Question

Let f(x) is a real valued function defined by $f\left(x\right)=x^2+x^2{\int }_{-1}^{1}tf\left(t\right)dt+x^3{\int }_{-1}^{1}f\left(t\right)dt$, then

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

$f\left(1\right)-f(-1)=\frac{20}{11}$

$f\left(x\right)=x^2+a{x}^2+b{x}^3=(a+1)x^2+b{x}^{3}a={\int }_{-1}^{1}tf\left(t\right)dt={\int }_{-1}^1(a+1)t^3+b{t}^{4}dt=2b{\int }_0^1{t}^{4}dt\therefore a=\frac{2b}{5}....\left(i\right)b={\int }_{-1}^{1}f\left(t\right)dt={\int }_{-1}^1(a+1)t^2+b{t}^{3}dt={\frac{(a+1)t^3}{3}|}_{-1}^1\Rightarrow b=(a+1)\frac{2}{3}......\left(ii\right)$
From (i)and (ii)
$a=\frac{4}{11}$ and $b=\frac{10}{11}$
$f\left(x\right)=\frac{15x^2}{11}+\frac{10x^3}{11}f\left(1\right)=\frac{25}{11}f(-1)=\frac{5}{11}$