Explanation:
Let the zeros of the given polynomial be p and q.
Given that q is the reciprocal of p, i.e. pq = 1.
Using the sum and product of zeros formula, we have:
p + q = 13/(k^2 - 4) and pq = 1/4k
Substituting pq = 1, we get:
p + q = 13/(k^2 - 4) and p = 1/4kq
Substituting p = 1/4kq in the first equation, we get:
1/4kq + q = 13/(k^2 - 4)
Multiplying both sides by 4kq(k^2 - 4), we get:
k^2q^2 + 4q(k^2 - 4) = 52kq
Rearranging the terms, we get:
k^2q^2 - 52kq + 4q(k^2 - 4) = 0
Dividing both sides by q, we get:
k^2q - 52k + 4(k^2 - 4) = 0
Simplifying, we get:
(k^2 - 16)(kq - 1) = 0
Since k cannot be equal to ±4, as it would make the denominator of p + q zero, we have:
kq - 1 = 0
i.e. q = 1/k
Since q is the reciprocal of p, we have:
p = 1/q = k
Therefore, the two zeros of the polynomial are p = k and q = 1/k.
Since k^2 - 4 ≠ 0, we have:
p + q = 13/(k^2 - 4) = k + 1/k
Multiplying both sides by k, we get:
k^2 + 1 = 13k/(k^2 - 4)
Multiplying both sides by (k^2 - 4), we get:
k^4 - 13k^2 + 4k + 16 = 0
This is a quadratic equation in k^2. Solving it using the quadratic formula, we get:
k^2 = (13 ± 5)/2
Since k^2 > 0, we have:
k^2 = (13 + 5)/2 = 9
Therefore, k = ±3
But k cannot be equal to -3, since it would make the denominator of p + q negative.
Therefore, k = 3.
Hence, the answer is option (B).

Let a=alpha..nd B=Bita.... NOW, a×B=1..As a(=1/B) ....(1) Also a×B=c/a=4k/(k^2+4).=1..(using1).... so,4k/(k^2+4)=1 ....4k=k^2+4... k^2-4k+4=0... k^2-2k-2k+4=0.... k(k-2)-2(k-2)=0... (k-2) (k-2)=0... so, k=2 or 2...