If x 0 3 + (x) + (x) + (1 - x)3 ^2(x) = 13, then 2 - is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2024Limits

If $\lim_{x\to 0} \frac{3 + \alpha \sin(x) + \beta \cos(x) + \log(1 - x)}{3\tan^2(x)} = \frac{1}{3}$ , then $2\alpha - \beta$ is equal to:

Answer

Correct answer:C

Standard Method

Given:

\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log(1-x)}{3\tan^2 x} = \frac{1}{3}

Find: $2\alpha - \beta$

Using the small-angle and Taylor expansions shown in the solution:

\sin x \approx x, \qquad \cos x \approx 1, \qquad \log(1-x) \approx -x, \qquad \tan x \approx x

So the numerator becomes

3 + \alpha x + \beta - x = 3 + \beta + (\alpha - 1)x

and the denominator becomes

3\tan^2 x \approx 3x^2

Hence the limit reduces to

\lim_{x \to 0} \frac{3 + \beta + (\alpha - 1)x}{3x^2}

For this limit to exist and be finite, the constant term and the coefficient of $x$ in the numerator must vanish.

Therefore,

3 + \beta = 0 \implies \beta = -3

and

\alpha - 1 = 0 \implies \alpha = 1

Now compute

2\alpha - \beta = 2(1) - (-3) = 5

Therefore, the correct option is C.

Taylor Expansion Method

Given:

\lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e(1-x)}{3\tan^2 x} = \frac{1}{3}

Find: $2\alpha - \beta$

Using the expansions stated in the solution:

\sin x \approx x, \qquad \cos x \approx 1 - \frac{x^2}{2}, \qquad \log_e(1-x) \approx -x - \frac{x^2}{2}

Substituting these into the numerator,

3 + \alpha x + \beta\left(1 - \frac{x^2}{2}\right) - x - \frac{x^2}{2}

which simplifies to

3 + \beta + (\alpha - 1)x + \left(-\frac{\beta}{2} - \frac{1}{2}\right)x^2

The solution states that for the limit to be finite, the lower-order terms must vanish. Thus,

3 + \beta = 0 \implies \beta = -3

and

\alpha - 1 = 0 \implies \alpha = 1

Now,

2\alpha - \beta = 2 \times 1 - (-3) = 5

Therefore, the value of $2\alpha - \beta$ is $5$ , so the correct option is C.

Setting only the constant term to zero and forgetting the coefficient of $x$ . Since the denominator behaves like $x^2$ , both the constant term and the linear term in the numerator must vanish to keep the limit finite.
Using only $\sin x \approx x$ and $\tan x \approx x$ but mishandling $\log(1-x)$ . Near $x=0$ , the correct first terms are $\log(1-x) \approx -x - \frac{x^2}{2}$ , not $+x$ .
Confusing the condition 'limit equals $\frac{1}{3}$ ' with directly equating the numerator to $1$ . The limit is determined by matching the order of the numerator with the denominator, not by comparing isolated constants.

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