Question

y = $\left[x\right(x-3){]}^2$ is increasing when-

• A. $0<x<\frac{3}{2}$
• B. $0<x<$ $\infty$
• C. $-\infty$ $<x<0$
• D. $1<x<3$

Answer 1

The correct option is A $0<x<\frac{3}{2}$

$y={\left[x(x-3)\right]}^2$ increases when For function to be increasing $\frac{dy}{dx}>0$

$\frac{dy}{dx}=2x(x-3)\times (2x-3)>0$

Now,

$x(x-3)(2x-3)>0$

Plotting on number line

(Image)

At a value less than zero, expression is negative

Therefore $y={\left[x(x-3)\right]}^2$ increases for

$xϵ(0,\frac{3}{2})\bigcup (3,\infty )$