Question

If $\int \frac{{\sec }^{2}x}{{\text{cosec}}^{2}x}dx=k\cdot \tan \left[f\right(x\left)\right]+lx+C,$ where $k,l$ are fixed constants and $C$ is arbitrary constant, then which of the following is/are true

• A. $k+l=2$
• B. The line $y=kx+l$ passes through $(5,4)$
• C. $f\left(x\right)$ is monotonically increasing function.
• D. $f\left(x\right)$ has no stationary points

Answer 1

Answer: (B), (C), (D)

$f\left(x\right)$ has no stationary points

$\int \frac{{\sec }^{2}x}{{\text{cosec}}^{2}x}dx=\int \frac{\frac{1}{{\cos }^{2}x}}{\frac{1}{{\sin }^{2}x}}dx=\int \frac{{\sin }^{2}x}{{\cos }^{2}x}dx$
$=\int {\tan }^{2}xdx$
Now $\{{\tan }^{2}x={\sec }^{2}x-1\}$
$=\int ({\sec }^{2}x-1)dx$
By taking terms separately, we get,
$=\int {\sec }^{2}xdx-\int 1dx$
Therefore, we get
$=\tan x-x+C$
$\therefore k=1,f\left(x\right)=x$ and $l=-1$
$\Rightarrow k+l=0$
The line $y=kx+l(i.e.)y=x-1$ passes through $(5,4)$
$f^{\prime }\left(x\right)=1>0$
$\Rightarrow f\left(x\right)$ monotonically increases
$f^{\prime }\left(x\right)=1\ne 0$
$\Rightarrow f\left(x\right)$ has no stationary points