Given that f(x)=10cosx+(13+2x)sinx . . . (i) On differentiating w.r.t. x, we get f′(x)=−10sinx+(13+2x)cosx+2sinx Again differentiating, we get f′′(x)=−10cosx−(13+2x)sinx+2cosx +2cosx... (ii) On adding Eqs. (i) and (ii), we get f′′(x)+f(x)=2cosx+2cosx=4cosx