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AllQuestion and Answers: Page 1409

Question Number 70108    Answers: 1   Comments: 0

Question Number 70103    Answers: 2   Comments: 0

if m^3 +2p^3 =3mn, a^3 +b^3 =p^3 and a^2 +b^2 =n then prove that a+b=m.

$$\mathrm{if}\:\mathrm{m}^{\mathrm{3}} +\mathrm{2p}^{\mathrm{3}} =\mathrm{3mn},\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{p}^{\mathrm{3}} \:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{n}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}=\mathrm{m}. \\ $$

Question Number 70075    Answers: 0   Comments: 3

Question Number 70074    Answers: 1   Comments: 1

∫_1 ^2 [3+(1/t^2 )]dt=

$$\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{3}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]{dt}= \\ $$

Question Number 70069    Answers: 1   Comments: 2

Π_(n=1) ^5 (((12n−2)^4 +18^2 )/((12n−8)^4 +18^2 )) =(((10^4 +324)(22^4 +324)(34^4 +324)(46^4 +324)(58^4 +324))/((4^4 +324)(16^4 +324)(28^4 +324)(40^4 +324)(52^4 +324)))

$$\underset{{n}=\mathrm{1}} {\overset{\mathrm{5}} {\prod}}\frac{\left(\mathrm{12}{n}−\mathrm{2}\right)^{\mathrm{4}} +\mathrm{18}^{\mathrm{2}} }{\left(\mathrm{12}{n}−\mathrm{8}\right)^{\mathrm{4}} +\mathrm{18}^{\mathrm{2}} } \\ $$$$=\frac{\left(\mathrm{10}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{22}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{34}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{46}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{58}^{\mathrm{4}} +\mathrm{324}\right)}{\left(\mathrm{4}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{16}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{28}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{40}^{\mathrm{4}} +\mathrm{324}\right)\left(\mathrm{52}^{\mathrm{4}} +\mathrm{324}\right)} \\ $$

Question Number 70066    Answers: 2   Comments: 0

Solve a) e^(2x) −e^(x+1) −e^x +e<0 b)4.2^(2x) −9.2^x <−2 c)9^x −4.3^(x+1) +27>0

$${Solve} \\ $$$$\left.\mathrm{a}\right)\:{e}^{\mathrm{2}{x}} −{e}^{{x}+\mathrm{1}} −{e}^{{x}} +{e}<\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{4}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{9}.\mathrm{2}^{{x}} <−\mathrm{2} \\ $$$$\left.{c}\right)\mathrm{9}^{{x}} −\mathrm{4}.\mathrm{3}^{{x}+\mathrm{1}} +\mathrm{27}>\mathrm{0} \\ $$$$ \\ $$

Question Number 70048    Answers: 1   Comments: 6

Question Number 70044    Answers: 0   Comments: 1

(√(2016 + 2007(√(2018 + 2009(√(2020 + 2011(√(2022 + …)))))))) = ...

$$\sqrt{\mathrm{2016}\:+\:\mathrm{2007}\sqrt{\mathrm{2018}\:+\:\mathrm{2009}\sqrt{\mathrm{2020}\:+\:\mathrm{2011}\sqrt{\mathrm{2022}\:+\:\ldots}}}}\:\:=\:\:... \\ $$

Question Number 70040    Answers: 1   Comments: 3

If, a^2 b^2 c^2 ((1/a^3 )+(1/b^3 )+(1/c^3 ))=a^3 +b^3 +c^3 than prove that a,b,c Successive Proportional.

$$\mathrm{If},\:\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:\mathrm{than} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{Successive}\:\mathrm{Proportional}. \\ $$

Question Number 70035    Answers: 0   Comments: 4

Question Number 70031    Answers: 0   Comments: 0

∫[x]dx

$$\int\left[{x}\right]{dx} \\ $$

Question Number 70030    Answers: 1   Comments: 0

Find the convergence of Σ_(n=1) ^∞ (((1/n) + 1)/(−n^2 ))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{of} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\frac{\mathrm{1}}{{n}}\:+\:\mathrm{1}}{−{n}^{\mathrm{2}} } \\ $$

Question Number 70052    Answers: 1   Comments: 0

sin A+sin B=n and cos A+cos B=m sin (A+B)=?

$$ \\ $$$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}={n}\:\:\:\:\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}={m} \\ $$$$\mathrm{sin}\:\left({A}+{B}\right)=? \\ $$

Question Number 70051    Answers: 2   Comments: 0

((a+b)/c)=((cos(((a−b)/2)))/(cos(c/2)))

$$\frac{{a}+{b}}{{c}}=\frac{{cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}{{cos}\frac{{c}}{\mathrm{2}}} \\ $$

Question Number 70025    Answers: 0   Comments: 1

use ε-δ defintion to prove that lim_(x→0) (((√(1+x))−(√(1−x)))/x)=1

$${use}\:\varepsilon-\delta\:{defintion}\:{to}\:{prove}\:{that} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{{x}}=\mathrm{1} \\ $$

Question Number 70022    Answers: 0   Comments: 1

Find all pairs of (p, q) integer(s) such that p^3 − q^5 = (p + q)^2

$${Find}\:\:\:{all}\:\:{pairs}\:\:{of}\:\:\:\left({p},\:{q}\right)\:\:{integer}\left({s}\right)\:\:{such}\:\:{that} \\ $$$${p}^{\mathrm{3}} \:−\:{q}^{\mathrm{5}} \:\:=\:\:\left({p}\:+\:{q}\right)^{\mathrm{2}} \\ $$

Question Number 70021    Answers: 0   Comments: 2

(((√(√(12345689654321233 + 5333334096 (√(12345679))))) − (√(√(12345689654321233 − 5333334096 (√(12345679)))))))^(1/3) = ...

$$\sqrt[{\mathrm{3}}]{\sqrt{\sqrt{\mathrm{12345689654321233}\:+\:\mathrm{5333334096}\:\sqrt{\mathrm{12345679}}}}\:−\:\sqrt{\sqrt{\mathrm{12345689654321233}\:−\:\mathrm{5333334096}\:\sqrt{\mathrm{12345679}}}}}\:\:=\:\:... \\ $$

Question Number 70008    Answers: 1   Comments: 0

Find the value of x. log_8 x + log_4 x + log_2 x = 11.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\mathrm{log}_{\mathrm{8}} \mathrm{x}\:+\:\mathrm{log}_{\mathrm{4}} \mathrm{x}\:+\:\mathrm{log}_{\mathrm{2}} \mathrm{x}\:=\:\mathrm{11}. \\ $$

Question Number 70017    Answers: 0   Comments: 2

Solution- log_8 x+log_4 x+log_2 x=11 ⇒(1/(log_x 8))+(1/(log_x 4))+(1/(log_x 2))=11 ⇒(1/(log_x 2^3 ))+(1/(log_x 2^2 ))+(1/(log_x 2))=11 ⇒(1/(3log_x 2))+(1/(2log_x 2))+(1/(log_x 2))=11 ⇒((1/3)+(1/2)+1)(1/(log_x 2))=11 ⇒((11)/6)×(1/(log_x 2))=11 ⇒(1/(log_x 2))=11×(6/(11)) ⇒log_2 x=6 ⇒x=2^6 ∴x=64 is this rule correct????

$$\mathrm{Solution}- \\ $$$$\mathrm{log}_{\mathrm{8}} \mathrm{x}+\mathrm{log}_{\mathrm{4}} \mathrm{x}+\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{8}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{4}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3log}_{\mathrm{x}} \mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2log}_{\mathrm{x}} \mathrm{2}}+\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{11}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{x}} \mathrm{2}}=\mathrm{11}×\frac{\mathrm{6}}{\mathrm{11}} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}^{\mathrm{6}} \\ $$$$\therefore\mathrm{x}=\mathrm{64} \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{rule}\:\mathrm{correct}???? \\ $$

Question Number 70003    Answers: 0   Comments: 3

Let f(x)=3^(x−1) , g(x)=3^x and h(x)=4 determine the values of x for which: f(x)+g(x)≥h(x).

$${Let}\:{f}\left({x}\right)=\mathrm{3}^{{x}−\mathrm{1}} ,\:{g}\left({x}\right)=\mathrm{3}^{{x}} \:{and}\:{h}\left({x}\right)=\mathrm{4} \\ $$$${determine}\:{the}\:{values}\:{of}\:\boldsymbol{{x}}\:{for}\:{which}: \\ $$$${f}\left({x}\right)+{g}\left({x}\right)\geqslant{h}\left({x}\right). \\ $$$$ \\ $$

Question Number 69995    Answers: 4   Comments: 0

1. { (((√(x^2 +y^2 ))+(√(x^2 −y^2 ))=a)),(((√(x+y))+(√(x−y))=b)) :} [a,b∈R^+ ] 2. { (((√((√x)+y))+(√(x+(√y)))=a)),(((√((√x)−y))+(√(x−(√y)))=b)) :} 3. { ((x^2 +y^2 =(a−b)xy)),((x^3 +y^3 =ab(x−y))) :} 4. { ((x+y+z=a)),((x^2 +y^2 +z^2 =b)),((x^3 +y^3 +z^3 =abxyz)) :} 5. (√(x^2 +(√x)))+(√(x^2 −(√x)))=2 6. (√(x^2 +x+1))+(√(x^2 +x−1))+(√(x^2 −x−1))=1 7. 16^(sin^2 x) +16^(cos^2 x) =10 8. 2^(lnx) =x.ln(x+(√2))

$$\:\:\mathrm{1}.\begin{cases}{\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases}\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}^{+} \right] \\ $$$$ \\ $$$$\:\:\:\mathrm{2}.\begin{cases}{\sqrt{\sqrt{\mathrm{x}}+\mathrm{y}}+\sqrt{\mathrm{x}+\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{a}}}\\{\sqrt{\sqrt{\mathrm{x}}−\mathrm{y}}+\sqrt{\mathrm{x}−\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\mathrm{3}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\boldsymbol{\mathrm{ab}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right)}\end{cases} \\ $$$$ \\ $$$$\:\:\mathrm{4}.\begin{cases}{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{a}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\boldsymbol{\mathrm{b}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} +\boldsymbol{\mathrm{z}}^{\mathrm{3}} =\boldsymbol{\mathrm{abxyz}}}\end{cases} \\ $$$$\:\:\:\mathrm{5}.\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\sqrt{\boldsymbol{\mathrm{x}}}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\sqrt{\boldsymbol{\mathrm{x}}}}=\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\mathrm{6}.\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{1}}+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}−\mathrm{1}}=\mathrm{1} \\ $$$$ \\ $$$$\:\mathrm{7}.\:\:\mathrm{16}^{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} +\mathrm{16}^{\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} =\mathrm{10} \\ $$$$ \\ $$$$\:\:\mathrm{8}.\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\boldsymbol{\mathrm{lnx}}} =\boldsymbol{\mathrm{x}}.\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{2}}\right) \\ $$

Question Number 69979    Answers: 0   Comments: 2

Using the definition calculate the derivative at point x=2 f(x)=2x^3

$${Using}\:{the}\:{definition}\:{calculate}\:{the} \\ $$$${derivative}\:{at}\:{point}\:{x}=\mathrm{2} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} \\ $$$$ \\ $$

Question Number 69974    Answers: 0   Comments: 3

Identify domain and range of this function that f(x)= ln((4−x)/(4+x)).

$$\mathrm{Identify}\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\mathrm{this}\: \\ $$$$\mathrm{function}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{ln}\frac{\mathrm{4}−\mathrm{x}}{\mathrm{4}+\mathrm{x}}. \\ $$

Question Number 69954    Answers: 0   Comments: 3

A= (((1 2)),((0 1)) ) find A^n

$$\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\mathrm{find}\:\mathrm{A}^{\mathrm{n}} \\ $$

Question Number 69953    Answers: 0   Comments: 5

if tan θ+sec θ= (√3) than find out the value of _ θ where 0^o ≤θ≤2π.

$$\mathrm{if}\:\mathrm{tan}\:\theta+\mathrm{sec}\:\theta=\:\sqrt{\mathrm{3}}\:\mathrm{than}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\underset{} {\:}\theta\:\mathrm{where}\:\mathrm{0}^{\mathrm{o}} \leqslant\theta\leqslant\mathrm{2}\pi. \\ $$

Question Number 69944    Answers: 0   Comments: 4

If 2^x =0 find x

$$\mathrm{If}\:\mathrm{2}^{\mathrm{x}} \:=\mathrm{0}\:\:\:\:\:\mathrm{find}\:\:\mathrm{x} \\ $$

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