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Question Number 31914 Answers: 0 Comments: 0

Question Number 31899 Answers: 1 Comments: 0

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Question Number 31868 Answers: 0 Comments: 0

Let a>b>1 be positive integers with b odd. Let n be a positive integer as well. If b^n divides a^n −1, prove that a^b > (3^n /n). Solution please. Thanks in advance!!

$${Let}\:{a}>{b}>\mathrm{1}\:{be}\:{positive}\:{integers}\:{with}\:{b}\:{odd}. \\ $$$${Let}\:{n}\:{be}\:{a}\:{positive}\:{integer}\:{as}\:{well}.\:{If}\:\:{b}^{{n}} \:{divides} \\ $$$${a}^{{n}} −\mathrm{1},\:{prove}\:{that}\:{a}^{{b}} \:>\:\frac{\mathrm{3}^{{n}} }{{n}}. \\ $$$${Solution}\:{please}.\:{Thanks}\:{in}\:{advance}!! \\ $$

Question Number 31865 Answers: 1 Comments: 6

Range of function : f(x)= 6^x +3^x +6^(−x) +3^(−x) +2.

$${Range}\:{of}\:{function}\:: \\ $$$${f}\left({x}\right)=\:\mathrm{6}^{{x}} +\mathrm{3}^{{x}} +\mathrm{6}^{−{x}} +\mathrm{3}^{−{x}} +\mathrm{2}. \\ $$

Question Number 31864 Answers: 1 Comments: 0

Let S_n = Σ_(k=1) ^(4n) (−1)^((k(k+1))/2) k^2 . Then S_n can take the value(s) 1) 1056 2) 1088 3) 1120 4) 1332.

$${Let}\:{S}_{{n}} =\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}{n}} {\sum}}\left(−\mathrm{1}\right)^{\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}} {k}^{\mathrm{2}} . \\ $$$${Then}\:{S}_{{n}} \:{can}\:{take}\:{the}\:{value}\left({s}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{1056} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{1088} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{1120} \\ $$$$\left.\mathrm{4}\right)\:\mathrm{1332}. \\ $$

Question Number 31863 Answers: 0 Comments: 9

Question Number 31858 Answers: 0 Comments: 1

∫((sinx)/x)dx

$$\int\frac{{sinx}}{{x}}{dx} \\ $$

Question Number 31846 Answers: 1 Comments: 2

25[(x−2)^2 +(y−3)^2 ]=(3x−4y+7)^2 is the equation of parabola.Find length of latus rectum

$$\mathrm{25}\left[\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} \right]=\left(\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{7}\right)^{\mathrm{2}} \\ $$$${is}\:{the}\:{equation}\:{of}\:{parabola}.{Find} \\ $$$${length}\:{of}\:{latus}\:{rectum} \\ $$

Question Number 31839 Answers: 0 Comments: 1

I = ∫ (√(x + (√(x^2 − 1)))) dx

$${I}\:=\:\int\:\sqrt{{x}\:+\:\sqrt{{x}^{\mathrm{2}} \:−\:\mathrm{1}}}\:{dx} \\ $$

Question Number 31838 Answers: 0 Comments: 0

Given f(x) = (3/(16)) (∫_0 ^1 f(x)dx)x^2 − (9/(10))(∫_0 ^2 f(x)dx)x + 2(∫_0 ^3 f(x)dx) + 4 Solve lim_(t→0) ((2t + (∫_(f(2) + 2) ^(f^(−1) (t)) [f ′(x)]^2 dx))/(1 − cos t cosh 2t cos 3t))

$$\mathrm{Given} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{16}}\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}\left({x}\right){dx}\right){x}^{\mathrm{2}} \:−\:\frac{\mathrm{9}}{\mathrm{10}}\left(\int_{\mathrm{0}} ^{\mathrm{2}} \:{f}\left({x}\right){dx}\right){x}\:+\:\mathrm{2}\left(\int_{\mathrm{0}} ^{\mathrm{3}} \:{f}\left({x}\right){dx}\right)\:+\:\mathrm{4} \\ $$$$\mathrm{Solve} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{t}\:+\:\left(\int_{{f}\left(\mathrm{2}\right)\:+\:\mathrm{2}} ^{{f}^{−\mathrm{1}} \left({t}\right)} \left[{f}\:'\left({x}\right)\right]^{\mathrm{2}} \:{dx}\right)}{\mathrm{1}\:−\:\mathrm{cos}\:{t}\:\mathrm{cosh}\:\mathrm{2}{t}\:\mathrm{cos}\:\mathrm{3}{t}} \\ $$

Question Number 31835 Answers: 0 Comments: 5