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The number of solutions of tan x = x - x3 with -1<= x <= 1 is

A. 1.

B. 2.

C. 3.

D. 4.

1
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A. 1.

B. 2.

C. 3.

D. 4.

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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.

============================

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.

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.

============================

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.