JEE MATHEMATICS MCQ: LIMITS

Solution approach

1. Correct Answer + Core Reason: 18. L'Hopital's rule is applied to evaluate the limit after expressing g(x) as a ratio of integrals.
   
   Step-by-Step Solution:
   
   Step 1: Express g(x) in terms of the integral.\nGiven $\int_6^{f(x)} 4t^3 dt = (x - 2)g(x)$.
   $g(x) = \frac{{\int_6^{f(x)} 4t^3 dt}}{{x - 2}}$.
   
   Step 2: Evaluate the limit as x approaches 2.
   $\lim\limits_{x \to 2} g(x) = \lim\limits_{x \to 2} \frac{{\int_6^{f(x)} 4t^3 dt}}{{x - 2}}$.\nThis is in the indeterminate form $\frac{0}{0}$ since $f(2) = 6$.
   
   Step 3: Apply L'Hopital's rule.
   $\lim\limits_{x \to 2} g(x) = \lim\limits_{x \to 2} \frac{{\frac{d}{{dx}}\int_6^{f(x)} 4t^3 dt}}{{\frac{d}{{dx}}(x - 2)}} = \lim\limits_{x \to 2} \frac{{4[f(x)]^3 f'(x)}}{1}$.
   
   Step 4: Substitute the given values.\nGiven $f(2) = 6$ and $f'(2) = \frac{1}{{{48}}}$.
   $\lim\limits_{x \to 2} g(x) = 4[f(2)]^3 f'(2) = 4(6)^3 \cdot \frac{1}{{{48}}} = 4(216) \cdot \frac{1}{{{48}}} = \frac{{864}}{{48}} = 18$.
   
   Why other options are wrong:
   - Option B: 24 is incorrect because the correct limit is 18.
   - Option C: 12 is incorrect because the correct limit is 18.
   - Option D: 36 is incorrect because the correct limit is 18.
   
   Key Concepts:
   - L'Hopital's Rule
   - Fundamental Theorem of Calculus
   
   Memory Aids:
   - Remember L'Hopital's rule for indeterminate forms.
   - Remember the Fundamental Theorem of Calculus for differentiating integrals.
   
   Quick Tips:
   - Check for indeterminate forms before applying L'Hopital's rule.
   - Use the Fundamental Theorem of Calculus to differentiate integrals.