Question and Answers Forum

AllQuestion and Answers: Page 1894

Question Number 13395 Answers: 2 Comments: 0

Find all positive integers n for which n^2 + 96 is a perfect square.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{n}\:\mathrm{for}\:\mathrm{which} \\ $$$${n}^{\mathrm{2}} \:+\:\mathrm{96}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Question Number 13394 Answers: 1 Comments: 0

Show that any circle with centre ((√2), (√3)) cannot pass through more than one lattice point. [Lattice points are points in cartesian plane, whose abscissa and ordinate both are integers.]

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{any}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:\left(\sqrt{\mathrm{2}},\:\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{cannot}\:\mathrm{pass}\:\mathrm{through}\:\mathrm{more}\:\mathrm{than}\:\mathrm{one} \\ $$$$\mathrm{lattice}\:\mathrm{point}.\:\left[\mathrm{Lattice}\:\mathrm{points}\:\mathrm{are}\:\mathrm{points}\right. \\ $$$$\mathrm{in}\:\mathrm{cartesian}\:\mathrm{plane},\:\mathrm{whose}\:\mathrm{abscissa}\:\mathrm{and} \\ $$$$\left.\mathrm{ordinate}\:\mathrm{both}\:\mathrm{are}\:\mathrm{integers}.\right] \\ $$

Question Number 13391 Answers: 1 Comments: 0

A four digit number has the following properties: (a) It is a perfect square (b) The first two digits are equal (c) The last two digits are equal. Find all such numbers.

$$\mathrm{A}\:\mathrm{four}\:\mathrm{digit}\:\mathrm{number}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{first}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{such}\:\mathrm{numbers}. \\ $$

Question Number 13389 Answers: 0 Comments: 5

Prove that [x] + [2x] + [4x] + [8x] + [16x] + [32x] = 12345 has no solution.

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[{x}\right]\:+\:\left[\mathrm{2}{x}\right]\:+\:\left[\mathrm{4}{x}\right]\:+\:\left[\mathrm{8}{x}\right]\:+\:\left[\mathrm{16}{x}\right]\:+\:\left[\mathrm{32}{x}\right]\:=\:\mathrm{12345} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$

Question Number 13388 Answers: 0 Comments: 5

Find the number of all rational numbers (m/n) such that (i) 0 < (m/n) < 1, (ii) m and n are relatively prime and (iii) mn = 25!.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{rational}\:\mathrm{numbers} \\ $$$$\frac{{m}}{{n}}\:\mathrm{such}\:\mathrm{that}\:\left(\mathrm{i}\right)\:\mathrm{0}\:<\:\frac{{m}}{{n}}\:<\:\mathrm{1},\:\left(\mathrm{ii}\right)\:{m}\:\mathrm{and} \\ $$$${n}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime}\:\mathrm{and}\:\left(\mathrm{iii}\right)\:{mn}\:=\:\mathrm{25}!. \\ $$

Question Number 13377 Answers: 1 Comments: 1

Question Number 13373 Answers: 0 Comments: 4

Question Number 13365 Answers: 2 Comments: 0

∫ e^((x + e^x )) dx

$$\int\:\:\mathrm{e}^{\left(\mathrm{x}\:+\:\mathrm{e}^{\mathrm{x}} \right)} \:\:\mathrm{dx} \\ $$

Question Number 13364 Answers: 2 Comments: 1

∫_( 0) ^( ((3(√3))/2)) (x^3 /(√(4x^2 − 9))) dx

$$\int_{\:\:\:\mathrm{0}} ^{\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:\:\frac{\mathrm{x}^{\mathrm{3}} }{\sqrt{\mathrm{4x}^{\mathrm{2}} \:−\:\mathrm{9}}}\:\:\mathrm{dx} \\ $$

Question Number 13362 Answers: 2 Comments: 0

∫ (x^3 /(√(16 − x^2 ))) dx

$$\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\sqrt{\mathrm{16}\:−\:\mathrm{x}^{\mathrm{2}} }}\:\:\mathrm{dx} \\ $$

Question Number 13360 Answers: 1 Comments: 0

∫ (x/(√(3 − 2x − x^2 ))) dx

$$\int\:\:\frac{\mathrm{x}}{\sqrt{\mathrm{3}\:−\:\mathrm{2x}\:−\:\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$

Question Number 13359 Answers: 0 Comments: 4

For two real and distinct solutions to : y=3 y=ax^2 +b As can be seen from graph in comment below, if a>0, b<3 while if a<0, b>3 .

$${For}\:{two}\:{real}\:{and}\:{distinct} \\ $$$${solutions}\:{to}\:: \\ $$$${y}=\mathrm{3} \\ $$$${y}={ax}^{\mathrm{2}} +{b} \\ $$$${As}\:{can}\:{be}\:{seen}\:{from}\:{graph}\:{in} \\ $$$${comment}\:{below}, \\ $$$${if}\:{a}>\mathrm{0},\:{b}<\mathrm{3} \\ $$$${while}\:{if}\:\:{a}<\mathrm{0},\:{b}>\mathrm{3}\:. \\ $$

Question Number 13349 Answers: 1 Comments: 0

Question Number 13346 Answers: 1 Comments: 2

Question Number 13345 Answers: 0 Comments: 0

Question Number 13336 Answers: 1 Comments: 2

A debating team is to be selected from a group of 8 boys and 6 girls. , in how many ways can this be done if the team (a) Must be either all boys or all girls (b) Must consist of two boys and two girls

$$\mathrm{A}\:\mathrm{debating}\:\mathrm{team}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{from}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{8}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{6}\:\mathrm{girls}.\:, \\ $$$$\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{done}\:\mathrm{if}\:\mathrm{the}\:\mathrm{team}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Must}\:\mathrm{be}\:\mathrm{either}\:\mathrm{all}\:\mathrm{boys}\:\mathrm{or}\:\mathrm{all}\:\mathrm{girls} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Must}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{two}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{two}\:\mathrm{girls} \\ $$

Question Number 13333 Answers: 0 Comments: 5

About the solution to question: For a,b,c>0 and abc=1, prove a^(b+c) b^(c+a) c^(a+b) ≤1 Way 1: Let′s say a≤b≤c. We can prove that a≤1: If a>1, we will get b≥a>1, c≥b>1, ⇒abc>1 but abc=1! so a>1 is not true, i.e. a≤1. Similarly we can also prove that c≥1: If c<1, we will get b≤c<1, a≤b<1, ⇒abc<1 but abc=1 so c<1 is not true, i.e. c≥1. We know also if p≤1, then p^x ≤1 for x≥0 if p≥1, then p^x ≥1 for x≥0 S=a^(b+c) b^(c+a) c^(a+b) =a^(b+c) ((1/(ac)))^(c+a) c^(a+b) =(a^(b−a) /c^(c−b) ) since a≤1 and b−a≥0, we have a^(b−a) ≤1 since c≥1 and c−b≥0, we have c^(b−a) ≥1 ⇒S= (a^(b−a) /c^(c−b) )=((≤1)/(≥1))≤1 Way 2: S=a^(b+c) b^(c+a) c^(a+b) =((a^(a+b+c) b^(c+a+b) c^(a+b+c) )/(a^a b^b c^c )) =(((abc)^(a+b+c) )/(a^a b^b c^c ))=(1/(a^a b^b c^c ))=(1/(a^a b^b ((1/(ab)))^(1/(ab)) )) =(((ab)^(1/(ab)) )/(a^a b^b )) let′s look at function F(x,y)=(((xy)^(1/(xy)) )/(x^x y^y )), the graph of F(x,y) see comment. It has a maximum at (1,1) which is F_(max) =1. Hence for x, y>0, 0<F(x,y)≤1 ⇒S=(((ab)^(1/(ab)) )/(a^a b^b ))=F(a,b)=F(b,a)≤1

$${About}\:{the}\:{solution}\:{to}\:{question}: \\ $$$${For}\:{a},{b},{c}>\mathrm{0}\:{and}\:{abc}=\mathrm{1},\:{prove} \\ $$$${a}^{{b}+{c}} {b}^{{c}+{a}} {c}^{{a}+{b}} \leqslant\mathrm{1} \\ $$$$ \\ $$$${Way}\:\mathrm{1}: \\ $$$${Let}'{s}\:{say}\:{a}\leqslant{b}\leqslant{c}. \\ $$$${We}\:{can}\:{prove}\:{that}\:{a}\leqslant\mathrm{1}: \\ $$$${If}\:{a}>\mathrm{1},\:{we}\:{will}\:{get}\:{b}\geqslant{a}>\mathrm{1},\:{c}\geqslant{b}>\mathrm{1}, \\ $$$$\Rightarrow{abc}>\mathrm{1} \\ $$$${but}\:{abc}=\mathrm{1}!\: \\ $$$${so}\:{a}>\mathrm{1}\:{is}\:{not}\:{true},\:{i}.{e}.\:{a}\leqslant\mathrm{1}. \\ $$$${Similarly}\:{we}\:{can}\:{also}\:{prove}\:{that}\:{c}\geqslant\mathrm{1}: \\ $$$${If}\:{c}<\mathrm{1},\:{we}\:{will}\:{get}\:{b}\leqslant{c}<\mathrm{1},\:{a}\leqslant{b}<\mathrm{1}, \\ $$$$\Rightarrow{abc}<\mathrm{1} \\ $$$${but}\:{abc}=\mathrm{1} \\ $$$${so}\:{c}<\mathrm{1}\:{is}\:{not}\:{true},\:{i}.{e}.\:{c}\geqslant\mathrm{1}. \\ $$$$ \\ $$$${We}\:{know}\:{also}\: \\ $$$${if}\:{p}\leqslant\mathrm{1},\:{then}\:{p}^{{x}} \leqslant\mathrm{1}\:{for}\:{x}\geqslant\mathrm{0} \\ $$$${if}\:{p}\geqslant\mathrm{1},\:{then}\:{p}^{{x}} \geqslant\mathrm{1}\:{for}\:{x}\geqslant\mathrm{0} \\ $$$$ \\ $$$${S}={a}^{{b}+{c}} {b}^{{c}+{a}} {c}^{{a}+{b}} ={a}^{{b}+{c}} \left(\frac{\mathrm{1}}{{ac}}\right)^{{c}+{a}} {c}^{{a}+{b}} \\ $$$$=\frac{{a}^{{b}−{a}} }{{c}^{{c}−{b}} } \\ $$$${since}\:{a}\leqslant\mathrm{1}\:{and}\:{b}−{a}\geqslant\mathrm{0},\:{we}\:{have} \\ $$$${a}^{{b}−{a}} \leqslant\mathrm{1} \\ $$$${since}\:{c}\geqslant\mathrm{1}\:{and}\:{c}−{b}\geqslant\mathrm{0},\:{we}\:{have} \\ $$$${c}^{{b}−{a}} \geqslant\mathrm{1} \\ $$$$\Rightarrow{S}=\:\frac{{a}^{{b}−{a}} }{{c}^{{c}−{b}} }=\frac{\leqslant\mathrm{1}}{\geqslant\mathrm{1}}\leqslant\mathrm{1} \\ $$$$ \\ $$$${Way}\:\mathrm{2}: \\ $$$${S}={a}^{{b}+{c}} {b}^{{c}+{a}} {c}^{{a}+{b}} =\frac{{a}^{{a}+{b}+{c}} {b}^{{c}+{a}+{b}} {c}^{{a}+{b}+{c}} }{{a}^{{a}} {b}^{{b}} {c}^{{c}} } \\ $$$$=\frac{\left({abc}\right)^{{a}+{b}+{c}} }{{a}^{{a}} {b}^{{b}} {c}^{{c}} }=\frac{\mathrm{1}}{{a}^{{a}} {b}^{{b}} {c}^{{c}} }=\frac{\mathrm{1}}{{a}^{{a}} {b}^{{b}} \left(\frac{\mathrm{1}}{{ab}}\right)^{\frac{\mathrm{1}}{{ab}}} } \\ $$$$=\frac{\left({ab}\right)^{\frac{\mathrm{1}}{{ab}}} }{{a}^{{a}} {b}^{{b}} } \\ $$$${let}'{s}\:{look}\:{at}\:{function}\:{F}\left({x},{y}\right)=\frac{\left({xy}\right)^{\frac{\mathrm{1}}{{xy}}} }{{x}^{{x}} {y}^{{y}} }, \\ $$$${the}\:{graph}\:{of}\:{F}\left({x},{y}\right)\:{see}\:{comment}. \\ $$$$ \\ $$$${It}\:{has}\:{a}\:{maximum}\:{at}\:\left(\mathrm{1},\mathrm{1}\right)\:{which} \\ $$$${is}\:{F}_{{max}} =\mathrm{1}. \\ $$$${Hence}\:{for}\:{x},\:{y}>\mathrm{0},\:\mathrm{0}<{F}\left({x},{y}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow{S}=\frac{\left({ab}\right)^{\frac{\mathrm{1}}{{ab}}} }{{a}^{{a}} {b}^{{b}} }={F}\left({a},{b}\right)={F}\left({b},{a}\right)\leqslant\mathrm{1} \\ $$

Question Number 13329 Answers: 0 Comments: 1

A surveyor standing at a point X, sites a post Y due east of him and a tower Z of a building on a bearing of 046°.After walking to a point W,a distance of 180m in the south-east direction he observes the bearing of Z and Y to be 337° and 50° respectively. calculate to the nearest meter (i)/XY/ (ii)/ZW/ (iii)if N is on /XY/ such that /XZ/=/ZN/ ,find the bearing of Z from N Please help . Thanks.

$$\mathrm{A}\:\mathrm{surveyor}\:\mathrm{standing}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{X}, \\ $$$$\mathrm{sites}\:\mathrm{a}\:\mathrm{post}\:\mathrm{Y}\:\mathrm{due}\:\mathrm{east}\:\mathrm{of}\:\mathrm{him}\:\mathrm{and}\: \\ $$$$\mathrm{a}\:\mathrm{tower}\:\mathrm{Z}\:\mathrm{of}\:\mathrm{a}\:\mathrm{building}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{bearing}\:\mathrm{of}\:\:\mathrm{046}°.\mathrm{After}\:\mathrm{walking}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{point}\:\mathrm{W},\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{180m}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{south}-\mathrm{east}\:\mathrm{direction}\:\mathrm{he}\:\mathrm{observes}\: \\ $$$$\mathrm{the}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{Z}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{to}\:\mathrm{be}\:\mathrm{337}° \\ $$$$\mathrm{and}\:\mathrm{50}°\:\mathrm{respectively}. \\ $$$$ \\ $$$$\mathrm{calculate}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{meter}\: \\ $$$$\left(\mathrm{i}\right)/\mathrm{XY}/ \\ $$$$\left(\mathrm{ii}\right)/\mathrm{ZW}/ \\ $$$$\left(\mathrm{iii}\right)\mathrm{if}\:\mathrm{N}\:\mathrm{is}\:\mathrm{on}\:/\mathrm{XY}/\:\mathrm{such}\:\mathrm{that} \\ $$$$/\mathrm{XZ}/=/\mathrm{ZN}/\:,\mathrm{find}\:\mathrm{the}\:\mathrm{bearing}\:\mathrm{of} \\ $$$$\mathrm{Z}\:\mathrm{from}\:\mathrm{N} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}\:.\:\mathrm{Thanks}. \\ $$$$ \\ $$

Question Number 13328 Answers: 1 Comments: 0

Can we define factorial for any real number???

$${Can}\:{we}\:{define}\:{factorial}\:{for}\:{any}\: \\ $$$${real}\:{number}??? \\ $$

Question Number 13327 Answers: 1 Comments: 0

how can we expand (1+x)^(1/2) ??

$${how}\:{can}\:{we}\:{expand}\:\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ?? \\ $$

Question Number 13316 Answers: 1 Comments: 0

{ ((x′(t)=4x(t)+5y(t))),((y′(t)=4y(t))) :}

$$\begin{cases}{{x}'\left({t}\right)=\mathrm{4}{x}\left({t}\right)+\mathrm{5}{y}\left({t}\right)}\\{{y}'\left({t}\right)=\mathrm{4}{y}\left({t}\right)}\end{cases} \\ $$

Question Number 13307 Answers: 1 Comments: 0

Question Number 13305 Answers: 1 Comments: 0

Does sin A=sin B imply sin (A/2)=sin(B/2)? For example A=π and B=−π?

$$\mathrm{Does} \\ $$$$\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\mathrm{B} \\ $$$$\mathrm{imply}\:\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{sin}\frac{\mathrm{B}}{\mathrm{2}}? \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{A}=\pi\:\mathrm{and}\:\mathrm{B}=−\pi? \\ $$

Question Number 13302 Answers: 2 Comments: 0

(a) A body of mass m initially at rest at a point O on a smooth horizontal surface. A horizontal force F is applied to the body and caused it to move in a straight line accross the surface. The magnitude of F is given by F = (1/(s + α)), Where S is the distance of the body from O and α is the positive constant. If v is the speed of the body at any moment, Show that S = α (e^((1/2)mv^2 ) − 1). (b) If F = 15s + 4 and m = 1 kg and the body is initially at rest at point O. Determine, (i) v when s = 2m (ii) s when v = 8m/s

Question Number 13888 Answers: 1 Comments: 3

Prove that if A′, B′ and C′ are the midpoints of the sides BC, CA and AB, respectively, then AA′ + BB′ + CC′ < AB + BC + CA

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}, \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$${AA}'\:+\:{BB}'\:+\:{CC}'\:<\:{AB}\:+\:{BC}\:+\:{CA} \\ $$

Question Number 13886 Answers: 1 Comments: 0

Pg 1889 Pg 1890 Pg 1891 Pg 1892 Pg 1893 Pg 1894 Pg 1895 Pg 1896 Pg 1897 Pg 1898

Contact: info@tinkutara.com