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AllQuestion and Answers: Page 181

Question Number 202497 Answers: 3 Comments: 0

If the difference of the roots of x^2 + 2px + q = 0 is equal to the difference of the roots of x^2 + 2qx + p = 0 [p ≠ q] then show that p + q + 1 = 0.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$

Question Number 202490 Answers: 2 Comments: 2

Question Number 202485 Answers: 0 Comments: 3

Question Number 202477 Answers: 3 Comments: 0

If x = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) then show that b^2 x^2 − 2a^2 x + b^2 = 0.

$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0}. \\ $$

Question Number 202473 Answers: 0 Comments: 0

Find: 1. Σ_(n=1) ^∞ ((2^n ∙ n!)/((2n)!)) 2. Σ_(n=1) ^∞ (((x + 5)^n )/(3^(n+1) ∙ n ∙ ln(n)))

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:\centerdot\:\mathrm{n}!}{\left(\mathrm{2n}\right)!}\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{5}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\centerdot\:\mathrm{n}\:\centerdot\:\mathrm{ln}\left(\mathrm{n}\right)} \\ $$

Question Number 202468 Answers: 1 Comments: 0

Find: 1. Σ_(n=1) ^( ∞) ((16)/(16n^2 − 8n − 3)) = ? 2. Σ_(n=1) ^( ∞) (((−1)^n )/(2n^3 )) = ?

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\mathrm{16}}{\mathrm{16n}^{\mathrm{2}} \:−\:\mathrm{8n}\:−\:\mathrm{3}}\:=\:?\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{2n}^{\mathrm{3}} }\:=\:? \\ $$

Question Number 202466 Answers: 1 Comments: 0

lim_(x→+∞) ((x(√(ln (x^2 +1))))/(1+e^(x−3) ))

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{1}+{e}^{{x}−\mathrm{3}} } \\ $$

Question Number 202464 Answers: 0 Comments: 0

Question Number 202462 Answers: 1 Comments: 0

Question Number 202459 Answers: 3 Comments: 0

If the difference of two roots of x^2 − lx + m = 0 is 1 then prove that l^2 + 4m^2 = (1 + 2m)^2 .

$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:{lx}\:+\:{m}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \:. \\ $$

Question Number 202449 Answers: 0 Comments: 0

Question Number 202448 Answers: 2 Comments: 0

Question Number 202447 Answers: 2 Comments: 0

rationnalise le denominateur de x = (((2)^(1/(3 )) −1)/(1−^3 (√2)+^3 (√4)))

$$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$

Question Number 202436 Answers: 2 Comments: 0

If α, β and γ are the roots of ax^3 + bx + c = 0 then frame an equation whose roots are α^2 , β^2 , γ^2 .

$$\mathrm{If}\:\alpha,\:\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{3}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{frame}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} ,\:\gamma^{\mathrm{2}} \:.\: \\ $$

Question Number 202419 Answers: 0 Comments: 0

Hard integral: Q202393

$$\mathrm{Hard}\:\mathrm{integral}:\:\mathrm{Q202393} \\ $$

Question Number 202418 Answers: 1 Comments: 0

Hard integral ∫∫∫∫∫∫∫∫∫ determinant ((a,b,c),(f,g,h),(j,k,l))dl dk dj dh dg df dc db da=

$$\mathrm{Hard}\:\mathrm{integral} \\ $$$$\int\int\int\int\int\int\int\int\int\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{f}}&{{g}}&{{h}}\\{{j}}&{{k}}&{{l}}\end{vmatrix}{dl}\:{dk}\:{dj}\:{dh}\:{dg}\:{df}\:{dc}\:{db}\:{da}= \\ $$

Question Number 202415 Answers: 2 Comments: 0

The value of ∫g′(x)f′(g(x))dx is...

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\int{g}'\left({x}\right){f}'\left({g}\left({x}\right)\right){dx}\:\mathrm{is}... \\ $$

Question Number 202408 Answers: 0 Comments: 0

Question Number 202406 Answers: 2 Comments: 0

Question Number 202400 Answers: 2 Comments: 1

Solve for a, b and c (1/a) + (1/(b + c)) = (1/2) (1/b) + (1/(c + a)) = (1/3) (1/c) + (1/(a + b)) = (1/4)

$$\mathrm{Solve}\:\mathrm{for}\:{a},\:{b}\:\mathrm{and}\:{c} \\ $$$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}\:+\:{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Question Number 202395 Answers: 0 Comments: 0

Question Number 202393 Answers: 2 Comments: 0

Question Number 202392 Answers: 2 Comments: 11

If the ratio of the roots of ax^2 + bx + b = 0 is p : q then show that (√(p/q)) + (√(q/p)) + (√(b/a)) = 0.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{b}\:=\:\mathrm{0} \\ $$$$\mathrm{is}\:{p}\::\:{q}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\sqrt{\frac{{p}}{{q}}}\:+\:\sqrt{\frac{{q}}{{p}}}\:+\:\sqrt{\frac{{b}}{{a}}}\:=\:\mathrm{0}. \\ $$

Question Number 202390 Answers: 0 Comments: 0

Question Number 202388 Answers: 1 Comments: 0

P rove that: ∫ (dx/(b^4 +2ax^2 +c))=((tan^(−1) ((((√2)(√a)x)/( (√(c+b^4 ))))))/( (√2)(√a)(√(c+b^4 ))))+C if a∙(c+b^4 )>0

$$\:\:\boldsymbol{{P}}\:\boldsymbol{{rove}}\:\boldsymbol{{that}}:\:\:\:\:\int\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{b}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{c}}}=\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}\right)}{\:\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}+\boldsymbol{{C}} \\ $$$$\boldsymbol{{if}}\:\:\boldsymbol{{a}}\centerdot\left(\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} \right)>\mathrm{0} \\ $$$$ \\ $$

Question Number 202383 Answers: 2 Comments: 0

Show that ((a(√b) − b(√a))/(a(√b) + b(√a))) = (1/(a − b))(a + b − 2(√(ab))).

$$\mathrm{Show}\:\mathrm{that}\:\frac{{a}\sqrt{{b}}\:−\:{b}\sqrt{{a}}}{{a}\sqrt{{b}}\:+\:{b}\sqrt{{a}}}\:=\:\frac{\mathrm{1}}{{a}\:−\:{b}}\left({a}\:+\:{b}\:−\:\mathrm{2}\sqrt{{ab}}\right). \\ $$

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