Let λ ∈ R and let the equation E be |x|2-2|x|+|λ-3|=0. Then the largest element in the set S= x+λ: x is an integer solution of E is

Question

Let λ ∈ R and let the equation E be |x|2-2|x|+|λ-3|=0. Then the largest element in the set S= x+λ: x is an integer solution of E is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

| x |2-2| x |+|λ-3|=0 | x |2-2| x |+|λ-3|-1=0 (| x |-1)2+|λ-3|=1 At λ=3, x =0 and 2 , at λ=4 or 2 , then x =1 or -1 So maximum value of x+λ=5

Q. Let $λ \in R$ and let the equation $E$ be $∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ = 0$ . Then the largest element in the set $S =$ ${x + λ : x$ is an integer solution of $E}$ is

$∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ = 0$
$∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ - 1 = 0$
$(∣ x ∣ - 1)^{2} + ∣ λ - 3∣ = 1$
At $λ = 3, x = 0$ and 2 ,
at $λ = 4$ or 2 ,
then $x = 1$ or $- 1$
So maximum value of $x + λ = 5$

Q. Let λ∈R and let the equation E be ∣x∣2−2∣x∣+∣λ−3∣=0. Then the largest element in the set S= {x+λ:x is an integer solution of E} is

∣x∣2−2∣x∣+∣λ−3∣=0 ∣x∣2−2∣x∣+∣λ−3∣−1=0 (∣x∣−1)2+∣λ−3∣=1 At λ=3,x=0 and 2 , at λ=4 or 2 , then x=1 or −1 So maximum value of x+λ=5

Q. Let $λ \in R$ and let the equation $E$ be $∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ = 0$ . Then the largest element in the set $S =$ ${x + λ : x$ is an integer solution of $E}$ is

$∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ = 0$
$∣ x ∣^{2} - 2∣ x ∣ + ∣ λ - 3∣ - 1 = 0$
$(∣ x ∣ - 1)^{2} + ∣ λ - 3∣ = 1$
At $λ = 3, x = 0$ and 2 ,
at $λ = 4$ or 2 ,
then $x = 1$ or $- 1$
So maximum value of $x + λ = 5$