InterviewSolution
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InterviewSolution
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Consider the equation sec theta +cosec theta=a, theta in (0, 2pi) -{pi//2, pi, 3pi//2} If the equation has two distinct real roots, then |
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Answer» `|a| ge 2sqrt(2)` 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)`. |
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