For any positive integer n,define f_n:(0, inf | vedclass

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(A) We have $f_n(x) = \sum_{j=1}^n 	an^{-1}\left(\frac{(x+j)-(x+j-1)}{1+(x+j)(x+j-1)}ight)$.Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-

Vedclass · Accepted Answer

$A, B, D$

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(A) We have $f_n(x) = \sum_{j=1}^n 	an^{-1}\left(\frac{(x+j)-(x+j-1)}{1+(x+j)(x+j-1)}ight)$.Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we get:$f_n(x) = \sum_{j=1}^n (	an^{-1}(x+j) - 	an^{-1}(x+j-1)) = 	an^{-1}(x+n) - 	an^{-1}(x)$.Thus,$	an(f_n(x)) = \frac{(x+n)-x}{1+(x+n)x} = \frac{n}{1+x^2+nx}$.For $(C)$,$\lim_{x ightarrow \infty} 	an(f_n(x)) = \lim_{x ightarrow \infty} \frac{n}{1+x^2+nx} = 0$. So $(C)$ is $FALSE$.For $(D)$,$\lim_{x ightarrow \infty} \sec^2(f_n(x)) = 1 + \lim_{x ightarrow \infty} 	an^2(f_n(x)) = 1 + 0 = 1$. So $(D)$ is $TRUE$.For $(A)$,$f_j(0) = 	an^{-1}(j) - 	an^{-1}(0) = 	an^{-1}(j)$. Thus $	an^2(f_j(0)) = j^2$. $\sum_{j=1}^5 j^2 = 1+4+9+16+25 = 55$. So $(A)$ is $TRUE$.For $(B)$,$f_j'(x) = \frac{1}{1+(x+j)^2} - \frac{1}{1+x^2}$. $f_j'(0) = \frac{1}{1+j^2} - 1 = \frac{-j^2}{1+j^2}$.Then $1+f_j'(0) = 1 - \frac{j^2}{1+j^2} = \frac{1}{1+j^2}$.Also $\sec^2(f_j(0)) = 1 + 	an^2(f_j(0)) = 1+j^2$.So $(1+f_j'(0)) \sec^2(f_j(0)) = \frac{1}{1+j^2} \cdot (1+j^2) = 1$.$\sum_{j=1}^{10} 1 = 10$. So $(B)$ is $TRUE$.The correct options are $(A), (B), (D)$.

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