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Question Number 156188 Answers: 0 Comments: 2

Given a rational function f(x) = ((ax^2 +bx+c)/(x+q)) has minimum point at (−2,9) and maximum point at (2,1) . Find value of a, b, c, and q .

$${Given}\:\:{a}\:\:{rational}\:\:{function} \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:=\:\:\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{x}+{q}} \\ $$$${has}\:\:{minimum}\:\:{point}\:\:{at}\:\left(−\mathrm{2},\mathrm{9}\right)\:\:{and}\:\:{maximum}\:\:{point}\:\:{at}\:\:\left(\mathrm{2},\mathrm{1}\right)\:. \\ $$$${Find}\:\:{value}\:\:{of}\:\:{a},\:{b},\:{c},\:\:{and}\:\:{q}\:. \\ $$

Question Number 156186 Answers: 1 Comments: 2

Question Number 156184 Answers: 2 Comments: 1

Question Number 155810 Answers: 1 Comments: 0

Verify the identity in Excercise below 1). cos θsec θ=1 2). (1+cos β)(1−cos β)=sin^2 β 3). cos^2 x(sec^2 x−1)=sin^2 x 4). ((sin t)/(cosec t))+((cos t)/(sec t))=1 5). ((cosec^2 θ)/(1+tan^2 θ)) = cot^2 θ

$$\mathrm{Verify}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{in}\:\mathrm{Excercise}\:\mathrm{below} \\ $$$$\left.\mathrm{1}\right).\:\mathrm{cos}\:\theta\mathrm{sec}\:\theta=\mathrm{1} \\ $$$$\left.\mathrm{2}\right).\:\left(\mathrm{1}+\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\beta\right)=\mathrm{sin}\:^{\mathrm{2}} \beta \\ $$$$\left.\mathrm{3}\right).\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)=\mathrm{sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\left.\mathrm{4}\right).\:\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{cosec}\:\mathrm{t}}+\frac{\mathrm{cos}\:\mathrm{t}}{\mathrm{sec}\:\mathrm{t}}=\mathrm{1} \\ $$$$\left.\mathrm{5}\right).\:\frac{\mathrm{cosec}\:^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}\:=\:\mathrm{cot}\:^{\mathrm{2}} \theta \\ $$

Question Number 155809 Answers: 1 Comments: 0

Given that f○g = (x^2 /(2x^2 − x + 4)) and g(x) = (x/(x − 2)), find f(x) ?

$$\:\:\mathrm{Given}\:\mathrm{that}\:{f}\circ{g}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{4}}\:\:\mathrm{and} \\ $$$$\:\:{g}\left({x}\right)\:=\:\frac{{x}}{{x}\:−\:\mathrm{2}},\:\mathrm{find}\:{f}\left({x}\right)\:? \\ $$$$ \\ $$

Question Number 155806 Answers: 1 Comments: 0

Question Number 155803 Answers: 2 Comments: 0

proof that Σ_(n=1) ^(10) n×n! = 11!−1

$$\mathrm{proof}\:\mathrm{that}\: \\ $$$$\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\mathrm{n}×\mathrm{n}!\:=\:\mathrm{11}!−\mathrm{1} \\ $$

Question Number 155802 Answers: 1 Comments: 0

find the value of 2×1!+4×2!+6×3!+...+200×100!

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{2}×\mathrm{1}!+\mathrm{4}×\mathrm{2}!+\mathrm{6}×\mathrm{3}!+...+\mathrm{200}×\mathrm{100}! \\ $$

Question Number 155801 Answers: 3 Comments: 0

{ ((m^2 =n+2)),((n^2 =m+2)) :} ⇒m≠n 4mn−m^3 −n^3 =?

$$\:\begin{cases}{\mathrm{m}^{\mathrm{2}} =\mathrm{n}+\mathrm{2}}\\{\mathrm{n}^{\mathrm{2}} =\mathrm{m}+\mathrm{2}}\end{cases}\:\Rightarrow\mathrm{m}\neq\mathrm{n} \\ $$$$\:\mathrm{4mn}−\mathrm{m}^{\mathrm{3}} −\mathrm{n}^{\mathrm{3}} =?\: \\ $$

Question Number 155797 Answers: 0 Comments: 6

a,b,c,d,e (kids) are in ascending order of heights. If they are to stand in a circle in a way so that the sum of ∣difference in heights of adjacent pairs∣ is a minimum, find this minimum.

$${a},{b},{c},{d},{e}\:\:\left({kids}\right)\:{are}\:{in}\:{ascending} \\ $$$${order}\:{of}\:{heights}.\:{If}\:{they}\:{are} \\ $$$${to}\:{stand}\:{in}\:{a}\:{circle}\:{in}\:{a}\:{way}\:{so} \\ $$$${that}\:{the}\:{sum}\:{of}\:\mid{difference}\:{in} \\ $$$${heights}\:{of}\:{adjacent}\:{pairs}\mid \\ $$$$\:{is}\:{a}\:{minimum},\:{find}\:{this} \\ $$$$\:{minimum}. \\ $$

Question Number 155775 Answers: 0 Comments: 0