Question

Let $f\left(x\right)=a{x}^7+b{x}^3+cx-5$, where $a,b$ and $c$ are constant. If $f(-7)=7$, then $f\left(7\right)$ equals$?$

• A. $-17$
• B. $7$
• C. $14$
• D. $21$

Answer 1

The correct option is A $-17$

Consider the given function.

$f\left(x\right)=a{x}^7+b{x}^3+cx-5$

Since, $f(-7)=7$

So,

$f(-7)=a(-7{)}^7+b(-7{)}^3+c(-7)-5=7$

$-7^{7}a-7^{3}b-7c-5=7$

$-(7^{7}a+7^{3}b+7c+5)=7$

$7^{7}a+7^{3}b+7c+5=-7$

$7^{7}a+7^{3}b+7c=-12$ $............\left(1\right)$

Put $x=7$

$f\left(7\right)=a(7{)}^7+b(7{)}^3+c\left(7\right)-5$

$f\left(7\right)=7^{7}a+7^{3}b+7c-5$

$f\left(7\right)=-12-5$ $fromequation\left(1\right)$

$f\left(7\right)=-17$

Hence, this is the answer.