Pnc, algebra,...

[languid ferry]

solve it fellas.... somehoe do it!!

[warped vapor]

which book ?

[languid ferry]

some jee level book fella

[zealous oracle]

Observe the expression $-2 \lambda ^2 + 11 \lambda - 8$

[shrewd jasper]

Hellohowareyou

[zealous oracle]

Judging by the fact that it is under a square root, what can you say about this expression?

[Use the fact that λ>0]

[languid ferry]

well i formed an inequality - (11x-2x^2-8)/(2x)>0

and after that determined that lambda can be 1,2,3,or 4

after that i am unable to do more stuff to somehow get lambda

[lucid topaz]

is the answer 6?

[languid ferry]

no ideas

ill verify and deploy tommorow

[lucid topaz]

oki

[zealous oracle]

?? Just place these values into the expression? See which returns integer?

[lucid topaz]

i tried that method first

but none of it works

so u like

find out for what values the expression is greater than 0

that was my initial thought

but then

then veritcal line after then lamda didnt make much sense to me

so i think it means that S is lamda

and u need lamda such that it satisfies that equation

and then i graphed it

turns out theres 4 values of lamda which satisty the equation

[languid ferry]

no man i used the inequality to the get the values of lambda

which i got as 1,2,3,or 4

[unborn reef]

S = {λ | √((11λ - 2λ² - 8)/(2λ)) is an integer, λ > 0}
√((11λ - 2λ² - 8)/(2λ)) = integer
(11λ - 2λ² - 8)/(2λ) = perfect square

simplify (11λ - 2λ² - 8)/(2λ) to 11/2 - λ - 4/λ
11/2 - λ - 4/λ = perfect square
5.5 - perfect square = λ + 4/λ

using this, go through the perfect squares to express the solutions in terms of a few quadratics

from AM-GM, (λ + 4/λ) / 2 ≥ √(λ * 4/λ) = 2
5.5 - perfect square = λ + 4/λ ≥ 2
3.5 - perfect square = λ - 2 + 4/λ ≥ 0
perfect square is 0 or 1
3.5 - (this is 0 or 1) = λ - 2 + 4/λ
(this is 3.5 or 2.5) = λ - 2 + 4/λ
0 = λ - 2 - (this is 3.5 or 2.5) + 4/λ
0 = λ - (this is 5.5 or 4.5) + 4/λ
0 = λ² - (this is 5.5 or 4.5) λ + 4

so the four solutions are contained in λ² - 5.5λ + 4 = 0 and λ² - 4.5λ + 4 = 0
note that λ² - 4λ + 4 = 0 has one solution, so the above with a higher - λ must have two solutions
so the four solutions are roots to (λ² - 5.5λ + 4)(λ² - 4.5λ + 4) = 0
(you know they dont share any roots because they equalling would mean λ = 0)

from there, you need to find a₁, a₂, a₃, a₄ of x⁴ + a₁ x³ + a₂ x² + a₃ x + a₄ = 0 where the roots are the four solutions of λ
so λ⁴ + a₁ λ³ + a₂ λ² + a₃ λ + a₄ = 0
however, we already know (λ² - 5.5λ + 4)(λ² - 4.5λ + 4) = 0
so λ⁴ + a₁ λ³ + a₂ λ² + a₃ λ + a₄ and (λ² - 5.5λ + 4)(λ² - 4.5λ + 4) are the same polynomial
match coefficients:
a₁ = (1)(-4.5) + (-5.5)(1) = -10
a₄ = (4)(4) = 16
a₁ + a₄ = 6

next,
"(n - 2) is the 3rd term of a GP which adds to an"
"2 is the 3rd term of a GP which adds to 16"
8 + 4 + 2 + ... = 16
8 = 2 √16 = 2 √a₄

if you need help understanding this, you can ask someone else about it, I gtg

[lucid topaz]

i suppose the second answer is 2√a4?

[unborn reef]

thats a typo, thanks

[languid ferry]

@lucid topaz @zealous oracle @unborn reef the answers are 1-a; 2-c

please do explain if any of you got this answer, and in detail too

[lucid topaz]

this is the actual method

i study for jee so i used to pretty wild approximations and assumptions

if you get lucky you get the answer quickly

🙏

[languid ferry]

me too

[lucid topaz]

first year or second year?

[languid ferry]

2

[lucid topaz]

sameee

next year saath m mrenge ❤️

[languid ferry]

maybe