Solutions of ex−1−k⋅arctan⁡x=0, consider h(x)=ex−1−k⋅arctan⁡x then find on which condition on k there will be two solutions for h(x)=0 (k is real).

Note that x=0 is the essential root of f(x)=0f(x)=ex−1−ktan−1x⇒f′(x)=ex−k(1+x2)⇒f″(x)=ex+2kx(1+x2)f(−∞)=1+kπ2 and f(∞)=+∞. So f(x)=0 will have even number (0,2,4,...) of real roots in (−∞,∞) if f(−∞)&gt;0⇒k&gt;2π. So this eq. will have 2 real roots if k&gt;2π

"Solutions of ex−1−k⋅arctan⁡x=0, consider h(x)=ex−1−k⋅arctan⁡x then find on..." was answered by our experts

pogonofor9z

Answered question

2022-01-16

Solutions of

e^{x} - 1 - k \cdot \arctan x = 0

,
consider

h (x) = e^{x} - 1 - k \cdot \arctan x

then find on which condition on k there will be two solutions for h(x)=0 (k is real).

Answer & Explanation

Charles Benedict

Beginner2022-01-17Added 32 answers

x=0 is always one of the solutions. Then the function

f (x) = \frac{e^{x} - 1}{\arctan x}

can be shown to be strictly increasing and has an horizontal asymptote

y = \frac{2}{π}

You can conclude.

Jordan Mitchell

Beginner2022-01-18Added 31 answers

Note that x=0 is the essential root of f(x)=0

f (x) = e^{x} - 1 - k \tan^{- 1} x \Rightarrow f^{'} (x) = e^{x} - \frac{k}{(1 + x^{2})} \Rightarrow f^{″} (x) = e^{x} + \frac{2 k x}{(1 + x^{2})}

f (- \infty) = 1 + \frac{k π}{2} and f (\infty) = + \infty

. So f(x)=0 will have even number (0,2,4,...) of real roots in

(- \infty, \infty)

f (- \infty) > 0 \Rightarrow k > \frac{2}{π}

. So this eq. will have 2 real roots if

k > \frac{2}{π}

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