Question

If $f\left(t\right)=t^2+\frac{3}{2}t,$ then calculate the value of $f(q-1)$.

• A. $q^2-\frac{3}{2}$
• B. $q^2-\frac{1}{2}q$
• C. $q^2-\frac{1}{2}q+\frac{1}{2}$
• D. $q^2-\frac{1}{2}q-\frac{1}{2}$
• E. $q^2-\frac{1}{2}q-1$

Answer 1

The correct option is D $q^2-\frac{1}{2}q-\frac{1}{2}$

Given, $f\left(t\right)=t^2+\frac{3t}{2}$
Therefore, the value of

$f(q-1)={(q-1)}^2+\frac{3(q-1)}{2}$

$=q^2-2q+1+\frac{3q}{2}-\frac{3}{2}$

$=q^2-\frac{q}{2}-\frac{1}{2}$