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XAT 2021 — QA & DI Question 8

Domain & RangeEasy

Let f(x) = x2+1x2-1 if x ≠ 1, -1, and 1 if x = 1, -1. Let g(x) = x+1x-1 if x ≠ 1, and 3 if x = 1. What is the minimum possible values of f(x)g(x)?

Answer & solution

  • A

    12

  • B

    -1

  • C

    14

  • 13

  • E

    1

Solution

f(x)g(x) = (x2+1)x2-1 . (x-1)x+1 = (x2+1)(x+1)2

This function is definitely greater than 0

let y = (x2+1)(x+1)2

⇒ x2 (y − 1) + 2yx + (y − 1) = 0 which is quadratic in x

Disctiminant should be greater than 0

4y2 - 4(y - 1)2 ≥ 0

⇒ y >= 1/2

When x =1, f(x)/g(x) = 1/3
Hence either the value should be greater than 1/2 or should be equal to 1/3

XAT 2021 QA & DI Q8: Let f(x) = x 2 + 1 x 2 - 1 if x ≠ 1, -1, and 1 if x = 1, -1. Let g(x) = x + 1 x - 1 if x ≠ 1, and 3 if x — Solution | TheCATExam