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1

### JEE Advanced 2021 Paper 1 Online

MCQ (More than One Correct Answer)
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?
A
$${S_{10}}(x) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$$, for all x > 0
B
$$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = x$$, for all x > 0
C
The equation $${S_3}(x) = {\pi \over 4}$$ has a root in (0, $$\infty$$)
D
$$tan({S_n}(x)) \le {1 \over 2}$$, 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.
2

### JEE Advanced 2018 Paper 2 Offline

MCQ (More than One Correct Answer)
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?
A
$$\sum\limits_{j = 1}^5 {{{\tan }^2}({f_j}(0)) = 55}$$
B
$$\sum\limits_{j = 1}^{10} {(1 + f{'_j}(0)){{\sec }^2}({f_j}(0)) = 10}$$
C
For any fixed positive integer n, $$\mathop {\lim }\limits_{x \to \infty } \tan ({f_n}(x)) = {1 \over n}$$
D
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.
3

### JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
If $$\alpha$$ $$= 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$ and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right),$$ where the inverse trigonimetric functions take only the principal values, then the correct options(s) is (are)
A
$$cos\beta > 0$$
B
$$\sin \beta < 0$$
C
$$\cos \left( {\alpha + \beta } \right) > 0$$
D
$$\cos \alpha < 0$$

## Explanation

Here, $$\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$

and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$ as $${6 \over {11}} > {1 \over 2}$$

$$\Rightarrow {\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 6}$$

$$\therefore$$ $$\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\pi \over 2} \Rightarrow \cos \alpha < 0$$

Now, $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$

As $${4 \over 9} < {1 \over 2} \Rightarrow {\cos ^{ - 1}}\left( {{4 \over 9}} \right) > {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$$

$$\therefore$$ $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right) > \pi$$

$$\therefore$$ $$\cos \beta < 0$$ and $$\sin \beta < 0$$

Now, $$\alpha + \beta$$ is slightly greater than $${{3\pi } \over 2}$$.

$$\therefore$$ $$\cos (\alpha + \beta ) > 0$$

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