Menu Close

Number-of-integers-n-for-which-3x-3-25x-n-0-has-three-real-roots-is-

Tinku Tara 0 Comments
Pin




Question Number 46974 by rahul 19 last updated on 03/Nov/18
Number of integers n for which 3x^3 −25x+n=0 has three real roots is ?
$${Number}\:{of}\:{integers}\:{n}\:{for}\:{which}\: \\ $$$$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{25}{x}+{n}=\mathrm{0}\:{has}\:{three}\:{real}\:{roots}\:{is}\:? \\ $$$$ \\ $$
Commented by rahul 19 last updated on 03/Nov/18
Ans. is 55 but i′m getting infinite... plss check my method.... f(x)= 3x^3 −25x+n=0 f′(x)= 9x^2 −25=0 ⇒ x= ∓ (√(5/3)) ⇒ f′(x) has 2 real roots ⇒ definitely f(x) will 3 real roots ... hence n can be any integer→infinite...
$${Ans}.\:{is}\:\mathrm{55}\:{but}\:{i}'{m}\:{getting}\:{infinite}… \\ $$$${plss}\:{check}\:{my}\:{method}…. \\ $$$${f}\left({x}\right)=\:\mathrm{3}{x}^{\mathrm{3}} −\mathrm{25}{x}+{n}=\mathrm{0} \\ $$$${f}'\left({x}\right)=\:\mathrm{9}{x}^{\mathrm{2}} −\mathrm{25}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\:\mp\:\:\sqrt{\frac{\mathrm{5}}{\mathrm{3}}}\:\:\:\Rightarrow\:{f}'\left({x}\right)\:{has}\:\mathrm{2}\:{real}\:{roots} \\ $$$$\Rightarrow\:{definitely}\:{f}\left({x}\right)\:{will}\:\mathrm{3}\:{real}\:{roots}\:… \\ $$$${hence}\:{n}\:{can}\:{be}\:{any}\:{integer}\rightarrow{infinite}… \\ $$
Answered by MJS last updated on 03/Nov/18
x^3 +px+q=0 D=(p^3 /(27))+(q^2 /4) D<0 ⇔ 3 real roots in this case x^3 −((25)/3)x+(n/3)=0 D=(n^2 /(36))−((15625)/(729))<0 ⇒ −((250)/9)≤n≤((250)/9) n∈Z ⇒ −27≤n≤27 ⇒ 55 integer values for n
$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${D}<\mathrm{0}\:\Leftrightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{roots} \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{25}}{\mathrm{3}}{x}+\frac{{n}}{\mathrm{3}}=\mathrm{0} \\ $$$${D}=\frac{{n}^{\mathrm{2}} }{\mathrm{36}}−\frac{\mathrm{15625}}{\mathrm{729}}<\mathrm{0}\:\Rightarrow\:−\frac{\mathrm{250}}{\mathrm{9}}\leqslant{n}\leqslant\frac{\mathrm{250}}{\mathrm{9}} \\ $$$${n}\in\mathbb{Z}\:\Rightarrow\:−\mathrm{27}\leqslant{n}\leqslant\mathrm{27}\:\Rightarrow\:\mathrm{55}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{for}\:{n} \\ $$
Commented by rahul 19 last updated on 04/Nov/18
thanks sir ����
Answered by ajfour last updated on 03/Nov/18
f(x)=3x^3 −25x f ′(x)=9x^2 −25 f ′(x)=0 ⇒ x= ±(5/3), 0 f((5/3))=3((5/3))^3 −25((5/3)) = ((125)/9)−((125)/3) = − ((250)/9) so having three real roots we can raise or lower f(x) by n< ((250)/9) ⇒ n∈ (−((250)/9) , ((250)/9)) ⇒ n∈ {−27,−26,....,−1,0, 1,..27} ⇒ 55 integral values.
$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{25}{x} \\ $$$${f}\:'\left({x}\right)=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{25}\: \\ $$$${f}\:'\left({x}\right)=\mathrm{0}\:\:\Rightarrow\:\:{x}=\:\pm\frac{\mathrm{5}}{\mathrm{3}},\:\mathrm{0} \\ $$$${f}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)=\mathrm{3}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{3}} −\mathrm{25}\left(\frac{\mathrm{5}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{125}}{\mathrm{9}}−\frac{\mathrm{125}}{\mathrm{3}}\:=\:−\:\frac{\mathrm{250}}{\mathrm{9}} \\ $$$${so}\:\:{having}\:{three}\:{real}\:{roots}\:{we} \\ $$$${can}\:{raise}\:{or}\:{lower}\:{f}\left({x}\right)\:{by} \\ $$$$\:\:{n}<\:\frac{\mathrm{250}}{\mathrm{9}}\: \\ $$$$\Rightarrow\:\:\:\:{n}\in\:\left(−\frac{\mathrm{250}}{\mathrm{9}}\:,\:\frac{\mathrm{250}}{\mathrm{9}}\right) \\ $$$$\Rightarrow\:{n}\in\:\left\{−\mathrm{27},−\mathrm{26},….,−\mathrm{1},\mathrm{0},\:\mathrm{1},..\mathrm{27}\right\} \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{55}\:{integral}\:{values}. \\ $$
Commented by rahul 19 last updated on 04/Nov/18
thanks sir �� ��

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *