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Question Number 16707    Answers: 1   Comments: 0

If x, y, z are pth, qth and rth terms respectively, of an AP and also of GP, then x^(y−z) y^(z−x) z^(x−y) is equal to

$$\mathrm{If}\:\:{x},\:{y},\:{z}\:\mathrm{are}\:{p}\mathrm{th},\:{q}\mathrm{th}\:\mathrm{and}\:{r}\mathrm{th}\:\mathrm{terms}\:\mathrm{respectively}, \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{and}\:\mathrm{also}\:\mathrm{of}\:\mathrm{GP},\:\mathrm{then}\: \\ $$$${x}^{{y}−{z}} \:{y}^{{z}−{x}} \:{z}^{{x}−{y}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 16703    Answers: 2   Comments: 0

Evaluate: ∫_0 ^(1/2) (dx/((1 + x^2 )(√(1 − x^2 ))))

$$\mathrm{Evaluate}:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }} \\ $$

Question Number 16701    Answers: 1   Comments: 0

Find distinct natural numbers from 1 to 9 such that these six equations are satisfied simultaneously: (1) a + bc = 20 (2) d + e + f = 20 (3) g − hi = −20 (4) adg = 20 (5) b + eh = 20 (6) c + f − i = 10

$$\mathrm{Find}\:\mathrm{distinct}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{1} \\ $$$$\mathrm{to}\:\mathrm{9}\:\mathrm{such}\:\mathrm{that}\:\mathrm{these}\:\mathrm{six}\:\mathrm{equations}\:\mathrm{are} \\ $$$$\mathrm{satisfied}\:\mathrm{simultaneously}: \\ $$$$\left(\mathrm{1}\right)\:{a}\:+\:{bc}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{2}\right)\:{d}\:+\:{e}\:+\:{f}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{3}\right)\:{g}\:−\:{hi}\:=\:−\mathrm{20} \\ $$$$\left(\mathrm{4}\right)\:{adg}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{5}\right)\:{b}\:+\:{eh}\:=\:\mathrm{20} \\ $$$$\left(\mathrm{6}\right)\:{c}\:+\:{f}\:−\:{i}\:=\:\mathrm{10} \\ $$

Question Number 16691    Answers: 1   Comments: 0

The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of horizontal motion of projectile, giving it a constant horizontal acceleration equal to g. Under the same conditions of projection, the new range will be (g = acceleration due to gravity) [Answer: R + 4H]

$$\mathrm{The}\:\mathrm{horizontal}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile} \\ $$$$\mathrm{is}\:{R}\:\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\mathrm{by}\:\mathrm{it}\:\mathrm{is}\:{H}.\:\mathrm{A}\:\mathrm{strong}\:\mathrm{wind}\:\mathrm{now}\:\mathrm{begins}\:\mathrm{to} \\ $$$$\mathrm{blow}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{horizontal} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{projectile},\:\mathrm{giving}\:\mathrm{it}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{horizontal}\:\mathrm{acceleration}\:\mathrm{equal}\:\mathrm{to}\:{g}. \\ $$$$\mathrm{Under}\:\mathrm{the}\:\mathrm{same}\:\mathrm{conditions}\:\mathrm{of}\:\mathrm{projection}, \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{range}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left({g}\:=\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\right) \\ $$$$\left[\boldsymbol{\mathrm{Answer}}:\:{R}\:+\:\mathrm{4}{H}\right] \\ $$

Question Number 16689    Answers: 1   Comments: 0

why any infinitely differentiable function is a power series?

$${why}\:{any}\:{infinitely}\:{differentiable}\:{function}\:{is}\:{a}\:{power}\:{series}? \\ $$

Question Number 16687    Answers: 0   Comments: 0

Three consecutive terms of an A.P form the three consecutive terms of a G.P, If the common ratio of the G.P forms the common difference of the A.P by adding the first term of the G.P to itself. Find the sum of the fifth term of the G.P.

$$\mathrm{Three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}\:\mathrm{form}\:\mathrm{the}\:\mathrm{three}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}, \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P}\:\mathrm{forms}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{A}.\mathrm{P}\:\mathrm{by} \\ $$$$\mathrm{adding}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P}\:\mathrm{to}\:\mathrm{itself}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fifth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{G}.\mathrm{P}. \\ $$

Question Number 16685    Answers: 1   Comments: 0

If cos (θ−α),cos θ and cos (θ+α) are in HP, then cos θsec (α/2) is equal to a)±(√2) b) ±(√3) c)±(1/(√2)) d)None of these

$$\mathrm{If}\:\mathrm{cos}\:\left(\theta−\alpha\right),\mathrm{cos}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\left(\theta+\alpha\right) \\ $$$$\:\mathrm{are}\:\mathrm{in}\:\mathrm{HP},\:\mathrm{then} \\ $$$$\mathrm{cos}\:\theta\mathrm{sec}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left.\mathrm{a}\right)\pm\sqrt{\mathrm{2}} \\ $$$$\left.{b}\right)\:\pm\sqrt{\mathrm{3}} \\ $$$$\left.{c}\right)\pm\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\left.{d}\right)\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$

Question Number 16682    Answers: 1   Comments: 0

Question Number 16681    Answers: 1   Comments: 0

If tan x+tan ((π/3)+x)+tan (((2π)/3)+x)=3 a) tan x=1 b) tan 2x=1 c) tan 3x=1 d) None of above

$$\mathrm{If} \\ $$$$\mathrm{tan}\:{x}+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}+{x}\right)+\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)=\mathrm{3} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:{x}=\mathrm{1} \\ $$$$\left.{b}\right)\:\mathrm{tan}\:\mathrm{2}{x}=\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{tan}\:\mathrm{3}{x}=\mathrm{1} \\ $$$$\left.{d}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{above} \\ $$

Question Number 16675    Answers: 2   Comments: 0

The number of intersecting points on the graph for sin x = (x/(10)) for x ∈ [−π, π] is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{graph}\:\mathrm{for}\:\mathrm{sin}\:{x}\:=\:\frac{{x}}{\mathrm{10}}\:\mathrm{for}\:{x}\:\in\:\left[−\pi,\:\pi\right] \\ $$$$\mathrm{is} \\ $$

Question Number 16699    Answers: 0   Comments: 8

Related to Q16675: Find the number of intersection points of graph sin x=(x/(10)). Let′s see sin x = (x/n) with n>1. For n≤1 there is one intersection point. Let x=2kπ+t with k∈N ∧ t∈[0,2π] sin x=sin t cos x=cos t we find the point on f(x)=sin x where its tangent is g(x)=(x/n). f′(x)=cos x=cos t g′(x)=(1/n) cos t=(1/n) t=cos^(−1) (1/n) sin t=(n/(√(n^2 +1))) so that f(x) intersects with g(x), ((sin x)/x)≥(1/n) ⇒n sin x≥x ⇒n sin t≥2kπ+t ⇒k≤((n sin t −t)/(2π))=(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π)) k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ number of intersecting points is m=2×2(k_(max) +1)−1=4k_(max) +3 for n=10 k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ =⌊((((10^2 )/(√(10^2 +1)))−cos^(−1) (1/(10)))/(2π))⌋=⌊1.35⌋=1 ⇒m=4×1+3=7 for n=20 k_(max) =⌊((((20^2 )/(√(20^2 +1)))−cos^(−1) (1/(20)))/(2π))⌋=⌊2.94⌋=2 ⇒m=4×2+3=11

$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16675}: \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{graph}\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{x}}{\mathrm{10}}. \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{see}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{x}}{\mathrm{n}}\:\mathrm{with}\:\mathrm{n}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{n}\leqslant\mathrm{1}\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{intersection}\:\mathrm{point}. \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{x}=\mathrm{2k}\pi+\mathrm{t}\:\mathrm{with}\:\mathrm{k}\in\mathbb{N}\:\wedge\:\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{sin}\:\mathrm{t} \\ $$$$\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\mathrm{x}\:\mathrm{where}\:\mathrm{its} \\ $$$$\mathrm{tangent}\:\mathrm{is}\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{n}}. \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{cos}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{t}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{sin}\:\mathrm{t}=\frac{\mathrm{n}}{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{intersects}\:\mathrm{with}\:\mathrm{g}\left(\mathrm{x}\right), \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\geqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{x}\geqslant\mathrm{x} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\geqslant\mathrm{2k}\pi+\mathrm{t} \\ $$$$\Rightarrow\mathrm{k}\leqslant\frac{\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\:−\mathrm{t}}{\mathrm{2}\pi}=\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$ \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{is} \\ $$$$\mathrm{m}=\mathrm{2}×\mathrm{2}\left(\mathrm{k}_{\mathrm{max}} +\mathrm{1}\right)−\mathrm{1}=\mathrm{4k}_{\mathrm{max}} +\mathrm{3} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{10} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$=\lfloor\frac{\frac{\mathrm{10}^{\mathrm{2}} }{\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{10}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{1}.\mathrm{35}\rfloor=\mathrm{1} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{1}+\mathrm{3}=\mathrm{7} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{20} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{20}^{\mathrm{2}} }{\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{20}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{2}.\mathrm{94}\rfloor=\mathrm{2} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{2}+\mathrm{3}=\mathrm{11} \\ $$

Question Number 16754    Answers: 2   Comments: 1

Question Number 16665    Answers: 0   Comments: 1

Question Number 16663    Answers: 0   Comments: 0

Question Number 16656    Answers: 1   Comments: 0

A particle moves with a speed of 10 ms^(−1) from the point (2, −2) in the direction 3i^∧ + 4j^∧ . The position vector after 3 s is

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{10} \\ $$$$\mathrm{ms}^{−\mathrm{1}} \:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\:−\mathrm{2}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{direction}\:\mathrm{3}\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}.\:\mathrm{The}\:\mathrm{position}\:\mathrm{vector} \\ $$$$\mathrm{after}\:\mathrm{3}\:\mathrm{s}\:\mathrm{is} \\ $$

Question Number 16647    Answers: 0   Comments: 4

A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height (d/2). Neglecting subsequent motion and air resistance its velocity V varies with height h above the ground is

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{dropped}\:\mathrm{vertically}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{height}\:{d}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{It}\:\mathrm{hits}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{and}\:\mathrm{bounces}\:\mathrm{up}\:\mathrm{vertically}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{height}\:\frac{{d}}{\mathrm{2}}.\:\mathrm{Neglecting}\:\mathrm{subsequent} \\ $$$$\mathrm{motion}\:\mathrm{and}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{its}\:\mathrm{velocity} \\ $$$${V}\:\mathrm{varies}\:\mathrm{with}\:\mathrm{height}\:{h}\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{is} \\ $$

Question Number 16645    Answers: 1   Comments: 1

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x_1 (t) after time ′t′ and that of the second body by x_2 (t) after the same time interval. Which of the following graphs correctly describes (x_1 − x_2 ) as a function of time ′t′?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:{x}\:=\:\mathrm{0}.\:\mathrm{At}\:{t}\:=\:\mathrm{0},\:\mathrm{it} \\ $$$$\mathrm{starts}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:{x}-\mathrm{direction} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{acceleration}.\:\mathrm{At}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{instant}\:\mathrm{another}\:\mathrm{body}\:\mathrm{passes} \\ $$$$\mathrm{through}\:{x}\:=\:\mathrm{0}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive} \\ $$$${x}-\mathrm{direction}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{speed}.\:\mathrm{The} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{body}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${x}_{\mathrm{1}} \left({t}\right)\:\mathrm{after}\:\mathrm{time}\:'{t}'\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{body}\:\mathrm{by}\:{x}_{\mathrm{2}} \left({t}\right)\:\mathrm{after}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}\:\mathrm{interval}.\:\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{graphs}\:\mathrm{correctly}\:\mathrm{describes}\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \right)\:\mathrm{as} \\ $$$$\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:'{t}'? \\ $$

Question Number 16641    Answers: 0   Comments: 0

Prove that p is a prime number if and only if every equiangular polygon with p sides of rational lengths is regular.

$$\mathrm{Prove}\:\mathrm{that}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\mathrm{every}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with} \\ $$$${p}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{lengths}\:\mathrm{is}\:\mathrm{regular}. \\ $$

Question Number 16638    Answers: 1   Comments: 0

∫_( 0) ^( (π/4)) tan^2 (3x) dx

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\mathrm{tan}^{\mathrm{2}} \left(\mathrm{3x}\right)\:\mathrm{dx} \\ $$

Question Number 16635    Answers: 0   Comments: 0

Solve for x, y, z (√x) + (√y) + (√z) = 4 x^2 + y^2 + z^2 = 40.125 e^(xyz) = 4.771

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y},\:\mathrm{z} \\ $$$$\sqrt{\mathrm{x}}\:+\:\sqrt{\mathrm{y}}\:+\:\sqrt{\mathrm{z}}\:=\:\mathrm{4} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{40}.\mathrm{125} \\ $$$$\mathrm{e}^{\mathrm{xyz}} \:=\:\mathrm{4}.\mathrm{771} \\ $$

Question Number 16634    Answers: 1   Comments: 2

Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as function of time will be

$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{rubber}\:\mathrm{ball}\:\mathrm{freely}\:\mathrm{falling} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:{h}\:=\:\mathrm{4}.\mathrm{9}\:\mathrm{m}\:\mathrm{onto}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{elastic}\:\mathrm{plate}.\:\mathrm{Assume}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{duration}\:\mathrm{of}\:\mathrm{collision}\:\mathrm{is}\:\mathrm{negligible} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{collision}\:\mathrm{with}\:\mathrm{the}\:\mathrm{plate}\:\mathrm{is} \\ $$$$\mathrm{totally}\:\mathrm{elastic}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{and}\:\mathrm{the}\:\mathrm{height}\:\mathrm{as} \\ $$$$\mathrm{function}\:\mathrm{of}\:\mathrm{time}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 16633    Answers: 0   Comments: 1

(√1) . (√2) . (√3). ... (√n) = S_n what is S_n ?

$$\sqrt{\mathrm{1}}\:.\:\sqrt{\mathrm{2}}\:.\:\sqrt{\mathrm{3}}.\:...\:\sqrt{\mathrm{n}}\:\:=\:\mathrm{S}_{\mathrm{n}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\:\mathrm{S}_{\mathrm{n}} \:? \\ $$

Question Number 16632    Answers: 0   Comments: 0

e^(xy) + y^2 = sin(x + y) Express the variable y interms of x only.

$$\mathrm{e}^{\mathrm{xy}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$$$\mathrm{Express}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{y}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{x}\:\mathrm{only}. \\ $$

Question Number 16623    Answers: 1   Comments: 0

A ball is thrown up in a lift with a velocity u relative to the lift. If it returns to the lift in time t, then acceleration of the lift is [Answer: ((2u − gt)/t) upwards]

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{up}\:\mathrm{in}\:\mathrm{a}\:\mathrm{lift}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}.\:\mathrm{If}\:\mathrm{it} \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{in}\:\mathrm{time}\:{t},\:\mathrm{then} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\left[\boldsymbol{\mathrm{Answer}}:\right. \\ $$$$\left.\frac{\mathrm{2}{u}\:−\:{gt}}{{t}}\:\mathrm{upwards}\right] \\ $$

Question Number 16621    Answers: 1   Comments: 1

A body moves in a straight line with a velocity whose square decreases linearly with displacement between two points A and B as shown. Determine the acceleration of the particle. [Answer: (8/3) ms^(−2) ]

$$\mathrm{A}\:\mathrm{body}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:\mathrm{whose}\:\mathrm{square}\:\mathrm{decreases} \\ $$$$\mathrm{linearly}\:\mathrm{with}\:\mathrm{displacement}\:\mathrm{between} \\ $$$$\mathrm{two}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}.\:\left[\boldsymbol{\mathrm{Answer}}:\:\frac{\mathrm{8}}{\mathrm{3}}\:\mathrm{ms}^{−\mathrm{2}} \right] \\ $$

Question Number 16600    Answers: 1   Comments: 8

Solve simultaneously x^2 + y^2 = 61 .............. equation (i) x^3 − y^3 = 91 .............. equation (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{61}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

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