If for p ≠ q ≠ 0, the function
\(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\)
is continuous at x = 0, then
If for p ≠ q ≠ 0, the function
\(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\)
is continuous at x = 0, then
\(7pq f(0)-1 = 0\)
\(63qf (0)-p^2 = 0\)
\(21qf(0) - p^2 = 0\)
\(7pqf(0)-9 = 0\)
The Correct Option is B
Solution and Explanation
The correct answer is (B) : \(63qf (0)-p^2 = 0\)
\(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\)
For continuity at x = 0, \(\lim_{{x \to 0}} f(x) = f(0)\)
Now,
\(\therefore \lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{{\sqrt[7]{p(729+x)} - 3}}{{\sqrt[3]{729 + qx }- 9}}\)
⇒ p = 3 (To make indeterminant form)
So,
\(\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{(3^7+3x)^{\frac{1}{7}}-3}{(729+qx)^{\frac{1}{3}}-9}\)
\(\lim_{{x \to 0}} \frac{3[(1+\frac{x}{3^6})^{\frac{1}{7}}-1]}{9[(1+\frac{q}{729})^{\frac{1}{3}}-1]}\)
\(= \frac{1}{3}.\frac{\frac{1}{7}.\frac{1}{3^6}}{\frac{1}{3}.\frac{q}{729}}\)
\(∴ f(0) = \frac{1}{7q}\)
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.