If the function
f(x)= (72ˣ-9ˣ-8ˣ+1)/(√2-√(1+cos x)), x ≠ 0 ; a log ₑ 2 log ₑ 3 , x=0
is continuous at x=0, then the value of a² is equal to

Solution

Maths sir · 4 steps

Factor the top into (9ˣ-1)(8ˣ-1) and rationalise the bottom. Both then behave like x².

1. Factor the numerator. Since 72 = 9 × 8, the numerator is 9ˣ 8ˣ - 9ˣ - 8ˣ + 1 = (9ˣ - 1)(8ˣ - 1). Using (cˣ - 1)/x → ln c, it behaves like x² ln 9 ln 8 = 6x² ln 2 ln 3.
2. Rationalise the denominator. Multiply by the conjugate: √2 - √(1+cos x) = (1 - cos x)/(√2 + √(1+cos x)). As x → 0, 1 - cos x ≈ x²/2 and the conjugate tends to 2√2, so the denominator behaves like x²/4√2.

   lim(x → 0) f(x) = (6x² ln 2 ln 3)/(x² / (4√2)) = 24√2 ln 2 ln 3

3. Match with f(0). Continuity needs a ln 2 ln 3 = 24√2 ln 2 ln 3, so a = 24√2.
4. So. a² = 576 × 2 = 1152, which is option (c).

The idea to keep. When you see (ab)ˣ - aˣ - bˣ + 1, factor it as (aˣ-1)(bˣ-1). When you see a difference of square roots, rationalise it so a standard limit like 1-cos x shows up.

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