# Doubt from quadratic equations

For 4) th one double derivative should have two roots

#4

double diffrenciate them D\ge 0

4 colliner point= 4 roots

#8

base change theoram

#5

(x-1)^2=4(1- cos(a+bx))

x=1 satisfying will turn our

cos(a+bx)=0

i m sorryy

cos(a+bx)=1

for which one ans is 3??

if it if for this then x=3 is also satisfying it two equation in at that time

cos(a+3b)=0 two equation two variable u will get reasult..

u r correct its caculation error of mine

(x-1)^2=-4(1+cos(a+bx) it was

ans is 3?? only

max(a+b)=3\pi see domain if given

@Achyut_2020

Please explain why you have not taken pie ,and you have taken 3 pie

pi is min value 3pi is max value

So why only 3 pie you have taken

a,b \in \mathcal 0,5

0<a+b<10

the max(a+b)=3\pi

3\pi \approx >9

now what so sophisticated to osberve why according to u \pi give max value value of a+b

@Achyut_2020

because 4 roots are their on 1 derivative their will be 3 roots and second derivative their will be

2 roots simple

its done in forum by jagdish sir before

question asked by chirag_hegde1

dnt remember now ask jagdish sir he can help u