Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1753

Question Number 25053    Answers: 2   Comments: 0

If 2A=3B=4C find the value of A:B:C

$${If}\:\mathrm{2}{A}=\mathrm{3}{B}=\mathrm{4}{C}\:{find}\:{the}\:{value}\:{of}\:{A}:{B}:{C} \\ $$

Question Number 25049    Answers: 1   Comments: 1

If I = Σ_(k=1) ^(98) ∫_k ^(k+1) ((k + 1)/(x(x + 1)))dx, then (1) I > ((49)/(50)) (2) I < ((49)/(50)) (3) I < log_e 99 (4) I > log_e 99

$$\mathrm{If}\:{I}\:=\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{98}} {\sum}}\underset{{k}} {\overset{{k}+\mathrm{1}} {\int}}\frac{{k}\:+\:\mathrm{1}}{{x}\left({x}\:+\:\mathrm{1}\right)}{dx},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{I}\:>\:\frac{\mathrm{49}}{\mathrm{50}} \\ $$$$\left(\mathrm{2}\right)\:{I}\:<\:\frac{\mathrm{49}}{\mathrm{50}} \\ $$$$\left(\mathrm{3}\right)\:{I}\:<\:\mathrm{log}_{{e}} \mathrm{99} \\ $$$$\left(\mathrm{4}\right)\:{I}\:>\:\mathrm{log}_{{e}} \mathrm{99} \\ $$

Question Number 25046    Answers: 1   Comments: 0

Show that (a) N=((10^(143) −1)/9) is composite, and (b) N has two factors each of which is a series of a G.P.

$${Show}\:{that} \\ $$$$\left({a}\right)\:{N}=\frac{\mathrm{10}^{\mathrm{143}} −\mathrm{1}}{\mathrm{9}}\:{is}\:{composite},\:{and} \\ $$$$\left({b}\right)\:{N}\:{has}\:{two}\:{factors}\:{each}\:{of}\:{which}\:{is} \\ $$$${a}\:{series}\:{of}\:{a}\:{G}.{P}. \\ $$

Question Number 25038    Answers: 1   Comments: 0

For a particle of a rotating rigid body, v = rω. So (1) ω ∝ (1/r) (2) ω ∝ v (3) v ∝ r (4) ω is independent of r

$$\mathrm{For}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rotating}\:\mathrm{rigid}\:\mathrm{body}, \\ $$$${v}\:=\:{r}\omega.\:\mathrm{So} \\ $$$$\left(\mathrm{1}\right)\:\omega\:\propto\:\left(\mathrm{1}/{r}\right) \\ $$$$\left(\mathrm{2}\right)\:\omega\:\propto\:{v} \\ $$$$\left(\mathrm{3}\right)\:{v}\:\propto\:{r} \\ $$$$\left(\mathrm{4}\right)\:\omega\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{r} \\ $$

Question Number 25034    Answers: 2   Comments: 0

f(x)=((sin x+sec x)/(1+xtan x)) find f′(x)

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}}{\mathrm{1}+\mathrm{xtan}\:\mathrm{x}} \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{f}'\left(\mathrm{x}\right) \\ $$

Question Number 25026    Answers: 2   Comments: 0

lim_(x→1) (((x^(1/3) −1)/(x^(1/4) −1))) Evaluate this

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{1}/\mathrm{4}} −\mathrm{1}}\right) \\ $$$${Evaluate}\:{this} \\ $$$$ \\ $$

Question Number 25025    Answers: 2   Comments: 0

If a^4 + b^4 + c^4 + d^4 = 16, prove that: a^5 + b^5 + c^5 + d^5 ≤ 32 for a, b, c, d ∈ R

$$\mathrm{If}\:\:\:\:\mathrm{a}^{\mathrm{4}} \:+\:\mathrm{b}^{\mathrm{4}} \:+\:\mathrm{c}^{\mathrm{4}} \:+\:\mathrm{d}^{\mathrm{4}} \:=\:\mathrm{16},\:\:\mathrm{prove}\:\mathrm{that}:\:\:\mathrm{a}^{\mathrm{5}} \:+\:\mathrm{b}^{\mathrm{5}} \:+\:\mathrm{c}^{\mathrm{5}} \:+\:\mathrm{d}^{\mathrm{5}} \:\leqslant\:\mathrm{32} \\ $$$$\mathrm{for}\:\:\mathrm{a},\:\mathrm{b},\:\mathrm{c},\:\mathrm{d}\:\in\:\mathbb{R} \\ $$

Question Number 25023    Answers: 1   Comments: 0

Consider the function f(x) which satisfying the functional equation 2f(x) + f(1 − x) = x^2 + 1, ∀ x ∈ R and g(x) = 3f(x) + 1. The range of φ(x) = g(x) + (1/(g(x) + 1)) is

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{which} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{functional}\:\mathrm{equation} \\ $$$$\mathrm{2}{f}\left({x}\right)\:+\:{f}\left(\mathrm{1}\:−\:{x}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{1},\:\forall\:{x}\:\in\:{R} \\ $$$$\mathrm{and}\:{g}\left({x}\right)\:=\:\mathrm{3}{f}\left({x}\right)\:+\:\mathrm{1}.\:\mathrm{The}\:\mathrm{range}\:\mathrm{of} \\ $$$$\phi\left({x}\right)\:=\:{g}\left({x}\right)\:+\:\frac{\mathrm{1}}{{g}\left({x}\right)\:+\:\mathrm{1}}\:\mathrm{is} \\ $$

Question Number 25013    Answers: 0   Comments: 3

With reference to figure of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube.) (1) The moment of inertia of cube about z-axis is, I_z = I_x + I_y (2) The moment of inertia of cube about z′ is, I_(z′) = I_z + ((ma^2 )/2) (3) The moment of inertia of cube about z′′ is, I_(z′) = I_z + ((ma^2 )/2) (4) I_x = I_y

$$\mathrm{With}\:\mathrm{reference}\:\mathrm{to}\:\mathrm{figure}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{of} \\ $$$$\mathrm{edge}\:{a}\:\mathrm{and}\:\mathrm{mass}\:{m},\:\mathrm{state}\:\mathrm{whether}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{are}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false}.\:\left(\mathrm{O}\:\mathrm{is}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}.\right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}-\mathrm{axis}\:\mathrm{is},\:{I}_{{z}} \:=\:{I}_{{x}} \:+\:{I}_{{y}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}'\:\mathrm{is},\:{I}_{{z}'} \:=\:{I}_{{z}} \:+\:\frac{{ma}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}''\:\mathrm{is},\:{I}_{{z}'} \:=\:{I}_{{z}} \:+\:\frac{{ma}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:{I}_{{x}} \:=\:{I}_{{y}} \\ $$

Question Number 25002    Answers: 1   Comments: 1

Question Number 25001    Answers: 0   Comments: 5

If x, y > 0, then the minimum value of 2x^2 + (2/x) − 2x + 2y^2 + (2/y) − 2y + 2 is equal to

$$\mathrm{If}\:{x},\:{y}\:>\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:+\:\frac{\mathrm{2}}{{x}}\:−\:\mathrm{2}{x}\:+\:\mathrm{2}{y}^{\mathrm{2}} \:+\:\frac{\mathrm{2}}{{y}}\:−\:\mathrm{2}{y}\:+\:\mathrm{2}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 25000    Answers: 1   Comments: 0

find x,y from the equation: (1/2)x−yi+(1/(1+i))=((√(1+ω^8 ))+(√(1+ω^(10) )))^4

$${find}\:{x},{y}\:{from}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}−{yi}+\frac{\mathrm{1}}{\mathrm{1}+{i}}=\left(\sqrt{\mathrm{1}+\omega^{\mathrm{8}} }+\sqrt{\mathrm{1}+\omega^{\mathrm{10}} }\right)^{\mathrm{4}} \\ $$$$ \\ $$

Question Number 24997    Answers: 0   Comments: 0

Question Number 24998    Answers: 0   Comments: 2

A fair coin is tossed 100 times. The probability of getting tails an odd number of times is

$$\mathrm{A}\:\mathrm{fair}\:\mathrm{coin}\:\mathrm{is}\:\mathrm{tossed}\:\mathrm{100}\:\mathrm{times}.\:\mathrm{The} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{tails}\:\mathrm{an}\:\mathrm{odd} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{times}\:\mathrm{is} \\ $$

Question Number 24985    Answers: 0   Comments: 0

Question Number 24981    Answers: 4   Comments: 1

Question Number 24974    Answers: 1   Comments: 0

If (x^(23) /x^m ) = x^(15) and (x^4 )^n =x^(20) , then mn=

$$\mathrm{If}\:\:\frac{{x}^{\mathrm{23}} }{{x}^{{m}} }\:=\:{x}^{\mathrm{15}} \:\mathrm{and}\:\left({x}^{\mathrm{4}} \right)^{{n}} ={x}^{\mathrm{20}} ,\:\mathrm{then}\:{mn}= \\ $$

Question Number 24973    Answers: 1   Comments: 2

If a sin^2 x+b cos^2 x=c, b sin^2 y+a cos^2 y=d and a tan x= b tan y then (a^2 /b^2 ) =? (in terms of a,b,c,d)

$$\mathrm{If}\:{a}\:\mathrm{sin}^{\mathrm{2}} {x}+{b}\:\mathrm{cos}^{\mathrm{2}} {x}={c},\:{b}\:\mathrm{sin}^{\mathrm{2}} {y}+{a}\:\mathrm{cos}^{\mathrm{2}} {y}={d} \\ $$$$\mathrm{and}\:\:{a}\:\mathrm{tan}\:{x}=\:{b}\:\mathrm{tan}\:{y}\:\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=? \\ $$$$\left(\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a},{b},{c},{d}\right) \\ $$

Question Number 24972    Answers: 1   Comments: 0

If α,β and γ are conected by the relation 2tan^2 α tan^2 β tan^2 γ+tan^2 α tan^2 β + tan^2 β tan^2 γ+tan^2 γ tan^2 α=1 then which of these are correct(multi correct) (A)sin^2 α+sin^2 β+ sin^2 γ=1 (B)cos^2 α+cos^2 β+cos^2 γ=2 (C)cos2α+ cos2β+ cos2γ=1 (D)cos(α+β) cos(α−β)= cos^2 γ

$$\mathrm{If}\:\alpha,\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{conected}\:\mathrm{by}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\mathrm{2tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:+ \\ $$$$\:\:\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \gamma\:\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{which}\:\mathrm{of}\:\mathrm{these}\:\mathrm{are}\:\mathrm{correct}\left(\mathrm{multi}\:\mathrm{correct}\right) \\ $$$$\left(\mathrm{A}\right)\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \beta+\:\mathrm{sin}^{\mathrm{2}} \gamma=\mathrm{1}\: \\ $$$$\left(\mathrm{B}\right)\mathrm{cos}^{\mathrm{2}} \alpha+\mathrm{cos}^{\mathrm{2}} \beta+\mathrm{cos}^{\mathrm{2}} \gamma=\mathrm{2} \\ $$$$\left(\mathrm{C}\right)\mathrm{cos2}\alpha+\:\mathrm{cos2}\beta+\:\mathrm{cos2}\gamma=\mathrm{1}\: \\ $$$$\left(\mathrm{D}\right)\mathrm{cos}\left(\alpha+\beta\right)\:\mathrm{cos}\left(\alpha−\beta\right)=\:\mathrm{cos}^{\mathrm{2}} \gamma \\ $$

Question Number 24971    Answers: 1   Comments: 0

Show that ((sin x)/(cos 3x))+((sin 3x)/(cos 9x))+((cos 9x)/(cos 27x))=(1/2)(tan 27x−tan x).

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:\mathrm{3}{x}}+\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{9}{x}}+\frac{\mathrm{cos}\:\mathrm{9}{x}}{\mathrm{cos}\:\mathrm{27}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:\mathrm{27}{x}−\mathrm{tan}\:{x}\right). \\ $$

Question Number 24966    Answers: 0   Comments: 1

find the focus of the hyperbola x^2 −16xy−11y^2 −12x+6y+21=0 ?

$$\mathrm{find}\:\mathrm{the}\:\mathrm{focus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{16}{xy}−\mathrm{11}{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{6}{y}+\mathrm{21}=\mathrm{0}\:? \\ $$

Question Number 24961    Answers: 0   Comments: 0

Question Number 24959    Answers: 2   Comments: 0

In a △ABC, ∠B=π/6, ∠C=π/4 and D divides BC internally in the ratio 1 : 3 then , ((sin ∠BAD)/(sin ∠CAD)) equal to__

$$\mathrm{In}\:\mathrm{a}\:\bigtriangleup\mathrm{ABC},\:\angle\mathrm{B}=\pi/\mathrm{6},\:\angle\mathrm{C}=\pi/\mathrm{4}\:\mathrm{and} \\ $$$$\mathrm{D}\:\mathrm{divides}\:\mathrm{BC}\:\mathrm{internally}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\: \\ $$$$\mathrm{1}\::\:\mathrm{3}\:\mathrm{then}\:,\:\:\frac{\mathrm{sin}\:\angle\mathrm{BAD}}{\mathrm{sin}\:\angle\mathrm{CAD}}\:\:\:\mathrm{equal}\:\mathrm{to\_\_} \\ $$

Question Number 24948    Answers: 1   Comments: 3

Assuming that the moon′s diameter subtends and angle (1/2)° at the eye of an observer, find how far from the eye of a coin of 10 cm diameter must be held so as just to hide moon ?

$$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{the}\:\mathrm{moon}'\mathrm{s}\:\mathrm{diameter}\: \\ $$$$\mathrm{subtends}\:\mathrm{and}\:\mathrm{angle}\:\left(\mathrm{1}/\mathrm{2}\right)°\:\mathrm{at}\:\mathrm{the}\:\mathrm{eye}\: \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{observer},\:\mathrm{find}\:\mathrm{how}\:\mathrm{far}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{eye}\:\mathrm{of}\:\mathrm{a}\:\mathrm{coin}\:\mathrm{of}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{diameter}\:\mathrm{must}\: \\ $$$$\mathrm{be}\:\mathrm{held}\:\mathrm{so}\:\mathrm{as}\:\mathrm{just}\:\mathrm{to}\:\mathrm{hide}\:\mathrm{moon}\:? \\ $$

Question Number 24944    Answers: 0   Comments: 4

Question Number 24945    Answers: 0   Comments: 4

A particle of mass m moving with a speed v hits elastically another stationary particle of mass 2m on a smooth horizontal circular tube of radius r. The time in which the next collision will take place is equal to

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{speed}\:{v}\:\mathrm{hits}\:\mathrm{elastically}\:\mathrm{another} \\ $$$$\mathrm{stationary}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}{m}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{circular}\:\mathrm{tube}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{r}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{in}\:\mathrm{which}\:\mathrm{the}\:\mathrm{next} \\ $$$$\mathrm{collision}\:\mathrm{will}\:\mathrm{take}\:\mathrm{place}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

  Pg 1748      Pg 1749      Pg 1750      Pg 1751      Pg 1752      Pg 1753      Pg 1754      Pg 1755      Pg 1756      Pg 1757   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com