Definite Integration Ques 75
- A cubic $f(x)$ vanishes at $x=-2$ and has relative minimum / maximum at $x=-1$ and $x=1 / 3$.
If $\int _{-1}^{1} f(x) d x=14 / 3$, find the cubic $f(x)$.
$(1992,4$ M)
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Answer:
Correct Answer: 75.$f(x)=x^{3}+x^{2}-x+2$
Solution:
- Since, $f(x)$ is a cubic polynomial. Therefore, $f^{\prime}(x)$ is a quadratic polynomial and $f(x)$ has relative maximum and minimum at $x=\frac{1}{3}$ and $x=-1$ respectively, therefore, -1 and $1 / 3$ are the roots of $f^{\prime}(x)=0$.
$$ \begin{aligned} \therefore \quad f^{\prime}(x) & =a(x+1) \quad x-\frac{1}{3}=a \quad x^{2}-\frac{1}{3} x+x-\frac{1}{3} \\ & =a x^{2}+\frac{2}{3} x-\frac{1}{3} \end{aligned} $$
Now, integrating w.r. t. $x$, we get
$$ f(x)=a \frac{x^{3}}{3}+\frac{x^{2}}{3}-\frac{x}{3}+c $$
where, $c$ is constant of integration.
Again, $f(-2)=0$
$$ \begin{aligned} & & f(-2) & =a-\frac{8}{3}+\frac{4}{3}+\frac{2}{3}+c \\ & & 0 & =a \frac{-8+4+2}{3}+c \\ & \Rightarrow & 0 & =\frac{-2 a}{3}+c \Rightarrow c=\frac{2 a}{3} \\ & \therefore & f(x) & =a \frac{x^{3}}{3}+\frac{x^{2}}{3}-\frac{x}{3}+\frac{2 a}{3}=\frac{a}{3}\left(x^{3}+x^{2}-x+2\right) \end{aligned} $$
Again, $\int _{-1}^{1} f(x) d x=\frac{14}{3}$
[given]
$$ \begin{aligned} & \Rightarrow \quad \int _{-1}^{1} \frac{a}{3}\left(x^{3}+x^{2}-x+2\right) d x=\frac{14}{3} \\ & \Rightarrow \quad \int _{-1}^{1} \frac{a}{3}\left(0+x^{2}+0+2\right) d x=\frac{14}{3} \end{aligned} $$
$\left[\because y=x^{3}\right.$ and $y=-x$ are odd functions $]$
$\Rightarrow \frac{a}{3} 2 \int _0^{1} x^{2} d x+4 \int _0^{1} 1 d x=\frac{14}{3}$
$\Rightarrow \quad \frac{a}{3} \quad \frac{2 x^{3}}{3}+4 x \quad=\frac{14}{3}$
$$ \begin{array}{rlrlrl} \Rightarrow & & \frac{a}{3} \frac{2}{3}+4 & =\frac{14}{3} \Rightarrow & \frac{a}{3} \frac{14}{3}=\frac{14}{3} \\ \Rightarrow & a & =3 \end{array} $$
Hence,
$$ f(x)=x^{3}+x^{2}-x+2 $$
BETA
