1.

In the given questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.I: 3x2 – 15x + 12 = 0II : 4y2 – 80y + 256 = 01. x > y2. x ≤ y3. x ≥ y4.  x < y

Answer» Correct Answer - Option 2 : x ≤ y

Calculation:

I: 3x2 – 15x + 12 = 0

⇒ 3x2 – 12x – 3x + 12 = 0

⇒ 3x(x – 4) – 3(x – 4) = 0

⇒ (x – 4) × (3x – 3) = 0

⇒ x = 4, 1

II: 4y2 – 80y + 256 = 0

Dividing the equation by 4 we'll get

⇒ y2 – 20y + 64 = 0

⇒ y2 – 4y – 16y + 64 = 0

⇒ y × (y – 4) – 16 × (y – 4) = 0

⇒ (y – 4) × (y – 16) = 0

⇒ y = 4 and 16

Comparing values of x and y,

Value of x

Relation

Value of y

4

=

4

4

< 

16

1

< 

4

1

< 

16


∴ From the table we can say, x ≤  y.


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