JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f:R \to R\) be a continuously differentiable function such that \(f\left( 2 \right) = 6\) and \(f'\left( 2 \right) = \frac{1}{{48}}\). If \(\int_6^{f\left( x \right)} {4{t^3}} \,dt = \left( {x - 2} \right)\,g\left( x \right)\), then \(\mathop {\lim }\limits_{x \to 2} \,g\left( x \right)\) is equal to
- A \(24\)
- B \(18\)
- C \(12\)
- D \(36\)
Answer & Solution
Correct Answer
(B) \(18\)
Step-by-step Solution
Detailed explanation
\(\mathop {\lim }\limits_{x \to 2} \frac{{\int\limits_6^{f\left( x \right)} {4{t^2}dt} }}{{x - 2}}\) \( = \mathop {\lim }\limits_{x \to 2} \frac{{4.{f^3}\left( x \right).f'\left( x \right)}}{1}\) \( = 4{f^3}\left( 2 \right)f'\left( 2 \right) = 18\)
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