Let a2, a3 ∈ R be such that |a2-a3|=6. Let f(x)=| 1 a3 a2 1 a3 2 a2-x 1 2 a3-x a2 |, x ∈ R The maximum value of f(x) is

Question

Let a2, a3 ∈ R be such that |a2-a3|=6. Let f(x)=| 1 a3 a2 1 a3 2 a2-x 1 2 a3-x a2 |, x ∈ R The maximum value of f(x) is (A) 6 (B) 9 (C) 12 (D) 36

Tardigrade Admin · Accepted Answer

Using R2 arrow R2-R1, R3 arrow R3-R1, we get f(x) =|1 a3 a2 0 0 a2-x 0 a3-x 0| =-(a2-x)(a3-x)=-[x2-(a2+a3) x+a2 a3] =(1/4)(a2-a3)2-(x-(a2+a3/2))2 ≤ 9 f(x) attains maximum value 9 when x=(1/2)(a2+a3)

Q. Let a2​,a3​∈R be such that ∣a2​−a3​∣=6. Let f(x)=∣∣​111​a3​a3​2a3​−x​a2​2a2​−xa2​​∣∣​,x∈R The maximum value of f(x) is

6

9

12

36

Using R2​→R2​−R1​,R3​→R3​−R1​, we get f(x)=∣∣​100​a3​0a3​−x​a2​a2​−x0​∣∣​ =−(a2​−x)(a3​−x)=−[x2−(a2​+a3​)x+a2​a3​] =41​(a2​−a3​)2−(x−2a2​+a3​​)2≤9 f(x) attains maximum value 9 when x=21​(a2​+a3​)

Q. Let $a_{2}, a_{3} \in R$ be such that $∣ a_{2} - a_{3} ∣ = 6$ .
Let $f (x) = ∣ ∣ 111 a_{3} a_{3} 2 a_{3} - x a_{2} 2 a_{2} - x a_{2} ∣ ∣, x \in R$
The maximum value of $f (x)$ is