If p(x) is a cubic polynomial with p(1)=3, p( | vedclass

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

Question

(D) Let $p(x) = ax^3 + bx^2 + cx + d$.Given $p(0) = 2$,we have $d = 2$.Given $p(1) = a + b + c + d = 3 \Rightarrow a + b + c = 1$ $(i)$.Given $p(-1) =

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let $p(x) = ax^3 + bx^2 + cx + d$.Given $p(0) = 2$,we have $d = 2$.Given $p(1) = a + b + c + d = 3 \Rightarrow a + b + c = 1$ $(i)$.Given $p(-1) = -a + b - c + d = 4 \Rightarrow -a + b - c = 2$ $(ii)$.Adding $(i)$ and $(ii)$,we get $2b = 3$,so $b = \frac{3}{2}$.We need to evaluate $I = \int_{-1}^1 (ax^3 + bx^2 + cx + d) dx$.Since $ax^3$ and $cx$ are odd functions,their integral over $[-1, 1]$ is $0$.Thus,$I = \int_{-1}^1 (bx^2 + d) dx = 2 \int_0^1 (bx^2 + d) dx$.$I = 2 \left[ \frac{bx^3}{3} + dx \right]_0^1 = 2 \left( \frac{b}{3} + d \right)$.Substituting $b = \frac{3}{2}$ and $d = 2$:$I = 2 \left( \frac{3/2}{3} + 2 \right) = 2 \left( \frac{1}{2} + 2 \right) = 2 \left( \frac{5}{2} \right) = 5$.

If $p(x)$ is a cubic polynomial with $p(1)=3, p(0)=2$ and $p(-1)=4$,then $\int_{-1}^1 p(x) dx$ is

Similar Questions

$\int_0^3 |x^2 - 3x + 2| dx = $

$\int_0^{2\pi} e^{x/2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx = $

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{5}{2}}} d x=$

If $I = \int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$, then $\alpha$ lies in the interval

$\int_0^{10} (5 - \sqrt{10x - x^2}) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module