Question

If $f\left(x\right)=2x^n+a$ for $n>0$, if $f\left(2\right)=26$ and $f\left(4\right)=138$, then $f\left(3\right)$ is equal to:

• A. $56$
• B. $82$
• C. $64$
• D. $122$

Answer 1

Answer: (C)

$64$

$f\left(2\right)=2(2{)}^n+a=26$...... (1)

$f\left(4\right)=2(4{)}^n+a=138$ .............. (2)

Subtracting (1) from (2) we get,

$138-26=2(4{)}^n-2(2{)}^n$

$112=2(4^n-2^n)$

$56=4^n-2^n$

$56=2^{2n}-2^n$ ...... (3)

Let say $2^n=y$

Then eq.(3) can be written as:

$y^2-y-56=0$

On solving for finding roots we get,

$y=\frac{+1\pm \sqrt{1+224}}{2}$

$y=\frac{1\pm 15}{2}$

Taking positive root because n is given positive in the question,

$y=8$

$2^n=8=2^3$

$n=3$ ...... (4)

Substituting eq. (4) in eq. (1) we get,

$26=2\times 8+a$

$a=10$

$f\left(3\right)=2(3{)}^n+a=2(3^3)+10=54+10=64$

Hence $f\left(3\right)=64$