Question

Let \(f:\mathbb{R}\to\mathbb{R}\) be a continuously differentiable function such that \(f(2)=6\) and \(f'(2)=\frac{1}{48}\). If
\(\int_{6}^{f(x)} 4t^{3}\,dt=(x-2)\,g(x)\),
then \(\lim_{x\to 2} g(x)\) is equal to :

• A. 18
• B. 36
• C. 12
• D. 24

Hint:

Evaluate the integral to get \(f(x)^4-6^4=(x-2)g(x)\); then \(g(2)\) is the derivative of \(f(x)^4\) at \(x=2\).

Answer 1

The Correct Option is A

The integral gives \(\int_6^{f(x)}4t^3dt=[t^4]_6^{f(x)}=f(x)^4-6^4\).
So \(g(x)=\frac{f(x)^4-f(2)^4}{x-2}\) (since \(f(2)=6\)).
As \(x\to2\), this is the derivative of \(f(x)^4\) at \(x=2\): \(g(2)=4f(2)^3f'(2)\).
\(=4\cdot 6^3\cdot\frac{1}{48}=\frac{4\cdot216}{48}=18\).
Answer: (1) 18