If lim x ? 0 3 + sin x + cos x + log e ( 1 - x ) 3 tan 2 x = 1 3 , then 2 is equal to [2024]

Question

If $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3},$ then $2 α - β$ is equal to [2024]

Smart Achievers · Accepted Answer

(1)

Given, $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3}$

$\Rightarrow \lim_{x \to 0} \frac{3 + α (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots) + β (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots) + (- x - \frac{x^{2}}{2!} - \frac{2 x^{3}}{3!} - \dots)}{3 {(\frac{\tan x}{x})}^{2} \cdot x^{2}} = \frac{1}{3}$

$\Rightarrow \lim_{x \to 0} \frac{3 + α (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots) + β (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots) + (- x - \frac{x^{2}}{2!} - \frac{2 x^{3}}{3!} - \dots)}{3 x^{2}} = \frac{1}{3}$

$Comparing the coefficient of x^{0}, we get$

$3 + β = 0 \Rightarrow β = - 3$

$Comparing the coefficient of x^{'}, we get$

$α - 1 = 0 \Rightarrow α = 1$

$∴$ $2 α - β = 2 (1) - (- 3) = 2 + 3 = 5$

Solution

Q.
If $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3},$ then $2 α - β$ is equal to [2024]

1 5

2 1

3 7

4 2

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Solution

Q. If limx→03+αsinx+βcosx+loge(1-x)3tan2x=13, then 2α-β is equal to [2024]

1 5

2 1

3 7

4 2

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Q.
If $\lim_{x \to 0} \frac{3 + α \sin x + β \cos x + \log_{e} (1 - x)}{3 \tan^{2} x} = \frac{1}{3},$ then $2 α - β$ is equal to [2024]