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I cudnt solve these Question ......can u help me out plzz:- Q.1 The Smallest Positive root of the equation tan(x) - x = 0, lies in :- a) (0,pi/2) b) (pi/2, pi) c) (pi, 3pi/2) d) (3pi/2, 2pi) Q.2 The number of roots of the equation x + 2tan (x) = pi/2 in the interval [0, 2pi] is a) 1 b)2 c)3 d) infinite Thanks!

```
9 years ago

```							Dear student,
You can show that x = tan(x) has a solution in every interval of the form ([2*k-1]*Pi/2, [2*k+1]*Pi/2), where k is any integer. That is because x is bounded in these intervals, but tan(x) is unbounded in both positive and negative directions.  Thus their graphs must cross.  If you look at the graph of y = tan(x), you will see that it has  some vertical asymptotes.  They occur at x = Pi/2, 3*Pi/2, 5*Pi/2,  7*Pi/2, ..., and also at x = -Pi/2, -3*Pi/2, ...  These are the odd  multiples of Pi/2.  That is why you look at the interval between two consecutive odd  multiples of Pi/2.  Two consecutive odd numbers have the form 2*k-1 and  2*k+1 for some integer k. Thus the interval between two asymptotes is  given by the form [2*k-1]*Pi/2 < x < [2*k+1]*Pi/2. Call that  interval I(k).  At the values of x where there is an asymptote, the value of tan(x) is  undefined. Just to the right of any asymptote, the function is positive  and very large in absolute value.  Just to the left of any asymptote,  the function is negative and very large in absolute value. The closer  you get to the asymptote, the larger the absolute value of tan(x). In  fact, you can make the absolute value of tan(x) as large as you like by  taking x close enough to the asymptote.  Now look at the graph of y = x.  This is a straight line, of course.  When x is in I(k), since x = y, it is also true that y is in I(k). That  means that |y| < (1+2*|k|)*Pi/2.  Putting this together, you can conclude the following.   For values of x close enough to the left end of I(k),  tan(x) < -(1+2*|k|)*Pi/2 < x. Thus tan(x) - x < 0.

All the best.
Win exciting gifts by                                                                                          answering     the           questions    on                 Discussion                Forum.      So           help                      discuss            any                       query     on                  askiitians          forum     and             become    an            Elite                      Expert         League                         askiitian.

Sagar Singh
B.Tech, IIT Delhi

```
9 years ago
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