Question

If $\int \frac{\operatorname{cosec}^{2} x-2010}{\cos ^{2010} x} d x=-\frac{f(x)}{(g(x))^{2010}}+C$; where $f(\frac{\pi}{4})=1 ;$ then the number of solutions of the equation $\frac{f(x)}{g(x)}=\{x\}$ in $[0,2 \pi]$ is/are : (where $\{.\}$ represents fractional part function)

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

$\int \sec ^{2010} x \operatorname{cosec}^{2} x d x-\int 2010 \sec ^{2010} x d x$
$=\int \sec ^{2010} x(-\cot x)-\int 2010 \sec ^{2010} x \cdot \tan x$
$\cdot(-\cot x)-\int 2010 \sec ^{2010} x d x$
$=-\frac{\cot x}{(\cos x)^{2010}}+2010 \int \sec ^{2010} x d x$
$-2010 \int \sec ^{2010} x d x +C$
$=\frac{-\cot x}{(\cos x)^{2010}}+C$
$\therefore \frac{f(x)}{g(x)}=\frac{1}{\sin x}=\{x\}$
No solution.