Messages posted by akki_0410

i am getting

u = v (sec@ - 1)

what's the ans..??

let v be the vel of mass at highest position around circle with centre Q.

at highest pt, tension t= 0

mg = mv²/(2a/3)

=> v²= 2ag/3

let u be initial required vel.

by energy conservation,

mu²/2 = mv²/2 + mg(5a/3)

u² = 4ag

=> u = 2 rt.(ag)

let, $I=\int_{2}^{3}\frac{dx}{lnx}$

put lnx = t

dx = e^t dt

so,the integral becomes $I=\int_{ln2}^{ln3}\frac{e^t}{t}dt$

now by using C-S inequility in integral form,

$\int_{ln2}^{ln3}\frac{e^t}{t}dt\le \sqrt{\int_{ln2}^{ln3}e^{2t}dt}.\sqrt{\int_{ln2}^{ln3}\frac{1}{t^2}dt}$

solving we get,

$\int_{ln2}^{ln3}\frac{e^t}{t}dt\le \sqrt{\frac{5}{2}(\frac{1}{ln2}-\frac{1}{ln3})}$

$\int_{ln2}^{ln3}\frac{e^t}{t}dt=I \le 1.1538$

so the value of integral is less than 2

$I=\int\sqrt{tanx}dx\\ let\ tanx=t^2\\ I=\int\frac{2t^2}{1+t^4}dt\\ I=\int\frac{2dt}{(t+\frac{1}{t})^2-2}\\ I=\frac{1}{\sqrt{2}}\int(\frac{t}{t^2-\sqrt{2}t+1}-\frac{t}{t^2+\sqrt{2}t+1})dt$

the further integrations is in standard form , hope u can solve it further...

it can be written as

(a+b+c) (1/a+1/b+1/c) - 3

by AM-HM ineq.

(a+b+c) (1/a+1/b+1/c) > 9

=> (a+b+c) (1/a+1/b+1/c) - 3 > 6

hence min value is 6

given,

F(1 ) = 2

F(-2) = -1

let q(x) be quotient when F(x) is divided by x^2+x-2

so, F(x) = (x²+x-2)q(x) + ax+b

-2a+b = -1

a+b = 2

we get, a=b=1

so the remainder will be x+1

@ sujit, u made a calculation mistake in writing I₁

$I=\int\frac{1-x^2}{x(1-2x)}dx=\int\frac{dx}{x(1-2x)}+\int\frac{(1-2x)-1}{2(1-2x)}dx$

$I=\int(\frac{1}{x}+\frac{1}{2}+\frac{3}{2(1-2x)})dx=lnx+\frac{x}{2}-\frac{3ln(1-2x)}{4}+C$

can u write the inequility by equn editor,

it is not clear where it is e^x+1 and where e^{x+1............}!!!!!!!!!!

clearly from figure,

area of tri. ABC = $A=ab(sin\alpha-\frac{sin2\alpha}{2})$

for max area A'=0,we get,

cos@ = -0.5 => @ = 120⁰

so,

$A_{max}=\frac{3\sqrt{3}ab}{4}$

we have,

$y=e^{-x^2}$

by the graph of the equn, we can see that it is symmetric about x=0

so, area of required rectangle will be $A=2xe^{-x^2}$

for max area A^'=0

we get

so max area $A_{max}=\sqrt{\frac{2}{e}}$

let v be the final pd across the combination,

900-400 = 3v+2v

v=100 volt.

so final charge on 3uF cap = 300 uC, &on 2uF=200uC

so, total charge passed = 900-300 = 600uC

let y²=4ax be general parabola

let the point of trisection of double ordinate be (h,k)

& (at²,2at) & (at²,-2at) will be the general points where doubleordinate cuts parabola

we have relation,

at²= h

2at = 3k

eleminating 't'

y²=a^'x will be the req. locus ....

to find,

$I=\int\frac{x^3}{(x+x^3)^\frac{2}{3}}dx$

$I=\int(x+x^3)^\frac{1}{3}dx-\int\frac{x^\frac{1}{3}}{(1+x^2)^\frac{2}{3}}dx$

I = g(x) - I₁

where $I_1=\int\frac{x^\frac{1}{3}}{(1+x^2)^\frac{2}{3}}dx$

put $x=tan\theta$

$I_1=\int\frac{\sqrt[3]{sin\theta}}{cos\theta}d\theta$

put $sin\theta = t^3$

$I_1=\int\frac{3t^3}{1-t^6}dt$

$I_1=\frac{3}{2}\int(\frac{1}{1-t^3}-\frac{1}{1+t^3})dt$

$I_1=\frac{1}{2}\int(\frac{dt}{1-t}-\frac{dt}{1+t})+\frac{1}{4}\int(\frac{2t+1}{t^2+t+1}+\frac{2t-1}{t^2-t+1})dt$ $+\frac{3}{4}\int(\frac{1}{t^2+t+1}-\frac{1}{t^2-t+1})dt$