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Only one person sits between K5 and the one who likes Lily. K1 sits second to the left of the one who likes Lily. If the one who likes Lily sits at place no.1, then K1 sits at place no. 7 and K5 sits at place no. 3. K5 sits second to the left of the one who likes Lavender. So, the one who likes Lavender at place no. 5. Only two persons sit between K7 and the one who has likes Lavender. From this statement, we will have two cases: K7 will sit either at place no. 2 or 8. K8 is an immediate neighbour of K3. K3 is not an immediate neighbour of K1 or K5 and he does not likes Lavender. The one who likes Rose sits second to the right of K8. K3 cannot sit at place no. 6, 5, 4 and 8. So, only place left for him is place no. 1. As K8 is an immediate neighbour of K3. So, K8 will sit either at place no. 8 or 2. The one who likes Rose sits either at place no. 4 or 2. Only three people sit between the one who likes Rose and the one who likes Marigold. So, the one who likes Marigold sits either at place no. 6 or 8. K6 sits second to the left of the one who likes Marigold. So, K6 sits either at place no. 4 or 6. Case-1
Case-2
K8 does not likes Daisy. The one who likes Daisy is not an immediate neighbor of K8. K4 sits on the immediate left of the one who likes Daisy. The one who likes Lris sits on the immediate right of the one who likes Pansy. Case 1 will get discarded as the one who likes Daisy cannot sit at place no. 7 and 8. If the one who likes Daisy sits either at place no. 4 or 3, then there is no place for K4 as K4 sits on the immediate left of the one who likes Daisy In case 2, if the one who likes Daisy sits at place no. 6, then K4 sits at place no. 5. The one who likes Lris and the one who likes Pansy sit at place no. 3 and 2 respectively Only person left for place no. 4 is K2. Only flowers left for K1 is Jasmine Case-1
Case-2
Final arrangement as shown below:
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If f(x) is periodic of period T, then:
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