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Question Number 11050 by lepan last updated on 09/Mar/17

The function f(x)=acosx+b where  a<0 has a maximum value of 8 and  a minimum value of −2.Find a+b.

$${The}\:{function}\:{f}\left({x}\right)={acosx}+{b}\:{where} \\ $$ $${a}<\mathrm{0}\:{has}\:{a}\:{maximum}\:{value}\:{of}\:\mathrm{8}\:{and} \\ $$ $${a}\:{minimum}\:{value}\:{of}\:−\mathrm{2}.\boldsymbol{{F}}{ind}\:{a}+{b}. \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

Answered by Mahmoud A.R last updated on 09/Mar/17

−1 ≤ cosx ≤ 1   − a ≥ acosx ≥ a  b −a ≥ acosx + b ≥ a  + b  minimum of f(x) = a + b = −2

$$−\mathrm{1}\:\leqslant\:\mathrm{cos}{x}\:\leqslant\:\mathrm{1}\: \\ $$ $$−\:{a}\:\geqslant\:{a}\mathrm{cos}{x}\:\geqslant\:{a} \\ $$ $${b}\:−{a}\:\geqslant\:{a}\mathrm{cos}{x}\:+\:{b}\:\geqslant\:{a}\:\:+\:{b} \\ $$ $${minimum}\:{of}\:{f}\left({x}\right)\:=\:{a}\:+\:{b}\:=\:−\mathrm{2} \\ $$

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