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Question Number 19358 Answers: 1 Comments: 1

In the arrangement shown in figure two beads slide along a smooth horizontal rod. The relation between v and v_0 in given position will be

$$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure} \\ $$$$\mathrm{two}\:\mathrm{beads}\:\mathrm{slide}\:\mathrm{along}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{rod}.\:\mathrm{The}\:\mathrm{relation}\:\mathrm{between} \\ $$$${v}\:\mathrm{and}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{given}\:\mathrm{position}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 19355 Answers: 0 Comments: 3

Two blocks are placed on a smooth horizontal surface and connected by a string pulley arrangement as shown. If a force F starts acting on block m_1 , then find the relation between acceleration of both masses and their values

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{and}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{string}\:\mathrm{pulley}\:\mathrm{arrangement}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{force}\:{F}\:\mathrm{starts}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{block}\:{m}_{\mathrm{1}} , \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{masses}\:\mathrm{and}\:\mathrm{their}\:\mathrm{values} \\ $$

Question Number 19352 Answers: 1 Comments: 0

Prove that ∣z_1 − z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 − 2∣z_1 ∣ ∣z_2 ∣ cos (θ_1 − θ_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$−\:\mathrm{2}\mid{z}_{\mathrm{1}} \mid\:\mid{z}_{\mathrm{2}} \mid\:\mathrm{cos}\:\left(\theta_{\mathrm{1}} \:−\:\theta_{\mathrm{2}} \right) \\ $$

Question Number 19351 Answers: 1 Comments: 2

Prove that ∣z_1 + z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 ⇔ (z_1 /z_2 ) is purely imaginary number.

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{number}. \\ $$

Question Number 19350 Answers: 1 Comments: 0

Prove that ∣z_1 + z_2 ∣ = ∣z_1 − z_2 ∣ ⇔ arg(z_1 ) − arg(z_2 ) = (π/2)

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid\:\Leftrightarrow \\ $$$$\mathrm{arg}\left({z}_{\mathrm{1}} \right)\:−\:\mathrm{arg}\left({z}_{\mathrm{2}} \right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Question Number 19349 Answers: 1 Comments: 2

Prove that ∣z_1 + z_2 + z_3 + .... + z_n ∣ ≤ ∣z_1 ∣ + ∣z_2 ∣ + ∣z_3 ∣ + .... + ∣z_n ∣

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:+\:....\:+\:{z}_{{n}} \mid\:\leqslant \\ $$$$\mid{z}_{\mathrm{1}} \mid\:+\:\mid{z}_{\mathrm{2}} \mid\:+\:\mid{z}_{\mathrm{3}} \mid\:+\:....\:+\:\mid{z}_{{n}} \mid \\ $$

Question Number 19452 Answers: 1 Comments: 0

Question Number 19345 Answers: 0 Comments: 0

A thin bi − convex lens rest on a plane mirror . it is found that a point objects placed 20cm above the object coincide with it own image. Determine the position and nature of the image when the object is placed (i) 8cm and (ii) 12 from the lens mirror combinatiom

$$\mathrm{A}\:\mathrm{thin}\:\mathrm{bi}\:−\:\mathrm{convex}\:\mathrm{lens}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{mirror}\:.\:\:\mathrm{it}\:\mathrm{is}\:\mathrm{found}\:\mathrm{that}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{objects}\:\mathrm{placed}\:\mathrm{20cm}\:\mathrm{above}\:\mathrm{the}\:\mathrm{object}\:\mathrm{coincide}\:\mathrm{with}\:\mathrm{it}\:\mathrm{own}\:\mathrm{image}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{position}\:\mathrm{and}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{image}\:\mathrm{when}\:\mathrm{the}\:\mathrm{object}\:\mathrm{is}\:\mathrm{placed} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{8cm}\:\:\mathrm{and}\:\:\left(\mathrm{ii}\right)\:\mathrm{12}\:\:\:\:\mathrm{from}\:\mathrm{the}\:\mathrm{lens}\:\mathrm{mirror}\:\mathrm{combinatiom} \\ $$

Question Number 19341 Answers: 0 Comments: 2

Question Number 19476 Answers: 1 Comments: 0

STATEMENT-1 : The graph between kinetic energy and vertical displacement is a straight line for a projectile. STATEMENT-2 : The graph between kinetic energy and horizontal displacement is a straight line for a projectile. STATEMENT-3 : The graph between kinetic energy and time is a parabola for a projectile.

$$\mathrm{STATEMENT}-\mathrm{1}\::\:\mathrm{The}\:\mathrm{graph}\:\mathrm{between} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{and}\:\mathrm{vertical}\:\mathrm{displacement} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{for}\:\mathrm{a}\:\mathrm{projectile}. \\ $$$$\mathrm{STATEMENT}-\mathrm{2}\::\:\mathrm{The}\:\mathrm{graph}\:\mathrm{between} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{and}\:\mathrm{horizontal} \\ $$$$\mathrm{displacement}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{for}\:\mathrm{a} \\ $$$$\mathrm{projectile}. \\ $$$$\mathrm{STATEMENT}-\mathrm{3}\::\:\mathrm{The}\:\mathrm{graph}\:\mathrm{between} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{and}\:\mathrm{time}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parabola} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{projectile}. \\ $$

Question Number 19332 Answers: 1 Comments: 1

Let S_n = n^2 + 20n + 12, n a positive integer. What is the sum of all possible values of n for which S_n is a perfect square?

$$\mathrm{Let}\:{S}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{20}{n}\:+\:\mathrm{12},\:{n}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{values}\:\mathrm{of}\:{n}\:\mathrm{for}\:\mathrm{which}\:{S}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect} \\ $$$$\mathrm{square}? \\ $$

Question Number 19405 Answers: 1 Comments: 0

Convert i(√((2(√2)−1)/2)) into polarform.

$${Convert}\:{i}\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}}\:{into}\:{polarform}. \\ $$

Question Number 19330 Answers: 1 Comments: 0

A triangle with perimeter 7 has integer side lengths. What is the maximum possible area of such a triangle?

$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{perimeter}\:\mathrm{7}\:\mathrm{has}\:\mathrm{integer} \\ $$$$\mathrm{side}\:\mathrm{lengths}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{possible}\:\mathrm{area}\:\mathrm{of}\:\mathrm{such}\:\mathrm{a}\:\mathrm{triangle}? \\ $$

Question Number 19329 Answers: 0 Comments: 5

Solve the equation y^3 = x^3 + 8x^2 − 6x + 8 for positive integers x and y.

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:{y}^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} \:+\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:{x}\:\mathrm{and}\:{y}. \\ $$

Question Number 19324 Answers: 0 Comments: 0

y=tan^(−1) 3a^2 x−x^3 /a(a^2 −3x^2 )

$$\mathrm{y}=\mathrm{tan}^{−\mathrm{1}} \mathrm{3a}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{3}} /\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{3x}^{\mathrm{2}} \right) \\ $$

Question Number 19323 Answers: 0 Comments: 0

y=sin (2tan^(−1) (√(1−x/1+x)))

$$\mathrm{y}=\mathrm{sin}\:\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\left.\mathrm{1}−\mathrm{x}/\mathrm{1}+\mathrm{x}\right)}\right. \\ $$

Question Number 19313 Answers: 1 Comments: 0

Prove that ∣z_1 ± z_2 ∣^2 = ∣z_2 ∣^2 + ∣z_1 ∣^2 ± 2Re(z_1 z_2 ^ ) = ∣z_1 ∣^2 + ∣z_2 ∣^2 ± 2Re(z_1 ^ .z_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:\pm\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:\pm \\ $$$$\mathrm{2Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\pm\:\mathrm{2Re}\left(\bar {{z}}_{\mathrm{1}} .{z}_{\mathrm{2}} \right) \\ $$

Question Number 19312 Answers: 1 Comments: 0

Product of n, n^(th) roots of unity = 1.α.α^2 .α^3 ..... α^(n−1) = (−1)^(n−1) Why? How to get RHS?

$$\mathrm{Product}\:\mathrm{of}\:{n},\:{n}^{\mathrm{th}} \:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity} \\ $$$$=\:\mathrm{1}.\alpha.\alpha^{\mathrm{2}} .\alpha^{\mathrm{3}} \:.....\:\alpha^{{n}−\mathrm{1}} \:=\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \\ $$$$\mathrm{Why}?\:\mathrm{How}\:\mathrm{to}\:\mathrm{get}\:\mathrm{RHS}? \\ $$

Question Number 19321 Answers: 1 Comments: 1

Parallel tangents to a circle at A and B are cut in the points C and D by a tangent to the circle at E. Prove that AD, BC and the line joining the middle points of AE and BE are concurrent.

$$\mathrm{Parallel}\:\mathrm{tangents}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{A} \\ $$$$\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{cut}\:\mathrm{in}\:\mathrm{the}\:\mathrm{points}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{at}\:\mathrm{E}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{AD},\:\mathrm{BC}\:\mathrm{and}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{joining}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{points}\:\mathrm{of}\:\mathrm{AE} \\ $$$$\mathrm{and}\:\mathrm{BE}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$

Question Number 19301 Answers: 1 Comments: 0

e^(iπ) +1=0

$${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$

Question Number 19325 Answers: 0 Comments: 0

xcos^(−1) x/(√(1−x^2 ))

$$\mathrm{xcos}^{−\mathrm{1}} \mathrm{x}/\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 19326 Answers: 0 Comments: 1

tan^(−1) ((√x)−x/1+x^(3/2) )

$$\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{x}}−\mathrm{x}/\mathrm{1}+\mathrm{x}^{\mathrm{3}/\mathrm{2}} \right) \\ $$

Question Number 19294 Answers: 1 Comments: 0

Question Number 19293 Answers: 1 Comments: 3

Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that ∠APB = ∠BQC = 120° and an isosceles triangle RAC on the other side of AC such that ∠ARC = 120°. Show that PQR is an equilateral triangle.

$$\mathrm{Let}\:{AC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{and}\:{B}\:\mathrm{a}\:\mathrm{point}\:\mathrm{between}\:{A}\:\mathrm{and}\:{C}. \\ $$$$\mathrm{Construct}\:\mathrm{isosceles}\:\mathrm{triangles}\:{PAB}\:\mathrm{and} \\ $$$${QBC}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment}\:{AC} \\ $$$$\mathrm{such}\:\mathrm{that}\:\angle{APB}\:=\:\angle{BQC}\:=\:\mathrm{120}°\:\mathrm{and} \\ $$$$\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:{RAC}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{of}\:{AC}\:\mathrm{such}\:\mathrm{that}\:\angle{ARC}\:=\:\mathrm{120}°. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{PQR}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$

Question Number 19292 Answers: 1 Comments: 1

Prove the equality sin (π/(2n)) sin ((2π)/(2n)) ... sin (((n − 1)π)/(2n)) = ((√n)/2^(n−1) ) .

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{equality} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{2}{n}}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{2}{n}}\:...\:\mathrm{sin}\:\frac{\left({n}\:−\:\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:=\:\frac{\sqrt{{n}}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$

Question Number 19291 Answers: 1 Comments: 0

Prove that (1/(cos 6°)) + (1/(sin 24°)) + (1/(sin 48°)) = (1/(sin 12°)) .

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{6}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{24}°}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{48}°}\:=\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}\:. \\ $$

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