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Question Number 21048    Answers: 0   Comments: 0

If the equation 2cos2x − (a + 7)cosx + 3a − 13 = 0 possesses atleast one real solution, then the maximum integral value of ′a′ can be

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2cos2}{x}\:−\:\left({a}\:+\:\mathrm{7}\right)\mathrm{cos}{x}\:+ \\ $$$$\mathrm{3}{a}\:−\:\mathrm{13}\:=\:\mathrm{0}\:\mathrm{possesses}\:\mathrm{atleast}\:\mathrm{one}\:\mathrm{real} \\ $$$$\mathrm{solution},\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{integral} \\ $$$$\mathrm{value}\:\mathrm{of}\:'{a}'\:\mathrm{can}\:\mathrm{be} \\ $$

Question Number 21038    Answers: 1   Comments: 1

Question Number 21031    Answers: 0   Comments: 0

if :∀ε>0, ∀(a,b)∈R^2 ,a<b+ε prove: a≤b

$${if}\::\forall\epsilon>\mathrm{0},\:\forall\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b}+\epsilon \\ $$$${prove}:\:{a}\leqslant{b} \\ $$

Question Number 21029    Answers: 0   Comments: 0

Question Number 21021    Answers: 1   Comments: 2

Question Number 21015    Answers: 0   Comments: 0

Question Number 21013    Answers: 1   Comments: 1

Question Number 21012    Answers: 1   Comments: 3

A spring with one end attached to a mass and the other to a rigid support is stretched and released. (a) Magnitude of acceleration, when just released is maximum. (b) Magnitude of acceleration, when at equilibrium position, is maximum. (c) Speed is maximum when mass is at equilibrium position. (d) Magnitude of displacement is always maximum whenever speed is minimum.

$$\mathrm{A}\:\mathrm{spring}\:\mathrm{with}\:\mathrm{one}\:\mathrm{end}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{mass}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{to}\:\mathrm{a}\:\mathrm{rigid}\:\mathrm{support}\:\mathrm{is} \\ $$$$\mathrm{stretched}\:\mathrm{and}\:\mathrm{released}. \\ $$$$\left({a}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{just}\:\mathrm{released}\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({b}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{at}\:\mathrm{equilibrium}\:\mathrm{position},\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({c}\right)\:\mathrm{Speed}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{equilibrium}\:\mathrm{position}. \\ $$$$\left({d}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{displacement}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{maximum}\:\mathrm{whenever}\:\mathrm{speed}\:\mathrm{is} \\ $$$$\mathrm{minimum}. \\ $$

Question Number 21009    Answers: 1   Comments: 0

Question Number 21006    Answers: 0   Comments: 0

Let z_1 and z_2 be two distinct complex numbers and let z = (1 − t)z_1 + tz_2 for some real number t with 0 < t < 1. If arg(w) denotes the principal argument of a non-zero complex number w, then (1) ∣z − z_1 ∣ + ∣z − z_2 ∣ = ∣z_1 − z_2 ∣ (2) Arg (z − z_1 ) = Arg (z − z_2 ) (3) determinant (((z − z_1 ),(z^ − z_1 ^ )),((z_2 − z_1 ),(z_2 ^ − z_1 ^ ))) = 0 (4) Arg (z − z_1 ) = Arg (z_2 − z_1 )

$$\mathrm{Let}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{two}\:\mathrm{distinct}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{and}\:\mathrm{let}\:{z}\:=\:\left(\mathrm{1}\:−\:{t}\right){z}_{\mathrm{1}} \:+\:{tz}_{\mathrm{2}} \:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{real}\:\mathrm{number}\:{t}\:\mathrm{with}\:\mathrm{0}\:<\:{t}\:<\:\mathrm{1}.\:\mathrm{If} \\ $$$$\mathrm{arg}\left({w}\right)\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{argument} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{complex}\:\mathrm{number}\:{w},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mid{z}\:−\:{z}_{\mathrm{1}} \mid\:+\:\mid{z}\:−\:{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Arg}\:\left({z}\:−\:{z}_{\mathrm{1}} \right)\:=\:\mathrm{Arg}\:\left({z}\:−\:{z}_{\mathrm{2}} \right) \\ $$$$\left(\mathrm{3}\right)\:\begin{vmatrix}{{z}\:−\:{z}_{\mathrm{1}} }&{\bar {{z}}\:−\:\bar {{z}}_{\mathrm{1}} }\\{{z}_{\mathrm{2}} \:−\:{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{2}} \:−\:\bar {{z}}_{\mathrm{1}} }\end{vmatrix}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Arg}\:\left({z}\:−\:{z}_{\mathrm{1}} \right)\:=\:\mathrm{Arg}\:\left({z}_{\mathrm{2}} \:−\:{z}_{\mathrm{1}} \right) \\ $$

Question Number 21005    Answers: 0   Comments: 0

If z_1 = a + ib and z_2 = c + id are complex numbers such that ∣z_1 ∣ = ∣z_2 ∣ = 1 and Re(z_1 z_2 ^ ) = 0, then the pair of complex numbers ω_1 = a + ic and ω_2 = b + id satisfy (1) ∣ω_1 ∣ = 1 (2) ∣ω_2 ∣ = 1 (3) Re(ω_1 ω_2 ^ ) = 0 (4) ∣ω_1 ∣ = 2∣ω_2 ∣

$$\mathrm{If}\:{z}_{\mathrm{1}} \:=\:{a}\:+\:{ib}\:\mathrm{and}\:{z}_{\mathrm{2}} \:=\:{c}\:+\:{id}\:\mathrm{are}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\omega_{\mathrm{1}} \:=\:{a}\:+\:{ic}\:\mathrm{and}\:\omega_{\mathrm{2}} \:=\:{b}\:+\:{id} \\ $$$$\mathrm{satisfy} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega_{\mathrm{1}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mid\omega_{\mathrm{2}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Re}\left(\omega_{\mathrm{1}} \bar {\omega}_{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mid\omega_{\mathrm{1}} \mid\:=\:\mathrm{2}\mid\omega_{\mathrm{2}} \mid \\ $$

Question Number 21001    Answers: 0   Comments: 0

determinant (((a 1 1)),((1 b 1)),((1 1 c)))>0 then showthat abc>−8−99

$$\begin{vmatrix}{{a}\:\mathrm{1}\:\mathrm{1}}\\{\mathrm{1}\:{b}\:\mathrm{1}}\\{\mathrm{1}\:\mathrm{1}\:{c}}\end{vmatrix}>\mathrm{0}\:{then}\:{showthat}\:{abc}>−\mathrm{8}−\mathrm{99} \\ $$

Question Number 20989    Answers: 1   Comments: 1

In the figure shown below, the block of mass 2 kg is at rest. If the spring constant of both the springs A and B is 100 N/m and spring B is cut at t = 0, then magnitude of acceleration of block immediately is

$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown}\:\mathrm{below},\:\mathrm{the}\:\mathrm{block}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{constant} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{the}\:\mathrm{springs}\:{A}\:\mathrm{and}\:{B}\:\mathrm{is}\:\mathrm{100}\:\mathrm{N}/\mathrm{m} \\ $$$$\mathrm{and}\:\mathrm{spring}\:{B}\:\mathrm{is}\:\mathrm{cut}\:\mathrm{at}\:{t}\:=\:\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{magnitude}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{block} \\ $$$$\mathrm{immediately}\:\mathrm{is} \\ $$

Question Number 20988    Answers: 0   Comments: 1

Question Number 20991    Answers: 0   Comments: 1

x^3 −12x

$${x}^{\mathrm{3}} −\mathrm{12}{x} \\ $$

Question Number 20992    Answers: 3   Comments: 1

Question Number 20986    Answers: 1   Comments: 1

A 50 kg log rest on the smooth horizontal surface. A motor deliver a towing force T as shown below. The momentum of the particle at t = 5 s is

$$\mathrm{A}\:\mathrm{50}\:\mathrm{kg}\:\mathrm{log}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{the}\:\mathrm{smooth}\:\mathrm{horizontal} \\ $$$$\mathrm{surface}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{deliver}\:\mathrm{a}\:\mathrm{towing}\:\mathrm{force} \\ $$$${T}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{below}.\:\mathrm{The}\:\mathrm{momentum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:{t}\:=\:\mathrm{5}\:\mathrm{s}\:\mathrm{is} \\ $$

Question Number 20984    Answers: 1   Comments: 1

A ball of mass m is moving with a velocity u rebounds from a wall with same speed. The collision is assumed to be elastic and the force of interaction between the ball and the wall varies as shown in the figure given below. The value of F_m is

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{rebounds}\:\mathrm{from}\:\mathrm{a}\:\mathrm{wall}\:\mathrm{with} \\ $$$$\mathrm{same}\:\mathrm{speed}.\:\mathrm{The}\:\mathrm{collision}\:\mathrm{is}\:\mathrm{assumed} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{elastic}\:\mathrm{and}\:\mathrm{the}\:\mathrm{force}\:\mathrm{of}\:\mathrm{interaction} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{and}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{varies}\:\mathrm{as} \\ $$$$\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{given}\:\mathrm{below}.\:\mathrm{The} \\ $$$$\mathrm{value}\:\mathrm{of}\:{F}_{{m}} \:\mathrm{is} \\ $$

Question Number 20983    Answers: 1   Comments: 0

Find the number of ordered triples (a, b, c) of positive integers such that abc = 108.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{triples} \\ $$$$\left({a},\:{b},\:{c}\right)\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$${abc}\:=\:\mathrm{108}. \\ $$

Question Number 20953    Answers: 0   Comments: 2

Question Number 20944    Answers: 1   Comments: 0

Question Number 20939    Answers: 0   Comments: 0

Demostration of the volume of an sphere V=((4πr^3 )/3) x^2 +y^2 +z^2 =r^2 We divide the sphere in 8 parts. So the volume of a part is ∫_0 ^( r) ∫_0 ^( (√(r^2 −x^2 ))) (√(r^2 −x^2 −y^2 ))∂y∂x Lets asumme a^2 =r^2 −x^2 ∫_0 ^( r) ∫_0 ^( a) (√(a^2 −y^2 ))∂y∂x ∫_0 ^( r) a∫_0 ^( a) (√(1−((y/a))^2 ))∂y∂x Lets assume (y/a)=sinθ⇒(∂y/a)=cosθ∂θ ∫(√(1−sin^2 θ))acosθ∂θ ∫acos^2 θ∂θ a((θ/2)−((sin2θ)/4)) a(((arcsin((y/a)))/2)−((y(√(a^2 −y^2 )))/(2a^2 ))) ∫_0 ^( r) a^2 (((arcsin((y/a)))/2)−((y(√(a^2 −y^2 )))/(2a^2 )))∣_0 ^a ∂x ∫_0 ^( r) (((a^2 arcsin((y/a))−y(√(a^2 −y^2 )))/2))∣_0 ^a ∂x ∫_0 ^( r) ((πa^2 )/4)∂x ∫_0 ^( r) ((π(r^2 −x^2 ))/4)∂x (((6πr^2 x−2πx^3 )/(24)))∣_0 ^r ((πr^3 )/6)=1/8Volume of the sphrere so... V=((4πr^3 )/3)

$$\mathrm{Demostration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{an}\:\mathrm{sphere}\:\mathrm{V}=\frac{\mathrm{4}\pi\mathrm{r}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \:\mathrm{We}\:\mathrm{divide}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{in}\:\mathrm{8}\:\mathrm{parts}.\:\mathrm{So}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{part}\:\mathrm{is} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \int_{\mathrm{0}} ^{\:\sqrt{\mathrm{r}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }} \sqrt{\mathrm{r}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\partial\mathrm{y}\partial\mathrm{x}\:\:\mathrm{Lets}\:\mathrm{asumme}\:\mathrm{a}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \int_{\mathrm{0}} ^{\:\mathrm{a}} \sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\partial\mathrm{y}\partial\mathrm{x} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \mathrm{a}\int_{\mathrm{0}} ^{\:\mathrm{a}} \sqrt{\mathrm{1}−\left(\frac{\mathrm{y}}{\mathrm{a}}\right)^{\mathrm{2}} }\partial\mathrm{y}\partial\mathrm{x}\:\mathrm{Lets}\:\mathrm{assume}\:\frac{\mathrm{y}}{\mathrm{a}}=\mathrm{sin}\theta\Rightarrow\frac{\partial\mathrm{y}}{\mathrm{a}}=\mathrm{cos}\theta\partial\theta \\ $$$$\int\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}\mathrm{acos}\theta\partial\theta \\ $$$$\int\mathrm{acos}^{\mathrm{2}} \theta\partial\theta \\ $$$$\mathrm{a}\left(\frac{\theta}{\mathrm{2}}−\frac{\mathrm{sin2}\theta}{\mathrm{4}}\right) \\ $$$$\mathrm{a}\left(\frac{\mathrm{arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right)}{\mathrm{2}}−\frac{\mathrm{y}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{2a}^{\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \mathrm{a}^{\mathrm{2}} \left(\frac{\mathrm{arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right)}{\mathrm{2}}−\frac{\mathrm{y}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{2a}^{\mathrm{2}} }\right)\mid_{\mathrm{0}} ^{\mathrm{a}} \partial\mathrm{x} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \left(\frac{\mathrm{a}^{\mathrm{2}} \mathrm{arcsin}\left(\frac{\mathrm{y}}{\mathrm{a}}\right)−\mathrm{y}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\mathrm{a}} \partial\mathrm{x} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\partial\mathrm{x} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{r}} \frac{\pi\left(\mathrm{r}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{4}}\partial\mathrm{x} \\ $$$$\left(\frac{\mathrm{6}\pi\mathrm{r}^{\mathrm{2}} \mathrm{x}−\mathrm{2}\pi\mathrm{x}^{\mathrm{3}} }{\mathrm{24}}\right)\mid_{\mathrm{0}} ^{\mathrm{r}} \\ $$$$\frac{\pi\mathrm{r}^{\mathrm{3}} }{\mathrm{6}}=\mathrm{1}/\mathrm{8Volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphrere}\:\mathrm{so}... \\ $$$$\mathbb{V}=\frac{\mathrm{4}\pi\mathrm{r}^{\mathrm{3}} }{\mathrm{3}}\: \\ $$$$ \\ $$

Question Number 20936    Answers: 1   Comments: 1

A graph of x versus t is shown in Figure. Choose correct alternatives from below. (a) The particle was released from rest at t = 0 (b) At B, the acceleration a > 0 (c) At C, the velocity and the acceleration vanish (d) Average velocity for the motion between A and D is positive (e) The speed at D exceeds that at E.

$$\mathrm{A}\:\mathrm{graph}\:\mathrm{of}\:{x}\:\mathrm{versus}\:{t}\:\mathrm{is}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{Figure}. \\ $$$$\mathrm{Choose}\:\mathrm{correct}\:\mathrm{alternatives}\:\mathrm{from}\:\mathrm{below}. \\ $$$$\left({a}\right)\:\mathrm{The}\:\mathrm{particle}\:\mathrm{was}\:\mathrm{released}\:\mathrm{from} \\ $$$$\mathrm{rest}\:\mathrm{at}\:{t}\:=\:\mathrm{0} \\ $$$$\left({b}\right)\:\mathrm{At}\:{B},\:\mathrm{the}\:\mathrm{acceleration}\:{a}\:>\:\mathrm{0} \\ $$$$\left({c}\right)\:\mathrm{At}\:{C},\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{vanish} \\ $$$$\left({d}\right)\:\mathrm{Average}\:\mathrm{velocity}\:\mathrm{for}\:\mathrm{the}\:\mathrm{motion} \\ $$$$\mathrm{between}\:{A}\:\mathrm{and}\:{D}\:\mathrm{is}\:\mathrm{positive} \\ $$$$\left({e}\right)\:\mathrm{The}\:\mathrm{speed}\:\mathrm{at}\:{D}\:\mathrm{exceeds}\:\mathrm{that}\:\mathrm{at}\:{E}. \\ $$

Question Number 20935    Answers: 1   Comments: 0

If ∣z + ω∣^2 = ∣z∣^2 + ∣ω∣^2 , where z and ω are complex numbers, then (1) (z/ω) is purely real (2) (z/ω) is purely imaginary (3) zω^ + z^ ω = 0 (4) amp((z/ω)) = (π/2)

$$\mathrm{If}\:\mid{z}\:+\:\omega\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mid\omega\mid^{\mathrm{2}} ,\:\mathrm{where}\:{z}\:\mathrm{and}\:\omega \\ $$$$\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{real} \\ $$$$\left(\mathrm{2}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\left(\mathrm{3}\right)\:{z}\bar {\omega}\:+\:\bar {{z}}\omega\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{amp}\left(\frac{{z}}{\omega}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Question Number 20934    Answers: 0   Comments: 1

If z satisfies ∣z − 1∣ < ∣z + 3∣, then ω = 2z + 3 − i satisfies (1) ∣ω − 5 − i∣ < ∣ω + 3 + i∣ (2) ∣ω − 5∣ < ∣ω + 3∣ (3) Im (iω) > 1 (4) ∣arg(ω − 1)∣ < (π/2)

$$\mathrm{If}\:{z}\:\mathrm{satisfies}\:\mid{z}\:−\:\mathrm{1}\mid\:<\:\mid{z}\:+\:\mathrm{3}\mid,\:\mathrm{then}\:\omega\:= \\ $$$$\mathrm{2}{z}\:+\:\mathrm{3}\:−\:{i}\:\mathrm{satisfies} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega\:−\:\mathrm{5}\:−\:{i}\mid\:<\:\mid\omega\:+\:\mathrm{3}\:+\:{i}\mid \\ $$$$\left(\mathrm{2}\right)\:\mid\omega\:−\:\mathrm{5}\mid\:<\:\mid\omega\:+\:\mathrm{3}\mid \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Im}\:\left({i}\omega\right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\mid\mathrm{arg}\left(\omega\:−\:\mathrm{1}\right)\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$

Question Number 20933    Answers: 1   Comments: 0

If z is a complex number satisfying z + z^(−1) = 1, then z^n + z^(−n) , n ∈ N, has the value (1) 2(−1)^n , when n is a multiple of 3 (2) (−1)^(n−1) , when n is not a multiple of 3 (3) (−1)^(n+1) , when n is a multiple of 3 (4) 0 when n is not a multiple of 3

$$\mathrm{If}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{satisfying} \\ $$$${z}\:+\:{z}^{−\mathrm{1}} \:=\:\mathrm{1},\:\mathrm{then}\:{z}^{{n}} \:+\:{z}^{−{n}} ,\:{n}\:\in\:{N},\:\mathrm{has} \\ $$$$\mathrm{the}\:\mathrm{value} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\left(−\mathrm{1}\right)^{{n}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of} \\ $$$$\mathrm{3} \\ $$$$\left(\mathrm{3}\right)\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{0}\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$

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