Concept: Inequalities & Modulus

CONTENTS

CALCULATING RANGE OF X

Consider a1 < a2 < a3 < ... < an

f(x) = (x - a1) × (x - a2) × (x - a3) × ... × (x - an)

If x > an,
then (x - a1), (x - a2), (x - a3), ..., (x - an) are all positive, hence f(x) > 0, hence x > an is acceptable value of x.

If an-1 < x < an,
then (x - a1), (x - a2), (x - a3), ..., (x - an-1) are positive but (x - an) < 0, hence f(x) < 0.

If an-2 < x < an-1,
then (x - a1), (x - a2), (x - a3), ..., (x - an-2) > 0 but (x - an-1), (x - an) < 0, hence f(x) > 0.

If an-3 < x < an-2,
then (x - a1), (x - a2), (x - a3), ..., (x - an-3) > 0 but (x - an-2), ..., (x - an) < 0, hence f(x) < 0.

∴ f(x) is alternately positive & negative starting with x > an

Here, a1, a2, a3, ..., an are called critical points.

Example: Find range of x for which (x - 2)(x - 5)(x - 9) > 0

Solution:
Let f(x) = (x - 2)(x - 5)(x - 9)

Here, critical points are 2, 5 & 9

If x > 9, (x - 2), (x - 5) & (x - 9) > 0, hence f(x) > 0.
∴ x > 9 is acceptable.

If 5 < x < 9, (x - 2) & (x - 5) > 0 but (x - 9) < 0, hence f(x) < 0.
∴ 5 < x < 9 is not acceptable.

If 2 < x < 5, (x - 2) > 0 but (x - 5) & (x - 9) < 0, hence f(x) > 0.
∴ 2 < x < 5 is acceptable.

If x < 2, (x - 2), (x - 5) & (x - 9) < 0, hence f(x) < 0.
∴ x < 2 is not acceptable.

⇒ f(x) > 0 for alternate ranges of x.

⇒ x ∈ (2, 5) ∪ (9, ∞)

Example: Find range of x for which (x + 1)(x - 5)(x - 7) > 0

Solution:
Let f(x) = (x + 1)(x - 5)(x - 7)

Here, the critical points are -1, 5 & 7.

f(x) > 0 for alternate ranges of x starting with x > 7.

⇒ x ∈ (-1, 5) ∪ (7, ∞)

Example: Find range of x for which (x + 1)(x + 3)(x - 4) < 0

Solution:
Let f(x) = (x + 1)(x + 3)(x - 4)

Here, the critical points are -1, -2 & 4.

f(x) > 0 for alternate ranges of x starting with x > 4.

∴ f(x) < 0 for alternate ranges of x starting with -2 < x < 4.

⇒ x ∈ (∞, -1) ∪ (-2, 4)

Example: Find range of x for which $\frac{1}{\left(\mathrm{x}-2\right)\left(\mathrm{x}-5\right)\left(\mathrm{x}-9\right)}$ > 0

Solution:
Let g(x) = $\frac{1}{\left(\mathrm{x}-2\right)\left(\mathrm{x}-5\right)\left(\mathrm{x}-9\right)}$, and
f(x) = (x - 2)(x - 5)(x - 9)

Now, g(x) > 0 if f(x) > 0.
Hence, required solution for g(x) > 0 is same as that for f(x) > 0. Hence, we effectively need to find range of x for which f(x) > 0.

Here, the critical points are 2, 5 & 7.

f(x) > 0 for alternate ranges of x starting with x > 7.

⇒ x ∈ (-1, 5) ∪ (7, ∞)

Example: Find range of x for which (x + 1)(x - 5)(x - 7) ≥ 0

Solution:
Here we will solve this question in 2 parts.

Part 1: Let f(x) = (x + 1)(x - 5)(x - 7) > 0 Here, the critical points are -1, 5 & 7.
f(x) > 0 for alternate ranges of x starting with x > 7.
⇒ x ∈ (-1, 5) ∪ (7, ∞)

Part 2: f(x) = (x + 1)(x - 5)(x - 7) = 0
This is true for x = -1, 5 & 7.

Now, we have possible solutions for x,
⇒ x ∈ (-1, 5) ∪ (7, ∞), or
⇒ x ∈ {-1, 5, 7}

∴ Combined range of x is
⇒ x ∈ [-1, 5] ∪ [7, ∞)
i.e., we include the critical points in our solution for f(x) > 0.

AM, GM RELATIONSHIP

Considering positive numbers, Arithmetic Mean of these numbers is always greater than or equal to their Geometric Mean.

Let AM and GM represent the Arithmetic and Geometric means of positive numbers a1, a2, ..., an. Then,
AM = (a1 + a2 + ... + an)/n
GM = $\sqrt[\mathrm{n}]{{\mathrm{a}}_{1}×{\mathrm{a}}_{2}×...×{\mathrm{a}}_{\mathrm{n}}}$

⇒ AM ≥ GM

Equality exists when all numbers are equal i.e., a1 = a2 = ... = an

Sum of a number and its reciprocal

Sum of a positive number and its reciprocal is always greater than or equal to 2. Sum is equal to 2 when the number is 1.

Let the number be x and its reciprocal be 1/x.

We know, AM ≥ GM
$\frac{\mathrm{x}+\frac{1}{\mathrm{x}}}{2}\ge \sqrt{\mathrm{x}×\frac{1}{\mathrm{x}}}$

$\frac{\mathrm{x}+\frac{1}{\mathrm{x}}}{2}\ge 1$

$\mathrm{x}+\frac{1}{\mathrm{x}}\ge 2$

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