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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

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

Question Number 20932    Answers: 0   Comments: 0

If a, b, c are real numbers and z is a complex number such that, a^2 + b^2 + c^2 = 1 and b + ic = (1 + a)z, then ((1 + iz)/(1 − iz)) equals. (1) ((b − ic)/(1 − ia)) (2) ((a + ib)/(1 + c)) (3) ((1 − c)/(a − ib)) (4) ((1 + a)/(b + ic))

$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that},\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \\ $$$$=\:\mathrm{1}\:\mathrm{and}\:{b}\:+\:{ic}\:=\:\left(\mathrm{1}\:+\:{a}\right){z},\:\mathrm{then}\:\frac{\mathrm{1}\:+\:{iz}}{\mathrm{1}\:−\:{iz}} \\ $$$$\mathrm{equals}. \\ $$$$\left(\mathrm{1}\right)\:\frac{{b}\:−\:{ic}}{\mathrm{1}\:−\:{ia}} \\ $$$$\left(\mathrm{2}\right)\:\frac{{a}\:+\:{ib}}{\mathrm{1}\:+\:{c}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}\:−\:{c}}{{a}\:−\:{ib}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}\:+\:{a}}{{b}\:+\:{ic}} \\ $$

Question Number 20926    Answers: 0   Comments: 0

Question Number 20925    Answers: 2   Comments: 0

In a △ABC, if a=2, b=60° and c=75°, then b =

$$\mathrm{In}\:\mathrm{a}\:\bigtriangleup{ABC},\:\mathrm{if}\:{a}=\mathrm{2},\:{b}=\mathrm{60}°\:\mathrm{and}\:{c}=\mathrm{75}°, \\ $$$$\mathrm{then}\:{b}\:= \\ $$

Question Number 20927    Answers: 1   Comments: 0

(a+b)×(a+b)

$$\left({a}+{b}\right)×\left({a}+{b}\right) \\ $$

Question Number 20916    Answers: 1   Comments: 0

A body starts rotating about a stationary axis with an angular acceleration b = 2t rad/s^2 . How soon after the beginning of rotation will the total acceleration vector of an arbitrary point on the body forms an angle of 60° with its velocity vector? (1) (2(√3))^(1/3) s (2) (2(√3))^(1/2) s (3) (2(√3)) s (4) (2(√3))^2 s

$$\mathrm{A}\:\mathrm{body}\:\mathrm{starts}\:\mathrm{rotating}\:\mathrm{about}\:\mathrm{a} \\ $$$$\mathrm{stationary}\:\mathrm{axis}\:\mathrm{with}\:\mathrm{an}\:\mathrm{angular} \\ $$$$\mathrm{acceleration}\:{b}\:=\:\mathrm{2}{t}\:\mathrm{rad}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{How}\:\mathrm{soon} \\ $$$$\mathrm{after}\:\mathrm{the}\:\mathrm{beginning}\:\mathrm{of}\:\mathrm{rotation}\:\mathrm{will}\:\mathrm{the} \\ $$$$\mathrm{total}\:\mathrm{acceleration}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{an}\:\mathrm{arbitrary} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{body}\:\mathrm{forms}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}° \\ $$$$\mathrm{with}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{vector}? \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} \:\mathrm{s} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{2}} \:\mathrm{s} \\ $$$$\left(\mathrm{3}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)\:\mathrm{s} \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\mathrm{s} \\ $$

Question Number 20915    Answers: 1   Comments: 0

Two shells are fired from a canon with speed u each, at angles of α and β respectively with the horizontal. The time interval between the shots is t. They collide in mid air after time T from the first shot. Which of the following conditions must be satisfied? (a) α > β (b) T cos α = (T − t) cos β (c) (T − t) cos α = T cos β (d) u sin α T − (1/2) g T^2 = u sin β (T − t) − (1/2) g (T − t)^2

$$\mathrm{Two}\:\mathrm{shells}\:\mathrm{are}\:\mathrm{fired}\:\mathrm{from}\:\mathrm{a}\:\mathrm{canon}\:\mathrm{with}\:\mathrm{speed}\:\mathrm{u}\:\mathrm{each},\:\mathrm{at} \\ $$$$\mathrm{angles}\:\mathrm{of}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{respectively}\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{The} \\ $$$$\mathrm{time}\:\mathrm{interval}\:\mathrm{between}\:\mathrm{the}\:\mathrm{shots}\:\mathrm{is}\:{t}.\:\mathrm{They}\:\mathrm{collide}\:\mathrm{in}\:\mathrm{mid} \\ $$$$\mathrm{air}\:\mathrm{after}\:\mathrm{time}\:{T}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{shot}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{conditions}\:\mathrm{must}\:\mathrm{be}\:\mathrm{satisfied}? \\ $$$$\left({a}\right)\:\alpha\:>\:\beta \\ $$$$\left({b}\right)\:{T}\:\mathrm{cos}\:\alpha\:=\:\left({T}\:−\:{t}\right)\:\mathrm{cos}\:\beta \\ $$$$\left({c}\right)\:\left({T}\:−\:{t}\right)\:\mathrm{cos}\:\alpha\:=\:{T}\:\mathrm{cos}\:\beta \\ $$$$\left({d}\right)\:{u}\:\mathrm{sin}\:\alpha\:{T}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:{g}\:{T}^{\mathrm{2}} \:=\:{u}\:\mathrm{sin}\:\beta\:\left({T}\:−\:{t}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:{g}\:\left({T}\:−\:{t}\right)^{\mathrm{2}} \\ $$

Question Number 20914    Answers: 1   Comments: 0

The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has (1) Only purely imaginary roots (2) All real roots (3) Two real and two purely imaginary roots (4) Neither real nor purely imaginary roots

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:{p}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{with} \\ $$$$\mathrm{real}\:\mathrm{coefficients}\:\mathrm{has}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{equation}\:{p}\left({p}\left({x}\right)\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Only}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{roots} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{real}\:\mathrm{roots} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Two}\:\mathrm{real}\:\mathrm{and}\:\mathrm{two}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Neither}\:\mathrm{real}\:\mathrm{nor}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\ $$

Question Number 20908    Answers: 1   Comments: 0

integrate with respect to x ∫(((2x+1)/(x^2 +4x+8)))dx

$${integrate}\:{with}\:{respect}\:{to}\:{x}\: \\ $$$$\int\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{8}}\right){dx} \\ $$

Question Number 20907    Answers: 1   Comments: 0

How many zeroes (0) there in 1×2×3×4..........99×100

$${How}\:{many}\:{zeroes}\:\left(\mathrm{0}\right)\:{there}\:{in}\:\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}..........\mathrm{99}×\mathrm{100}\: \\ $$

Question Number 20905    Answers: 1   Comments: 1

∫e^x^2 dx

$$\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 20903    Answers: 1   Comments: 0

Question Number 20891    Answers: 0   Comments: 11

The Figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearing. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle 30° with the horizontal. Then with respect to the horizontal surface (a) The point O has a linear velocity 3Rωi^∧ (b) The point P has a linear velocity ((11)/4)Rωi^∧ + ((√3)/4)Rωk^∧

$$\mathrm{The}\:\mathrm{Figure}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{system}\:\mathrm{consisting} \\ $$$$\mathrm{of}\:\left({i}\right)\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{radius}\:\mathrm{3}{R}\:\mathrm{rolling} \\ $$$$\mathrm{clockwise}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{speed} \\ $$$$\omega\:\mathrm{and}\:\left({ii}\right)\:\mathrm{an}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}{R} \\ $$$$\mathrm{rotating}\:\mathrm{anti}-\mathrm{clockwise}\:\mathrm{with}\:\mathrm{angular} \\ $$$$\mathrm{speed}\:\omega/\mathrm{2}.\:\mathrm{The}\:\mathrm{ring}\:\mathrm{and}\:\mathrm{disc}\:\mathrm{are} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{frictionless}\:\mathrm{ball}\:\mathrm{bearing}. \\ $$$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:{x}-{z}\:\mathrm{plane}.\:\mathrm{The} \\ $$$$\mathrm{point}\:{P}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance} \\ $$$${R}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{where}\:{OP}\:\mathrm{makes}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{30}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{Then} \\ $$$$\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{surface} \\ $$$$\left({a}\right)\:\mathrm{The}\:\mathrm{point}\:{O}\:\mathrm{has}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{velocity} \\ $$$$\mathrm{3}{R}\omega\overset{\wedge} {{i}} \\ $$$$\left({b}\right)\:\mathrm{The}\:\mathrm{point}\:{P}\:\mathrm{has}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{velocity} \\ $$$$\frac{\mathrm{11}}{\mathrm{4}}{R}\omega\overset{\wedge} {{i}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{R}\omega\overset{\wedge} {{k}} \\ $$

Question Number 20886    Answers: 1   Comments: 0

if sin x=msin y so proof that tan (1/2)(x−y)=((m−1)/(m+1))tan (1/2)(x+y)

$${if}\:\mathrm{sin}\:{x}={m}\mathrm{sin}\:{y} \\ $$$${so}\:{proof}\:{that} \\ $$$$\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)=\frac{{m}−\mathrm{1}}{{m}+\mathrm{1}}\mathrm{tan}\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right) \\ $$

Question Number 20885    Answers: 0   Comments: 0

if (θ−ϕ)subtle and sin θ+sin ϕ= (cos ϕ−cos θ)(√3) so proof sin 3θ+sin 3ϕ=0

$${if}\:\left(\theta−\varphi\right){subtle}\:{and}\:\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:\varphi= \\ $$$$\left(\mathrm{cos}\:\varphi−\mathrm{cos}\:\theta\right)\sqrt{\mathrm{3}} \\ $$$${so}\:{proof}\:\mathrm{sin}\:\mathrm{3}\theta+\mathrm{sin}\:\mathrm{3}\varphi=\mathrm{0} \\ $$

Question Number 20884    Answers: 1   Comments: 0

2cos (π/3)cos ((9π)/(13))+cos ((3π)/(13))+cos ((5π)/(13))=0

$$\mathrm{2cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}}=\mathrm{0} \\ $$

Question Number 20882    Answers: 1   Comments: 0

Question Number 20881    Answers: 2   Comments: 0

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