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`|a| gt 2sqrt(2)``|a| lt 2sqrt(2)``a ge -2sqrt(2)`none of these

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ASolution :

We have equation `sec x + cosec x =a` <br> 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. <br> Period of `y=sec x + cosec x` is `2pi`. <br> So, we draw the graph of the function for `x in [0, 2pi]`. <br> 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. <br> For example, in first quadrant, `sec x, cosec x gt 0`. <br> Also, when x approaches to zero, `cosec x` approaches to infinity. <br> So, `f(x)` approaches to infinity. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_TRI_C04_E12_004_S01.png" width="80%"> <br> Similarly, when x approaches to `pi//2 sec x` approaches to infinity. <br> So, `f(x)` approaches to infinity. <br> At `x=pi//4, f(x)` attains its least value which is `2sqrt(2)`.<br> With similar arguments, we can draw the graph of `y=f(x)` in intervals `(pi//2, pi), (pi, 3pi//2)` and `(3pi//2, 2pi)` <br> We have following graph of `y=f(x)`. <br> 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)`. <br> 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)`.