JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 1)
For any positive integer n, define
$${f_n}:(0,\infty ) \to R$$ as
$${f_n} = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)$$
for all x$$ \in $$(0, $$\infty $$). (Here, the inverse trigonometric function tan$$-$$1 x assumes values in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$). Then, which of the following statement(s) is (are) TRUE?
$${f_n}:(0,\infty ) \to R$$ as
$${f_n} = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)$$
for all x$$ \in $$(0, $$\infty $$). (Here, the inverse trigonometric function tan$$-$$1 x assumes values in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$). Then, which of the following statement(s) is (are) TRUE?
$$\sum\limits_{j = 1}^5 {{{\tan }^2}({f_j}(0)) = 55} $$
$$\sum\limits_{j = 1}^{10} {(1 + f{'_j}(0)){{\sec }^2}({f_j}(0)) = 10} $$
For any fixed positive integer n, $$\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = {1 \over n}$$
For any fixed positive integer n, $$\mathop {\lim }\limits_{x \to \infty } {\sec ^2}({f_n}(x)) = 1$$
Explanation
We have,
$${f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)$$
$$ \Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{{(x + j) - (x + j - 1)} \over {1 + (x + j)(x + j - 1)}}} \right)$$
$$ \Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {[{{\tan }^{ - 1}}} (x + j) - {\tan ^{ - 1}}(x + j - 1)]$$
for all $$x \in (0,\infty )$$
$$ \Rightarrow {f_n}(x) = ({\tan ^{ - 1}}(x + 1) - {\tan ^{ - 1}}x) + (ta{n^{ - 1}}(x + 2) - {\tan ^{ - 1}}(x + 1)) + ({\tan ^{ - 1}}(x + 3) - {\tan ^{ - 1}}(x + 2)) + ... + ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}(x + n - 1))$$
$$ \Rightarrow {f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x$$
This statement is false as $$x \ne 0$$. i.e., $$x \in (0,\infty )$$
(b) This statement is also false as 0$$ \notin $$(0, $$\infty $$)
(c) $${f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x$$
$$\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x)$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}{n \over {1 + nx + {x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } {n \over {1 + nx + {x^2}}} = 0$$
$$ \therefore $$ (c) statement is false.
(d) $$\mathop {\lim }\limits_{x \to \infty } {\sec ^2}({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } (1 + {\tan ^2}{f_n}(x)) = 1 + \mathop {\lim }\limits_{x \to \infty } {\tan ^2}({f_n}(x)) = 1 + 0 = 1$$
$$ \therefore $$ (d) statement is true.
$${f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{1 \over {1 + (x + j)(x + j - 1)}}} \right)$$
$$ \Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {{{\tan }^{ - 1}}} \left( {{{(x + j) - (x + j - 1)} \over {1 + (x + j)(x + j - 1)}}} \right)$$
$$ \Rightarrow {f_n}(x) = \sum\limits_{j = 1}^n {[{{\tan }^{ - 1}}} (x + j) - {\tan ^{ - 1}}(x + j - 1)]$$
for all $$x \in (0,\infty )$$
$$ \Rightarrow {f_n}(x) = ({\tan ^{ - 1}}(x + 1) - {\tan ^{ - 1}}x) + (ta{n^{ - 1}}(x + 2) - {\tan ^{ - 1}}(x + 1)) + ({\tan ^{ - 1}}(x + 3) - {\tan ^{ - 1}}(x + 2)) + ... + ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}(x + n - 1))$$
$$ \Rightarrow {f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x$$
This statement is false as $$x \ne 0$$. i.e., $$x \in (0,\infty )$$
(b) This statement is also false as 0$$ \notin $$(0, $$\infty $$)
(c) $${f_n}(x) = {\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x$$
$$\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan ({\tan ^{ - 1}}(x + n) - {\tan ^{ - 1}}x)$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } \tan \left( {{{\tan }^{ - 1}}{n \over {1 + nx + {x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } {n \over {1 + nx + {x^2}}} = 0$$
$$ \therefore $$ (c) statement is false.
(d) $$\mathop {\lim }\limits_{x \to \infty } {\sec ^2}({f_n}(x)) = \mathop {\lim }\limits_{x \to \infty } (1 + {\tan ^2}{f_n}(x)) = 1 + \mathop {\lim }\limits_{x \to \infty } {\tan ^2}({f_n}(x)) = 1 + 0 = 1$$
$$ \therefore $$ (d) statement is true.
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