Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x) = \begin{cases} \frac{x^{32}-2^{32}+4^{16}-1}{(x-1)^2} & , x \neq 1 \\ =k & , x=1 \end{cases}$, the value of $k$ so that the function is continuous at $x = 1$ is:
$2^{63}-2^{31}$
$2^{65}-2^{33}$
$(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2^{32}+2^{31})$
$(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2^{33}+2^{31})$
Step-by-Step Solution
Key Concept: Expand $x^n - 1$ as a geometric series and use polynomial division to identify the coefficient structure of the limit.
With $2^{32} = n$, rewrite the limit as $k = \lim_{x\to n}\frac{x^n - 1 - n(x-1)}{(x-1)^2}$. Expand the numerator using the binomial pattern: $x^n - 1 = (x-1)(1 + x + \cdots + x^{n-1})$. After canceling $(x-1)$ and simplifying, the sum becomes $(n-1) + (n-2) + \cdots + 1 = \frac{n(n-1)}{2} = \frac{2^{32}(2^{32}-1)}{2} = 2^{63} - 2^{31}$.
Correct Answer: 1,3