Q.
If the function f(x)=(m2−3m+2)cosx+(m−1)sinx+2(n−3),(m=1) is an odd function then the value of sinmθ+cosnθ is always
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Relations and Functions - Part 2
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Solution:
Θf(−x)=−f(x),∀x∈R ⇒(m2−3m+2)cosx−(m−1)sinx+2(n−3)=−(m2−3m+2)cosx−(m−1)sinx−2(n−3) ∴m2−3m+2=0⇒m=1 or 2 but m=1 ∴m=2&n−3=0⇒n=3 ∴sinmθ+cosnθ=sin2θ+cos3θ≤sin2θ+cos2θ≤1