Theory of EquationsmediumFree

Theory of Equations: Set Values Holds True

JEE Maths question with a full step-by-step solution.

Question

The set of values of

a

for which

(a-1)x^2 - (a+1)x + (a-1) \geq 0

holds true for all

x \geq 2

(1, \frac{7}{3}]

(-6,1)

[\frac{7}{3},\infty)

correct

(-3,-2)

Solution

Step 1: Rewrite the left side

(a-1)x^2 - (a+1)x + (a-1) = a(x^2 - x + 1) - (x^2 + x + 1)

Step 2: Since

x^2 - x + 1 > 0

for all real

x

, the condition becomes:

a \geq \frac{x^2 + x + 1}{x^2 - x + 1} \quad \text{for all } x \geq 2

Step 3: Let

g(x) = \dfrac{x^2+x+1}{x^2-x+1}

. Computing

g'(x)

g'(x) = \frac{-2(x^2-1)}{(x^2-x+1)^2} < 0 \quad \text{for } x \geq 2

g

is strictly decreasing on

[2, \infty)

. Its maximum is at

x = 2

g(2) = \frac{4+2+1}{4-2+1} = \frac{7}{3}

Step 4: The condition holds for all

x \geq 2

if and only if

a \geq \dfrac{7}{3}

a \in \left[\frac{7}{3},\, +\infty\right)

Correct answer:

a \geq \dfrac{7}{3}

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