Correct Answer - Option 2 : 6
Calculation:
Consider an,
If a ≠ + 1, if a ≠ -1 and a ≠ 0
Then, an is always 1
When, n = 0 ----(1)
If a = 1, an is always 1 ----(2)
If a = -1, n should be even for an to be 1 ----(3)
According to the question:
(x2 - 7x + 11)(x2- 13x + 42)
⇒ (x2 – 13x + 42) = 0
⇒ (x2 – 6x – 7x + 42) = 0
⇒ x(x – 6) – 7(x – 6) = 0
⇒ (x – 6)(x – 7) = 0
⇒ x = 6 or x = 7
If x = 6,
(x2 – 7x + 11)
⇒ 62 – 7 × 6 + 11 ≠ 0
⇒ 1 ≠ 0
x = 6, is one solution
If x = 7,
(x2 – 7x + 11)
⇒ 72 – 7 × 7 + 11 ≠ 0
⇒ 11 ≠ 0
x = 7, is one solution
Again,
If (x2 – 7x + 11) = 1
⇒ (x2 – 7x + 10) = 0
⇒ (x – 2)(x – 5) = 0
⇒ x = 2 or x = 5 are another 2 solutions
If (x2 – 7x + 11) = -1
⇒ (x2 – 7x + 12) = 0
⇒ (x – 3)(x – 4) = 0
⇒ x = 3 or x = 4,
If x = 3, (x2 – 13x + 42)
⇒ 32 – 13 × 3 + 42 = 12, which is even
x = 3, is one solution
If x = 4, (x2 – 13x + 42)
⇒ 42 – 13 × 4 + 42 = 6, which is even
x = 4, is one solution
Now,
x = 2, 3, 4, 5, 6, and 7
∴ There are 6 distinct positive integer-valued solutions exist to the equation.