Question

if $f$ and $g$ are real function defined by $f\left(x\right)=2x-1$ and $g\left(x\right)=x^2$, then

• A. $(3f-2g)\left(1\right)=1$
• B. $\left(fg\right)\left(2\right)=10$
• C. $g^3\left(2\right)=128$
• D. $\left(\frac{\sqrt{f}}{g}\right)\left(2\right)=\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $(3f-2g)\left(1\right)=1$

Solution:-

$f\left(x\right)=2x-1g\left(x\right)=x^2$

$\left(i\right)(3f-2g)1=3f\left(1\right)-2g\left(1\right)$

$=3(2-1)-2\left(1\right)$

$=3-2=1$

$\left(ii\right)\left(fg\right)\left(2\right)=f\left(g\right(2\left)\right)$

$=f\left(2^2\right)=f\left(4\right)$

$=2\times 4-1=9$

$\left(iii\right)g^3\left(2\right)=g\left(g\right(g\left(2\right))$

$=g\left(g\right(4\left)\right)$

$=g\left(16\right)$

$=256$

$\left(iv\right)\left(\frac{\sqrt{f}}{g}\right)g\left(2\right)=\frac{\sqrt{f}\left(2\right)}{g\left(2\right)}$

$=\frac{\sqrt{2\left(2\right)-1}}{4}$

$=\frac{\sqrt{3}}{4}$