Let P(x) be a real polynomial of degree 3 whi | vedclass

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

Question

(B) Let $P'(x) = k(x - 1)(x + 1) = k(x^2 - 1)$ for some constant $k$.Integrating $P'(x)$,we get $P(x) = k(\frac{x^3}{3} - x) + C$.Given $P(-3) = 0$,we

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $P'(x) = k(x - 1)(x + 1) = k(x^2 - 1)$ for some constant $k$.Integrating $P'(x)$,we get $P(x) = k(\frac{x^3}{3} - x) + C$.Given $P(-3) = 0$,we have $k(\frac{-27}{3} - (-3)) + C = 0$,which simplifies to $k(-9 + 3) + C = 0$,so $-6k + C = 0$,or $C = 6k$.Given $\int_{-1}^{1} P(x) dx = 18$,we calculate $\int_{-1}^{1} (k(\frac{x^3}{3} - x) + C) dx = 18$.Since $k(\frac{x^3}{3} - x)$ is an odd function,its integral over $[-1, 1]$ is $0$. Thus,$\int_{-1}^{1} C dx = 18$,which gives $2C = 18$,so $C = 9$.Using $C = 6k$,we find $9 = 6k$,so $k = \frac{9}{6} = \frac{3}{2}$.Thus,$P(x) = \frac{3}{2}(\frac{x^3}{3} - x) + 9 = \frac{1}{2}x^3 - \frac{3}{2}x + 9$.The sum of the coefficients of $P(x)$ is $P(1) = \frac{1}{2}(1)^3 - \frac{3}{2}(1) + 9 = \frac{1}{2} - \frac{3}{2} + 9 = -1 + 9 = 8$.

Let $P(x)$ be a real polynomial of degree $3$ which vanishes at $x = -3$. Let $P(x)$ have local minima at $x = 1$,local maxima at $x = -1$,and $\int_{-1}^{1} P(x) dx = 18$. Then the sum of all the coefficients of the polynomial $P(x)$ is equal to ....... .

Similar Questions

The number of integers $n$ for which $3x^3-25x+n=0$ has three real roots is

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$,then $p^{\prime}(0)$ is

The minimum value of $2x^2 + x - 1$ is

If $PQ$ and $PR$ are the two sides of a triangle,then the angle between them which gives the maximum area of the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module