Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma926c: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma926c Solution: For any positive integer n, define as for all x (0, ). (Here,…

For any positive integer n, define

fn:(0,∞)→R{f_n}:(0,\infty ) \to R as

fn=∑j=1ntan⁡−1(11+(x+j)(x+j−1)){f_n} = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)

for all x ∈\in (0, ∞\infty ). (Here, the inverse trigonometric function tan −- 1 x assumes values in (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right) ). Then, which of the following statement(s) is (are) TRUE?

Multiple Choice Options

  • A ∑j=15tan⁡2(fj(0))=55\sum\limits_{j = 1}^5 {{{\tan }^2}({f_j}(0)) = 55}
  • B ∑j=110(1+fj′(0))sec⁡2(fj(0))=10\sum\limits_{j = 1}^{10} {(1 + f{'_j}(0)){{\sec }^2}({f_j}(0)) = 10}
  • C For any fixed positive integer n, lim⁡x→∞tan⁡(fn(x))=1n\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = {1 \over n}
  • D For any fixed positive integer n, lim⁡x→∞sec⁡2(fn(x))=1\mathop {\lim }\limits_{x \to \infty } {\sec ^2}({f_n}(x)) = 1

Step-by-Step Verified Solution

We have,

fn(x)=∑j=1ntan⁡−1(11+(x+j)(x+j−1)){f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)

⇒fn(x)=∑j=1ntan⁡−1((x+j)−(x+j−1)1+(x+j)(x+j−1))\Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{{(x + j) - (x + j - 1)} \over {1 + (x + j)(x + j - 1)}}} \right)

⇒fn(x)=∑j=1n[tan⁡−1(x+j)−tan⁡−1(x+j−1)]\Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {[{{\tan }^{ - 1}}} (x + j) - {\tan ^{ - 1}}(x + j - 1)]

for all x∈(0,∞)x \in (0,\infty )

⇒fn(x)=(tan⁡−1(x+1)−tan⁡−1x)+(tan−1(x+2)−tan⁡−1(x+1))+(tan⁡−1(x+3)−tan⁡−1(x+2))+...+(tan⁡−1(x+n)−tan⁡−1(x+n−1))\Rightarrow {f_n}(x) = ({\tan ^{ - 1}}(x + 1) - {\tan ^{ - 1}}x) + (ta{n^{ - 1}}(x + 2) - {\tan ^{ - 1}}(x + 1)) + ({\tan ^{ - 1}}(x + 3) - {\tan ^{ - 1}}(x + 2)) + ... + ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}(x + n - 1))

⇒fn(x)=tan⁡−1(x+n)−tan⁡−1x\Rightarrow {f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x

This statement is false as x≠0x \ne 0 . i.e., x∈(0,∞)x \in (0,\infty )

(b) This statement is also false as 0 ∉\notin (0, ∞\infty )

(c) fn(x)=tan⁡−1(x+n)−tan⁡−1x{f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x

lim⁡x→∞tan⁡(fn(x))=lim⁡x→∞tan⁡(tan⁡−1(x+n)−tan⁡−1x)\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x)

⇒lim⁡x→∞tan⁡(fn(x))=lim⁡x→∞tan⁡(tan⁡−1n1+nx+x2)=lim⁡x→∞n1+nx+x2=0\Rightarrow \mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}{n \over {1 + nx + {x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } {n \over {1 + nx + {x^2}}} = 0

∴\therefore (c) statement is false.

(d) lim⁡x→∞sec⁡2(fn(x))=lim⁡x→∞(1+tan⁡2fn(x))=1+lim⁡x→∞tan⁡2(fn(x))=1+0=1\mathop {\lim }\limits_{x \to \infty } {\sec ^2}({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } (1 + {\tan ^2}{f_n}(x)) = 1 + \mathop {\lim }\limits_{x \to \infty } {\tan ^2}({f_n}(x)) = 1 + 0 = 1

∴\therefore (d) statement is true.

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