Anyone know how the Multiple choice for that how many number of root for cx^2 +ax+b works I got 2 real root cause it’s just guess work lmao
0 roots i think? i remember b=c coming up, and then doing some ******** to deduce a^2 - 4bc was negative. swear i remember a = -(b+c)/2 as well but idk lmao
pretty sure the key idea was just to utilise the fact that cos(90-x) = sin(x).
i dont remember the exact expression, but there were two brackets multiplied iirc, one was (1 - cos^2(2x-60)) and the other (3-cos(150-2x)) by applying the identity to the left bracket, you get (1-sin^2(150-2x)) = cos^2(150-2x) which leaves the expression in a form previous to the prior part, but with something to the power of 4 instead?
anyways i got an answer of 16^4 = 4^8 = 2^16. If you know the expression, I could explain it probably, but i'm pretty confident it was maximised when cos(150-2x) = -1.
omd yeah that makes sense i shioldve seen that. i got the right answer just by inspection p much but idt they'll give me many marks. at least 1 out of the 6 but idk it could also be 6/6 if I'm v lucky.
0 roots i think? i remember b=c coming up, and then doing some ******** to deduce a^2 - 4bc was negative. swear i remember a = -(b+c)/2 as well but idk lmao
yh me too but did you guess? i guessed it looked like the only good option
nah i looked at the last digit. either 1,4,5,6,9 or an even number of zeroes at the end. all the options can be eliminated except the one ending with 5