I am unable to find the answer of this question. Any help will be appreciated

If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.

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- Jul 30th 2013, 02:58 AMvarunkanpurQuadratic Problem
I am unable to find the answer of this question. Any help will be appreciated

If A^2 + B^2 =1 then find the range of (a+b) where a and b is real. - Jul 30th 2013, 08:23 AMHallsofIvyRe: Quadratic Problem
Are you aware that the graph of $\displaystyle x^2+ y^2= 1$ is a circle with center at (0, 0) and radius 1? So that the circle extends from x= -1 to x= 1 horizontally, and y= -1 to 1 vertically.

- Jul 30th 2013, 11:24 AMtopsquarkRe: Quadratic Problem
- Jul 30th 2013, 04:28 PMShakarriRe: Quadratic Problem
Assuming you haven't done geometry of the circle you can think of it this way.

$\displaystyle A^2+B^2=1$

$\displaystyle A^2=1-B^2$

$\displaystyle A=\sqrt{1-B^2}$

Think of the maximum and minimum vales of B so that A is real. Then do the same thing for A instead of B. - Aug 2nd 2013, 09:48 PMsmatikRe: Quadratic Problem
hmm if you've done trigonometry,try this. Let A=sin x and B=cos x[Parametric equation of circle]. Now range of A+B is Range of sin x + cos x i.e [-1/sqrt(2),1/sqrt(2)].

- Aug 3rd 2013, 06:59 PMjohngRe: Quadratic Problem
Hi,

smatak has a good idea, but unfortunately he gave the wrong answer.

$\displaystyle sin(x)+cos(x)=\sqrt2sin(x+\pi/4)$

So the range of sin(x)+cos(x) is $\displaystyle [-\sqrt2,\sqrt2]$ - Aug 3rd 2013, 11:37 PMsmatikRe: Quadratic Problem
oh yea.i calculated wrong.... :( btw its smatik