If set of all values of x∈(−π2,π2) satisfying |4sinx+√2|<√6 is (aπ24,bπ24), then the value of ∣∣∣a−b3∣∣∣=
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Solution
|4sinx+√2|<√6 ⇒−√6<4sinx+√2<√6⇒−√2(√3+1)<4sinx<√2(√3−1) ⇒−(√3+1)2√2<sinx<(√3−1)2√2⇒−sin5π12<sinx<sinπ12⇒−5π12<x<π12 for x∈(−π2,π2)
Comparing with aπ24<x<bπ24, we get, a=−10,b=2 ∴∣∣∣a−b3∣∣∣=∣∣∣−10−23∣∣∣=4