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Question Number 180856 by Shrinava last updated on 18/Nov/22

find all values of  m∈R  such that the equation:                         ∫_0 ^( x)  ((arctany)/y) dy = mx  has two real roots:   x_1 ∈(−∞;0) , x_2 ∈(0;∞)

$$\mathrm{find}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\:\mathrm{m}\in\mathbb{R}\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\boldsymbol{\mathrm{x}}} \:\frac{\mathrm{arctan}\boldsymbol{\mathrm{y}}}{\mathrm{y}}\:\mathrm{dy}\:=\:\mathrm{mx} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{real}\:\mathrm{roots}:\:\:\:\mathrm{x}_{\mathrm{1}} \in\left(−\infty;\mathrm{0}\right)\:,\:\mathrm{x}_{\mathrm{2}} \in\left(\mathrm{0};\infty\right) \\ $$

Answered by mr W last updated on 19/Nov/22

f(x)=∫_0 ^x ((tan^(−1) t)/t)dt  g(x)=mx  we can see f(x) is an odd function,  i.e. f(−x)=−f(x).  f′(x)=((tan^(−1) x)/x)  f′(0)=lim_(x→0) ((tan^(−1) x)/x)=1  i.e. the tangent of y=f(x) at x=0 is   y=kx with k=1.  lim_(x→+∞) f′(x)=lim_(x→+∞) ((tan^(−1) x)/x)=0  so we can know how the graph of  y=f(x) nearly looks like. the line   y=mx always intersects the curve   y=f(x) at x=0 and additionally at   two points P and P′ with x=±a  if y=mx is flater than the tangent  line at x=0, i.e. if 0<m<k=1.  that means if 0<m<1, the eqn.  f(x)=g(x) will have two roots:  x=±a. so the answer is 0<m<1.

$${f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{tan}^{−\mathrm{1}} {t}}{{t}}{dt} \\ $$$${g}\left({x}\right)={mx} \\ $$$${we}\:{can}\:{see}\:{f}\left({x}\right)\:{is}\:{an}\:{odd}\:{function}, \\ $$$${i}.{e}.\:{f}\left(−{x}\right)=−{f}\left({x}\right). \\ $$$${f}'\left({x}\right)=\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}} \\ $$$${f}'\left(\mathrm{0}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}=\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{tangent}\:{of}\:{y}={f}\left({x}\right)\:{at}\:{x}=\mathrm{0}\:{is}\: \\ $$$${y}={kx}\:{with}\:{k}=\mathrm{1}. \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}'\left({x}\right)=\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}=\mathrm{0} \\ $$$${so}\:{we}\:{can}\:{know}\:{how}\:{the}\:{graph}\:{of} \\ $$$${y}={f}\left({x}\right)\:{nearly}\:{looks}\:{like}.\:{the}\:{line}\: \\ $$$${y}={mx}\:{always}\:{intersects}\:{the}\:{curve}\: \\ $$$${y}={f}\left({x}\right)\:{at}\:{x}=\mathrm{0}\:{and}\:{additionally}\:{at}\: \\ $$$${two}\:{points}\:{P}\:{and}\:{P}'\:{with}\:{x}=\pm{a} \\ $$$${if}\:{y}={mx}\:{is}\:{flater}\:{than}\:{the}\:{tangent} \\ $$$${line}\:{at}\:{x}=\mathrm{0},\:{i}.{e}.\:{if}\:\mathrm{0}<{m}<{k}=\mathrm{1}. \\ $$$${that}\:{means}\:{if}\:\mathrm{0}<{m}<\mathrm{1},\:{the}\:{eqn}. \\ $$$${f}\left({x}\right)={g}\left({x}\right)\:{will}\:{have}\:{two}\:{roots}: \\ $$$${x}=\pm{a}.\:{so}\:{the}\:{answer}\:{is}\:\mathrm{0}<{m}<\mathrm{1}. \\ $$

Commented by mr W last updated on 19/Nov/22

Commented by mr W last updated on 19/Nov/22

for the solition of the question, we  don′t need to solve the integral  ∫_0 ^x  ((tan^(−1) t)/t)dt really. to be honest, i can′t  solve it. but it′s enough to know how  its graph nearly looks like.

$${for}\:{the}\:{solition}\:{of}\:{the}\:{question},\:{we} \\ $$$${don}'{t}\:{need}\:{to}\:{solve}\:{the}\:{integral} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{tan}^{−\mathrm{1}} {t}}{{t}}{dt}\:{really}.\:{to}\:{be}\:{honest},\:{i}\:{can}'{t} \\ $$$${solve}\:{it}.\:{but}\:{it}'{s}\:{enough}\:{to}\:{know}\:{how} \\ $$$${its}\:{graph}\:{nearly}\:{looks}\:{like}. \\ $$

Commented by Shrinava last updated on 19/Nov/22

perfect dear professor thank you

$$\mathrm{perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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