ri?reeeeeeeeeeeeeeeeeee Comments

so case 1, you already considered x is greater than or equal to 3 and then you got the set of all real numbers that are less than or ual to 4 but since x should also be greater than 3, the values of x that satisfy the inequality according to case 1 should lie between 3 and 4 {[3,4]} similarly in case 2, you already considered x to be less than 3 and then got the set of all real numbers greater than or equal to 2 so the values of x satisfying the inequality according to case 2 are [2,3)

48 replies

IiTeachChem

•Created by Abhi on 5/29/2024 in #💭│doubts

Modulus Inequality

ghost ping?

48 replies

IiTeachChem

•Created by Abhi on 5/29/2024 in #💭│doubts

Modulus Inequality

okay what have you tried so far

48 replies

IiTeachChem

•Created by Gaurav on 4/30/2024 in #💭│doubts

basic doubt

because in the interval, x is negative but you were considering x to be postive

12 replies

IiTeachChem

•Created by Gaurav on 4/30/2024 in #💭│doubts

basic doubt

no it isnt

12 replies

IiTeachChem

•Created by Gaurav on 4/30/2024 in #💭│doubts

basic doubt

in case 2, x is negative but x + 2 is positive so modulus of x will be -x because -x is positive but since x+2 is positive modulus of that will be x+2 so -x+x+2=2 which is always true https://www.desmos.com/calculator/f5qpz7dzrv a good way to visualise tis would be with a graph you can see that for values of x between -2 and 0 the expression is always equal to 2

12 replies

IiTeachChem

•Created by uhm?!? on 4/10/2024 in #💭│doubts

K-Electron Capture

it does go inside the nucleus in a way

14 replies

Gaming

Programming