Aspire Faculty ID #13744 · Topic: JAMIA MILLIA ISLAMIA MCA 2022 · Just now
JAMIA MILLIA ISLAMIA MCA 2022

If $f(x)=ax^7+bx^3+cx-5$, where $a,b,c$ are real constants, and $f(-7)=7$, then the range of $f(7)+17\cos x$ is:

Solution

$f(7)=-(f(-7))= -7 + 10 = 27$ (since odd-powered terms change sign). Actually, $f(7)=-f(-7)-10=...$ simplifying gives $f(7)=17$. Then $f(7)+17\cos x = 17 + 17\cos x$. Range = $[17-17,17+17]=[0,34]$.

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