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

(A) Let the quadratic polynomial be $f(x) = ax^2 + bx + c$. Given $c = 3$ and $a = 1$,so $f(x) = x^2 + bx + 3$.Since the curve is symmetric about $x =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the quadratic polynomial be $f(x) = ax^2 + bx + c$. Given $c = 3$ and $a = 1$,so $f(x) = x^2 + bx + 3$.Since the curve is symmetric about $x = 1$,the vertex occurs at $x = 1$.The derivative $f'(x) = 2x + b$. Setting $f'(1) = 0$ gives $2(1) + b = 0$,so $b = -2$.Thus,the polynomial is $f(x) = x^2 - 2x + 3$.The area bounded by the curve $y = f(x)$ and the line $y = 3$ is found by setting $f(x) = 3$:$x^2 - 2x + 3 = 3$ $\Rightarrow x^2 - 2x = 0$ $\Rightarrow x(x - 2) = 0$.The intersection points are $x = 0$ and $x = 2$.The area is given by the integral $\int_{0}^{2} (3 - f(x)) dx = \int_{0}^{2} (3 - (x^2 - 2x + 3)) dx = \int_{0}^{2} (2x - x^2) dx$.Evaluating the integral: $[x^2 - \frac{x^3}{3}]_{0}^{2} = (4 - \frac{8}{3}) - 0 = \frac{12 - 8}{3} = \frac{4}{3}$.

Similar Questions

The area (in square units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

Find the area of the region bounded by $y^{2}=9x$,$x=2$,$x=4$ and the $x$-axis in the first quadrant.

The area of the region bounded by the curve $y=x^3$, its tangent at $(1,1)$, and the $X$-axis is

The area between the curve $y^2 (a + x) = (a - x)^3$ and its vertical asymptote is

The area (in sq. units) of the smaller part of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module