SATHEE: Binomial Theorem 1 Question 6

Binomial Theorem 1 Question 6

7.

If the fourth term in the binomial expansion of $(\sqrt{x^{(} \frac{1}{1 + \log_{10} x})} + x^{\frac{1}{12}})^{6}$ is equal to $200$ , and $x > 1$ , then the

value of $x$ is

(a) $100$

(b) $10^{4}$

(d) $10^{3}$

Show Answer

Answer:

Correct Answer: 7. (c)

Solution:

Given binomial is $(\sqrt{x^{(} \frac{1}{1 + \log_{10} x}}) + x^{\frac{1}{12}})^{6}$

Since, the fourth term in the given expansion is 200.

$∴^{6} C_{3} (\frac{1}{x^{1 + \log_{10} x}})^{\frac{3}{2}} (x^{\frac{1}{12}})^{3} = 200$

$\begin{array}{lc} \Rightarrow & 20 \times x [^{\frac{3}{2 (1 + \log_{10} x)} + \frac{1}{4}}] = 200 \\ \Rightarrow & x^{\frac{3}{2 (1 + \log_{10} x)}} + \frac{1}{4} = 10 \\ \Rightarrow & [\frac{3}{2 (1 + \log_{10} x)} + \frac{1}{4}] \log_{10} x = 1 \end{array}$

[applying $\log_{10}$ both sides]

$\begin{aligned} \Rightarrow [6 + (1 + \log_{10} x)] \log_{10} x = 4 (1 + \log_{10} x) \\ \Rightarrow (7 + \log_{10} x) \log_{10} x = 4 + 4 \log_{10} x \\ \Rightarrow t^{2} + 7 t = 4 + 4 t [let \log_{10} x = t] \\ \Rightarrow t^{2} + 3 t - 4 = 0 \\ \Rightarrow t = 1, - 4 = \log_{10} x \\ \Rightarrow x = 10, 10^{- 4} \\ Since, x > 1 x = 10 \end{aligned}$

Binomial Theorem 1 Question 6

7.

Answer:

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Binomial Theorem 1 Question 6

7.

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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