Let f be an odd function defined on the real number such that f(x)=3sinx+4cosx
A
−3sinx+4cosx
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B
−3sinx−4cosx
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C
3sinx+4cosx
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D
3sinx−4cosx
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Solution
The correct option is D3sinx−4cosx Given, f(x)=3sinx+4cosx When, x<0 ∴−x>0 ∴f(−x)=3sin(−x)+4cos(−x)....(i) But f(x) is odd function ∴f(−x)=−f(x) when x<0 ⇒f(x)=−f(−x)=−[3sin(−x)+4cos(−x)]=3sinx−4cosx