Question

If $S_n=\sum _{r=1}^{n}r$ ! then for $n>6($ given $\sum _{r=1}^{6}r!=873)$

Column I		Column II
A	${\sin }^{-1}\left(\sin \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	P	$5-2\pi$
B	${\cos }^{-1}\left(\cos \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	Q	$2\pi -5$
C	${\tan }^{-1}\left(\tan \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	R	$6-2\pi$
D	${\cot }^{-1}\left(\cot \left(S_n-7\left[\frac{S_n}{7}\right]\right)\right)$	S	$5-\pi$
		T	$\pi -4$

(where [ ] denotes greatest integer function)

• A. (A) P; (B) Q ; (C) P ; (D) S
• B. (A) P; (B) Q ; (C)Q; (D) S
• C. (A) P; (B) Q ; (C) R; (D) S
• D. (A) P; (B) Q ; (C) S; (D) S

Answer 1

Answer: (A)

$S_n=\underline{1!+2!+3!+4!+5!+6!}+71$ where I be an integer
$S_n=873+7I$
$\frac{S_n}{7}=124.71+I\Rightarrow \left[\frac{S_n}{7}\right]=124+I$
$\therefore 7\left[\frac{S_n}{7}\right]=868+7I\Rightarrow S_n-7\left[\frac{S_n}{7}\right]=(873+7I)-(868+7I)=5$
$\text{ now }{\sin }^{-1}(\sin 5)={\sin }^{-1}(\sin (5-2\pi ))=5-2\pi \Rightarrow \text{ (P) }$
${\cos }^{-1}(\cos 5)={\cos }^{-1}(\cos (2\pi -5))=2\pi -5\Rightarrow \text{ (Q) }$
${\tan }^{-1}(\tan 5)={\tan }^{-1}(\tan (5-2\pi ))=5-2\pi \Rightarrow (P)$
$\text{ and }{\cot }^{-1}(\cot 5)={\cot }^{-1}(\cot (5-\pi ))=5-\pi \Rightarrow \text{ (R) }$