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AlgebraQuestion and Answers: Page 226

Question Number 130362 Answers: 3 Comments: 0

f(x)=((2x+1)/( (√(x^2 −∣2x−3∣)))) Domain D_f = ?

$$\mathrm{f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mid\mathrm{2}{x}−\mathrm{3}\mid}} \\ $$$$\mathrm{Domain}\:\mathrm{D}_{\mathrm{f}} \:=\:? \\ $$

Question Number 130354 Answers: 5 Comments: 0

f(x)=((4x^2 +1)/(2x^2 +1)) prove that 1 ≤ f(x) ≤ 2

$$\mathrm{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{1}\:\leqslant\:\mathrm{f}\left({x}\right)\:\leqslant\:\mathrm{2} \\ $$

Question Number 130290 Answers: 1 Comments: 0

Σ_(i=0) ^∞ Σ_(j=0) ^∞ Σ_(k=0) ^∞ (1/3^(i+j+k) ) ? where i≠j≠k

$$\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{j}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{i}+\mathrm{j}+\mathrm{k}} }\:?\:\mathrm{where}\:\mathrm{i}\neq\mathrm{j}\neq\mathrm{k} \\ $$

Question Number 130334 Answers: 2 Comments: 0

Complete the square in the expression y^2 +8y+9k and hence find the value of k that makes it a perfect square.

$$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{y}^{\mathrm{2}} \:+\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Question Number 130331 Answers: 0 Comments: 0

f(x)=ax^3 +bx^2 +c f ′(x)=3ax^2 +2bx f(0)=−2 ⇒a(0)^3 +b(0)^2 +c=−2 ⇒c=−2 f(−2)=2 f ′(−2)=0 ⇒ { ((a(−2)^3 +b(−2)^2 −2=2)),((3a(−2)^2 +2b(−2)=0)) :} ⇒ { ((−8a+4b=4)),((12a−4b=0)) :} −8a+12a+4b−4b=4 ⇒4a=4 ⇒a=1 −8a+4b=4 ⇒b=1+2a ⇒b=1+2 ⇒b=3 f(x)=x^3 +3x^2 −2 f ′(x)=3x^2 +6x f(0)=(0)^3 +3(0)^2 −2=−2 vrai f(−2)=(−2)^3 +3(−2)^2 −2=−8+12−2=2 vrai f ′(−2)=3(−2)^2 +6(−2)=12−12=0 vrai

$$\mathrm{f}\left(\mathrm{x}\right)={a}\mathrm{x}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{c} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{3}{a}\mathrm{x}^{\mathrm{2}} +\mathrm{2bx} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=−\mathrm{2} \\ $$$$\Rightarrow{a}\left(\mathrm{0}\right)^{\mathrm{3}} +\mathrm{b}\left(\mathrm{0}\right)^{\mathrm{2}} +\mathrm{c}=−\mathrm{2} \\ $$$$\Rightarrow\mathrm{c}=−\mathrm{2} \\ $$$$\mathrm{f}\left(−\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{f}\:'\left(−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{a}\left(−\mathrm{2}\right)^{\mathrm{3}} +{b}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2}}\\{\mathrm{3}{a}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{b}\left(−\mathrm{2}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{−\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{4}}\\{\mathrm{12}{a}−\mathrm{4}{b}=\mathrm{0}}\end{cases} \\ $$$$−\mathrm{8}{a}+\mathrm{12}{a}+\mathrm{4}{b}−\mathrm{4}{b}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}{a}=\mathrm{4} \\ $$$$\Rightarrow{a}=\mathrm{1} \\ $$$$−\mathrm{8}{a}+\mathrm{4}{b}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{1}+\mathrm{2}{a} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{1}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{b}=\mathrm{3} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} +\mathrm{6x} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\left(\mathrm{0}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{0}\right)^{\mathrm{2}} −\mathrm{2}=−\mathrm{2}\:\:\:\:\:\:\mathrm{vrai} \\ $$$$\mathrm{f}\left(−\mathrm{2}\right)=\left(−\mathrm{2}\right)^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=−\mathrm{8}+\mathrm{12}−\mathrm{2}=\mathrm{2}\:\:\:\:\mathrm{vrai} \\ $$$$\mathrm{f}\:'\left(−\mathrm{2}\right)=\mathrm{3}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{6}\left(−\mathrm{2}\right)=\mathrm{12}−\mathrm{12}=\mathrm{0}\:\:\:\:\:\mathrm{vrai} \\ $$

Question Number 130257 Answers: 0 Comments: 2

n is an integer prove algebraically that the sum of (1/2)n(n+1) and (1/2)(n+1)(n+2) is always a square number. note: write your expression in a form that clearly shows a square number.

$${n}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer} \\ $$$$\mathrm{prove}\:\mathrm{algebraically}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:\mathrm{is}\:\mathrm{always} \\ $$$$\mathrm{a}\:\mathrm{square}\:\mathrm{number}. \\ $$$$\mathrm{note}:\:{write}\:{your}\:{expression}\:{in}\:{a}\:{form} \\ $$$${that}\:{clearly}\:{shows}\:{a}\:{square}\:{number}. \\ $$

Question Number 130242 Answers: 0 Comments: 0

Find m n=(p_1 p_2 )^m 2n=(p_1 ^(1+log_(p_1 p_2 ) n) )(p_2 ^(1+log_(p_1 p_2 ) n) )

$$\mathrm{Find}\:\:\:{m} \\ $$$${n}=\left({p}_{\mathrm{1}} {p}_{\mathrm{2}} \right)^{{m}} \\ $$$$\mathrm{2}{n}=\left({p}_{\mathrm{1}} ^{\mathrm{1}+{log}_{{p}_{\mathrm{1}} {p}_{\mathrm{2}} } {n}} \right)\left({p}_{\mathrm{2}} ^{\mathrm{1}+{log}_{{p}_{\mathrm{1}} {p}_{\mathrm{2}} } {n}} \right) \\ $$

Question Number 130092 Answers: 1 Comments: 1

Question Number 130089 Answers: 1 Comments: 2

To solve x^3 =x+c Let y=(x−p)(x^3 −x−c) (dy/dx)=(x^3 −x−c)+(3x^2 −1)(x−p) let (dy/dx)=mx ⇒ (3x^2 −1)(x−p)=mx 3x^3 −3px^2 −(m+1)x+p=0 3(x+c)−3px^2 −(m+1)x+p=0 3px^2 +(m−2)x−(3c+p)=0 and since x^3 =x+c (m−2)x^2 +(2p−3c)x+3cp=0 [9cp^2 +(3c+p)(m−2)]x +[3cp(m−2)+(3c+p)(2p−3c)] =0 x=−(((3c+p)(2p−3c)+3cp(m−2))/(9cp^2 +(3c+p)(m−2))) (m−2)x^2 +(2p−3c)x+(3cp)=0 Now choosing suitable ′p′ value we find ′m′; and then x=f(p,m) ★

$${To}\:{solve}\:\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${Let}\:\:{y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−{c}\right) \\ $$$$\:\:\frac{{dy}}{{dx}}=\left({x}^{\mathrm{3}} −{x}−{c}\right)+\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right) \\ $$$$\:\:{let}\:\:\:\frac{{dy}}{{dx}}={mx} \\ $$$$\Rightarrow\:\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right)={mx} \\ $$$$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{3}{px}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+{p}=\mathrm{0} \\ $$$$\mathrm{3}\left({x}+{c}\right)−\mathrm{3}{px}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+{p}=\mathrm{0} \\ $$$$\mathrm{3}{px}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}−\left(\mathrm{3}{c}+{p}\right)=\mathrm{0} \\ $$$${and}\:{since}\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$$\left({m}−\mathrm{2}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{3}{c}\right){x}+\mathrm{3}{cp}=\mathrm{0} \\ $$$$\left[\mathrm{9}{cp}^{\mathrm{2}} +\left(\mathrm{3}{c}+{p}\right)\left({m}−\mathrm{2}\right)\right]{x} \\ $$$$+\left[\mathrm{3}{cp}\left({m}−\mathrm{2}\right)+\left(\mathrm{3}{c}+{p}\right)\left(\mathrm{2}{p}−\mathrm{3}{c}\right)\right] \\ $$$$=\mathrm{0} \\ $$$${x}=−\frac{\left(\mathrm{3}{c}+{p}\right)\left(\mathrm{2}{p}−\mathrm{3}{c}\right)+\mathrm{3}{cp}\left({m}−\mathrm{2}\right)}{\mathrm{9}{cp}^{\mathrm{2}} +\left(\mathrm{3}{c}+{p}\right)\left({m}−\mathrm{2}\right)} \\ $$$$\left({m}−\mathrm{2}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{3}{c}\right){x}+\left(\mathrm{3}{cp}\right)=\mathrm{0} \\ $$$${Now}\:{choosing}\:{suitable}\:'{p}'\:{value} \\ $$$${we}\:{find}\:'{m}'; \\ $$$${and}\:{then}\:{x}={f}\left({p},{m}\right)\:\bigstar \\ $$

Question Number 130036 Answers: 2 Comments: 0

If { ((a+b+c = 1)),((a^3 +b^3 +c^3 = 4)) :} find (1/(a+bc)) + (1/(b+ca)) + (1/(c+ab)).

$$\mathrm{If}\:\begin{cases}{{a}+{b}+{c}\:=\:\mathrm{1}}\\{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:\mathrm{4}}\end{cases} \\ $$$$\:\mathrm{find}\:\frac{\mathrm{1}}{{a}+{bc}}\:+\:\frac{\mathrm{1}}{{b}+{ca}}\:+\:\frac{\mathrm{1}}{{c}+{ab}}. \\ $$

Question Number 130006 Answers: 2 Comments: 0

Question Number 129979 Answers: 3 Comments: 0

If { ((∣x∣ + x + y = 5)),((x + ∣y∣ −y = 10)) :} then x+y ?

$$\mathrm{If}\:\begin{cases}{\mid\mathrm{x}\mid\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}}\\{\mathrm{x}\:+\:\mid\mathrm{y}\mid\:−\mathrm{y}\:=\:\mathrm{10}}\end{cases}\:\mathrm{then}\:\mathrm{x}+\mathrm{y}\:? \\ $$

Question Number 129841 Answers: 1 Comments: 2

{_((2^(2x) )^(1/3) +(2^(4x) )^(1/3) +4^x =5) ^(((2^x ))^(1/3) −2^x =2) }⇒2^x −8^x =? pleas solve thes

$$\left\{_{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2x}} }+\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{4x}} }+\mathrm{4}^{\mathrm{x}} =\mathrm{5}} ^{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{x}} \:}−\mathrm{2}^{\mathrm{x}} =\mathrm{2}} \right\}\Rightarrow\mathrm{2}^{\mathrm{x}} −\mathrm{8}^{\mathrm{x}} =? \\ $$$$\mathrm{pleas}\:\mathrm{solve}\:\mathrm{thes} \\ $$

Question Number 129832 Answers: 1 Comments: 2

Question Number 129687 Answers: 1 Comments: 0

x^3 −x−c=0 (solve for x even if 0<c<(2/(3(√3))))

$$\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$$\left({solve}\:{for}\:{x}\:\:{even}\:{if}\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$

Question Number 129677 Answers: 0 Comments: 2

Question Number 129663 Answers: 0 Comments: 4

(√(14+(√3) +((26−5(√3)))^(1/3) )) =?

$$\:\sqrt{\mathrm{14}+\sqrt{\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{\mathrm{26}−\mathrm{5}\sqrt{\mathrm{3}}}}\:=?\: \\ $$

Question Number 129622 Answers: 1 Comments: 2

Let f(x)=(x+1)^2 for all x≥−1. If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y=x, then g(x) is equal to...

$$\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\geqslant−\mathrm{1}.\:\mathrm{If}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{function}\:\mathrm{whose}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{x},\:\mathrm{then}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}... \\ $$

Question Number 129607 Answers: 2 Comments: 0

salom

$${salom} \\ $$

Question Number 129595 Answers: 2 Comments: 0

y=(√(sin x+(√(sin x+(√(sin x+.............∞)))))) (dy/dx)=?

$$\mathrm{y}=\sqrt{\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{sin}\:\mathrm{x}+.............\infty}}} \\ $$$$ \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=? \\ $$

Question Number 129489 Answers: 2 Comments: 0

if p(x+2)−2p(x)=x^2 −5x−3 find p(x)

$${if}\:{p}\left({x}+\mathrm{2}\right)−\mathrm{2}{p}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3} \\ $$$${find}\:{p}\left({x}\right) \\ $$

Question Number 129394 Answers: 3 Comments: 0

{ ((x^3 −3x^2 +5x+17=0)),((y^3 −3y^2 +5y+11=0)) :} find x+y .