Question

If $\int \frac{{\mathrm{cosec}}^{2}x-2010}{{\cos }^{2010}x}dx=-\frac{f(x)}{(g(x){)}^{2010}}+C$; 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)

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