Question

limits(38)

Answer 1

lol wut?

Answer 2

Let \(\alpha\) and \(\beta\) be the distinct roots of \(\ ax^2+bx+c=0\) then
\[\Huge \lim_{x \rightarrow \alpha} \frac{1-\cos(ax^2+bx+c)}{(x-\alpha)^2}=?\]

Answer 3

Attempt:
\[\Large \frac{1-cosa(x-\alpha)(x-\beta)}{(x-\alpha)^2}\]

Answer 4

\[\lim_{x \rightarrow \alpha}\Large \frac{1-cosa(x-\beta)}{(x-\alpha)}\]

Answer 5

1-cos(2x)=2 sin^2 x laga.. quadratic ko khol mat.

Answer 6

without opening the quadratic it wont solve :/ (x-alpha)^2 wont cancel i guess

Answer 7

do it.. sin^2 something aayega.. (x-alpha)^2 hai.. so [ sin(something)/(x-alpha)] ^2 ho jayega..

Answer 8

\[\Large \frac{ 2\sin^{2} (ax^2+bx+c)}{(x-\alpha)^2}\]
this?

Answer 9

sorry upon 2 too

Answer 10

\[\Large \frac{ 2\sin^{2} \frac{ (ax^2+bx+c)}{2}}{(x-\alpha)^2}\]

Answer 11

yup. ab quadratic ko (x-alpha)(x-beta) likh.. multiply divide by (x-beta)^2 ka whole square..
[sin(x-alpha)(x-beta)/(x-alpha)(x-beta)]^2 *2(x-beta)^2..
pehli term sinx/x ke form me hai to 1.. bach gya 2(x-beta)^2..
to aa gya. 2*(alpha-beta)^2 ??

Answer 12

answer to galat hai khair :P

Answer 13

pata nhi..

Answer 14

\[(a)0\]
\[(b) \frac{1}{2}a^(\alpha-\beta)^2\]
\[(c) \frac{1}{2}^(\alpha-\beta)^2\]
\[(d) \frac{-1}{2}a^2(\alpha-\beta)^2\]

Answer 15

ohho.. ax^2 + bx + c/2 tha.. :/

Answer 16

C hai?

Answer 17

nah :P

Answer 18

pta nhi.. aapko kya lgta hai?

Answer 19

x-beta wala step (2nd step) samaj nhi aya

Answer 20

(x-beta)^2 se multiply divide kr diya..

Answer 21

kyunkiya?

Answer 22

taaki sinx/x vali form ban sake.. [sin(x-alpha)(x-beta)/2]^2 hai.. to niche [(x-alpha)(x-beta)/2]^2 chahiye..
to (x-beta)^2/4 se multiply krenge..

Answer 23

bus ab rukja :P

Answer 24

@yrelhan4 (x-alpha) nahi use kar sakte (x-beta) ki jagah?

Answer 25

kaise? sin me x-alpha aur x- beta dono hain na.. to dono lane pdenge denominator me..

Answer 26

OH :O ab bus rukja :P

Answer 27

yo aagaya
im the winner :P
B hai

Answer 28

\[\LARGE \frac{2(\sin \frac{(a(x-\alpha)(x-\beta)}{2})^2}{(x-\alpha)^2} \times \frac{(\frac{a}{2})^2 \times (x-\beta)^2}{ (\frac{a}{2})^2 \times (x-\beta)^2}\]

Answer 29

u forgot about a :P

Answer 30

ah.. samajh gya. :/

Answer 31

:D

Answer 32

hmm iit. hmm

Answer 33

:P

Answer 34

1 ghanta ek question me lag raha hai :| ye 5minute ka tha,shubham ko ek minute me strike kar jata :|

Answer 35

ok bye.

Answer 36

thanks for the help :)

Answer 37

haha. :')