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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!

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!

Grade:11

1 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 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.

 

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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