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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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