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Question Number 186292    Answers: 0   Comments: 0

(x−p)(x^3 −x−(1/3))=0 x^4 −px^3 −x^2 +(p−(1/3))x+(p/3)=0 (x^2 +ax+h)(x^2 +bx+k)=0 a+b=−p h+k+ab=−1 bh+ak=p−(1/3) hk=(p/3) −−−−−− say ab=t −−−−−− ah+bk+p−(1/3)=p(1+t) bh+ak−(p−(1/3))=0 ⇒ (a−b)(h−k)=pt−p+(2/3) squaring (p^2 −4t){(1+t)^2 −((4p)/3)}=(pt−p+(2/3))^2 say t+1=z (p^2 +4−4z)(z^2 −((4p)/3))=(pz−2p+(2/3))^2 ⇒ −4z^3 +(p^2 +4)z^2 +((16pz)/3)−((4p)/3)(p^2 +4) =p^2 z^2 −4p(p−(1/3))z+4(p−(1/3))^2 ⇒ z^3 −z^2 −p(p+1)z+(p−(1/3))^2 +(p/3)(p^2 +4)=0 .....

$$\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{px}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){x}+\frac{{p}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{h}\right)\left({x}^{\mathrm{2}} +{bx}+{k}\right)=\mathrm{0} \\ $$$${a}+{b}=−{p} \\ $$$${h}+{k}+{ab}=−\mathrm{1} \\ $$$${bh}+{ak}={p}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${hk}=\frac{{p}}{\mathrm{3}} \\ $$$$−−−−−− \\ $$$${say}\:{ab}={t} \\ $$$$−−−−−− \\ $$$${ah}+{bk}+{p}−\frac{\mathrm{1}}{\mathrm{3}}={p}\left(\mathrm{1}+{t}\right) \\ $$$${bh}+{ak}−\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\left({a}−{b}\right)\left({h}−{k}\right)={pt}−{p}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${squaring} \\ $$$$\left({p}^{\mathrm{2}} −\mathrm{4}{t}\right)\left\{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} −\frac{\mathrm{4}{p}}{\mathrm{3}}\right\}=\left({pt}−{p}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${say}\:\:{t}+\mathrm{1}={z} \\ $$$$\left({p}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}{z}\right)\left({z}^{\mathrm{2}} −\frac{\mathrm{4}{p}}{\mathrm{3}}\right)=\left({pz}−\mathrm{2}{p}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$−\mathrm{4}{z}^{\mathrm{3}} +\left({p}^{\mathrm{2}} +\mathrm{4}\right){z}^{\mathrm{2}} +\frac{\mathrm{16}{pz}}{\mathrm{3}}−\frac{\mathrm{4}{p}}{\mathrm{3}}\left({p}^{\mathrm{2}} +\mathrm{4}\right) \\ $$$$\:\:\:\:={p}^{\mathrm{2}} {z}^{\mathrm{2}} −\mathrm{4}{p}\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){z}+\mathrm{4}\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{z}^{\mathrm{3}} −{z}^{\mathrm{2}} −{p}\left({p}+\mathrm{1}\right){z}+\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{{p}}{\mathrm{3}}\left({p}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$$.....\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 186285    Answers: 0   Comments: 2

Prove that: R (m , n) ≤ C_(m+n) ^m Here R states the Ramsey theory

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathrm{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$\mathrm{Here}\:\:\mathrm{R}\:\:\mathrm{states}\:\mathrm{the}\:\:\mathrm{Ramsey}\:\:\mathrm{theory} \\ $$

Question Number 186283    Answers: 0   Comments: 0

Question Number 186282    Answers: 1   Comments: 0

Question Number 186271    Answers: 2   Comments: 0

Question Number 186267    Answers: 1   Comments: 1

lim_(x→0) (((1+2016x)^(2017) −(1+2017x)^(2016) )/x^2 ) without L′H or series

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{2016}{x}\right)^{\mathrm{2017}} −\left(\mathrm{1}+\mathrm{2017}{x}\right)^{\mathrm{2016}} }{{x}^{\mathrm{2}} } \\ $$$${without}\:{L}'{H}\:{or}\:{series} \\ $$

Question Number 186265    Answers: 1   Comments: 0

f(x)= (√( x −a)) + (√(3a −x)) with ( a>0) is given .If , f_( max) . f_( min) = (√(32)) find , ” a ” = ?

$$ \\ $$$$\:\:\:{f}\left({x}\right)=\:\sqrt{\:{x}\:−{a}}\:\:+\:\sqrt{\mathrm{3}{a}\:−{x}}\:\:\:{with}\:\left(\:{a}>\mathrm{0}\right) \\ $$$$\:\:\:\:{is}\:\:{given}\:.{If}\:\:,\:\:{f}_{\:{max}} \:.\:{f}_{\:{min}} \:=\:\sqrt{\mathrm{32}} \\ $$$$\:\:\:\:\:{find}\:\:,\:\:\:\:\:\:\:''\:\:\:{a}\:\:''\:\:=\:? \\ $$

Question Number 186263    Answers: 2   Comments: 0

∫_0 ^( 1) (√(1+(1/(4x)))) dx

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}}}\:{dx} \\ $$$$ \\ $$

Question Number 186256    Answers: 0   Comments: 1

Prove that: R (m , n) ≤ C_(m+n) ^m

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathbb{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathbb{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$ \\ $$

Question Number 186253    Answers: 1   Comments: 1

Question Number 186249    Answers: 0   Comments: 0

Question Number 186248    Answers: 0   Comments: 0

Question Number 186247    Answers: 0   Comments: 0

Question Number 186246    Answers: 1   Comments: 0

Question Number 186241    Answers: 2   Comments: 3

x+(1/x)=((−1+(√5))/2) x?

$${x}+\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}? \\ $$

Question Number 186236    Answers: 1   Comments: 0

Question Number 186352    Answers: 1   Comments: 0

∫_1 ^( 2) ((tan^(−1) (x) + 2)/x^2 ) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\:\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \:\left(\boldsymbol{{x}}\right)\:+\:\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{2}} }\:\:\boldsymbol{{dx}}\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 186232    Answers: 1   Comments: 0

1+(x/(1+(x/(1+(x/∙_∙_∙ )))))=5 x=?

$$\mathrm{1}+\frac{{x}}{\mathrm{1}+\frac{{x}}{\mathrm{1}+\frac{{x}}{\centerdot_{\centerdot_{\centerdot} } }}}=\mathrm{5}\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Question Number 186230    Answers: 2   Comments: 0

Question Number 186231    Answers: 0   Comments: 0

Question Number 186223    Answers: 1   Comments: 0

∫_o ^(+oo) e^(−E(x)dx)

$$\int_{{o}} ^{+{oo}} {e}^{−{E}\left({x}\right){dx}} \\ $$

Question Number 186214    Answers: 0   Comments: 12

if S_a =cos(a)+sin(x+a) then ∫(S_1 /S_2 )−((x+S_1 )/(x−S_3 ))dx=?

$${if}\:{S}_{{a}} =\mathrm{cos}\left({a}\right)+\mathrm{sin}\left({x}+{a}\right) \\ $$$${then}\:\int\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }−\frac{{x}+{S}_{\mathrm{1}} }{{x}−{S}_{\mathrm{3}} }{dx}=? \\ $$

Question Number 186198    Answers: 1   Comments: 0

∫^3 _2 ((x^2 − 1)/(1 + ^x^2 (√(2 ln(x))))) dx

$$ \\ $$$$\:\:\:\:\:\underset{\mathrm{2}} {\int}^{\mathrm{3}} \:\:\:\frac{\boldsymbol{{x}}^{\mathrm{2}} \:\:−\:\:\mathrm{1}}{\mathrm{1}\:\:\:+\:\:\:^{{x}^{\mathrm{2}} } \sqrt{\mathrm{2}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)}}\:\:\:\boldsymbol{{dx}}\:\:\:\: \\ $$$$ \\ $$

Question Number 186196    Answers: 3   Comments: 2

∫_1 ^2 (((√(1 )) + cos (x))/( (√1) − cos (x))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\:\frac{\sqrt{\mathrm{1}\:}\:\:+\:\:\boldsymbol{{cos}}\:\left(\boldsymbol{{x}}\right)}{\:\sqrt{\mathrm{1}}\:\:\:−\:\:\boldsymbol{{cos}}\:\left(\boldsymbol{{x}}\right)}\:\:\boldsymbol{{dx}}\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 186195    Answers: 1   Comments: 0

∫_1 ^2 ((1/2 ∙(x^2 ) )/(x (√(x^2 + 2)))) dx

$$ \\ $$$$\:\:\:\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\:\:\frac{\mathrm{1}/\mathrm{2}\:\centerdot\left(\boldsymbol{{x}}^{\mathrm{2}} \right)\:}{\boldsymbol{{x}}\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} \:+\:\:\mathrm{2}}}\:\:\boldsymbol{{dx}}\:\:\:\: \\ $$

Question Number 186194    Answers: 1   Comments: 2

∫_2 ^4 ((2x^2 − 1)/(1 + (√x^2 ) − 2)) dx

$$ \\ $$$$\:\:\:\:\:\:\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:\:\frac{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{1}}{\mathrm{1}\:+\:\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} }\:\:−\:\:\mathrm{2}}\:\:\boldsymbol{{dx}}\:\:\:\: \\ $$$$ \\ $$

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