Q.

The values of f(x)=3sin(π216−x2)  lie in the interval

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a

(0,  32)

b

(−32,  32)

c

[0,  32]

d

[−32,  32]

answer is C.

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Detailed Solution

π216−x2≥0       ⇒−π4≤×≤π4 ∴Df=[−π4,π4]f(0)=3sinπ4=12 fπ4=3sin0=0 range is 0, 12
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