Question

(i) If x² – 6x + 1 = 0 then `x^2 + 1/x^2 = 34`

(ii) x³ + y² = (x + y)(x² + xy + y²)

विकल्प

• Only (i)

• Only (ii)

• Both (i) and (ii)

• Neither (i) nor (ii)

Answer 1

Only (i)

Explanation:

Let’s analyse the statements:

(i) Given: x² – 6x + 1 = 0

We want to find `x^2 + 1/x^2`.

Divide the equation by x assuming x ≠ 0:

`x - 6 + 1/x = 0`

⇒ `x + 1/x = 6`

Square both sides:

`(x + 1/x)^2 = 6^2`

`(x + 1/x)^2 = 36`

`x^2 + 2 + 1/x^2 = 36`

⇒ `x^2 + 1/x^2 = 34`

So the statement (i) is true.

(ii) The identity:

x³ + y³ = (x + y)(x² – xy + y²)   ...(Standard algebraic identity)

However, in the question, it is stated:

x³ + y³ = (x + y)(x² + xy + y²) 

This is incorrect because the correct factorisation has a minus sign for the middle term:

x³ + y³ = (x + y)(x² – xy + y²)

Therefore, statement (ii) is false.