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Question Number 18261    Answers: 1   Comments: 0

What is the maximum angle to the horizontal at which a stone can be thrown and always be moving away from the thrower?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{angle}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{at}\:\mathrm{which}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{thrown}\:\mathrm{and}\:\mathrm{always}\:\mathrm{be}\:\mathrm{moving}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{thrower}? \\ $$

Question Number 18258    Answers: 0   Comments: 0

Question Number 18168    Answers: 1   Comments: 0

If the coefficient of the middle term in the expansion of the (1+x)^(2n+2) is p and the coefficients of middle terms in the expansion of (1+x)^(2n+1) are q and r, then

$$\mathrm{If}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{the}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}+\mathrm{2}} \:\mathrm{is}\:{p}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{middle}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{are}\:{q}\:\mathrm{and}\:{r},\:\mathrm{then} \\ $$

Question Number 18153    Answers: 0   Comments: 1

Question Number 18152    Answers: 0   Comments: 0

Question Number 18149    Answers: 1   Comments: 2

Question Number 18146    Answers: 0   Comments: 1

Question Number 18142    Answers: 1   Comments: 1

An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0, the plank starts moving along the x- direction with an acceleration 1.5 ms^(−2) . At the same instant a stone is projected from the origin with a velocity u^→ as shown. A stationary person on the ground observe the stone hitting the object during its downward motion at an angle of 45° with the horizontal. Take g = 10 m/s^2 and consider all motions in the x-y plane. 1. The time after which the stone hits the object is 2. The initial velocity (u^→ ) of the particle is

$$\mathrm{An}\:\mathrm{object}\:{A}\:\mathrm{is}\:\mathrm{kept}\:\mathrm{fixed}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point} \\ $$$${x}\:=\:\mathrm{3}\:\mathrm{m}\:\mathrm{and}\:{y}\:=\:\mathrm{1}.\mathrm{25}\:\mathrm{m}\:\mathrm{on}\:\mathrm{a}\:\mathrm{plank}\:{P} \\ $$$$\mathrm{raised}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{At}\:\mathrm{time}\:{t}\:=\:\mathrm{0}, \\ $$$$\mathrm{the}\:\mathrm{plank}\:\mathrm{starts}\:\mathrm{moving}\:\mathrm{along}\:\mathrm{the}\:{x}- \\ $$$$\mathrm{direction}\:\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}\:\mathrm{1}.\mathrm{5}\:\mathrm{ms}^{−\mathrm{2}} . \\ $$$$\mathrm{At}\:\mathrm{the}\:\mathrm{same}\:\mathrm{instant}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\overset{\rightarrow} {{u}}\:\mathrm{as} \\ $$$$\mathrm{shown}.\:\mathrm{A}\:\mathrm{stationary}\:\mathrm{person}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{observe}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{hitting}\:\mathrm{the} \\ $$$$\mathrm{object}\:\mathrm{during}\:\mathrm{its}\:\mathrm{downward}\:\mathrm{motion}\:\mathrm{at} \\ $$$$\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{45}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}. \\ $$$$\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{consider}\:\mathrm{all} \\ $$$$\mathrm{motions}\:\mathrm{in}\:\mathrm{the}\:{x}-{y}\:\mathrm{plane}. \\ $$$$\mathrm{1}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{after}\:\mathrm{which}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{object}\:\mathrm{is} \\ $$$$\mathrm{2}.\:\mathrm{The}\:\mathrm{initial}\:\mathrm{velocity}\:\left(\overset{\rightarrow} {{u}}\right)\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{is} \\ $$

Question Number 18140    Answers: 1   Comments: 0

From a tower of height H, a particle is thrown vertically upward with speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (1) 2gH = n^2 u^2 (2) gH = (n − 2)^2 u^2 (3) 2gH = nu^2 (n − 2) (4) gH = (n − 2)u^2

$$\mathrm{From}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{of}\:\mathrm{height}\:{H},\:\mathrm{a}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{speed} \\ $$$${u}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{taken}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle},\:\mathrm{to} \\ $$$$\mathrm{hit}\:\mathrm{the}\:\mathrm{ground},\:\mathrm{is}\:{n}\:\mathrm{times}\:\mathrm{that}\:\mathrm{taken}\:\mathrm{by} \\ $$$$\mathrm{it}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{highest}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}. \\ $$$$\mathrm{The}\:\mathrm{relation}\:\mathrm{between}\:{H},\:{u}\:\mathrm{and}\:{n}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}{gH}\:=\:{n}^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right)^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}{gH}\:=\:{nu}^{\mathrm{2}} \left({n}\:−\:\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:{gH}\:=\:\left({n}\:−\:\mathrm{2}\right){u}^{\mathrm{2}} \\ $$

Question Number 18135    Answers: 0   Comments: 6

Let x be the LCM of 3^(2002) − 1 and 3^(2002) + 1. Find the last digit of x.

$$\mathrm{Let}\:{x}\:\mathrm{be}\:\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{3}^{\mathrm{2002}} \:−\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{3}^{\mathrm{2002}} \:+\:\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:{x}. \\ $$

Question Number 18131    Answers: 0   Comments: 3

A stone is projected from a point on the ground in such a direction so as to hit a bird on the top of a telegraph post of height h, and then attain a height 2h above the ground. If, at an instant of projection, the bird were to fly away horizontal with a uniform speed, find the ratio of the horizontal velocities of the bird and the stone, if the stone still hits the bird.

$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{hit}\:\mathrm{a} \\ $$$$\mathrm{bird}\:\mathrm{on}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{telegraph}\:\mathrm{post}\:\mathrm{of} \\ $$$$\mathrm{height}\:{h},\:\mathrm{and}\:\mathrm{then}\:\mathrm{attain}\:\mathrm{a}\:\mathrm{height}\:\mathrm{2}{h} \\ $$$$\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{If},\:\mathrm{at}\:\mathrm{an}\:\mathrm{instant}\:\mathrm{of} \\ $$$$\mathrm{projection},\:\mathrm{the}\:\mathrm{bird}\:\mathrm{were}\:\mathrm{to}\:\mathrm{fly}\:\mathrm{away} \\ $$$$\mathrm{horizontal}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{speed},\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{velocities}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{bird}\:\mathrm{and}\:\mathrm{the}\:\mathrm{stone},\:\mathrm{if}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{still} \\ $$$$\mathrm{hits}\:\mathrm{the}\:\mathrm{bird}. \\ $$

Question Number 18133    Answers: 1   Comments: 0

A 4×4×4 wooden cube is painted so that one pair of opposite faces is blue, one pair green and one pair red. The cube is now sliced into 64 cubes of side 1 unit each. (i) How many of the smaller cubes have no painted face? (ii) How many of the smaller cubes have exactly one painted face? (iii) How many of the smaller cubes have exactly two painted face? (iv) How many of the smaller cubes have exactly three painted face? (v) How many of the smaller cubes have exactly one face painted blue and one face painted green?

$$\mathrm{A}\:\mathrm{4}×\mathrm{4}×\mathrm{4}\:\mathrm{wooden}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{painted}\:\mathrm{so} \\ $$$$\mathrm{that}\:\mathrm{one}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{faces}\:\mathrm{is}\:\mathrm{blue}, \\ $$$$\mathrm{one}\:\mathrm{pair}\:\mathrm{green}\:\mathrm{and}\:\mathrm{one}\:\mathrm{pair}\:\mathrm{red}.\:\mathrm{The} \\ $$$$\mathrm{cube}\:\mathrm{is}\:\mathrm{now}\:\mathrm{sliced}\:\mathrm{into}\:\mathrm{64}\:\mathrm{cubes}\:\mathrm{of}\:\mathrm{side} \\ $$$$\mathrm{1}\:\mathrm{unit}\:\mathrm{each}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{no}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{two}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{iv}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{three}\:\mathrm{painted}\:\mathrm{face}? \\ $$$$\left(\mathrm{v}\right)\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{cubes}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{face}\:\mathrm{painted}\:\mathrm{blue}\:\mathrm{and}\:\mathrm{one} \\ $$$$\mathrm{face}\:\mathrm{painted}\:\mathrm{green}? \\ $$

Question Number 18167    Answers: 0   Comments: 3

Question Number 18123    Answers: 0   Comments: 2

Question Number 18114    Answers: 0   Comments: 0

(√(((1+cosθ)(1+cosθ))/(1−cosθ)(1+cosθ)))) ⇒(√(((1+cosθ)^2 )/(1−cos^2 θ))) recall that:cos^2 θ−1=sin^2 θ ((1+cosθ)/(sin^2 θ)) (1/(sinθ))+((cosθ)/(sinθ)) =cosecθ+cotθ

$$\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}{\left.\mathrm{1}−\mathrm{cos}\theta\right)\left(\mathrm{1}+\mathrm{cos}\theta\right)}} \\ $$$$\Rightarrow\sqrt{\frac{\left(\mathrm{1}+\mathrm{cos}\theta\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \theta}} \\ $$$$\mathrm{recall}\:\mathrm{that}:\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{1}=\mathrm{sin}^{\mathrm{2}} \theta \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\theta}{\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\theta}+\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta} \\ $$$$=\mathrm{cosec}\theta+\mathrm{cot}\theta \\ $$

Question Number 18111    Answers: 1   Comments: 3

Question Number 18109    Answers: 0   Comments: 4

Question Number 18107    Answers: 0   Comments: 0

The first and second ionization potentials of helium atoms are 24.58 eV and 54.4 eV per mole respectively. Calculate the energy in kJ required to produce 1 mole of He^(2+) ions.

$$\mathrm{The}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{ionization} \\ $$$$\mathrm{potentials}\:\mathrm{of}\:\mathrm{helium}\:\mathrm{atoms}\:\mathrm{are}\:\mathrm{24}.\mathrm{58}\:\mathrm{eV} \\ $$$$\mathrm{and}\:\mathrm{54}.\mathrm{4}\:\mathrm{eV}\:\mathrm{per}\:\mathrm{mole}\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{kJ}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{produce}\:\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{He}^{\mathrm{2}+} \:\mathrm{ions}. \\ $$

Question Number 18106    Answers: 0   Comments: 0

The ionization potential of hydrogen is 13.60 eV/mole. Calculate the energy in kJ required to produce 0.1 mole of H^+ ions. Given, 1 eV = 96.49 kJ mol^(−1) )

$$\mathrm{The}\:\mathrm{ionization}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{hydrogen}\:\mathrm{is} \\ $$$$\mathrm{13}.\mathrm{60}\:\mathrm{eV}/\mathrm{mole}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{in} \\ $$$$\mathrm{kJ}\:\mathrm{required}\:\mathrm{to}\:\mathrm{produce}\:\mathrm{0}.\mathrm{1}\:\mathrm{mole}\:\mathrm{of}\:\mathrm{H}^{+} \\ $$$$\left.\mathrm{ions}.\:\mathrm{Given},\:\mathrm{1}\:\mathrm{eV}\:=\:\mathrm{96}.\mathrm{49}\:\mathrm{kJ}\:\mathrm{mol}^{−\mathrm{1}} \right) \\ $$

Question Number 18095    Answers: 1   Comments: 0

A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.

$$\mathrm{A}\:\mathrm{boy}\:\mathrm{travelling}\:\mathrm{in}\:\mathrm{an}\:\mathrm{open}\:\mathrm{car}\:\mathrm{moving} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{levelled}\:\mathrm{road}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed} \\ $$$$\mathrm{tosses}\:\mathrm{a}\:\mathrm{ball}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{in}\:\mathrm{the}\:\mathrm{air}\:\mathrm{and} \\ $$$$\mathrm{catches}\:\mathrm{it}\:\mathrm{back}.\:\mathrm{Sketch}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{ball}\:\mathrm{as}\:\mathrm{observed}\:\mathrm{by}\:\mathrm{a}\:\mathrm{boy}\:\mathrm{standing} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{footpath}.\:\mathrm{Give}\:\mathrm{explanation}\:\mathrm{to} \\ $$$$\mathrm{support}\:\mathrm{your}\:\mathrm{diagram}. \\ $$

Question Number 18094    Answers: 0   Comments: 3

A^→ , B^(→) and C^(→) are three non-collinear, non co-planar vectors. What can you say about direction of A^(→) ×(B^(→) ×C^(→) )?

$$\overset{\rightarrow} {{A}},\:\overset{\rightarrow} {{B}}\:\mathrm{and}\:\overset{\rightarrow} {{C}}\:\mathrm{are}\:\mathrm{three}\:\mathrm{non}-\mathrm{collinear}, \\ $$$$\mathrm{non}\:\mathrm{co}-\mathrm{planar}\:\mathrm{vectors}.\:\mathrm{What}\:\mathrm{can}\:\mathrm{you} \\ $$$$\mathrm{say}\:\mathrm{about}\:\mathrm{direction}\:\mathrm{of}\:\overset{\rightarrow} {{A}}×\left(\overset{\rightarrow} {{B}}×\overset{\rightarrow} {{C}}\right)? \\ $$

Question Number 18093    Answers: 1   Comments: 1

The equation sinx + sin2x + 2sinxsin2x = 2cosx + cos2x is satisfied by values of x for which (1) x = nπ + (−1)^n (π/6) , n ∈ I (2) x = 2nπ + ((2π)/3) , n ∈ I (3) x = 2nπ − ((2π)/3) , n ∈ I (4) x = 2nπ − (π/2) , n ∈ I

$$\mathrm{The}\:\mathrm{equation}\:\mathrm{sin}{x}\:+\:\mathrm{sin2}{x}\:+\:\mathrm{2sin}{x}\mathrm{sin2}{x} \\ $$$$=\:\mathrm{2cos}{x}\:+\:\mathrm{cos2}{x}\:\mathrm{is}\:\mathrm{satisfied}\:\mathrm{by}\:\mathrm{values} \\ $$$$\mathrm{of}\:{x}\:\mathrm{for}\:\mathrm{which} \\ $$$$\left(\mathrm{1}\right)\:{x}\:=\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{\pi}{\mathrm{6}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{2}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:+\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{3}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:−\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:,\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{4}\right)\:{x}\:=\:\mathrm{2}{n}\pi\:−\:\frac{\pi}{\mathrm{2}}\:,\:{n}\:\in\:{I} \\ $$

Question Number 18092    Answers: 1   Comments: 0

A value of θ satisfying 4cos^2 θsinθ − 2sin^2 θ = 3sinθ is (1) ((9π)/(10)) (2) (π/(10)) (3) −((13π)/(10)) (4) −((17π)/(10))

$$\mathrm{A}\:\mathrm{value}\:\mathrm{of}\:\theta\:\mathrm{satisfying} \\ $$$$\mathrm{4cos}^{\mathrm{2}} \theta\mathrm{sin}\theta\:−\:\mathrm{2sin}^{\mathrm{2}} \theta\:=\:\mathrm{3sin}\theta\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{9}\pi}{\mathrm{10}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\pi}{\mathrm{10}} \\ $$$$\left(\mathrm{3}\right)\:−\frac{\mathrm{13}\pi}{\mathrm{10}} \\ $$$$\left(\mathrm{4}\right)\:−\frac{\mathrm{17}\pi}{\mathrm{10}} \\ $$

Question Number 18091    Answers: 1   Comments: 0

Which of the following statement(s) is/are correct? (1) cos(sin1) > sin(cos1) (2) cos(sin1.5) > sin(cos1.5) (3) cos(sin((7π)/(18))) > sin(cos((7π)/(18))) (4) cos(sin((5π)/(18))) > sin(cos((5π)/(18)))

$$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{statement}\left(\mathrm{s}\right) \\ $$$$\mathrm{is}/\mathrm{are}\:\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{cos}\left(\mathrm{sin1}\right)\:>\:\mathrm{sin}\left(\mathrm{cos1}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}\left(\mathrm{sin1}.\mathrm{5}\right)\:>\:\mathrm{sin}\left(\mathrm{cos1}.\mathrm{5}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{cos}\left(\mathrm{sin}\frac{\mathrm{7}\pi}{\mathrm{18}}\right)\:>\:\mathrm{sin}\left(\mathrm{cos}\frac{\mathrm{7}\pi}{\mathrm{18}}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{cos}\left(\mathrm{sin}\frac{\mathrm{5}\pi}{\mathrm{18}}\right)\:>\:\mathrm{sin}\left(\mathrm{cos}\frac{\mathrm{5}\pi}{\mathrm{18}}\right) \\ $$

Question Number 18077    Answers: 1   Comments: 0

A mango in a tree is located (30,40) from the point of projection of stone.Find the minimum speed and the angle of projevtion of the stone so as to hit the mango

$${A}\:{mango}\:{in}\:{a}\:{tree}\:{is}\:{located}\:\left(\mathrm{30},\mathrm{40}\right)\:{from}\:{the} \\ $$$${point}\:{of}\:{projection}\:{of}\:{stone}.{Find}\:{the}\: \\ $$$${minimum}\:{speed}\:{and}\:{the}\:{angle}\:{of}\:{projevtion} \\ $$$${of}\:{the}\:{stone}\:{so}\:{as}\:{to}\:{hit}\:{the}\:{mango} \\ $$

Question Number 18069    Answers: 0   Comments: 1

ai) If θ is the angle in the fourth quadrant satisfying the equation : cot^2 θ = 4 find the value of the function: f(θ) = (1/(√5)) (secθ − cosecθ) aii) Prove that: (√((1 + cosθ)/(1 − cosθ))) = cosecθ + cotθ, if cosθ ≠ 1 (b) Let R be a positive real number and let α satisfy the inequality 0 < α < 360. express the function 2sinθ + cosθ in the form Rsin(θ + α). Hence, find the value of θ between 0 and 360 which satisfy the equation. 3cosθ + 6sinθ = 1

$$\left.\mathrm{ai}\right)\:\:\mathrm{If}\:\theta\:\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{fourth}\:\mathrm{quadrant}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{equation}\::\:\mathrm{cot}^{\mathrm{2}} \theta\:=\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}:\:\:\mathrm{f}\left(\theta\right)\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:\left(\mathrm{sec}\theta\:−\:\mathrm{cosec}\theta\right) \\ $$$$\left.\mathrm{aii}\right)\:\:\mathrm{Prove}\:\mathrm{that}:\:\:\:\sqrt{\frac{\mathrm{1}\:+\:\mathrm{cos}\theta}{\mathrm{1}\:−\:\mathrm{cos}\theta}}\:\:=\:\:\mathrm{cosec}\theta\:+\:\mathrm{cot}\theta,\:\:\:\:\:\:\:\:\mathrm{if}\:\:\mathrm{cos}\theta\:\neq\:\mathrm{1} \\ $$$$\left(\mathrm{b}\right)\:\:\:\mathrm{Let}\:\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}\:\mathrm{and}\:\mathrm{let}\:\alpha\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{inequality}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:<\:\alpha\:<\:\mathrm{360}.\:\mathrm{express}\:\mathrm{the}\:\mathrm{function}\:\:\mathrm{2sin}\theta\:+\:\mathrm{cos}\theta\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\:\mathrm{Rsin}\left(\theta\:+\:\alpha\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Hence},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta\:\mathrm{between}\:\mathrm{0}\:\mathrm{and}\:\mathrm{360}\:\mathrm{which}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3cos}\theta\:+\:\mathrm{6sin}\theta\:=\:\mathrm{1} \\ $$

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