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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

જો \(p \neq q \neq 0\) માટે વિધેય  \(f(x)=\frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+q x}-9}\) એ  \(x=0\) આગળ સતત હોય તો .. . .  

  1. A \(7 pq f (0)-1=0\)
  2. B \(63 q f(0)- p ^{2}=0\)
  3. C \(21 q f(0)-p^{2}=0\)
  4. D \(7 pq f (0)-9=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(63 q f(0)- p ^{2}=0\)

Step-by-step Solution

Detailed explanation

\(f(0)=\lim _{x \rightarrow 0} f(x)\) Limit should be \(\frac{0}{0}\) form So, \(\sqrt[7]{ p .729}-3=0 \Rightarrow p \cdot 3^{6}=3^{7} \Rightarrow p =3\) Now, \(f(0)=\lim _{x \rightarrow 0} \frac{\sqrt[7]{3\left(3^{6}+x\right)}-3}{\sqrt[3]{3^{6}+q x}-9}\)…
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