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Question Number 205490 by mnjuly1970 last updated on 22/Mar/24
      If,f(x)= (√(2 + x)) + a (√(x − 1))       is monotone function .      find the range of  ” a ”
$$ \\ $$$$\:\:\:\:{If},{f}\left({x}\right)=\:\sqrt{\mathrm{2}\:+\:{x}}\:+\:{a}\:\sqrt{{x}\:−\:\mathrm{1}}\: \\ $$$$\:\:\:\:{is}\:{monotone}\:{function}\:. \\ $$$$\:\:\:\:{find}\:{the}\:{range}\:{of}\:\:''\:{a}\:'' \\ $$$$ \\ $$
Answered by mr W last updated on 22/Mar/24
x≥1  f(x)=(√(3+x−1))+a(√(x−1))  let t=x−1≥0  f(t)=(√(3+t^2 ))+at  f′(t)=(t/( (√(3+t^2 ))))+a=(1/( (√(1+(3/t^2 )))))+a  such that the function is monotone  for t≥0, (1/( (√(1+(3/t^2 )))))+a=0 should have  no solution. this is the case:  1) if a≥0  2) if a≤−1  i.e. the range of a is a≥0 or a≤−1  or  function is not monotone if  −1<a<0
$${x}\geqslant\mathrm{1} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{3}+{x}−\mathrm{1}}+{a}\sqrt{{x}−\mathrm{1}} \\ $$$${let}\:{t}={x}−\mathrm{1}\geqslant\mathrm{0} \\ $$$${f}\left({t}\right)=\sqrt{\mathrm{3}+{t}^{\mathrm{2}} }+{at} \\ $$$${f}'\left({t}\right)=\frac{{t}}{\:\sqrt{\mathrm{3}+{t}^{\mathrm{2}} }}+{a}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{t}^{\mathrm{2}} }}}+{a} \\ $$$${such}\:{that}\:{the}\:{function}\:{is}\:{monotone} \\ $$$${for}\:{t}\geqslant\mathrm{0},\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{t}^{\mathrm{2}} }}}+{a}=\mathrm{0}\:{should}\:{have} \\ $$$${no}\:{solution}.\:{this}\:{is}\:{the}\:{case}: \\ $$$$\left.\mathrm{1}\right)\:{if}\:{a}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{if}\:{a}\leqslant−\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{range}\:{of}\:{a}\:{is}\:{a}\geqslant\mathrm{0}\:{or}\:{a}\leqslant−\mathrm{1} \\ $$$${or} \\ $$$${function}\:{is}\:{not}\:{monotone}\:{if} \\ $$$$−\mathrm{1}<{a}<\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 22/Mar/24
thanks alot sir ...
$${thanks}\:{alot}\:{sir}\:… \\ $$

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