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Question Number 20187    Answers: 0   Comments: 1

t_n =(t_(n−1) /n^2 ), t_1 =3;t_2 ,t_3 ,(n≥2)

$${t}_{{n}} =\frac{{t}_{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} },\:{t}_{\mathrm{1}} =\mathrm{3};{t}_{\mathrm{2}} ,{t}_{\mathrm{3}} ,\left({n}\geqslant\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 20177    Answers: 0   Comments: 0

But ans is (1/(108))

$${But}\:{ans}\:{is}\:\frac{\mathrm{1}}{\mathrm{108}} \\ $$

Question Number 20174    Answers: 2   Comments: 0

(0.1^− )^2 {1−9(0.16^− )^2 }

$$\left(\mathrm{0}.\overset{−} {\mathrm{1}}\right)^{\mathrm{2}} \left\{\mathrm{1}−\mathrm{9}\left(\mathrm{0}.\mathrm{1}\overset{−} {\mathrm{6}}\right)^{\mathrm{2}} \right\} \\ $$

Question Number 20167    Answers: 1   Comments: 0

please solve it integrate with respect to x ∫((5x−2)/(3x^2 +2x+1))

$${please}\:{solve}\:{it} \\ $$$${integrate}\:{with}\:{respect}\:{to}\:{x} \\ $$$$\int\frac{\mathrm{5}{x}−\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$

Question Number 20166    Answers: 1   Comments: 0

∫cosec^2 xdx

$$\int\mathrm{cosec}\:^{\mathrm{2}} {xdx} \\ $$

Question Number 20164    Answers: 1   Comments: 0

∫(e^(tan^(−1) x) /(1+x^2 ))

$$\int\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} {x}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Question Number 20157    Answers: 1   Comments: 1

Question Number 20156    Answers: 1   Comments: 0

Solve: inverse laplace. L^(−1) ((s/(s^(2 ) + 6s + 25)))

$$\mathrm{Solve}:\:\mathrm{inverse}\:\mathrm{laplace}.\:\:\:\:\mathrm{L}^{−\mathrm{1}} \left(\frac{\mathrm{s}}{\mathrm{s}^{\mathrm{2}\:} +\:\mathrm{6s}\:+\:\mathrm{25}}\right) \\ $$

Question Number 20149    Answers: 2   Comments: 1

Question Number 20162    Answers: 1   Comments: 0

Compute the volume of a solid bounded by a surface with equation (x^2 +y^2 +z^2 )^2 =a^3 x .

$${Compute}\:{the}\:{volume}\:{of}\:{a}\:{solid} \\ $$$${bounded}\:{by}\:{a}\:{surface}\:{with}\:{equation} \\ $$$$\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{3}} {x}\:. \\ $$

Question Number 20138    Answers: 1   Comments: 0

lim_(x→0) ((1−cos ax)/(1−cos bx))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{ax}}{\mathrm{1}−\mathrm{cos}\:{bx}} \\ $$$$ \\ $$

Question Number 20118    Answers: 1   Comments: 0

The quadratic equations x^2 − 6x + a = 0 and x^2 − cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then, find the common root.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equations}\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:{a}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} \:−\:{cx}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\mathrm{have}\:\mathrm{one}\:\mathrm{root}\:\mathrm{in} \\ $$$$\mathrm{common}.\:\mathrm{The}\:\mathrm{other}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{4}\::\:\mathrm{3}.\:\mathrm{Then},\:\mathrm{find}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{root}. \\ $$

Question Number 20116    Answers: 1   Comments: 0

If a and b (≠ 0) are the roots of the equation x^2 + ax + b = 0, then find the least value of x^2 + ax + b (x ∈ R).

$$\mathrm{If}\:{a}\:\mathrm{and}\:{b}\:\left(\neq\:\mathrm{0}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\left({x}\:\in\:{R}\right). \\ $$

Question Number 20115    Answers: 1   Comments: 0

The value of a for which the equation (1 − a^2 )x^2 + 2ax − 1 = 0 has roots belonging to (0, 1) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$

Question Number 20110    Answers: 0   Comments: 1

Question Number 20102    Answers: 2   Comments: 0

Solve the equation: (log _(sin x) cos x)^2 =1

$${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{log}\:_{\mathrm{sin}\:{x}} \mathrm{cos}\:{x}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Question Number 20091    Answers: 1   Comments: 0

Prove that Σ_(n=0) ^3 tan^2 (((2n + 1)π)/(16)) = 28.

$$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\mathrm{tan}^{\mathrm{2}} \:\frac{\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi}{\mathrm{16}}\:=\:\mathrm{28}. \\ $$

Question Number 20079    Answers: 0   Comments: 2

Question Number 20068    Answers: 1   Comments: 1

Question Number 20058    Answers: 1   Comments: 0

What is the difference between ∮ and ∫? Where is ∮ used?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{between}\:\oint\:\mathrm{and} \\ $$$$\int?\:\mathrm{Where}\:\mathrm{is}\:\oint\:\mathrm{used}? \\ $$

Question Number 20054    Answers: 1   Comments: 0

If α and β (α < β) are the roots of the equation x^2 + bx + c = 0, where c < 0 < b, then (1) 0 < α < β (2) α < 0 < β < ∣α∣ (3) α < β < 0 (4) α < 0 < ∣α∣ < β

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\left(\alpha\:<\:\beta\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{where} \\ $$$${c}\:<\:\mathrm{0}\:<\:{b},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{0}\:<\:\alpha\:<\:\beta \\ $$$$\left(\mathrm{2}\right)\:\alpha\:<\:\mathrm{0}\:<\:\beta\:<\:\mid\alpha\mid \\ $$$$\left(\mathrm{3}\right)\:\alpha\:<\:\beta\:<\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\alpha\:<\:\mathrm{0}\:<\:\mid\alpha\mid\:<\:\beta \\ $$

Question Number 20053    Answers: 1   Comments: 0

If (4a + c)^2 ≤ 4b^2 then one root of ax^2 + bx + c = 0 lies in (1) (−2, 2) (2) (−1, 1) (3) (−∞, −2) (4) (2, ∞)

$$\mathrm{If}\:\left(\mathrm{4}{a}\:+\:{c}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{4}{b}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{lies}\:\mathrm{in} \\ $$$$\left(\mathrm{1}\right)\:\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\left(−\infty,\:−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2},\:\infty\right) \\ $$

Question Number 20052    Answers: 1   Comments: 0

If the roots α and β of the equation ax^2 + bx + c = 0 are real and of opposite sign then the roots of the equation α(x − β)^2 + β(x − α)^2 is/are (1) Positive (2) Negative (3) Real and opposite sign (4) Imaginary

$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{real}\:\mathrm{and}\:\mathrm{of}\:\mathrm{opposite} \\ $$$$\mathrm{sign}\:\mathrm{then}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\alpha\left({x}\:−\:\beta\right)^{\mathrm{2}} \:+\:\beta\left({x}\:−\:\alpha\right)^{\mathrm{2}} \:\mathrm{is}/\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Positive} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Negative} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Real}\:\mathrm{and}\:\mathrm{opposite}\:\mathrm{sign} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Imaginary} \\ $$

Question Number 20049    Answers: 0   Comments: 0

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Question Number 20047    Answers: 1   Comments: 0

Solve for x: ((√(x + 1))/x) + (√(x/(x + 1))) = ((13)/6)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{x}}\:+\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$

Question Number 20042    Answers: 0   Comments: 3

In the situation given, all surfaces are frictionless, pulley is ideal and string is light, F = ((mg)/2) , find the acceleration of block 2.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{situation}\:\mathrm{given},\:\mathrm{all}\:\mathrm{surfaces}\:\mathrm{are} \\ $$$$\mathrm{frictionless},\:\mathrm{pulley}\:\mathrm{is}\:\mathrm{ideal}\:\mathrm{and}\:\mathrm{string}\:\mathrm{is} \\ $$$$\mathrm{light},\:{F}\:=\:\frac{{mg}}{\mathrm{2}}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of} \\ $$$$\mathrm{block}\:\mathrm{2}. \\ $$

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