Question

Let $f$ be a real valued function defined as $f\left(x\right)=x^2+x^2\underset{-1}{\overset{1}{\int }}t\cdot f\left(t\right)dt+x^3\underset{-1}{\overset{1}{\int }}f\left(t\right)dt.$ Then which of the following hold(s) good ?

• A. $\underset{-1}{\overset{1}{\int }}t\cdot f\left(t\right)dt=\frac{10}{11}$
• B. $f\left(1\right)+f(-1)=\frac{30}{11}$
• C. $\underset{-1}{\overset{1}{\int }}t\cdot f\left(t\right)dt>\underset{-1}{\overset{1}{\int }}f\left(t\right)dt$
• D. $f\left(1\right)-f(-1)=\frac{20}{11}$

Answer 1

Answer: (B), (D)

The correct options are
B $f\left(1\right)+f(-1)=\frac{30}{11}$
D $f\left(1\right)-f(-1)=\frac{20}{11}$

We have $f\left(x\right)=x^2+a{x}^2+b{x}^3$
where $a=\underset{-1}{\overset{1}{\int }}t\cdot f\left(t\right)dt$ and $b=\underset{-1}{\overset{1}{\int }}f\left(t\right)dt$

Now, $a=\underset{-1}{\overset{1}{\int }}t\left[\right(a+1)t^2+b{t}^3]dt$
$\Rightarrow a=(a+1)\underset{-1}{\overset{1}{\int }}{t}^{3}dt+b\underset{-1}{\overset{1}{\int }}{t}^{4}dt$
$\Rightarrow a=0+2b\underset{0}{\overset{1}{\int }}{t}^{4}dt$
$\Rightarrow a=\frac{2b}{5}\dots \left(1\right)$

Again $b=\underset{-1}{\overset{1}{\int }}\left(\right(a+1)t^2+b{t}^3)dt$
$\Rightarrow b=2\underset{0}{\overset{1}{\int }}(a+1)t^{2}dt$
$\Rightarrow b=\frac{2(a+1)}{3}\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$\frac{5a}{2}=\frac{2(a+1)}{3}$
$\Rightarrow a=\frac{4}{11}$ and $b=\frac{10}{11}$

$\therefore \underset{-1}{\overset{1}{\int }}t\cdot f\left(t\right)dt=\frac{4}{11}$ and $\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=\frac{10}{11}$

$f\left(x\right)=(a+1)x^2+b{x}^3$
Now, $f\left(1\right)+f(-1)=2(a+1)=\frac{30}{11}$
And $f\left(1\right)-f(-1)=2b=\frac{20}{11}$