Given 

⇒ 

⇒ 

⇒ (x + 4) (x – 4) (x – 3) ≤ 0

The critical points are – 4, – 1, 3 and 4

When x = 0, the inequality is not true

Hence when x Î [-4, -1] or (3, 4) the inequality is true.

∴ The required solution set [-4, -1] ∪ (3, 4)

Given, 3 – x < 4

⇒ – x < 4 – 3 ⇒ – x < 1

⇒ x > – 1

6 < x < 8

Or        x ϵ [6, 8]

⇒        (x – 6) (x – 8) < 0

Given 4x² + 5x – 9< 0 Þ 4x² + 9x – 4x – 9 < 0

x (4x + 9) – 1(4x + 9) < 0

(4x + 9) (x – 1) < 0

∴ The critical points are 

When x = 0, the given in equation is true

Hence, the solution set is x ϵ 

Given

           6x  + 9 < 3x + 5          :           4x + 7 > 2x – 5

           6x – 3x < 5 – 9           :           4x – 2x > -5 – 7

           3x < – 4                      :           2x > – 12

           ⇒ x < –  ….(A)    :           x > – 6….. (B)

           ∴ From (A) and (B)

           We have x ϵ 

Choice (A) is a standard result

Consider choice (B)

4 > 2 and – 3 > -4

But 4(-3) < 2 (-4)

Consider choice (C),

4 > 2 and -2 > -5

But 4 + 2 < 2 + 5

I. x² +  = 2 or x⁴ – 2x² + 1 = 0 is not a quadratic polynomial.

II. x +  = x² + 3 = x³ ⇒ x³ – x² – 3 = 0 is not a quadratic polynomial.

III. 2x² – x + 2 = x² + 4x – 4

      Or 2x² – x + 2 – x² – 4x + 4 = 0

      Or x² – 5x + 6 = 0 is a quadratic polynomial.

IV.  x³ + 6x² + 2x – 1 = 0 is not a quadratic polynomial.

Given 3x + 4 < 6 and 4x + 3 > 6

3x < 2 and 4x > 3

x <  and x >  ………. (2)

There is no real value of x satisfying the above In equations simultaneously

∴ The solution set is empty set.

Given 4x + 3 > 6x + 7

⇒ 4x  – 6x > 7 – 3

(-5, 2] (By definition)