If ^ -1 ( )= ^ -1 2+ ^ -1 8+ ^ -1 18 + ^ -1 32+ . . upto 100 term… - Vidyadip

MCQ

If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18$ $+\cot ^{-1} 32+\ldots . .$ upto $100$ terms, then $\alpha$ is

✓
$1.01$
B
$1.00$
C
$1.02$
D
$1.03$

✓

Answer

Correct option: A.

$1.01$

a
$\operatorname{Cot}^{-1}(\alpha)=\cot ^{-1}(2)+\cot ^{-1}(8)+\cot ^{-1}(18)+\ldots .$

$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{2}{4 n^{2}}\right)$

$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}\right)$

$=\sum_{n=1}^{100} \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$

$=\tan ^{-1} 201-\tan ^{-1} 1$

$=\tan ^{-1}\left(\frac{200}{202}\right)$

$\therefore \cot ^{-1}(\alpha)=\cot ^{-1}\left(\frac{202}{200}\right)$

$\alpha=1.01$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let the solution curve $y = y ( x )$ of the differential equation $\quad \frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{\frac{3}{2}}} y=2 x$ $\exp \frac{x^3-\tan ^{-1} x^3}{\sqrt{(1+x)^6}}$ pass through the origin. Then $y (1)$ is equal to:

The mean of $10$ terms is $3$ . If the first term is increased by $1$ , second by $2$ and so on, then the new mean is

If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ and $y=9 x^2 f(x),$ then $y$ is strictly increasing in :

The equation of latus rectum of a parabola is $x + y = 8$ and the equation of the tangent at the vertex is $x + y = 12$, then length of the latus rectum is

The equation of normal to the circle $2{x^2} + 2{y^2} - 2x - 5y + 3 = 0$ at $(1, 1)$ is

Consider the line $\mathrm{L}$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right)$ from the line $\mathrm{L}$ along the line $\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2}$ is equal to :

The area of the circle whose centre is at $(1, 2)$ and which passes through the point $(4, 6)$ is

${\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }} = $

Let $T > 0$ be a fixed number. Suppose $f$ is a continuous function such that for all $x \in R,\,f(x + T) = f(x)$. If $I = \int_{\,0}^{\,T} {f(x)\,dx} $ then the value of $\int_{\,3}^{\,3 + 3T} {f(2x)\,dx,} $ is

If the solution curve of the differential equation $\left(y-2 \log _e x\right) d x+\left(x \log _e x^2\right) d y=0, x > 1$ passes through the points $\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$, then $\alpha$ is equal to $................$.