Doubt from quadratic equations

q4,5,8

For 4) th one double derivative should have two roots

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#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

@Sneha_2021 but the answer is given 3 please also explain 8 th one i tried but i am confused

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..

@Sneha_2021 on putting x=3 we get no solution

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

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@Bhuvanitha_2020 could you pls explain why is it so..? ?

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

Post the link @Sneha_2021

dnt remember now ask jagdish sir he can help u

@pushkar_2020 creadit to jagdish sir

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