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Test 1. Solve equation (k^2 −1)x^2 +(k−1)x+(k+1)=0 k∈R (30) 2. Prove ((sin(x))/(1+cos(x)))=((1−cos(x))/(sin(x))) (35) 3.P(x)=−2x^3 −2x^2 −x+2409 Find P(−11) (35) Evaluate other answers and give marks I want to see how math teachers evaluate in other countries Sorry foy my english

$${Test} \\ $$$$\mathrm{1}.\:{Solve}\:{equation} \\ $$$$\left({k}^{\mathrm{2}} −\mathrm{1}\right){x}^{\mathrm{2}} +\left({k}−\mathrm{1}\right){x}+\left({k}+\mathrm{1}\right)=\mathrm{0}\:\:{k}\in\mathbb{R} \\ $$$$\left(\mathrm{30}\right) \\ $$$$\mathrm{2}.\:{Prove} \\ $$$$\frac{{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}=\frac{\mathrm{1}−{cos}\left({x}\right)}{{sin}\left({x}\right)} \\ $$$$\left(\mathrm{35}\right) \\ $$$$\mathrm{3}.{P}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2409} \\ $$$${Find}\:{P}\left(−\mathrm{11}\right) \\ $$$$\left(\mathrm{35}\right) \\ $$$$ \\ $$$${Evaluate}\:{other}\:{answers}\:{and}\:{give}\:{marks} \\ $$$${I}\:{want}\:{to}\:{see}\:{how}\:{math}\:{teachers}\:{evaluate}\:{in}\:{other}\:{countries} \\ $$$${Sorry}\:{foy}\:{my}\:{english} \\ $$

Question Number 13438 Answers: 1 Comments: 0

7x^5 −4x^4 +9x^3 +12x^2 +5x−9=0 How many roots of this equation are Negative?

$$\mathrm{7}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{9}=\mathrm{0} \\ $$$${How}\:{many}\:{roots}\:{of}\:{this}\:{equation} \\ $$$${are}\:{Negative}? \\ $$$$ \\ $$

Question Number 13412 Answers: 1 Comments: 0

e^(−kN) − kN − 1 = 0 Find the value of N

$$\mathrm{e}^{−\mathrm{kN}} \:−\:\mathrm{kN}\:\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{N} \\ $$

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 13349 Answers: 1 Comments: 0

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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 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 13294 Answers: 0 Comments: 3

If a, b and c are the sides of a triangle and a + b + c = 2, then prove that a^2 + b^2 + c^2 + 2abc < 2

$$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{and}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:\mathrm{2}{abc}\:<\:\mathrm{2} \\ $$

Question Number 13295 Answers: 1 Comments: 3

For n ∈ N, n > 1, show that (1/n) + (1/(n + 1)) + (1/(n + 2)) + ... + (1/n^2 ) > 1

$$\mathrm{For}\:{n}\:\in\:\mathbb{N},\:{n}\:>\:\mathrm{1},\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{n}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{2}}\:+\:...\:+\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:>\:\mathrm{1} \\ $$

Question Number 17713 Answers: 0 Comments: 2

Evaluate: (−(√3))^((−(√2)))

$$\mathrm{Evaluate}:\:\:\:\:\left(−\sqrt{\mathrm{3}}\right)^{\left(−\sqrt{\mathrm{2}}\right)} \\ $$

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