Question

1. Let y = f(x) be any curve and P be any point on it then the slope of the curve at P is $\frac{dy}{dx}=m=tan\theta$
2. Equation of the tangent at P is $y-y_1=m(x-x_1)$
3. Equation of the normal at P is $y-y_1=\frac{-1}{m}(x-x_1)$
On the basis of these 3 points answer the following questions:

If the curve $y=a{x}^2+6x+b$ passes through (0, 2) and has its tangent parallel to x-axis at $x=\frac{3}{2}$, then the value of a and b are respectively

• A. 2 and 2
• B. -2 and -2
• C. -2 and 2
• D. 2 and -2

Answer 1

The correct option is C -2 and 2

$\frac{dy}{dx}=\pm 1$
$\therefore 2x^2+x-1=0$
$x=-1,\frac{1}{2}$
$\therefore$ The points are
$(-1,-\frac{1}{6}),(\frac{1}{2},\frac{5}{24})$
$y=a{x}^2+6x+b...\left(1\right)$
In equation (1) by putting x = 0, y = 2 We get $\Rightarrow b=2$
Equation (2) ${\left|\frac{dy}{dx}\right|}_{atx=\frac{3}{2}}=0\Rightarrow 3a+6=0\Rightarrow a=-2$.