If for p ≠ q ≠ 0, the function f(x)=(⁷√p(729+x)-3)/(∛(729+q x)-9) is continuous at x=0, then :

Solution

Maths sir · 4 steps

The denominator goes to 0 at x=0, so the numerator must too. That pins down p before you do anything else.

1. Find p first. At x=0 the denominator is ∛729-9 = 0. For the limit to be finite, the numerator must also vanish: (729p)^(1/7) = 3, so 729p = 3⁷ = 2187 and p = 3.
2. Use the standard limit. Use lim(t→ a)(tⁿ-aⁿ)/(t-a) = naⁿ⁻¹. In the numerator, with u = (2187+3x)^(1/7), we have u⁷-3⁷ = 3x, so u-3 ≈ 3x/(7· 3⁶). In the denominator, with v = (729+qx)^(1/3), we have v³-9³ = qx, so v-9 ≈ qx/(3· 9²).

   f(0)=lim(x→ 0)(3x/(7× 729))/(qx/243) = (3× 243)/(7× 729 q) = 1/7q

3. Test the options. 63q f(0) = 63q/7q = 9, and p² = 9, so (b) holds. Also 7pq f(0) = 3 and 21q f(0) = 3, which rule out (a), (c) and (d).
4. So. With p = 3 and f(0) = 1/7q, the true relation is 63q f(0) - p² = 0, which is option (b).

The idea to keep. When a 0/0 form must give a finite limit, first force the numerator to vanish to find the hidden constant. Then turn each root into (tⁿ-aⁿ)/(t-a) → naⁿ⁻¹.

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