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Prove that: (a) If ∣z_1 + z_2 ∣ = ∣z_1 − z_2 ∣, the difference of the arguements of z_1 and z_2 is (π/2) (b) If arg{((z_1 + z_2 )/(z_1 − z_2 ))} = (π/2) , then ∣z_1 ∣ = ∣z_2 ∣

Question Number 51163 Answers: 0 Comments: 0

Question Number 51156 Answers: 2 Comments: 0

Show that the equation of tangent to the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 at the end of lactus rectum which lie in the 1^(st) quadrant is xe+y−a=0 ∗merry X−mas and happy new year∗

$${Show}\:{that}\:{the}\:{equation} \\ $$$${of}\:{tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:{at}\:{the}\:{end}\:{of} \\ $$$${lactus}\:{rectum}\:{which} \\ $$$${lie}\:{in}\:{the}\:\mathrm{1}^{{st}} {quadrant}\:{is} \\ $$$${xe}+{y}−{a}=\mathrm{0} \\ $$$$ \\ $$$$\ast{merry}\:{X}−{mas}\:{and}\:{happy}\:{new}\:{year}\ast \\ $$

Question Number 51153 Answers: 3 Comments: 1

Question Number 51214 Answers: 3 Comments: 0

1.lim_(x→−(i/2)) (((z−i)^2 )/((2z−i)(3−z))) 2.lim_(x→e^((πi)/4) ) ((2z^2 )/(z^3 −z−1)) 3.lim_(x→2i) ((2z^2 +8)/((√z^4 )−^3 (√(64)))) 4.lim_(x→0) ((cos 4z−1)/(z sin z))

$$\mathrm{1}.\underset{{x}\rightarrow−\frac{\mathrm{i}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\left({z}−{i}\right)^{\mathrm{2}} }{\left(\mathrm{2}{z}−{i}\right)\left(\mathrm{3}−{z}\right)} \\ $$$$\mathrm{2}.\underset{{x}\rightarrow{e}^{\frac{\pi{i}}{\mathrm{4}}} } {\mathrm{lim}}\:\frac{\mathrm{2}{z}^{\mathrm{2}} }{{z}^{\mathrm{3}} −{z}−\mathrm{1}} \\ $$$$\mathrm{3}.\underset{{x}\rightarrow\mathrm{2}{i}} {\mathrm{lim}}\:\frac{\mathrm{2}{z}^{\mathrm{2}} +\mathrm{8}}{\sqrt{{z}^{\mathrm{4}} }−^{\mathrm{3}} \sqrt{\mathrm{64}}} \\ $$$$\mathrm{4}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4}{z}−\mathrm{1}}{{z}\:\mathrm{sin}\:{z}} \\ $$$$ \\ $$

Question Number 51151 Answers: 1 Comments: 0

Find the equation of tangent to the ellipse x^2 +4y^2 =4 which are perpendicular to the line 2x−3y=1 ∗merry X−mas and happy new year∗

$${Find}\:{the}\:{equation}\:{of} \\ $$$${tangent}\:{to}\:{the}\:\:{ellipse} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{4}\:{which}\:{are} \\ $$$${perpendicular}\:{to}\:{the}\: \\ $$$${line}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1} \\ $$$$ \\ $$$$\ast{merry}\:{X}−{mas}\:{and}\:{happy}\:{new}\:{year}\ast \\ $$

Question Number 51150 Answers: 2 Comments: 0

from left hand sides prove that ((sinαsin β)/(cos α+cos β))=((2tan(α/2) tan (β/2))/(1−tan^2 (α/2)tan^2 (β/2)))

$${from}\:{left}\:{hand}\:{sides} \\ $$$${prove}\:{that} \\ $$$$\frac{{sin}\alpha\mathrm{sin}\:\beta}{\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta}=\frac{\mathrm{2tan}\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}\mathrm{tan}^{\mathrm{2}} \:\frac{\beta}{\mathrm{2}}} \\ $$

Question Number 51148 Answers: 1 Comments: 0

Two hot cubes are at same temperature and both are cooled by forced convetion the cube are made from the same material but one has side of L metre and other one 2L metre. which cube is at quicker rate of cooling

$${Two}\:{hot}\:{cubes}\:{are}\:\:{at}\:{same}\: \\ $$$${temperature}\:{and} \\ $$$${both}\:{are}\:{cooled}\:{by}\:{forced}\: \\ $$$${convetion}\:{the}\:{cube}\:{are}\:{made} \\ $$$${from}\:{the}\:{same}\:{material} \\ $$$${but}\:{one}\:{has}\:{side}\:{of}\:{L}\:{metre} \\ $$$${and}\:{other}\:{one}\:\mathrm{2}{L}\:{metre}. \\ $$$${which}\:{cube}\:{is}\:{at}\:{quicker}\: \\ $$$${rate}\:{of}\:{cooling} \\ $$$$ \\ $$

Question Number 51141 Answers: 1 Comments: 0

Question Number 51134 Answers: 1 Comments: 1

Question Number 51122 Answers: 1 Comments: 1

∫ (dx/(tgx−(√(tgx))))=?

$$\int\:\:\:\:\frac{\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{tgx}}−\sqrt{\boldsymbol{\mathrm{tgx}}}}=? \\ $$

Question Number 51107 Answers: 2 Comments: 1

A ball is bouncing down a flight of stairs.The coefficiate of restitution is e.The height of each step d and the ball descends one step at each bounce. After each bounce it rebounds to heigt h above the next lower step.The height h is/ large enough compare with width of a step that the empacts are effectively head on.show that h=(d/(1−e^2 ))

$${A}\:{ball}\:{is}\:{bouncing}\:{down} \\ $$$${a}\:{flight}\:{of}\:{stairs}.{The}\: \\ $$$${coefficiate}\:{of}\:{restitution} \\ $$$${is}\:{e}.{The}\:{height}\:{of}\:{each}\:{step} \\ $$$${d}\:{and}\:{the}\:{ball}\:{descends} \\ $$$${one}\:{step}\:{at}\:{each}\:{bounce}. \\ $$$${After}\:{each}\:{bounce}\:{it}\:{rebounds} \\ $$$${to}\:{heigt}\:\:{h}\:{above}\:{the}\:{next} \\ $$$${lower}\:{step}.{The}\:{height}\:{h}\:{is}/ \\ $$$${large}\:{enough}\:{compare}\:{with} \\ $$$${width}\:{of}\:{a}\:{step}\:{that} \\ $$$${the}\:{empacts}\:{are}\:{effectively}\: \\ $$$${head}\:{on}.{show}\:{that} \\ $$$${h}=\frac{{d}}{\mathrm{1}−{e}^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 51109 Answers: 0 Comments: 1

Question Number 51098 Answers: 1 Comments: 0

Two similar spherical bodies of radius R and 2R are initially are at same temperature.if they are kept to cool under the same condition.show qualitatively which of the two spherical body will cool faster.

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Question Number 51077 Answers: 0 Comments: 0

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Question Number 51069 Answers: 3 Comments: 0

If sinA+cos2A =(1/2) and cosA+sin2A=(1/3) , then find the value of sin3A.

$${If}\:\mathrm{sin}{A}+\mathrm{cos2}{A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{and} \\ $$$$\mathrm{cos}{A}+\mathrm{sin2}{A}=\frac{\mathrm{1}}{\mathrm{3}}\:,\:{then}\:{find}\:{the}\:{value} \\ $$$${of}\:\mathrm{sin3}{A}. \\ $$

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