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Question

Consider the four sets A,B,C and D defined as A={a:a=8(log4(x))3+4log√2(x4), x>0} B={b:b=20+13(logx2)2, x>0} C={c∈N:c∈A∩B for same x>0} D={x∈Z:(x−2)2(15x2−56x+17)<0} Then the number of elements in C×D is

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Solution

C denotes the common elements (natural numbers) in A and B. 8(log4(x))3+4log√2(x4)=20+13(logx2)2 ⇒8(log4x)3+4log√2(x4)−20−13(log2x)2=0 ; x≠1 ⇒8×(12)3(log2x)3+4×2×4log2x−13(log2x)2−20=0 ⇒(log2x)3−13(log2x)2+32log2x−20=0 Let log2x=t Then t3−13t2+32t−20=0 By trial and error method, we find that t=1 is a root. Applying synthetic division t=11−13322001−12−201−12−200 ⇒t3−13t2+32t−20=(t−1)(t2−12−20)=0 ⇒(t−1)(t−2)(t−10)=0 ⇒t=1,2,10 ⇒x=21,22,210=2,4,1024 ∴C={2,4,1024} (x−2)2(15x2−56x+17)<0 ⇒15x2−56x+17<0 ;x≠2 ⇒15x2−5x−51x+17<0 ⇒(3x−1)(5x−17)<0 ⇒(x−13)(x−175)<0 ⇒x∈(13,175)−{2} ∴D={1,3} ∴n(C×D)=3×2=6

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