Here, x = √√5 + 1/√5 - 1
On rationalising the terms given in square root, we get
x = √√5 + 1/√5 - 1 x √5 + 1/√5 + 1
√5 + 1/2
Now, substituting the value of x in x² - x - 1.
∴ x² - x - 1 = (√5 + 1/2)² - (√5 + 1/2) - 1
= 5 + 1 + 2√5/4 - √5 + 1/2 - 1
= 6 + 2√5 - 2√5 - 2 - 4 /4 = 0

Correct Option: A

Here, x = √√5 + 1/√5 - 1
On rationalising the terms given in square root, we get
x = √√5 + 1/√5 - 1 x √5 + 1/√5 + 1
√5 + 1/2
Now, substituting the value of x in x² - x - 1.
∴ x² - x - 1 = (√5 + 1/2)² - (√5 + 1/2) - 1
= 5 + 1 + 2√5/4 - √5 + 1/2 - 1
= 6 + 2√5 - 2√5 - 2 - 4 /4 = 0

Given, α and β are the roots of the equation ax² + bx + c = 0
∴ Sum of two roots = -b/a
α + β = -b/a
and product of two roots = c/a
α β = c/a
∴ α²/β + β²/α = ( α³ + β³) / α β
= [(α + β) ³ - 3 α β (α + β)] / αβ
[∵ (a + b)³ = a³ + b³ + 3ab(a + b) ]

Correct Option: D

Given, α and β are the roots of the equation ax² + bx + c = 0
∴ Sum of two roots = -b/a
α + β = -b/a
and product of two roots = c/a
α β = c/a
∴ α²/β + β²/α = ( α³ + β³) / α β
= [(α + β) ³ - 3 α β (α + β)] / αβ
[∵ (a + b)³ = a³ + b³ + 3ab(a + b) ]
= (-b/a)³ - [3c/a(-b/a)/c/a = -b³/a³ + 3bc/a²]/(c/a)
= 3abc - b³/a²c

Given equation is
x² - 11x + 24 = 0
Then, required equation is
(x - 2)² - 11(x - 2) + 24 = 0

Correct Option: E

Given equation is
x² - 11x + 24 = 0
Then, required equation is
(x - 2)² - 11(x - 2) + 24 = 0
⇒ x² - 4x + 4 - 11x + 22 + 24 = 0
⇒ x² - 15x + 50 = 0

Here, x² = 6 + √6 + √6 + √6 + .... ∞ ,
So, x² = 6 + √x²

Correct Option: D

Here, x² = 6 + √6 + √6 + √6 + .... ∞ ,
So, x² = 6 + √x²
⇒ x² = 6 + x
⇒ x² - x - 6 = 0
⇒ x² + 2x - 3x - 6 = 0
⇒ x(x + 2) - 3(x + 2) = 0
⇒ (x - 3) (x + 2) = 0
∴ x = 3

ab/(a + b) = [p/(p + q)] x [q/(p - q)] / [p/(p + q) + q/(p - q)]
= pq/p² - pq + pq + q²

Correct Option: A

ab/(a + b) = [p/(p + q)] x [q/(p - q)] / [p/(p + q) + q/(p - q)]
= pq/p² - pq + pq + q²
= pq/(p² + q²)