Question and Answers Forum

AllQuestion and Answers: Page 1830

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

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

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

Question Number 24938 Answers: 0 Comments: 1

3^y =2 2^x =3 4^(xy+1) ×9^(xy) = 24^(xy+1) =

$$\mathrm{3}^{{y}} =\mathrm{2} \\ $$$$\mathrm{2}^{{x}} =\mathrm{3} \\ $$$$\mathrm{4}^{{xy}+\mathrm{1}} ×\mathrm{9}^{{xy}} = \\ $$$$\mathrm{24}^{{xy}+\mathrm{1}} = \\ $$

Question Number 24937 Answers: 0 Comments: 1

Question Number 24933 Answers: 0 Comments: 2

if 6sin^4 θ+3cos^4 θ=2 then find the value of (7cosec^6 θ+8sec^6 θ)^(1/3)

$$\mathrm{if}\:\mathrm{6sin}^{\mathrm{4}} \theta+\mathrm{3cos}^{\mathrm{4}} \theta=\mathrm{2}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\left(\mathrm{7cosec}^{\mathrm{6}} \theta+\mathrm{8sec}^{\mathrm{6}} \theta\right)^{\mathrm{1}/\mathrm{3}} \\ $$

Question Number 24934 Answers: 0 Comments: 3

A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle xyz with its vertices on the perimeter of the disc. One second after applying the forces, the angular speed of the disc in rad/s is :

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{circular}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}.\mathrm{5}\:\mathrm{kg} \\ $$$$\mathrm{and}\:\mathrm{radius}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{is}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{frictionless}\:\mathrm{surface}.\:\mathrm{Three} \\ $$$$\mathrm{forces}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitude}\:{F}\:=\:\mathrm{0}.\mathrm{5}\:\mathrm{N} \\ $$$$\mathrm{are}\:\mathrm{applied}\:\mathrm{simultaneously}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{three}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$${xyz}\:\mathrm{with}\:\mathrm{its}\:\mathrm{vertices}\:\mathrm{on}\:\mathrm{the}\:\mathrm{perimeter} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{disc}.\:\mathrm{One}\:\mathrm{second}\:\mathrm{after}\:\mathrm{applying} \\ $$$$\mathrm{the}\:\mathrm{forces},\:\mathrm{the}\:\mathrm{angular}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{disc} \\ $$$$\mathrm{in}\:\mathrm{rad}/\mathrm{s}\:\mathrm{is}\:: \\ $$

Question Number 24930 Answers: 1 Comments: 0

a+a=

$${a}+{a}= \\ $$

Question Number 24913 Answers: 0 Comments: 2

find the derivative of (sec(√(1+cos x)))

$$\mathrm{find}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of} \\ $$$$\left(\mathrm{sec}\sqrt{\left.\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)}\right. \\ $$

Question Number 24912 Answers: 1 Comments: 0

Question Number 24909 Answers: 0 Comments: 0

Question Number 24908 Answers: 0 Comments: 0

If a, b, c are the sides of a triangle prove the following inequality: (a/(c + a − b)) + (b/(a + b − c)) + (c/(b + c − a)) ≥ 3.

$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{prove} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}: \\ $$$$\frac{{a}}{{c}\:+\:{a}\:−\:{b}}\:+\:\frac{{b}}{{a}\:+\:{b}\:−\:{c}}\:+\:\frac{{c}}{{b}\:+\:{c}\:−\:{a}}\:\geqslant\:\mathrm{3}. \\ $$

Question Number 24893 Answers: 0 Comments: 2

About a collision which of the following are not correct a. Physical touch is a must b. Particles cannot change c. Effect of external force is not considered d. Momentum may or may not change multi−correct question

$$\mathrm{About}\:\mathrm{a}\:\mathrm{collision}\:\mathrm{which}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{are}\:\mathrm{not}\:\mathrm{correct} \\ $$$$\mathrm{a}.\:\mathrm{Physical}\:\mathrm{touch}\:\mathrm{is}\:\mathrm{a}\:\mathrm{must} \\ $$$$\mathrm{b}.\:\mathrm{Particles}\:\mathrm{cannot}\:\mathrm{change} \\ $$$$\mathrm{c}.\:\mathrm{Effect}\:\mathrm{of}\:\mathrm{external}\:\mathrm{force}\:\mathrm{is}\:\mathrm{not}\:\mathrm{considered} \\ $$$$\mathrm{d}.\:\mathrm{Momentum}\:\mathrm{may}\:\mathrm{or}\:\mathrm{may}\:\mathrm{not}\:\mathrm{change} \\ $$$$\mathrm{multi}−\mathrm{correct}\:\mathrm{question} \\ $$

Question Number 24879 Answers: 0 Comments: 4

A particle of mass m is fixed to one end of a light rigid rod of length l and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{is}\:\mathrm{fixed}\:\mathrm{to}\:\mathrm{one}\:\mathrm{end} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{light}\:\mathrm{rigid}\:\mathrm{rod}\:\mathrm{of}\:\mathrm{length}\:{l}\:\mathrm{and}\:\mathrm{rotated} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{about}\:\mathrm{its} \\ $$$$\mathrm{other}\:\mathrm{end}.\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{at}\:\mathrm{its}\:\mathrm{highest}\:\mathrm{point}\:\mathrm{must}\:\mathrm{be} \\ $$

Question Number 24877 Answers: 0 Comments: 7

Question Number 24866 Answers: 2 Comments: 0

3^x =2 2^y =4 (((3^x )^(4y−1) ×(3^(2x) )^(y+1) )/((3^(3y+2) )^x ))=

$$\mathrm{3}^{{x}} =\mathrm{2} \\ $$$$\mathrm{2}^{{y}} =\mathrm{4} \\ $$$$\frac{\left(\mathrm{3}^{{x}} \right)^{\mathrm{4}{y}−\mathrm{1}} ×\left(\mathrm{3}^{\mathrm{2}{x}} \right)^{{y}+\mathrm{1}} }{\left(\mathrm{3}^{\mathrm{3}{y}+\mathrm{2}} \right)^{{x}} }= \\ $$

Question Number 24865 Answers: 1 Comments: 0

((27^(x−1) +81^x )/3^(3x) )=(4/(27))

$$\frac{\mathrm{27}^{{x}−\mathrm{1}} +\mathrm{81}^{{x}} }{\mathrm{3}^{\mathrm{3}{x}} }=\frac{\mathrm{4}}{\mathrm{27}} \\ $$

Question Number 24864 Answers: 1 Comments: 0

Solve the following trigonometric limit: lim_(x → (π/4)) (5tg(x)) =

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trigonometric}\:\mathrm{limit}: \\ $$$$ \\ $$$$\underset{\mathrm{x}\:\rightarrow\:\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\left(\mathrm{5tg}\left(\mathrm{x}\right)\right)\:=\: \\ $$

Question Number 24858 Answers: 2 Comments: 0

If three positive numbers a, b, c are in A.P. and (1/a^2 ), (1/b^2 ), (1/c^2 ) also in A.P., then (1) a = b = c (2) 2b = 3a + c (3) b^2 = ((ac)/8) (4) 2c = 2b + a

$$\mathrm{If}\:\mathrm{three}\:\mathrm{positive}\:\mathrm{numbers}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{in} \\ $$$$\mathrm{A}.\mathrm{P}.\:\mathrm{and}\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} },\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} },\:\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:\mathrm{also}\:\mathrm{in}\:\mathrm{A}.\mathrm{P}.,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{a}\:=\:{b}\:=\:{c} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}{b}\:=\:\mathrm{3}{a}\:+\:{c} \\ $$$$\left(\mathrm{3}\right)\:{b}^{\mathrm{2}} \:=\:\frac{{ac}}{\mathrm{8}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}{c}\:=\:\mathrm{2}{b}\:+\:{a} \\ $$

Question Number 24852 Answers: 0 Comments: 1

please prove that∫_0 ^(π/2) log(sinx)dx=−(π/2)log2 or ∫_0 ^((π )/2) log(cosx)dx=−(π/2)log2

$$\mathrm{please}\:\mathrm{prove}\:\mathrm{that}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sinx}\right)\mathrm{dx}=−\frac{\pi}{\mathrm{2}}\mathrm{log2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi\:\:}{\mathrm{2}}} \mathrm{log}\left(\mathrm{cosx}\right)\mathrm{dx}=−\frac{\pi}{\mathrm{2}}\mathrm{log2} \\ $$

Pg 1825 Pg 1826 Pg 1827 Pg 1828 Pg 1829 Pg 1830 Pg 1831 Pg 1832 Pg 1833 Pg 1834

Contact: info@tinkutara.com