Question

Prove that for every `x>0,x/(1+x^2)<tan^(-1)(x)<x`?

Answer 1

`(1) If,f'(x)>0 ,AAin(a,b)` then `f` `"is "color(blue)"Strictly Increasing function"` on `(a,b)"`
`(2)f` `"is "color(blue)"Strictly Increasing function"` on `(a,b)"`,then
`x_1>x_2=>f(x_1)>f(x_2)`

Explanation:

We have, `x>0=>x in(0,00)` given to prove,

`x/(1+x^2) < tan^-1x < x`

Let,

`color(red)(f(x)=tan^-1x-x/(1+x^2)....to(A)`

`=>f^'(x)=1/(1+x^2)-((1+x^2)*1-x(2x))/((1+x^2)^2`

#=>f^'(x)=1/(1+x^2)-(1-x^2)/((1+x^2)^2)= (2x^2)/((1+x^2)^2)>0#

i.e. `x>0=>f^'(x)>0=>f,` #"is "color(blue)"Strictly Increasing function"#

`:.x>0=>f(x)>f(0)`,where,`f(0)=tan^-1 0-0/1=0`

`=>f(x)>0..............to [ Use (A)` ]

`=>tan^-1x-x/(1+x^2)>0=>color(blue)(tan^-1x>x/(1+x^2)...to(M)`

Again let,

`color(red)(g(x)=x-tan^-1x....to(B)`

`=>g^'(x)=1-1/(1+x^2)=x^2/(1+x^2)> 0`

i.e. `x>0=>g^'(x)>0=>g,` #"is "color(blue)"Strictly Increasing function"#

`:.x>0=>g(x)>g(0),where, g(0)=0-tan^-1 0=0`

`=>g(x)>0................to [ Use (B) ]`

`=>x-tan^-1x>0`

`=>color(blue)(x>tan^-1x...to(N)`

From `(N) and (M)`

`x >tan^-1x> x/(1+x^2)`

i.e. `x/(1+x^2) < tan^-1x < x`

Answer 2

Kindly refer to the Explanation.

Explanation:

We use the following Form of the Mean Value Theorem (MVT) :

If `f` is continuous on `[0,x]` and derivable on `(0,x)`, then,

`(f(x)-f(0))/x=f'(thetax)` for some `theta in (0,1)`.

This version of MVT can be easily proved from its usual version.

So, let us skip it it and proceed to prove the desired inequality.

Consider `f(x)=tan^-1x, :. f'(x)=1/(1+x^2)`.

We know that the conditions of MVT are fulfilled by `f`.

`:. {tan^-1 x-tan^-1 0}/x=1/{1+(thetax)^2}; x>0, 0ltthetalt 1`.

`i.e., tan^-1 x/x=1/(1+theta^2x^2); x>0, 0 < theta <1`.

`:. x/tan^-1x-1=theta^2x^2; x>0, 0 < theta <1......(star)`.

Now multiplying `0ltthetalt1" by "x^2gt0`, we have,

`0lttheta^2x^2ltx^2`.

Utilising this in `(star)`, we get,

`0 lt x/tan^-1x-1 ltx^2`.

Multiplying this by `1/x gt 0`, we get, `0 lt 1/tan^-1 x-1/x lt x`.

Adding `1/x` we get, `1/x lt 1/tan^-1 x lt x+1/x=(1+x^2)/x`.

Inverting, `x gt tan^-1x gt x/(1+x^2)`, or what is the same as to say,

`x/(1+x^2) lt tan^-1x lt x," for "x gt 0`.

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