Consider the equation
`sec theta +cosec theta=a, theta in (0, 2pi) -{pi//2, pi, 3pi//2}`
If the equation has four distinct real roots, then (a) `|a| gt 2sqrt(2)` (b) `|a| lt 2sqrt(2)` (c) `a ge -2sqrt(2)` (d) none of these

Options

A. `|a| ge 2sqrt(2)`
B. `a lt 2sqrt(2)`
C. `|a| lt 2sqrt(2)`
D. none of these

Answer

C

Explanation

We have equation `sec x + cosec x =a`
To analyze the roots of the equation, we draw the graph of function `y= sec x+ cosec x` and check how many times line `y=a` intersects this graph.
Period of `y=sec x + cosec x` is `2pi`.
So, we draw the graph of the function for `x in [0, 2pi]`.
The garph of function can be easily drawn by drawing the graph of `y=sec x` and `y=cosec x` and then adding the values of `sec x` and `cosec x` by inspection.
For example, in first quadrant, `sec x, cosec x gt 0`.
Also, when x approaches to zero, `cosec x` approaches to infinity.
So, `f(x)` approaches to infinity.

Similarly, when x approaches to `pi//2 sec x` approaches to infinity.
So, `f(x)` approaches to infinity.
At `x=pi//4, f(x)` attains its least value which is `2sqrt(2)`.
With similar arguments, we can draw the graph of `y=f(x)` in intervals `(pi//2, pi), (pi, 3pi//2)` and `(3pi//2, 2pi)`
We have following graph of `y=f(x)`.
From the figure, we can say that `f(x)=a` has two distinct solution if line `y=a` cuts the graph `y=f(x)` between `y=2sqrt(2)` and `y=-2sqrt(2)` i.e., `|a| lt 2sqrt(2)`.
If line `y=a`, cuts the graph of `y=f(x)` above `y=2sqrt(2)` and below `y=-2sqrt(2)`, then `f(x)=a` has four distinct solutions. So, `|a| gt 2sqrt(2)`.