# The number of solutions of tan x = x - x3 with -1<= x <= 1 is A. 1. B. 2. C. 3. D. 4.

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2015-06-18T21:37:41+05:30

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Tan x =  x - x³
Let Y = tan x - x + x³
we want solutions of  Y = Tan x - x + x³ = 0

We can see that  at x = 0 ,  the equation is satisfied as Tan x = x - x³.
So x = 0 is one solution.

Derivative of Tan x = Sec² x  > 0 for  0 <= x <= 1.
So Tan x is rising continuously in the interval  0<= x <= 1.

Derivative of x - x³ = 1 - 3 x²  > 0  for  0<= x < 1/√3.
< 0  for  1/√3 < x <= 1
so during the interval 1/√3 < x < 1,  x - x³ decreases.

At   x = 1/√3,  Tan x = 0.6518
x - x³ =  0.3849

Thus we find that value of Tan x >    x - x³    for  0< x <= 1

we apply similar principle for the 3rd quadrant too.
so,    Tan x  <  x - x³   for  -1 <= x < 0.

Hence, there is only one solution at x = 0.
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Derivatives of  tan x  and  x - x³  we find.
d tan x  / dx = sec² x    > 1      we know that secant is always > 1.
So y = tan x curve is above  y = x.

y =  x - x³
slope of the curve =  d y / dx = 1 - 3 x²    < 1

Hence,   tan x always rises faster than  x - x³  for x > 0  or  falls faster than x - x³ for x < 0.  Hence they meet only at  x = 0.

There is only one solution for the above equation.

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