If lim x → 0 (3+α sin x+β cos x+log ₑ(1-x))/(3 tan ² x)=1/3 , then 2 α-β is equal to :

Solution

Maths sir · 4 steps

The denominator starts at x², so the numerator's constant and x terms must vanish. Kill them one by one.

1. Expand everything. Near x = 0, sin x ≈ x, cos x ≈ 1 - x²/2 and logₑ(1-x) ≈ -x - x²/2. The denominator is 3 tan² x ≈ 3x².

   Numerator = (3+β) + (α - 1) x + (-β/2 - 1/2)x² + …

2. Finite limit. Dividing by 3x² gives a finite limit only if the constant and x terms are zero. So 3 + β = 0, giving β = -3, and α - 1 = 0, giving α = 1.
3. Check the value. The x² coefficient is 3/2 - 1/2 = 1, so the limit is 1/3, as given. The values are consistent.
4. So. 2α - β = 2(1) - (-3) = 5, which is option (c).

The idea to keep. When a limit is finite but the denominator behaves like xⁿ, every numerator term below xⁿ must be zero. Series expansion gives all the conditions at once, faster than repeated L'Hopital.

Answer from the NTA final answer key. Solution by PrepWiser.