Question

If y = $ta{n}^{-1}\left(\frac{1}{1+x+x^2}\right)+ta{n}^{-1}\left(\frac{1}{x^2+3x+3}\right)+ta{n}^{-1}\left(\frac{1}{x^2+5x+7}\right)$ + ........+

upto n terms they y' (o) is equal to

• A.
• B.
• C.
• D. 1+n

Answer 1

The correct option is B

y = $ta{n}^{-1}\left\{\frac{1}{1+x(1+x)}\right\}+ta{n}^{-1}\left\{\frac{1}{1+(x+1)(x+2)}\right\}$

+ $ta{n}^{-1}\left\{\frac{1}{1+(x+2)(x+3)}\right\}$

+ ........ + $ta{n}^{-1}\left\{\frac{1}{1+(x+n-1)(x+n)}\right\}$

= $\sum _{r=1}^n$ $ta{n}^{-1}\left\{\frac{1}{1+(x+r-1)(x+r)}\right\}$

= $\sum _{r=1}^n$ $ta{n}^{-1}\left\{\frac{(x+r)-(x+r-1)}{1+(x+r-1)(x+r)}\right\}$

= $\sum _{r=1}^n$ { \( tan^-1 (x+r) - tan^{-1} ( x + r -1 ) }

= $ta{n}^{-1}(x+n)-ta{n}^{-1}x$

y' = $\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$

$\Rightarrow$ y'(0) = $\frac{1}{1+n^2}-1=\frac{-n^2}{1+n^2}$