If lim(x → 0) (α eˣ+β e⁻ˣ+γ sin x)/(x sin ² x)=2/3, where α, β, γ ∈ R, then which of the following is NOT correct?

Solution

Maths sir · 4 steps

The denominator behaves like x³, so expand the top and kill every power below x³.

1. Size of denominator. Since sin x ≈ x near 0, x sin² x ≈ x³. For a finite non-zero limit, the numerator must also start at x³.

   α eˣ+β e⁻ˣ+γ sin x=(α+β)+(α-β+γ)x+(α+β)/2x²+(α-β-γ)/6x³+…

2. Match coefficients. The constant and x terms must vanish, so α+β=0 and α-β+γ=0. The x² term then vanishes too. The x³ coefficient gives the limit, so (α-β-γ)/6=2/3, that is α-β-γ=4.
3. Solve. Adding α-β+γ=0 and α-β-γ=4 gives α-β=2 and γ=-2. With α+β=0 this gives α=1 and β=-1.

   Option	Value with α=1, β=-1, γ=-2	Holds?
   (a)	1+1+4=6	yes
   (b)	-1+2-2+1=0	yes
   (c)	1-4-2+3=-2	no
   (d)	1-1+4=4	yes

4. So. Only (c) fails, because αβ²+βγ²+γα²+3=-2 ≠ 0. The answer is (c).

The idea to keep. When the denominator behaves like xⁿ, expand the numerator in a series. Every coefficient below xⁿ must be zero, and the xⁿ coefficient gives the limit.

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