}$

Binomial Theorem 1 Question 12

13.

Answer:

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Binomial Theorem 1 Question 12

13.

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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