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Limits Continuity and Differentiability question

2021 · 16 Mar · Shift 1 · Q24
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  5. /2021 · 16 Mar · Shift 1 · Q24

Limits Continuity and Differentiability question

2021 · 16 Mar · Shift 1 · Q24

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let Sk=∑r=1ktan⁡−1(6r22r+1+32r+1){S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)}Sk​=r=1∑k​tan−1(22r+1+32r+16r​). Then lim⁡k→∞Sk\mathop {\lim }\limits_{k \to \infty } {S_k}k→∞lim​Sk​ is equal to :
  1. A
    cot⁡−1(32){\cot ^{ - 1}}\left( {{3 \over 2}} \right)cot−1(23​)
  2. B
    π2{\pi \over 2}2π​
  3. C
    tan −-− 1 (3)
  4. D
    tan⁡−1(32){\tan ^{ - 1}}\left( {{3 \over 2}} \right)tan−1(23​)
View written solutionFree

Correct answer: A

  1. We need to evaluate
Sk=∑r=1ktan⁡−1 ⁣(6r22r+1+32r+1).S_k=\sum_{r=1}^k \tan^{-1}\! \left(\frac{6^r}{2^{2r+1}+3^{2r+1}}\right).Sk​=r=1∑k​tan−1(22r+1+32r+16r​).
  1. Rewrite the denominator:
22r+1+32r+1=2⋅4r+3⋅9r.2^{2r+1}+3^{2r+1}=2\cdot 4^r+3\cdot 9^r.22r+1+32r+1=2⋅4r+3⋅9r.

Also,

6r=(2⋅3)r.6^r=(2\cdot 3)^r.6r=(2⋅3)r.
  1. Observe the standard 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 angles lie in the principal range.

We try

a=(32)r,b=(32)r−1.a=\left(\frac{3}{2}\right)^r,\qquad b=\left(\frac{3}{2}\right)^{r-1}.a=(23​)r,b=(23​)r−1.

Then

a−b=(32)r−1(32−1)=12(32)r−1.a-b=\left(\frac{3}{2}\right)^{r-1}\left(\frac32-1\right)=\frac12\left(\frac{3}{2}\right)^{r-1}.a−b=(23​)r−1(23​−1)=21​(23​)r−1.

And

1+ab=1+(32)2r−1.1+ab=1+\left(\frac{3}{2}\right)^{2r-1}.1+ab=1+(23​)2r−1.

So

a−b1+ab=12(32)r−11+(32)2r−1.\frac{a-b}{1+ab} =\frac{\frac12\left(\frac{3}{2}\right)^{r-1}}{1+\left(\frac{3}{2}\right)^{2r-1}}.1+aba−b​=1+(23​)2r−121​(23​)r−1​.

Multiply numerator and denominator by 22r2^{2r}22r:

a−b1+ab=3r−12r22r+2⋅32r−1.\frac{a-b}{1+ab} =\frac{3^{r-1}2^r}{2^{2r}+2\cdot 3^{2r-1}}.1+aba−b​=22r+2⋅32r−13r−12r​.

Now simplify denominator:

22r+2⋅32r−1=22r+2332r=22r+2⋅32r−1.2^{2r}+2\cdot 3^{2r-1}=2^{2r}+\frac{2}{3}3^{2r}=2^{2r}+2\cdot 3^{2r-1}.22r+2⋅32r−1=22r+32​32r=22r+2⋅32r−1.

Instead, let us verify directly with a better choice.

  1. Take
a=(32)r,b=(32)r+1.a=\left(\frac{3}{2}\right)^r, \qquad b=\left(\frac{3}{2}\right)^{r+1}.a=(23​)r,b=(23​)r+1.

Then

tan⁡−1b−tan⁡−1a=tan⁡−1(b−a1+ab).\tan^{-1}b-\tan^{-1}a =\tan^{-1}\left(\frac{b-a}{1+ab}\right).tan−1b−tan−1a=tan−1(1+abb−a​).

Now

b−a=(32)r(32−1)=12(32)rb-a=\left(\frac{3}{2}\right)^r\left(\frac32-1\right)=\frac12\left(\frac{3}{2}\right)^rb−a=(23​)r(23​−1)=21​(23​)r

and

1+ab=1+(32)2r+1.1+ab=1+\left(\frac{3}{2}\right)^{2r+1}.1+ab=1+(23​)2r+1.

Hence

b−a1+ab=12(32)r1+(32)2r+1.\frac{b-a}{1+ab} =\frac{\frac12\left(\frac{3}{2}\right)^r}{1+\left(\frac{3}{2}\right)^{2r+1}}.1+abb−a​=1+(23​)2r+121​(23​)r​.

Multiply numerator and denominator by 22r+12^{2r+1}22r+1:

b−a1+ab=12⋅3r2r+122r+1+32r+1=3r2r22r+1+32r+1=6r22r+1+32r+1.\frac{b-a}{1+ab} =\frac{\frac12\cdot 3^r 2^{r+1}}{2^{2r+1}+3^{2r+1}} =\frac{3^r2^r}{2^{2r+1}+3^{2r+1}} =\frac{6^r}{2^{2r+1}+3^{2r+1}}.1+abb−a​=22r+1+32r+121​⋅3r2r+1​=22r+1+32r+13r2r​=22r+1+32r+16r​.

Therefore,

tan⁡−1(6r22r+1+32r+1)=tan⁡−1((32)r+1)−tan⁡−1((32)r).\tan^{-1}\left(\frac{6^r}{2^{2r+1}+3^{2r+1}}\right) =\tan^{-1}\left(\left(\frac32\right)^{r+1}\right)-\tan^{-1}\left(\left(\frac32\right)^r\right).tan−1(22r+1+32r+16r​)=tan−1((23​)r+1)−tan−1((23​)r).
  1. So the sum telescopes:
Sk=∑r=1k[tan⁡−1((32)r+1)−tan⁡−1((32)r)].S_k=\sum_{r=1}^k \left[\tan^{-1}\left(\left(\frac32\right)^{r+1}\right)-\tan^{-1}\left(\left(\frac32\right)^r\right)\right].Sk​=r=1∑k​[tan−1((23​)r+1)−tan−1((23​)r)].

Hence,

Sk=tan⁡−1((32)k+1)−tan⁡−1(32).S_k=\tan^{-1}\left(\left(\frac32\right)^{k+1}\right)-\tan^{-1}\left(\frac32\right).Sk​=tan−1((23​)k+1)−tan−1(23​).
  1. Now take limit as k→∞k\to\inftyk→∞:
(32)k+1→∞\left(\frac32\right)^{k+1}\to\infty(23​)k+1→∞

so

tan⁡−1((32)k+1)→π2.\tan^{-1}\left(\left(\frac32\right)^{k+1}\right)\to \frac\pi2.tan−1((23​)k+1)→2π​.

Thus,

lim⁡k→∞Sk=π2−tan⁡−1(32).\lim_{k\to\infty} S_k=\frac\pi2-\tan^{-1}\left(\frac32\right).k→∞lim​Sk​=2π​−tan−1(23​).

Using

π2−tan⁡−1x=cot⁡−1x(x>0),\frac\pi2-\tan^{-1}x=\cot^{-1}x \quad (x>0),2π​−tan−1x=cot−1x(x>0),

we get

lim⁡k→∞Sk=cot⁡−1(32).\lim_{k\to\infty} S_k=\cot^{-1}\left(\frac32\right).k→∞lim​Sk​=cot−1(23​).
  1. Checking options:
  • A: cot⁡−1(32)\cot^{-1}\left(\frac32\right)cot−1(23​) ✅
  • B: π2\frac\pi22π​ ❌
  • C: tan⁡−1(3)\tan^{-1}(3)tan−1(3) ❌
  • D: tan⁡−1(32)\tan^{-1}\left(\frac32\right)tan−1(23​) ❌

Therefore, the correct answer is A.

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