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Mathematics 9 Online
OpenStudy (jiteshmeghwal9):

If \(\LARGE{\color{red}{f(x)=ax^2+bx+c}}\). Which one of the following is true :- a. \(\LARGE{f(x_1/x_2)=f(x_1)f(x_2)}\) b. \(\LARGE{f(x_1+x_2)=f(x_1)+f(x_2)}\) c. \(\LARGE{f(x_1/x_2)=f(x_1)/f(x_2)}\) d. none of these.

OpenStudy (jiteshmeghwal9):

@hartnn @mathslover @satellite73 plz help :)

OpenStudy (jiteshmeghwal9):

& explain with reason :)

OpenStudy (anonymous):

Try plugging it in.

OpenStudy (anonymous):

\[\Large f(x_1/x_2)=a(x_1/x_2)^2+b(x_1/x_2) +c \]Simplify can see where it goes.

OpenStudy (anonymous):

If you cant get it in the form of: \[ \Large f(x_1)f(x_2) = [a(x_1)^2+b(x_1)+c] [a(x_2)^2+b(x_2)+c] \]Then it is true.

OpenStudy (anonymous):

It might actually be better to try going backwards.

OpenStudy (anonymous):

if you CAN get it in the form^

OpenStudy (anonymous):

@jiteshmeghwal9 Does it make sense?

OpenStudy (jiteshmeghwal9):

Dude, i'm totally confused :(

OpenStudy (jiteshmeghwal9):

I must try this a bit later well u did ur best :)

OpenStudy (jiteshmeghwal9):

u deserve a medal for all this stuff :)

OpenStudy (anonymous):

I can give you a hint. Consider what happens to \(c\).

OpenStudy (anonymous):

Or even better, try doing \(x_1 = 0\) and \(x_2=1\) See if it holds true. If it doesn't, you have a counter example. So you know it would be false.

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