Which of the following number is the coefficient of in the expansion of , ?

0.01 Explanation: Step 1: Standard Logarithmic Expansion Formula Recall the standard Maclaurin series expansion for when : Step 2: Simplify the Given Logarithmic Expression Using the logarithmic property , we can break down the expression: Step 3: Expand Both Terms Independently Expanding : The term containing in this expansion corresponds to : Expanding : Substitute into the standard expansion formula: We want the power of to be , so we set : Step 4: Calculate the Combined Coefficient of Now combine the components from both expansions according to our simplified expression: Substitute the calculated coefficients: Find a common denominator: Convert the fraction into its decimal representation: Conclusion The coefficient of in the given expansion is 0.01 . The correct option is A) 0.01 .

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NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

Which of the following number is the coefficient of $x^{100}$ in the expansion of $lo g_{e} (\frac{1 + x}{1 + x ^{2}})$ , $|x|<1$ ?

Choose the correct answer:

Explanation

Step 1: Standard Logarithmic Expansion Formula

Recall the standard Maclaurin series expansion for $lo g_{e} (1 + t)$ when $|t| < 1$ :

$lo g_{e} (1 + t) = t - \frac{t ^{2}}{2} + \frac{t ^{3}}{3} - \frac{t ^{4}}{4} + \dots + (- 1)^{n - 1} \frac{t ^{n}}{n} + \dots$

Step 2: Simplify the Given Logarithmic Expression

Using the logarithmic property $lo g_{e} (\frac{M}{N}) = lo g_{e} M - lo g_{e} N$ , we can break down the expression:

$lo g_{e} (\frac{1 + x}{1 + x ^{2}}) = lo g_{e} (1 + x) - lo g_{e} (1 + x^{2})$

Step 3: Expand Both Terms Independently

Expanding $lo g_{e} (1 + x)$ :
$lo g_{e} (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots + (- 1)^{n - 1} \frac{x ^{n}}{n} + \dots$
The term containing $x^{100}$ in this expansion corresponds to $n = 100$ :
$Term = (- 1)^{100 - 1} \frac{x ^{100}}{100} = (- 1)^{99} \frac{x ^{100}}{100} = - \frac{x ^{100}}{100}$
Expanding $lo g_{e} (1 + x^{2})$ :
Substitute $t = x^{2}$ into the standard expansion formula:
$lo g_{e} (1 + x^{2}) = (x^{2}) - \frac{( x ^{2} ) ^{2}}{2} + \frac{( x ^{2} ) ^{3}}{3} - \dots + (- 1)^{k - 1} \frac{( x ^{2} ) ^{k}}{k} + \dots$
$lo g_{e} (1 + x^{2}) = x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{3} - \dots + (- 1)^{k - 1} \frac{x ^{2 k}}{k} + \dots$
We want the power of $x$ to be $100$ , so we set $2 k = 100 ⟹ k = 50$ :
$Term = (- 1)^{50 - 1} \frac{x ^{100}}{50} = (- 1)^{49} \frac{x ^{100}}{50} = - \frac{x ^{100}}{50}$

Step 4: Calculate the Combined Coefficient of $x^{100}$

Now combine the components from both expansions according to our simplified expression:

$Coefficient of x^{100} = (Coefficient from lo g_{e} (1 + x)) - (Coefficient from lo g_{e} (1 + x^{2}))$

Substitute the calculated coefficients:

$Coefficient of x^{100} = (- \frac{1}{100}) - (- \frac{1}{50})$

$Coefficient of x^{100} = - \frac{1}{100} + \frac{1}{50}$

Find a common denominator:

$Coefficient of x^{100} = \frac{- 1 + 2}{100} = \frac{1}{100}$

Convert the fraction into its decimal representation:

$\frac{1}{100} = 0.01$

Conclusion

The coefficient of $x^{100}$ in the given expansion is 0.01.

The correct option is A) 0.01.