Messages posted by Pankti Kansara

fringe width = La/d

so no. of fringes per metre = d/aL

the first step....

log _x a = 1 / log _a x

log _ax a = 1 / log _aax

= 1 / ( log ax / log a)

= 1 / [ (log a + log x) / log a ]

= 1 / [ 1 + (log x / log a) ]

= 1 / ( 1 + log _a x )

similary log _a^2/x a = 1 / ( 2 + log _a x )

( log x * log 3 * log 3 ) / ( log 3 * log x/3 * log x/81 ) = 0

for the numerator to be 0, log x should be 0.....implies x = 1

u can derive them from the basic formulae...

but as u'll keep practising u'll eventually learn them by heart...so dont worry bout it!

jus keep practising....

wers d figure???

which chp dear???...

by applying Heisenberg's uncertainity principle

if we assume the electron to be inside the nucleus and calculate its velocity inside the nucleus it comes out to b more than the vel of light which is not possible. so it cannot exist inside the nucleus

we calc the vel of electron in the foll way:

as e^-is assumed inside the nucleus delta x = 10^-15m

the product delta x. delta v = h/(4. pi. m)

therefore delta v = h/(4. pi. m. delta x)

subs the values of h, pi, m and delta x to get the ans as 5.8 x 10¹⁰m/s

can u pls give d diagram...

(x+5)²- (x-5)² = 9

(x+5+x-5)(x+5-x+5) = 9 .........using a²-b² = (a+b)(a-b)

(2x)(10) = 9

20x=9

x=9/20

if v have 2 chk d continuity at x=1 then f(x) is defined as 4x+1

is not considered at all.....so it makes no difference wat value takes.....f(x) will always be continuous at x=1.

obviously there will be a change in velocity.....its direction changes on every instant n so does d velocity

v've been told dat d 1st one is correct n its explanation has been given in resnick halliday krane...

if u want d expln then temme......i'll post it

ders no fixed value of t

the meaning of the notation g(x) = max f(x) ;

is that g(x) attains the maximum value of f(t) wer t is the local maxima of f(x) and it should lie between x-1 and x+1.....n the range of x is already given. So u can find the value of g(x)

hope its clear....if not then u can nudge me

pls post the full ques dear.......

neways i read ur 1st half of d ques from d previous msg....d ans 2 ur ques is

¹²C₃ - ⁴C₃- ⁵C₃ - 1 = 205

point of infection is where f "(x) i.e. the second derivative of the function is either 0 or does not exist......

it can be defined as the point where the graph of f'(x) changes from increasing to decreasing or vice-versa......

no.......coz the very first condition for differentiability is dat it shud b a continuous function

yeah.....even in a ring i think the c.o.m. lies at the centre of the circle dats outside the ring

v_com = (m₁v₁+ m₂v₂) / m₁+ m₂

= (4+40) / 6 = 44/6 m/s

f(x) = x² - x sin x

for f(x) to be an increasing function f'(x) 0

f'(x) = 2x - x cos x - sin x

T.P.T: f'(x) 0

f"(x) = 2 + x sin x - 2 cos x

since

f''(x) > 0 (x sin x is always +ve in 0 to pi/2 and 2 - 2cos x is also >=0)

thus f'(x) is an increasing function

for an increasing function, if x>0 then g(x) > g(0)

but here f'(0) = 0

so f'(x) > f'(0) i.e. 0.........f'(x) >0

since f'(x) > 0, f(x) is an increasing function

hence proved

ya dats d rite ans......

( - infinity , - sqrt 2] U [ sqrt 2 , 2 ) U ( 3 , infinity )

sry for d wrong ans b4.....actually i had made a mistk in considering s step function

bt dis is d rite ans now....

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