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Question Number 21097    Answers: 1   Comments: 0

Suppose in the plane 10 pairwise nonparallel lines intersect one another. What is the maximum possible number of polygons (with finite areas) that can be formed?

$$\mathrm{Suppose}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{10}\:\mathrm{pairwise} \\ $$$$\mathrm{nonparallel}\:\mathrm{lines}\:\mathrm{intersect}\:\mathrm{one}\:\mathrm{another}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{possible}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{polygons}\:\left(\mathrm{with}\:\mathrm{finite}\:\mathrm{areas}\right)\:\mathrm{that}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{formed}? \\ $$

Question Number 21071    Answers: 2   Comments: 0

The values of ′k′ for which the equation ∣x∣^2 (∣x∣^2 − 2k + 1) = 1 − k^2 , has repeated roots, when k belongs to (1) {1, −1} (2) {0, 1} (3) {0, −1} (4) {2, 3}

$$\mathrm{The}\:\mathrm{values}\:\mathrm{of}\:'{k}'\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mid{x}\mid^{\mathrm{2}} \left(\mid{x}\mid^{\mathrm{2}} \:−\:\mathrm{2}{k}\:+\:\mathrm{1}\right)\:=\:\mathrm{1}\:−\:{k}^{\mathrm{2}} ,\:\mathrm{has} \\ $$$$\mathrm{repeated}\:\mathrm{roots},\:\mathrm{when}\:{k}\:\mathrm{belongs}\:\mathrm{to} \\ $$$$\left(\mathrm{1}\right)\:\left\{\mathrm{1},\:−\mathrm{1}\right\} \\ $$$$\left(\mathrm{2}\right)\:\left\{\mathrm{0},\:\mathrm{1}\right\} \\ $$$$\left(\mathrm{3}\right)\:\left\{\mathrm{0},\:−\mathrm{1}\right\} \\ $$$$\left(\mathrm{4}\right)\:\left\{\mathrm{2},\:\mathrm{3}\right\} \\ $$

Question Number 21070    Answers: 1   Comments: 2

Let us consider an equation f(x) = x^3 − 3x + k = 0. Then the values of k for which the equation has 1. Exactly one root which is positive, then k belongs to 2. Exactly one root which is negative, then k belongs to 3. One negative and two positive root if k belongs to

$$\mathrm{Let}\:\mathrm{us}\:\mathrm{consider}\:\mathrm{an}\:\mathrm{equation}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \\ $$$$−\:\mathrm{3}{x}\:+\:{k}\:=\:\mathrm{0}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{k}\:\mathrm{for} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has} \\ $$$$\mathrm{1}.\:\mathrm{Exactly}\:\mathrm{one}\:\mathrm{root}\:\mathrm{which}\:\mathrm{is}\:\mathrm{positive}, \\ $$$$\mathrm{then}\:{k}\:\mathrm{belongs}\:\mathrm{to} \\ $$$$\mathrm{2}.\:\mathrm{Exactly}\:\mathrm{one}\:\mathrm{root}\:\mathrm{which}\:\mathrm{is}\:\mathrm{negative}, \\ $$$$\mathrm{then}\:{k}\:\mathrm{belongs}\:\mathrm{to} \\ $$$$\mathrm{3}.\:\mathrm{One}\:\mathrm{negative}\:\mathrm{and}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{root} \\ $$$$\mathrm{if}\:{k}\:\mathrm{belongs}\:\mathrm{to} \\ $$

Question Number 21067    Answers: 2   Comments: 0

∀n∈N, prove 9∣[n^3 +(n+1)^3 +(n+2)^3 ]

$$\forall{n}\in\mathbb{N},\:{prove}\:\mathrm{9}\mid\left[{n}^{\mathrm{3}} +\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\left({n}+\mathrm{2}\right)^{\mathrm{3}} \right] \\ $$

Question Number 21076    Answers: 1   Comments: 0

integrate with respect to x ∫x^(sinx)

$${integrate}\:{with}\:{respect}\:{to}\:{x} \\ $$$$\int{x}^{{sinx}} \\ $$

Question Number 21060    Answers: 0   Comments: 8

write sin 1° in surd form please show workings.

$$\mathrm{write}\:\mathrm{sin}\:\mathrm{1}°\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{workings}. \\ $$

Question Number 21140    Answers: 1   Comments: 1

if tan β=((2sin αsin γ)/(sin (α+γ))) so proof cot γ+cot α=2cot β

$${if}\:\mathrm{tan}\:\beta=\frac{\mathrm{2sin}\:\alpha\mathrm{sin}\:\gamma}{\mathrm{sin}\:\left(\alpha+\gamma\right)} \\ $$$${so}\:{proof}\:\mathrm{cot}\:\gamma+\mathrm{cot}\:\alpha=\mathrm{2cot}\:\beta \\ $$

Question Number 21053    Answers: 0   Comments: 0

Question Number 21050    Answers: 1   Comments: 0

The most general solution of the equation sinx + cosx = min_(a∈R) {1, a^2 − 4a + 6} is

$$\mathrm{The}\:\mathrm{most}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{sin}{x}\:+\:\mathrm{cos}{x}\:=\:\underset{{a}\in{R}} {\mathrm{min}}\left\{\mathrm{1},\:{a}^{\mathrm{2}} \:−\:\mathrm{4}{a}\:+\:\mathrm{6}\right\} \\ $$$$\mathrm{is} \\ $$

Question Number 21048    Answers: 0   Comments: 0

If the equation 2cos2x − (a + 7)cosx + 3a − 13 = 0 possesses atleast one real solution, then the maximum integral value of ′a′ can be

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2cos2}{x}\:−\:\left({a}\:+\:\mathrm{7}\right)\mathrm{cos}{x}\:+ \\ $$$$\mathrm{3}{a}\:−\:\mathrm{13}\:=\:\mathrm{0}\:\mathrm{possesses}\:\mathrm{atleast}\:\mathrm{one}\:\mathrm{real} \\ $$$$\mathrm{solution},\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{integral} \\ $$$$\mathrm{value}\:\mathrm{of}\:'{a}'\:\mathrm{can}\:\mathrm{be} \\ $$

Question Number 21038    Answers: 1   Comments: 1

Question Number 21031    Answers: 0   Comments: 0

if :∀ε>0, ∀(a,b)∈R^2 ,a<b+ε prove: a≤b

$${if}\::\forall\epsilon>\mathrm{0},\:\forall\left({a},{b}\right)\in\mathbb{R}^{\mathrm{2}} ,{a}<{b}+\epsilon \\ $$$${prove}:\:{a}\leqslant{b} \\ $$

Question Number 21029    Answers: 0   Comments: 0

Question Number 21021    Answers: 1   Comments: 2

Question Number 21015    Answers: 0   Comments: 0

Question Number 21013    Answers: 1   Comments: 1

Question Number 21012    Answers: 1   Comments: 3

A spring with one end attached to a mass and the other to a rigid support is stretched and released. (a) Magnitude of acceleration, when just released is maximum. (b) Magnitude of acceleration, when at equilibrium position, is maximum. (c) Speed is maximum when mass is at equilibrium position. (d) Magnitude of displacement is always maximum whenever speed is minimum.

$$\mathrm{A}\:\mathrm{spring}\:\mathrm{with}\:\mathrm{one}\:\mathrm{end}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{mass}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{to}\:\mathrm{a}\:\mathrm{rigid}\:\mathrm{support}\:\mathrm{is} \\ $$$$\mathrm{stretched}\:\mathrm{and}\:\mathrm{released}. \\ $$$$\left({a}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{just}\:\mathrm{released}\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({b}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{acceleration},\:\mathrm{when} \\ $$$$\mathrm{at}\:\mathrm{equilibrium}\:\mathrm{position},\:\mathrm{is}\:\mathrm{maximum}. \\ $$$$\left({c}\right)\:\mathrm{Speed}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{equilibrium}\:\mathrm{position}. \\ $$$$\left({d}\right)\:\mathrm{Magnitude}\:\mathrm{of}\:\mathrm{displacement}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{maximum}\:\mathrm{whenever}\:\mathrm{speed}\:\mathrm{is} \\ $$$$\mathrm{minimum}. \\ $$

Question Number 21009    Answers: 1   Comments: 0

Question Number 21006    Answers: 0   Comments: 0

Let z_1 and z_2 be two distinct complex numbers and let z = (1 − t)z_1 + tz_2 for some real number t with 0 < t < 1. If arg(w) denotes the principal argument of a non-zero complex number w, then (1) ∣z − z_1 ∣ + ∣z − z_2 ∣ = ∣z_1 − z_2 ∣ (2) Arg (z − z_1 ) = Arg (z − z_2 ) (3) determinant (((z − z_1 ),(z^ − z_1 ^ )),((z_2 − z_1 ),(z_2 ^ − z_1 ^ ))) = 0 (4) Arg (z − z_1 ) = Arg (z_2 − z_1 )

$$\mathrm{Let}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{two}\:\mathrm{distinct}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{and}\:\mathrm{let}\:{z}\:=\:\left(\mathrm{1}\:−\:{t}\right){z}_{\mathrm{1}} \:+\:{tz}_{\mathrm{2}} \:\mathrm{for} \\ $$$$\mathrm{some}\:\mathrm{real}\:\mathrm{number}\:{t}\:\mathrm{with}\:\mathrm{0}\:<\:{t}\:<\:\mathrm{1}.\:\mathrm{If} \\ $$$$\mathrm{arg}\left({w}\right)\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{argument} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{complex}\:\mathrm{number}\:{w},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mid{z}\:−\:{z}_{\mathrm{1}} \mid\:+\:\mid{z}\:−\:{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Arg}\:\left({z}\:−\:{z}_{\mathrm{1}} \right)\:=\:\mathrm{Arg}\:\left({z}\:−\:{z}_{\mathrm{2}} \right) \\ $$$$\left(\mathrm{3}\right)\:\begin{vmatrix}{{z}\:−\:{z}_{\mathrm{1}} }&{\bar {{z}}\:−\:\bar {{z}}_{\mathrm{1}} }\\{{z}_{\mathrm{2}} \:−\:{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{2}} \:−\:\bar {{z}}_{\mathrm{1}} }\end{vmatrix}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Arg}\:\left({z}\:−\:{z}_{\mathrm{1}} \right)\:=\:\mathrm{Arg}\:\left({z}_{\mathrm{2}} \:−\:{z}_{\mathrm{1}} \right) \\ $$

Question Number 21005    Answers: 0   Comments: 0

If z_1 = a + ib and z_2 = c + id are complex numbers such that ∣z_1 ∣ = ∣z_2 ∣ = 1 and Re(z_1 z_2 ^ ) = 0, then the pair of complex numbers ω_1 = a + ic and ω_2 = b + id satisfy (1) ∣ω_1 ∣ = 1 (2) ∣ω_2 ∣ = 1 (3) Re(ω_1 ω_2 ^ ) = 0 (4) ∣ω_1 ∣ = 2∣ω_2 ∣

$$\mathrm{If}\:{z}_{\mathrm{1}} \:=\:{a}\:+\:{ib}\:\mathrm{and}\:{z}_{\mathrm{2}} \:=\:{c}\:+\:{id}\:\mathrm{are}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\omega_{\mathrm{1}} \:=\:{a}\:+\:{ic}\:\mathrm{and}\:\omega_{\mathrm{2}} \:=\:{b}\:+\:{id} \\ $$$$\mathrm{satisfy} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega_{\mathrm{1}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mid\omega_{\mathrm{2}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Re}\left(\omega_{\mathrm{1}} \bar {\omega}_{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mid\omega_{\mathrm{1}} \mid\:=\:\mathrm{2}\mid\omega_{\mathrm{2}} \mid \\ $$

Question Number 21001    Answers: 0   Comments: 0

determinant (((a 1 1)),((1 b 1)),((1 1 c)))>0 then showthat abc>−8−99

$$\begin{vmatrix}{{a}\:\mathrm{1}\:\mathrm{1}}\\{\mathrm{1}\:{b}\:\mathrm{1}}\\{\mathrm{1}\:\mathrm{1}\:{c}}\end{vmatrix}>\mathrm{0}\:{then}\:{showthat}\:{abc}>−\mathrm{8}−\mathrm{99} \\ $$

Question Number 20989    Answers: 1   Comments: 1

In the figure shown below, the block of mass 2 kg is at rest. If the spring constant of both the springs A and B is 100 N/m and spring B is cut at t = 0, then magnitude of acceleration of block immediately is

$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown}\:\mathrm{below},\:\mathrm{the}\:\mathrm{block}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{constant} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{the}\:\mathrm{springs}\:{A}\:\mathrm{and}\:{B}\:\mathrm{is}\:\mathrm{100}\:\mathrm{N}/\mathrm{m} \\ $$$$\mathrm{and}\:\mathrm{spring}\:{B}\:\mathrm{is}\:\mathrm{cut}\:\mathrm{at}\:{t}\:=\:\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{magnitude}\:\mathrm{of}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{block} \\ $$$$\mathrm{immediately}\:\mathrm{is} \\ $$

Question Number 20988    Answers: 0   Comments: 1

Question Number 20991    Answers: 0   Comments: 1

x^3 −12x

$${x}^{\mathrm{3}} −\mathrm{12}{x} \\ $$

Question Number 20992    Answers: 3   Comments: 1

Question Number 20986    Answers: 1   Comments: 1

A 50 kg log rest on the smooth horizontal surface. A motor deliver a towing force T as shown below. The momentum of the particle at t = 5 s is

$$\mathrm{A}\:\mathrm{50}\:\mathrm{kg}\:\mathrm{log}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{the}\:\mathrm{smooth}\:\mathrm{horizontal} \\ $$$$\mathrm{surface}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{deliver}\:\mathrm{a}\:\mathrm{towing}\:\mathrm{force} \\ $$$${T}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{below}.\:\mathrm{The}\:\mathrm{momentum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{at}\:{t}\:=\:\mathrm{5}\:\mathrm{s}\:\mathrm{is} \\ $$

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