UPSC NDA Math Model Paper 4

Using the definition of modulus function, we have the following two cases:
CASE 1: If 2−sin‌θ≥0⇒2≥sin‌θ, then:
|2−sin‌θ|=2−sin‌θ
In order to maximize 2−sin‌θ,sin‌θ should be minimum.
The minimum value of sin‌θ is -1 .
∴ The maximum value of 2−sin‌θ is 2−(−1)=3.
CASE 2: If 2−sin‌θ<0⇒2<sin‌θ, which is not possible.Therefore, the maximum value of |2−sin‌θ| is |3|=3.