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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If ∑_ r=1 ^ 50 ^ -1 (1)/(2 r^2)=p then p is

If ∑r=150tan⁡−112r2=p\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}=p then tan⁡p\tan p is

Options

  1. A100101\frac {100}{101}
  2. B5150\frac {51}{50}
  3. C5051\frac {50}{51}
  4. D101102\frac {101}{102}

Correct answer

C. 5051\frac {50}{51}

Step-by-step solution

∑r=150tan⁡−112r2=p⇒∑r=150tan⁡−1(24r2)=p⇒∑r=150tan⁡−1[(2r+1)−(2r−1)1+(2r+1)(2r−1)]=p⇒∑r=150[tan⁡−1(2r+1)−tan⁡−1(2r−1)]=p⇒tan⁡−1(101)−tan⁡−1(1)=p⇒tan⁡−1(101−11+101)=p⇒100102=tan⁡p⇒tan⁡p=5051\begin{aligned} & \sum_{\mathrm{r}=1}^{50} \tan ^{-1} \frac{1}{2 \mathrm{r}^2}=\mathrm{p} \\ & \Rightarrow \sum_{\mathrm{r}=1}^{50} \tan ^{-1}\left(\frac{2}{4 \mathrm{r}^2}\right)=\mathrm{p} \\ & \Rightarrow \sum_{\mathrm{r}=1}^{50} \tan ^{-1}\left[\frac{(2 \mathrm{r}+1)-(2 \mathrm{r}-1)}{1+(2 \mathrm{r}+1)(2 \mathrm{r}-1)}\right]=\mathrm{p} \\ & \Rightarrow \sum_{\mathrm{r}=1}^{50}\left[\tan ^{-1}(2 \mathrm{r}+1)-\tan ^{-1}(2 \mathrm{r}-1)\right]=\mathrm{p} \\ & \Rightarrow \tan ^{-1}(101)-\tan ^{-1}(1)=\mathrm{p} \\ & \Rightarrow \tan ^{-1}\left(\frac{101-1}{1+101}\right)=\mathrm{p} \\ & \Rightarrow \frac{100}{102}=\tan \mathrm{p} \\ & \Rightarrow \tan \mathrm{p}=\frac{50}{51}\end{aligned}
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