Question

The point on the curve $y=12x-x^2$, where the slope of the tangent is zero will be

• A. $(0,0)$
• B. $(2,16)$
• C. $(3,9)$
• D. $(6,36)$

Answer 1

The correct option is D $(6,36)$

Let the point be $P(x,y)$

$y=12x-x^2$

$\frac{dy}{dx}=12-2x$

${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=12-2x_1$

since slope of tangent is zero

so ${\left(\frac{dy}{dx}\right)}_{x_1,y_1}=0$

$12-2x_1=0$

$2x_1=12$

$x_1=6$

Also curve passing through tangent

$y_1=12x_1-x_1^2$

$y_1=12\times 6-36$

$y_1=72-36$

$y_1=36$

The points are $(6,36)$.