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Question Number 140974 Answers: 2 Comments: 3

If the equations x^2 −3x+a=0 and x^2 +ax−3=0 have a common root, then a possible value of a is (A) 3 (B) 1 (C) −2 (D) 2

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+{a}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} +{a}\mathrm{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{then}\:\mathrm{a}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{2} \\ $$

Question Number 140973 Answers: 1 Comments: 0

If α and β are roots of the equation 2x^2 +ax+b=0, then one of the roots of the equation 2(αx+β)^2 + a(αx+β)+b=0 is (A) 0 (B) ((α+2b)/α^2 ) (C) ((aα+b)/(2α^2 )) (D) ((aα−2b)/(2α^2 ))

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2x}^{\mathrm{2}} +{a}\mathrm{x}+{b}=\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}\left(\alpha\mathrm{x}+\beta\right)^{\mathrm{2}} + \\ $$$${a}\left(\alpha\mathrm{x}+\beta\right)+{b}=\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\alpha+\mathrm{2}{b}}{\alpha^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:\frac{{a}\alpha+{b}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{a}\alpha−\mathrm{2}{b}}{\mathrm{2}\alpha^{\mathrm{2}} } \\ $$

Question Number 140971 Answers: 0 Comments: 1

Question Number 140969 Answers: 0 Comments: 0

The roots of the equation x^2 −(m−3)x+m=0 are such that exactly one of them lies in the interval (1, 2). Then (A) 5<m<7 (B) m<10 (C) 2<m<5 (D) m>10

$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} −\left({m}−\mathrm{3}\right)\mathrm{x}+{m}=\mathrm{0}\:\mathrm{are} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right).\:\mathrm{Then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{5}<{m}<\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{m}<\mathrm{10} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{2}<{m}<\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{m}>\mathrm{10} \\ $$

Question Number 140965 Answers: 0 Comments: 0

If a≠0 and a(l+m)^2 +2blm+c=0 and a(l+n)^2 + 2bln+c=0, then (A) mn=l^2 +c/a (B) lm=n^2 +c/a (C) ln=m^2 +c/a (D) mn=l^2 +bc/a

$$\mathrm{If}\:{a}\neq\mathrm{0}\:\mathrm{and}\:{a}\left({l}+{m}\right)^{\mathrm{2}} +\mathrm{2}{blm}+{c}=\mathrm{0}\:\mathrm{and}\:{a}\left({l}+{n}\right)^{\mathrm{2}} + \\ $$$$\mathrm{2}{bln}+{c}=\mathrm{0},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:{mn}={l}^{\mathrm{2}} +{c}/{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{lm}={n}^{\mathrm{2}} +{c}/{a} \\ $$$$\left(\mathrm{C}\right)\:{ln}={m}^{\mathrm{2}} +{c}/{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{mn}={l}^{\mathrm{2}} +{bc}/{a} \\ $$

Question Number 140961 Answers: 2 Comments: 0

∫_0 ^( 1) ((ln (x+(√(1−x^2 ))))/x) dx =?

$$\:\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\:\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}\:{dx}\:=?\: \\ $$

Question Number 140959 Answers: 1 Comments: 0

Let a,b ≥ 0. Prove that (1/4)∙(((2+a)(2+b))/((1+a)(1+b))) ≥ ((4−a−b)/(4+a+b))

$$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}}{\mathrm{4}+{a}+{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140958 Answers: 0 Comments: 0

find e^ (((−1 1)),((2 −1)) )

$$\mathrm{find}\:\mathrm{e}^{\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}} \\ $$

Question Number 140956 Answers: 1 Comments: 0

.....advanced......calculus..... prove that: 𝛗:= ∫_(−∞) ^( ∞) ((sin^4 (x).cos^4 (x))/x^2 )dx=(π/(16)) m.n

$$\:\:\:\:\:\:\:\:\:\:.....{advanced}......{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\boldsymbol{\phi}:=\:\int_{−\infty} ^{\:\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right).{cos}^{\mathrm{4}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{16}} \\ $$$$\:\:{m}.{n} \\ $$

Question Number 140966 Answers: 1 Comments: 0

.......nice......calculus..... if Σ_(n=0) ^∞ (((√(cos (nπ))) )/((2n)!!)) = ω then Re(ω):=??

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{nice}......{calculus}..... \\ $$$$\:\:\:\:\:{if}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\sqrt{{cos}\:\left({n}\pi\right)}\:}{\left(\mathrm{2}{n}\right)!!}\:=\:\omega \\ $$$$\:\:\:\:\:\:\:{then}\:\:\:{Re}\left(\omega\right):=?? \\ $$

Question Number 140941 Answers: 1 Comments: 2

Question Number 140939 Answers: 1 Comments: 0

x^(sgn (x^3 −x)) = x^2 − (4/9)

$$\:{x}^{\mathrm{sgn}\:\left({x}^{\mathrm{3}} −{x}\right)} \:=\:{x}^{\mathrm{2}} \:−\:\frac{\mathrm{4}}{\mathrm{9}}\: \\ $$

Question Number 140938 Answers: 0 Comments: 0

Question Number 140937 Answers: 0 Comments: 0

can someone please share maple with me or show me the site to download free?

$${can}\:{someone}\:{please}\:{share}\:{maple}\:{with}\:{me}\:{or}\:{show}\:{me}\:{the}\:{site}\:{to}\:{download}\:{free}? \\ $$

Question Number 140930 Answers: 3 Comments: 0

∫_0 ^π (dx/( (√2) −cos x))

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{2}}\:−\mathrm{cos}\:\mathrm{x}} \\ $$

Question Number 141002 Answers: 1 Comments: 0

If a>0 and one root of ax^2 +bx+c=0 is less than −2 and the other is greater than 2, then (A) 4a+2∣b∣+c<0 (B) 4a+2∣b∣+c>0 (C) 4a+2∣b∣+c=0 (D) a+b=c

$$\mathrm{If}\:{a}>\mathrm{0}\:\mathrm{and}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{a}\mathrm{x}^{\mathrm{2}} +{b}\mathrm{x}+\mathrm{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}>\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}=\mathrm{0} \\ $$$$\left(\mathrm{D}\right)\:{a}+{b}={c} \\ $$

Question Number 140915 Answers: 1 Comments: 2

Question Number 142212 Answers: 0 Comments: 0

Question Number 140916 Answers: 0 Comments: 4

Question Number 140908 Answers: 0 Comments: 0

ab=c let (a−p)(b−q)=0 ⇒ c−(aq+bp)+pq=0 q=((bp−c)/(p−a)) say 4×2=8 (4−3)(2−((6−8)/(−1)))=1×0=0 And if q=b p=((ab−c)/(2b))=0 And if p+q=c (p−a)(c−p)=bp−c p^2 −(a+c−b)p+c(1−a)=0 2p=(a+c−b)±(√((a+c−b)^2 −4c(1−a))) ________________________

$$\:\:{ab}={c} \\ $$$${let}\:\:\:\left({a}−{p}\right)\left({b}−{q}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{c}−\left({aq}+{bp}\right)+{pq}=\mathrm{0} \\ $$$${q}=\frac{{bp}−{c}}{{p}−{a}} \\ $$$${say}\:\:\:\mathrm{4}×\mathrm{2}=\mathrm{8} \\ $$$$\:\:\:\:\:\left(\mathrm{4}−\mathrm{3}\right)\left(\mathrm{2}−\frac{\mathrm{6}−\mathrm{8}}{−\mathrm{1}}\right)=\mathrm{1}×\mathrm{0}=\mathrm{0} \\ $$$${And}\:\:{if}\:\:{q}={b} \\ $$$$\:\:{p}=\frac{{ab}−{c}}{\mathrm{2}{b}}=\mathrm{0} \\ $$$${And}\:{if}\:{p}+{q}={c} \\ $$$$\left({p}−{a}\right)\left({c}−{p}\right)={bp}−{c} \\ $$$${p}^{\mathrm{2}} −\left({a}+{c}−{b}\right){p}+{c}\left(\mathrm{1}−{a}\right)=\mathrm{0} \\ $$$$\mathrm{2}{p}=\left({a}+{c}−{b}\right)\pm\sqrt{\left({a}+{c}−{b}\right)^{\mathrm{2}} −\mathrm{4}{c}\left(\mathrm{1}−{a}\right)} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

Question Number 140907 Answers: 1 Comments: 0

(√(sin x)) cos x −(√(sin^5 x)) cos x = cos^3 x (√(sin x))

$$\:\sqrt{\mathrm{sin}\:\mathrm{x}}\:\mathrm{cos}\:\mathrm{x}\:−\sqrt{\mathrm{sin}\:^{\mathrm{5}} \mathrm{x}}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:\sqrt{\mathrm{sin}\:\mathrm{x}} \\ $$

Question Number 140905 Answers: 2 Comments: 0

lim_(x→0) ((1−cos (1−cos x))/(x (tan x−x))) =?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}\:\left(\mathrm{tan}\:\mathrm{x}−\mathrm{x}\right)}\:=? \\ $$

Question Number 140900 Answers: 1 Comments: 1

{ ((x^3 =xyz+1)),((y^3 =xyz+2)),((z^3 =xyz−3)) :}

$$\:\begin{cases}{\mathrm{x}^{\mathrm{3}} =\mathrm{xyz}+\mathrm{1}}\\{\mathrm{y}^{\mathrm{3}} =\mathrm{xyz}+\mathrm{2}}\\{\mathrm{z}^{\mathrm{3}} =\mathrm{xyz}−\mathrm{3}}\end{cases} \\ $$

Question Number 140899 Answers: 0 Comments: 2