JEE Advance - Mathematics (2021 - Paper 1 Online - No. 15)
For any positive integer n, let Sn : (0, $$\infty$$) $$\to$$ R be defined by $${S_n}(x) = \sum\nolimits_{k = 1}^n {{{\cot }^{ - 1}}\left( {{{1 + k(k + 1){x^2}} \over x}} \right)} $$, where for any x $$\in$$ R, $${\cot ^{ - 1}}(x) \in (0,\pi )$$ and $${\tan ^{ - 1}}(x) \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$. Then which of the following statements is (are) TRUE?
$${S_{10}}(x) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$$, for all x > 0
$$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = x$$, for all x > 0
The equation $${S_3}(x) = {\pi \over 4}$$ has a root in (0, $$\infty$$)
$$tan({S_n}(x)) \le {1 \over 2}$$, for all n $$\ge$$ 1 and x > 0
Explanation
For option (a) $${S_n}(x) = \sum\limits_{k = 1}^n {{{\cot }^{ - 1}}\left[ {{{1 + k(k + 1){x^2}} \over x}} \right]} $$ can be written as
$${S_n}(x) = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}\left[ {{{(k + 1)x - kx} \over {1 + kx\,.\,(k + 1)x}}} \right]} $$
$$ = \sum\limits_{k = 1}^n {[{{\tan }^{ - 1}}(k + 1)x - {{\tan }^{ - 1}}(kx)]} $$
$$ = {\tan ^{ - 1}}(n + 1)x - {\tan ^{ - 1}}x$$
$$ = {\tan ^{ - 1}}\left( {{{nx} \over {1 + (n + 1){x^2}}}} \right)$$
Now, $${S_{10}}(x) = ta{n^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right)$$
$$ = {\pi \over 2} - {\cot ^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$$
Option (a) is correct.
For option (b)
$$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = \cot \left( {{{\tan }^{ - 1}}\left( {{x \over {{x^2}}}} \right)} \right)$$
$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{1 \over x}} \right)} \right) = \cot (co{t^{ - 1}}x) = x,\,x > 0$$
Option (b) is correct.
For option (c)
$${S_3}(x) = {\pi \over 4} \Rightarrow {{3x} \over {1 + 4{x^2}}} = 1$$
$$ \Rightarrow 4{x^2} - 3x + 1 = 0$$ has no real root.
Option (c) is incorrect.
For option (d)
For $$x = 1,\,\tan ({S_n}(x)) = {n \over {n + 2}}$$ which is greater than $${1 \over 2}$$ for n $$\ge$$ 3
Option (d) is incorrect.
$${S_n}(x) = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}\left[ {{{(k + 1)x - kx} \over {1 + kx\,.\,(k + 1)x}}} \right]} $$
$$ = \sum\limits_{k = 1}^n {[{{\tan }^{ - 1}}(k + 1)x - {{\tan }^{ - 1}}(kx)]} $$
$$ = {\tan ^{ - 1}}(n + 1)x - {\tan ^{ - 1}}x$$
$$ = {\tan ^{ - 1}}\left( {{{nx} \over {1 + (n + 1){x^2}}}} \right)$$
Now, $${S_{10}}(x) = ta{n^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right)$$
$$ = {\pi \over 2} - {\cot ^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$$
Option (a) is correct.
For option (b)
$$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = \cot \left( {{{\tan }^{ - 1}}\left( {{x \over {{x^2}}}} \right)} \right)$$
$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{1 \over x}} \right)} \right) = \cot (co{t^{ - 1}}x) = x,\,x > 0$$
Option (b) is correct.
For option (c)
$${S_3}(x) = {\pi \over 4} \Rightarrow {{3x} \over {1 + 4{x^2}}} = 1$$
$$ \Rightarrow 4{x^2} - 3x + 1 = 0$$ has no real root.
Option (c) is incorrect.
For option (d)
For $$x = 1,\,\tan ({S_n}(x)) = {n \over {n + 2}}$$ which is greater than $${1 \over 2}$$ for n $$\ge$$ 3
Option (d) is incorrect.
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