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Question Number 54506 Answers: 3 Comments: 0

L′Hopital rule lim_(x→α) ((f(x))/(g(x)))=lim_(x→α) ((f ′(x))/(g′(x)))= ((f ′(α))/(g′(α))) f ′(x)=(d/dx)f(x) , g′(x)=(d/dx)g(x) differential What is it? Proof of the rule.. plz :)

$${L}'{Hopital}\:{rule} \\ $$$$\underset{{x}\rightarrow\alpha} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}=\underset{{x}\rightarrow\alpha} {\mathrm{lim}}\:\frac{{f}\:'\left({x}\right)}{{g}'\left({x}\right)}=\:\frac{{f}\:'\left(\alpha\right)}{{g}'\left(\alpha\right)} \\ $$$${f}\:'\left({x}\right)=\frac{{d}}{{dx}}{f}\left({x}\right)\:,\:{g}'\left({x}\right)=\frac{{d}}{{dx}}{g}\left({x}\right)\:{differential} \\ $$$$\left.{What}\:{is}\:{it}?\:{Proof}\:{of}\:{the}\:{rule}..\:\mathrm{plz}\::\right) \\ $$

Question Number 54504 Answers: 1 Comments: 0

please,Sir. Would u explain to me how ? if f((x/(x − 1))) + 2f(((x − 1)/x)) = (x/(x − 1)) , then f(x) is ....

$$\boldsymbol{\mathrm{please}},\boldsymbol{\mathrm{Sir}}.\:\:\:\boldsymbol{\mathrm{Would}}\:\boldsymbol{\mathrm{u}}\:\boldsymbol{\mathrm{explain}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{how}}\:? \\ $$$$\boldsymbol{\mathrm{if}}\:\:\boldsymbol{{f}}\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}\:−\:\mathrm{1}}\right)\:\:+\:\:\mathrm{2}\boldsymbol{{f}}\left(\frac{\boldsymbol{{x}}\:−\:\mathrm{1}}{\boldsymbol{{x}}}\right)\:\:=\:\:\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}\:−\:\mathrm{1}}\:\:\:, \\ $$$$\boldsymbol{\mathrm{then}}\:\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:\:\boldsymbol{\mathrm{is}}\:.... \\ $$

Question Number 54502 Answers: 0 Comments: 5

In △ABC cos A+cos B+cos C=(3/2) prove that trianle is equilateral

$${In}\:\bigtriangleup{ABC}\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}+\mathrm{cos}\:{C}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${prove}\:{that}\:{trianle}\:{is}\:{equilateral} \\ $$

Question Number 54480 Answers: 2 Comments: 0

show that tan α +tan (α+((2Λ^− )/5)) +tan (α+((4Λ^− )/5)) +tan (α+((6Λ^− )/5)) + tan (α+((8Λ^− )/5)) = 5tan 5α

$$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}\:\alpha\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{4}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{6}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\:\mathrm{tan}\:\left(\alpha+\frac{\mathrm{8}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:=\:\mathrm{5tan}\:\mathrm{5}\alpha \\ $$

Question Number 54481 Answers: 1 Comments: 1

Question Number 54473 Answers: 1 Comments: 0

Solve for x: 2^(2x − 4) = x^2

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\mathrm{2}^{\mathrm{2x}\:−\:\mathrm{4}} \:\:=\:\:\mathrm{x}^{\mathrm{2}} \\ $$

Question Number 54472 Answers: 1 Comments: 0

If A, B, C are angle of a triangle, show that tanA + tanB + tanC = tanA tanB tanC

$$\mathrm{If}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{are}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle},\:\:\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\:\mathrm{tanA}\:+\:\mathrm{tanB}\:+\:\mathrm{tanC}\:\:=\:\:\mathrm{tanA}\:\mathrm{tanB}\:\mathrm{tanC} \\ $$

Question Number 54468 Answers: 0 Comments: 0

First three terms of the sequence given by a_1 =1, a_n =a_(n−1) +2a_(n−2) are in

$$\mathrm{First}\:\mathrm{three}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{given} \\ $$$$\mathrm{by}\:\:{a}_{\mathrm{1}} =\mathrm{1},\:{a}_{{n}} ={a}_{{n}−\mathrm{1}} +\mathrm{2}{a}_{{n}−\mathrm{2}} \:\mathrm{are}\:\mathrm{in} \\ $$

Question Number 54466 Answers: 1 Comments: 1

((6+(√((6)^2 −4(1)(10))))/2)

$$\frac{\mathrm{6}+\sqrt{\left(\mathrm{6}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{10}\right)}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 54462 Answers: 1 Comments: 1

Question Number 54460 Answers: 1 Comments: 0

Solve (dy/dx)+3x=5

$${Solve}\:\frac{{dy}}{{dx}}+\mathrm{3}{x}=\mathrm{5} \\ $$

Question Number 54457 Answers: 0 Comments: 0

Question Number 54447 Answers: 2 Comments: 1

Question Number 54487 Answers: 0 Comments: 0

Question Number 54422 Answers: 1 Comments: 4

The median AD of the triangle ABC is bisected at E, BE meets AC in F, then AF : AC =

$$\mathrm{The}\:\mathrm{median}\:{AD}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:{ABC} \\ $$$$\mathrm{is}\:\mathrm{bisected}\:\mathrm{at}\:{E},\:{BE}\:\mathrm{meets}\:{AC}\:\mathrm{in}\:{F}, \\ $$$$\mathrm{then}\:{AF}\::\:{AC}\:= \\ $$

Question Number 54421 Answers: 2 Comments: 2

If the vector c, a=xi+yj+zk and b=j are such that a, c and b form a right handed system, then c is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{\mathrm{c}},\:\boldsymbol{\mathrm{a}}={x}\boldsymbol{\mathrm{i}}+{y}\boldsymbol{\mathrm{j}}+{z}\boldsymbol{\mathrm{k}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{j}} \\ $$$$\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{c}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}\:\mathrm{form}\:\mathrm{a}\:\mathrm{right} \\ $$$$\mathrm{handed}\:\mathrm{system},\:\mathrm{then}\:\boldsymbol{\mathrm{c}}\:\mathrm{is} \\ $$

Question Number 54420 Answers: 1 Comments: 0

Let a=i+j and b=2i−k, the point of intersection of the lines r×a=b×a and r×b=a×b is

$$\mathrm{Let}\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=\mathrm{2}\boldsymbol{\mathrm{i}}−\boldsymbol{\mathrm{k}},\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:\boldsymbol{\mathrm{r}}×\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}×\boldsymbol{\mathrm{a}} \\ $$$$\mathrm{and}\:\boldsymbol{\mathrm{r}}×\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{a}}×\boldsymbol{\mathrm{b}}\:\mathrm{is} \\ $$

Question Number 54419 Answers: 1 Comments: 0

If a=i+j−k, b=i−j+k and c is a unit vector ⊥ to the vector a and coplanar with a and b, then a unit vector d ⊥ to both a and c is

Question Number 54418 Answers: 1 Comments: 0

The projection of the vector a=4i−3j+2k on the axis making equal acute angles with the coordinate axes is

$$\mathrm{The}\:\mathrm{projection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{\mathrm{a}}=\mathrm{4}\boldsymbol{\mathrm{i}}−\mathrm{3}\boldsymbol{\mathrm{j}}+\mathrm{2}\boldsymbol{\mathrm{k}} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{making}\:\mathrm{equal}\:\mathrm{acute}\:\mathrm{angles} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{axes}\:\mathrm{is} \\ $$

Question Number 54417 Answers: 0 Comments: 0

A force of 39 kg weight is acting at a point P(−4, 2, 5) in the direction 12i−4j−3k. The moment of this force about a line through the origin having the direction of 2i−2j+k is

$$\mathrm{A}\:\mathrm{force}\:\mathrm{of}\:\mathrm{39}\:\mathrm{kg}\:\mathrm{weight}\:\mathrm{is}\:\mathrm{acting}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{point}\:{P}\left(−\mathrm{4},\:\mathrm{2},\:\mathrm{5}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction}\: \\ $$$$\mathrm{12}\boldsymbol{\mathrm{i}}−\mathrm{4}\boldsymbol{\mathrm{j}}−\mathrm{3}\boldsymbol{\mathrm{k}}.\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{this}\:\mathrm{force} \\ $$$$\mathrm{about}\:\mathrm{a}\:\mathrm{line}\:\mathrm{through}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{having} \\ $$$$\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{2}\boldsymbol{\mathrm{i}}−\mathrm{2}\boldsymbol{\mathrm{j}}+\boldsymbol{\mathrm{k}}\:\mathrm{is} \\ $$

Question Number 54416 Answers: 0 Comments: 0

Question Number 54413 Answers: 0 Comments: 8

Question Number 54409 Answers: 2 Comments: 1

lim_(x→∞) (√((x^2 +x+1)))−x=? pls solve this

$${li}\underset{{x}\rightarrow\infty} {{m}}\sqrt{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−{x}=? \\ $$$${pls}\:{solve}\:{this} \\ $$$$ \\ $$

Question Number 54396 Answers: 1 Comments: 1

Question Number 54388 Answers: 0 Comments: 0

Question Number 54378 Answers: 0 Comments: 3

∫_(π/3) ^(3π/2) [ 2 cos x ] dx =

$$\underset{\pi/\mathrm{3}} {\overset{\mathrm{3}\pi/\mathrm{2}} {\int}}\:\:\left[\:\mathrm{2}\:\mathrm{cos}\:{x}\:\right]\:{dx}\:= \\ $$

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