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DLS
 one year ago
Best ResponseYou've already chosen the best response.1Let \(\alpha\) and \(\beta\) be the distinct roots of \(\ ax^2+bx+c=0\) then \[\Huge \lim_{x \rightarrow \alpha} \frac{1\cos(ax^2+bx+c)}{(x\alpha)^2}=?\]

DLS
 one year ago
Best ResponseYou've already chosen the best response.1Attempt: \[\Large \frac{1cosa(x\alpha)(x\beta)}{(x\alpha)^2}\]

DLS
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{x \rightarrow \alpha}\Large \frac{1cosa(x\beta)}{(x\alpha)}\]

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.01cos(2x)=2 sin^2 x laga.. quadratic ko khol mat.

DLS
 one year ago
Best ResponseYou've already chosen the best response.1without opening the quadratic it wont solve :/ (xalpha)^2 wont cancel i guess

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0do it.. sin^2 something aayega.. (xalpha)^2 hai.. so [ sin(something)/(xalpha)] ^2 ho jayega..

DLS
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{ 2\sin^{2} (ax^2+bx+c)}{(x\alpha)^2}\] this?

DLS
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{ 2\sin^{2} \frac{ (ax^2+bx+c)}{2}}{(x\alpha)^2}\]

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0yup. ab quadratic ko (xalpha)(xbeta) likh.. multiply divide by (xbeta)^2 ka whole square.. [sin(xalpha)(xbeta)/(xalpha)(xbeta)]^2 *2(xbeta)^2.. pehli term sinx/x ke form me hai to 1.. bach gya 2(xbeta)^2.. to aa gya. 2*(alphabeta)^2 ??

DLS
 one year ago
Best ResponseYou've already chosen the best response.1answer to galat hai khair :P

DLS
 one year ago
Best ResponseYou've already chosen the best response.1\[(a)0\] \[(b) \frac{1}{2}a^(\alpha\beta)^2\] \[(c) \frac{1}{2}^(\alpha\beta)^2\] \[(d) \frac{1}{2}a^2(\alpha\beta)^2\]

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0ohho.. ax^2 + bx + c/2 tha.. :/

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0pta nhi.. aapko kya lgta hai?

DLS
 one year ago
Best ResponseYou've already chosen the best response.1xbeta wala step (2nd step) samaj nhi aya

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0(xbeta)^2 se multiply divide kr diya..

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0taaki sinx/x vali form ban sake.. [sin(xalpha)(xbeta)/2]^2 hai.. to niche [(xalpha)(xbeta)/2]^2 chahiye.. to (xbeta)^2/4 se multiply krenge..

DLS
 one year ago
Best ResponseYou've already chosen the best response.1@yrelhan4 (xalpha) nahi use kar sakte (xbeta) ki jagah?

yrelhan4
 one year ago
Best ResponseYou've already chosen the best response.0kaise? sin me xalpha aur x beta dono hain na.. to dono lane pdenge denominator me..

DLS
 one year ago
Best ResponseYou've already chosen the best response.1yo aagaya im the winner :P B hai

DLS
 one year ago
Best ResponseYou've already chosen the best response.1\[\LARGE \frac{2(\sin \frac{(a(x\alpha)(x\beta)}{2})^2}{(x\alpha)^2} \times \frac{(\frac{a}{2})^2 \times (x\beta)^2}{ (\frac{a}{2})^2 \times (x\beta)^2}\]

DLS
 one year ago
Best ResponseYou've already chosen the best response.11 ghanta ek question me lag raha hai : ye 5minute ka tha,shubham ko ek minute me strike kar jata :
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