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Inverse Trigonometric Functions question

2018 · Shift 2 · Q19
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Inverse Trigonometric Functions question

2018 · Shift 2 · Q19

JEE AdvancedMathematicsInverse Trigonometric FunctionsMultiple correct+4 / −2
For any positive integer n, define fn:(0,∞)→R{f_n}:(0,\infty ) \to Rfn​:(0,∞)→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)fn​=j=1∑n​tan−1(1+(x+j)(x+j−1)1​) 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)(−2π​,2π​)). Then, which of the following statement(s) is (are) TRUE?
  1. A
    ∑j=15tan⁡2(fj(0))=55\sum\limits_{j = 1}^5 {{{\tan }^2}({f_j}(0)) = 55}j=1∑5​tan2(fj​(0))=55
  2. 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}j=1∑10​(1+fj′​(0))sec2(fj​(0))=10
  3. 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}x→∞lim​tan(fn​(x))=n1​
  4. 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)) = 1x→∞lim​sec2(fn​(x))=1
View written solutionFree

Correct answer: A, B, D

  1. Simplify the summand

We are given

fn(x)=∑j=1ntan⁡−1 ⁣(11+(x+j)(x+j−1)),x>0. f_n(x)=\sum_{j=1}^n \tan^{-1}\! \left(\frac{1}{1+(x+j)(x+j-1)}\right), \qquad x>0.fn​(x)=j=1∑n​tan−1(1+(x+j)(x+j−1)1​),x>0.

First simplify the denominator:

1+(x+j)(x+j−1)=1+(x+j)2−(x+j)=(x+j)2−(x+j)+1.1+(x+j)(x+j-1)=1+(x+j)^2-(x+j)= (x+j)^2-(x+j)+1.1+(x+j)(x+j−1)=1+(x+j)2−(x+j)=(x+j)2−(x+j)+1.

Now recall the identity

tan⁡−1a−tan⁡−1b=tan⁡−1(a−b1+ab)\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)tan−1a−tan−1b=tan−1(1+aba−b​)

when the principal values are compatible. Take

a=x+j,b=x+j−1.a=x+j,\qquad b=x+j-1.a=x+j,b=x+j−1.

Then

a−b1+ab=11+(x+j)(x+j−1).\frac{a-b}{1+ab}=\frac{1}{1+(x+j)(x+j-1)}.1+aba−b​=1+(x+j)(x+j−1)1​.

Hence

tan⁡−1(11+(x+j)(x+j−1))=tan⁡−1(x+j)−tan⁡−1(x+j−1).\tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) =\tan^{-1}(x+j)-\tan^{-1}(x+j-1).tan−1(1+(x+j)(x+j−1)1​)=tan−1(x+j)−tan−1(x+j−1).

So the sum telescopes:

fn(x)=∑j=1n[tan⁡−1(x+j)−tan⁡−1(x+j−1)]. f_n(x)=\sum_{j=1}^n \big[\tan^{-1}(x+j)-\tan^{-1}(x+j-1)\big].fn​(x)=j=1∑n​[tan−1(x+j)−tan−1(x+j−1)].

Therefore,

fn(x)=tan⁡−1(x+n)−tan⁡−1(x).\boxed{f_n(x)=\tan^{-1}(x+n)-\tan^{-1}(x)}.fn​(x)=tan−1(x+n)−tan−1(x)​.
  1. Useful formula for tan⁡fn(x)\tan f_n(x)tanfn​(x)

Using

tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B,\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},tan(A−B)=1+tanAtanBtanA−tanB​,

with A=tan⁡−1(x+n)A=\tan^{-1}(x+n)A=tan−1(x+n) and B=tan⁡−1(x)B=\tan^{-1}(x)B=tan−1(x),

tan⁡(fn(x))=(x+n)−x1+x(x+n)=n1+x2+nx.\tan(f_n(x))=\frac{(x+n)-x}{1+x(x+n)}=\frac{n}{1+x^2+nx}.tan(fn​(x))=1+x(x+n)(x+n)−x​=1+x2+nxn​.

Thus

tan⁡(fn(x))=nx2+nx+1.\boxed{\tan(f_n(x))=\frac{n}{x^2+nx+1}}.tan(fn​(x))=x2+nx+1n​​.

Also,

sec⁡2(fn(x))=1+tan⁡2(fn(x))=1+(nx2+nx+1)2.\sec^2(f_n(x))=1+\tan^2(f_n(x)) =1+\left(\frac{n}{x^2+nx+1}\right)^2.sec2(fn​(x))=1+tan2(fn​(x))=1+(x2+nx+1n​)2.
  1. Check option C

For fixed positive integer nnn,

tan⁡(fn(x))=nx2+nx+1.\tan(f_n(x))=\frac{n}{x^2+nx+1}.tan(fn​(x))=x2+nx+1n​.

As x→∞x\to\inftyx→∞,

tan⁡(fn(x))→0.\tan(f_n(x))\to 0.tan(fn​(x))→0.

So option C, which claims the limit is 1n\frac1nn1​, is false.

  1. Check option D

Since

sec⁡2(fn(x))=1+(nx2+nx+1)2,\sec^2(f_n(x))=1+\left(\frac{n}{x^2+nx+1}\right)^2,sec2(fn​(x))=1+(x2+nx+1n​)2,

as x→∞x\to\inftyx→∞,

sec⁡2(fn(x))→1.\sec^2(f_n(x))\to 1.sec2(fn​(x))→1.

Hence option D is true.

  1. Check option A

We need fj(0)f_j(0)fj​(0). From the telescoping form,

fj(0)=tan⁡−1(j)−tan⁡−1(0)=tan⁡−1(j).f_j(0)=\tan^{-1}(j)-\tan^{-1}(0)=\tan^{-1}(j).fj​(0)=tan−1(j)−tan−1(0)=tan−1(j).

Therefore,

tan⁡(fj(0))=j⇒tan⁡2(fj(0))=j2.\tan(f_j(0))=j \quad \Rightarrow \quad \tan^2(f_j(0))=j^2.tan(fj​(0))=j⇒tan2(fj​(0))=j2.

So

∑j=15tan⁡2(fj(0))=∑j=15j2=1+4+9+16+25=55.\sum_{j=1}^5 \tan^2(f_j(0))=\sum_{j=1}^5 j^2=1+4+9+16+25=55.j=1∑5​tan2(fj​(0))=j=1∑5​j2=1+4+9+16+25=55.

Thus option A is true.

  1. Check option B

We need

∑j=110(1+fj′(0))sec⁡2(fj(0)).\sum_{j=1}^{10}(1+f_j'(0))\sec^2(f_j(0)).j=1∑10​(1+fj′​(0))sec2(fj​(0)).

From

fj(x)=tan⁡−1(x+j)−tan⁡−1(x),f_j(x)=\tan^{-1}(x+j)-\tan^{-1}(x),fj​(x)=tan−1(x+j)−tan−1(x),

we differentiate:

fj′(x)=11+(x+j)2−11+x2.f_j'(x)=\frac{1}{1+(x+j)^2}-\frac{1}{1+x^2}.fj′​(x)=1+(x+j)21​−1+x21​.

Thus at x=0x=0x=0,

fj′(0)=11+j2−1=−j21+j2.f_j'(0)=\frac{1}{1+j^2}-1=-\frac{j^2}{1+j^2}.fj′​(0)=1+j21​−1=−1+j2j2​.

Hence

1+fj′(0)=1−j21+j2=11+j2.1+f_j'(0)=1-\frac{j^2}{1+j^2}=\frac{1}{1+j^2}.1+fj′​(0)=1−1+j2j2​=1+j21​.

Also, from fj(0)=tan⁡−1(j)f_j(0)=\tan^{-1}(j)fj​(0)=tan−1(j),

sec⁡2(fj(0))=1+tan⁡2(fj(0))=1+j2.\sec^2(f_j(0))=1+\tan^2(f_j(0))=1+j^2.sec2(fj​(0))=1+tan2(fj​(0))=1+j2.

Therefore,

(1+fj′(0))sec⁡2(fj(0))=11+j2(1+j2)=1.(1+f_j'(0))\sec^2(f_j(0))=\frac{1}{1+j^2}(1+j^2)=1.(1+fj′​(0))sec2(fj​(0))=1+j21​(1+j2)=1.

So

∑j=110(1+fj′(0))sec⁡2(fj(0))=∑j=1101=10.\sum_{j=1}^{10}(1+f_j'(0))\sec^2(f_j(0))=\sum_{j=1}^{10}1=10.j=1∑10​(1+fj′​(0))sec2(fj​(0))=j=1∑10​1=10.

Thus option B is true.

  1. Final conclusion

The true statements are

A, B, D.\boxed{A,\ B,\ D}.A, B, D​.

Option C is false.

  1. Comparison with stored answer

Stored correct answer: DDD

But from the complete evaluation above, AAA, BBB, and DDD are all true. Hence I disagree with the stored answer.

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