A quadratic polynomial y = f(x) with absolute | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The quadratic polynomial is of the form $y = ax^2 + bx + c$. Given the absolute term $c = 3$ and the leading coefficient $a = 1$,we have $f(x) = x

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) The quadratic polynomial is of the form $y = ax^2 + bx + c$. Given the absolute term $c = 3$ and the leading coefficient $a = 1$,we have $f(x) = x^2 + bx + 3$.Since the curve is symmetric about the line $x = 1$,the vertex occurs at $x = 1$. Thus,the derivative $f'(x) = 2x + b$ must be zero at $x = 1$.$2(1) + b = 0 \Rightarrow b = -2$.So,the polynomial is $f(x) = x^2 - 2x + 3$.For point $A$,$x_1 = 1$,so $y_1 = f(1) = 1^2 - 2(1) + 3 = 2$. Thus,$A = (1, 2)$.For point $B$,$y_2 = 11$,so $11 = x^2 - 2x + 3 \Rightarrow x^2 - 2x - 8 = 0$.$(x - 4)(x + 2) = 0$. Since $B$ is in the first quadrant,$x_2 = 4$. Thus,$B = (4, 11)$.The vectors are $\vec{OA} = \hat{i} + 2\hat{j}$ and $\vec{OB} = 4\hat{i} + 11\hat{j}$.The scalar product $\vec{OA} \cdot \vec{OB} = (1)(4) + (2)(11) = 4 + 22 = 26$.

Similar Questions

For the hyperbola $\frac{x^2}{9} - \frac{y^2}{3} = 1$,the incorrect statement is:

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=4ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $2e^2=$

If the curves $2x^2 + ky^2 = 30$ and $3y^2 = 28x$ cut each other orthogonally,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module