Limits, Continuity & Differentiability
General
Grade 12
Question:
<p>For any positive integer n, define
fn(x) =
n
X
j=1
tan−1
1
1 + (x + j)(x + j −1)
,
x ∈(0, ∞).
Then which of the following statement(s) is(are) true?</p>
Step-by-Step Solution
Key Concept: Use the identity
tan-1
1
1 + uv
= tan-1(u) -tan-1(v)
with u = x + j and v = x + j -1.
<p>We have</p> tan-1 1 1 + (x + j)(x + j -1) = tan-1(x + j) -tan-1(x + j -1). Therefore the sum telescopes: fn(x) = tan-1(x + n) -tan-1(x). Hence, for fixed n, fn(x) \to 0 (x \to \infty). So sec2(fn(x)) \to sec2(0) = 1. That is exactly the true statement recorded by the official key. Shortcut / Fast View If a tan-1-sum has the pattern 1 + (x + j)(x + j -1) in the denominator, telescoping is the intended move.
Correct Answer: 4