Let f(x) is a continuous function for all real values of x and satisfies $\int\limits_0^x f(t) d t=\int\limits_x^1 t^2 . f(t) d t+\frac{x^{16}}{8}+\frac{x^6}{3}+a+a$ then value of ‘a’ is equal to

None of these

$\int\limits_0^x f(t) d t=\int\limits_x^1 t^2 \cdot f(t) d t+\frac{x^{16}}{8}+\frac{x^6}{3}+a$       .....(i)

For $x=1, \int\limits_0^1 f(t) d t=0+\frac{1}{8}+\frac{1}{3}+a=\frac{11}{24}+a$

Diff. both sides of (i) w. r. t. x we get;

$f(x)=0-x^2 f(x)+2 x^{15}+2 x^5$

$\Rightarrow 2 \int\limits_0^1 \frac{x^{15}+x^5}{1+x^2} d x=\frac{11}{24}+a$

$\Rightarrow 2 \int\limits_0^1\left(x^{13}-x^{11}+x^9-x^7+x^5\right) d x=\frac{11}{24}+a$

$\Rightarrow 2 \int\limits_0^1\left(\frac{1}{14}-\frac{1}{12}+\frac{1}{10}-\frac{1}{8}+\frac{1}{6}\right)=\frac{11}{24}+a \Rightarrow a=-\frac{167}{840}$