For any positive integer n, let Sn : (0, \infty ) \to R be defined by S_n(x) = \sum olimits_k = 1^n \cot ^ - 1( 1 + k(k + 1)x^2 \over x ) , where for any x \in R, \cot ^ - 1(x) \in (0,\pi ) and \tan ^ - 1(x) \in ( - \pi \over 2,\pi \over 2 ) . Then which of the following statements is (are) TRUE?

S_10(x) = \pi \over 2 - \tan ^ - 1( 1 + 11x^2 \over 10x ) , for all x > 0, \mathop \lim \limits_n \to \infty \cot (S_n(x)) = x , for all x > 0

S_10(x) = \pi \over 2 - \tan ^ - 1( 1 + 11x^2 \over 10x ) , 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

Sequences and Series PYQ — JEE Advanced 2021 · Shift 1 · Q34 with Solution

View written solutionFree

Correct answer: A, B

Rewrite each term using an arctangent identity

Let

\theta_k=\cot^{-1}\!\left(\frac{1+k(k+1)x^2}{x}\right),\qquad x>0.

Since the argument of $\cot^{-1}$ is positive, we have $\theta_k\in(0,\pi/2)$ , so

\theta_k=\tan^{-1}\!\left(\frac{x}{1+k(k+1)x^2}\right).

Thus

S_n(x)=\sum_{k=1}^n \tan^{-1}\!\left(\frac{x}{1+k(k+1)x^2}\right).

Now observe the key factorization:

\frac{x}{1+k(k+1)x^2} =\frac{kx-(k-1)x}{1+k(k+1)x^2}.

More usefully,

\tan^{-1}(kx)-\tan^{-1}((k-1)x) =\tan^{-1}\!\left(\frac{kx-(k-1)x}{1+k(k-1)x^2}\right) =\tan^{-1}\!\left(\frac{x}{1+k(k-1)x^2}\right).

So instead shift index properly:

\tan^{-1}((k+1)x)-\tan^{-1}(kx) =\tan^{-1}\!\left(\frac{x}{1+k(k+1)x^2}\right).

Because all quantities are positive, there is no branch issue. Therefore,

\cot^{-1}\!\left(\frac{1+k(k+1)x^2}{x}\right)=\tan^{-1}((k+1)x)-\tan^{-1}(kx).

Hence the sum telescopes:

S_n(x)=\sum_{k=1}^n \big[\tan^{-1}((k+1)x)-\tan^{-1}(kx)\big] =\tan^{-1}((n+1)x)-\tan^{-1}(x).

Check Option A

For $n=10$ ,

S_{10}(x)=\tan^{-1}(11x)-\tan^{-1}(x).

Using

\tan^{-1}a-\tan^{-1}b=\tan^{-1}\!\left(\frac{a-b}{1+ab}\right)

(for positive values here),

S_{10}(x)=\tan^{-1}\!\left(\frac{10x}{1+11x^2}\right).

Now for $t>0$ ,

\tan^{-1}(t)=\frac{\pi}{2}-\tan^{-1}\!\left(\frac{1}{t}\right).

S_{10}(x)=\frac{\pi}{2}-\tan^{-1}\!\left(\frac{1+11x^2}{10x}\right).

Thus A is true.

Check Option B

From

S_n(x)=\tan^{-1}((n+1)x)-\tan^{-1}(x),

let

A=\tan^{-1}((n+1)x),\qquad B=\tan^{-1}(x).

Then

\tan S_n(x)=\tan(A-B)=\frac{(n+1)x-x}{1+(n+1)x^2}=\frac{nx}{1+(n+1)x^2}.

Therefore

\cot S_n(x)=\frac{1+(n+1)x^2}{nx}=\frac{1}{nx}+\frac{n+1}{n}x.

Taking limit as $n\to\infty$ ,

\lim_{n\to\infty}\cot(S_n(x))=0+x=x.

Thus B is true.

Check Option C

For $n=3$ ,

S_3(x)=\tan^{-1}(4x)-\tan^{-1}(x).

If $S_3(x)=\pi/4$ , then

\tan(S_3(x))=1.

But

\tan(S_3(x))=\frac{4x-x}{1+4x^2}=\frac{3x}{1+4x^2}.

So we need

\frac{3x}{1+4x^2}=1

which gives

4x^2-3x+1=0.

Discriminant:

\Delta=9-16=-7<0.

So there is no real root, hence no root in $(0,\infty)$ . Thus C is false.

Check Option D

We already found

\tan(S_n(x))=\frac{nx}{1+(n+1)x^2}.

We need to test whether

\frac{nx}{1+(n+1)x^2}\le \frac12

for all $n\ge1$ , $x>0$ .

Consider

2nx\le 1+(n+1)x^2

which is equivalent to

(n+1)x^2-2nx+1\ge0.

Its discriminant is

(2n)^2-4(n+1)=4(n^2-n-1).

For large $n$ this is positive, so the quadratic is negative for some $x$ . Hence the inequality is not true for all $x>0$ .

A specific counterexample: take $n=2$ , $x=1$ . Then

\tan(S_2(1))=\frac{2}{1+3}=\frac12.

This satisfies equality, so not a counterexample. Take $n=3$ , $x=1/2$ :

\tan(S_3(1/2))=\frac{3\cdot \frac12}{1+4\cdot \frac14} =\frac{3/2}{2}=\frac34>\frac12.

So D is false.

Final conclusion

The true statements are:

\boxed{\text{A, B}}

This matches the stored correct answer.

More from Sequences and Series

View written solutionFree

Correct answer: A, B

Rewrite each term using an arctangent identity

Let

\theta_k=\cot^{-1}\!\left(\frac{1+k(k+1)x^2}{x}\right),\qquad x>0.

Since the argument of $\cot^{-1}$ is positive, we have $\theta_k\in(0,\pi/2)$ , so

\theta_k=\tan^{-1}\!\left(\frac{x}{1+k(k+1)x^2}\right).

Thus

S_n(x)=\sum_{k=1}^n \tan^{-1}\!\left(\frac{x}{1+k(k+1)x^2}\right).

Now observe the key factorization:

\frac{x}{1+k(k+1)x^2} =\frac{kx-(k-1)x}{1+k(k+1)x^2}.

More usefully,

\tan^{-1}(kx)-\tan^{-1}((k-1)x) =\tan^{-1}\!\left(\frac{kx-(k-1)x}{1+k(k-1)x^2}\right) =\tan^{-1}\!\left(\frac{x}{1+k(k-1)x^2}\right).

So instead shift index properly:

\tan^{-1}((k+1)x)-\tan^{-1}(kx) =\tan^{-1}\!\left(\frac{x}{1+k(k+1)x^2}\right).

Because all quantities are positive, there is no branch issue. Therefore,

\cot^{-1}\!\left(\frac{1+k(k+1)x^2}{x}\right)=\tan^{-1}((k+1)x)-\tan^{-1}(kx).

Hence the sum telescopes:

S_n(x)=\sum_{k=1}^n \big[\tan^{-1}((k+1)x)-\tan^{-1}(kx)\big] =\tan^{-1}((n+1)x)-\tan^{-1}(x).

Check Option A

For $n=10$ ,

S_{10}(x)=\tan^{-1}(11x)-\tan^{-1}(x).

Using

\tan^{-1}a-\tan^{-1}b=\tan^{-1}\!\left(\frac{a-b}{1+ab}\right)

(for positive values here),

S_{10}(x)=\tan^{-1}\!\left(\frac{10x}{1+11x^2}\right).

Now for $t>0$ ,

\tan^{-1}(t)=\frac{\pi}{2}-\tan^{-1}\!\left(\frac{1}{t}\right).

S_{10}(x)=\frac{\pi}{2}-\tan^{-1}\!\left(\frac{1+11x^2}{10x}\right).

Thus A is true.

Check Option B

From

S_n(x)=\tan^{-1}((n+1)x)-\tan^{-1}(x),

let

A=\tan^{-1}((n+1)x),\qquad B=\tan^{-1}(x).

Then

\tan S_n(x)=\tan(A-B)=\frac{(n+1)x-x}{1+(n+1)x^2}=\frac{nx}{1+(n+1)x^2}.

Therefore

\cot S_n(x)=\frac{1+(n+1)x^2}{nx}=\frac{1}{nx}+\frac{n+1}{n}x.

Taking limit as $n\to\infty$ ,

\lim_{n\to\infty}\cot(S_n(x))=0+x=x.

Thus B is true.

Check Option C

For $n=3$ ,

S_3(x)=\tan^{-1}(4x)-\tan^{-1}(x).

If $S_3(x)=\pi/4$ , then

\tan(S_3(x))=1.

But

\tan(S_3(x))=\frac{4x-x}{1+4x^2}=\frac{3x}{1+4x^2}.

So we need

\frac{3x}{1+4x^2}=1

which gives

4x^2-3x+1=0.

Discriminant:

\Delta=9-16=-7<0.

So there is no real root, hence no root in $(0,\infty)$ . Thus C is false.

Check Option D

We already found

\tan(S_n(x))=\frac{nx}{1+(n+1)x^2}.

We need to test whether

\frac{nx}{1+(n+1)x^2}\le \frac12

for all $n\ge1$ , $x>0$ .

Consider

2nx\le 1+(n+1)x^2

which is equivalent to

(n+1)x^2-2nx+1\ge0.

Its discriminant is

(2n)^2-4(n+1)=4(n^2-n-1).

For large $n$ this is positive, so the quadratic is negative for some $x$ . Hence the inequality is not true for all $x>0$ .

A specific counterexample: take $n=2$ , $x=1$ . Then

\tan(S_2(1))=\frac{2}{1+3}=\frac12.

This satisfies equality, so not a counterexample. Take $n=3$ , $x=1/2$ :

\tan(S_3(1/2))=\frac{3\cdot \frac12}{1+4\cdot \frac14} =\frac{3/2}{2}=\frac34>\frac12.

So D is false.

Final conclusion

The true statements are:

\boxed{\text{A, B}}

This matches the stored correct answer.

Sequences and Series question

Sequences and Series question

More from Sequences and Series

Sequences and Series question

Sequences and Series question

More from Sequences and Series