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

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Limits

lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}

, then

2 α - β

is equal to :

Answer: (C)

x \to 0 lim \frac{3 + α sin x + β cos x + lo g _{e} ( 1 - x )}{3 tan ^{2} x} = \frac{1}{3} .

Expand each term to

O (x^{3})

sin x = x - \frac{x ^{3}}{6} + \dots

cos x = 1 - \frac{x ^{2}}{2} + \dots

lo g_{e} (1 - x) = - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{3} - \dots

tan^{2} x = x^{2} + \dots

Numerator

= 3 + α x + β (1 - \frac{x ^{2}}{2}) + (- x - \frac{x ^{2}}{2}) + O (x^{3})

= (3 + β) + (α - 1) x - (\frac{β}{2} + \frac{1}{2}) x^{2} + O (x^{3}) .

The denominator behaves like

3 x^{2}

, so for the limit to be finite, the constant and

x

-coefficients must vanish:

3 + β = 0 \Rightarrow β = - 3, α - 1 = 0 \Rightarrow α = 1.

Then the coefficient of

x^{2}

- \frac{( - 3 )}{2} - \frac{1}{2} = 1

, giving the limit

\frac{1}{3}

, matching the given value.

2 α - β = 2 (1) - (- 3) = 5.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option