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Question

If θϵR, maximum value of
Δ=∣ ∣11111+sinθ1111+cosθ∣ ∣ is

A
12
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B
32
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C
2
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D
324
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Solution

The correct option is A 12
Δ=∣ ∣11111+sinθ1111+cosθ∣ ∣
Applying R2R2R1,R3R3R1
Δ=∣ ∣1110sinθ000cosθ∣ ∣=1(sinθcosθ)=sin2θ2
Therefore, max of Δ is 12

Hence, option 'A' is correct.

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