IIT JEE 1983 Maths - 'Adapted' - FIB to Single-Digit Integer Q1

Find the range of the real function $f(x) = 3 \sin \sqrt{ \dfrac{\pi^2}{16} - x^2}$ .

If the range can be expressed as the closed interval $[p,q]$ , submit your answer as $3\Big(\big\lvert\lfloor p\rfloor+\lfloor q\rfloor\big\rvert\Big)-1$ .

If the range can be expressed as the open interval $(p,q)$ , submit your answer as $3\Big(\big\lvert\lfloor p\rfloor+\lfloor q\rfloor\big\rvert\Big)+1$ .

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Notations:

• $\lfloor \cdot \rfloor$ denotes the floor function .

• $| \cdot |$ denotes the absolute value function .

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