Question

Let $f\left(x\right)$ is a real valued function defined by
$f\left(x\right)=x^2+x^2{\int }_{-1}^{1}t.f\left(t\right)dt+x^3{\int }_{-1}^{1}f\left(t\right)dt$, then which of the following hold(s) good?

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

Let

${\int }_{-1}^{1}tf\left(t\right)dt=a,$ and ${\int }_{-1}^{1}f\left(t\right)dt=b$

$\Rightarrow f\left(x\right)=x^2+a{x}^2+b{x}^3$

$\Rightarrow f\left(t\right)=t^2(1+a)+b{t}^3$

$\Rightarrow tf\left(t\right)=t^3(1+a)+b{t}^4$
Now

$a={\int }_{-1}^{1}tf\left(t\right)dt={\int }_{-1}^1(t^3(1+a)+b{t}^4)dt$

$=2{\left[\frac{b{t}^5}{5}\right]}_0^1=\frac{2b}{5}$ ...(1)
(using property of integral of odd and even function).

Similarly $b={\int }_{-1}^{1}f\left(t\right)dt={\int }_{-1}^1(t^2(1+a)+b{t}^3)dt$

$=2{\left[\frac{(1+a)t^3}{3}\right]}_0^1=\frac{2}{3}(1+a)$ ....(2)
Solving the equations

$a=\frac{2b}{5}$ and $b=\frac{2(1+a)}{3}$ simultaneously, we get

$a=\frac{4}{11}$ and $b=\frac{10}{11}$

Hence, options A and C are incorrect.
Substituting $x=1$ in the original equation we get,
$f\left(1\right)=1+a+b$ ..(3)
Similarly, substituting $x=-1$, we get
$f(-1)=1+a-b$ ...(4)
Adding $\left(3\right)$ and $\left(4\right)$ we get ,
$f\left(1\right)+f(-1)=2(1+a)=2\times \frac{15}{11}=\frac{30}{11}$
Similarly subtracting $\left(4\right)$ from $\left(3\right)$ we get,
$f\left(1\right)-f(-1)=2b=\frac{20}{11}$

Hence, options B and D are correct.