Question

$f\left(x\right)$ is a continuous function for all real values of $x$ and satisfies ${\int }_0^{x}f\left(t\right)dt={\int }_x^1{t}^{2}f\left(t\right)dt+\frac{x^{16}}{8}+\frac{x^6}{3}+k$. The value of $k$ is

• A. $\frac{167}{840}$
• B. $-\frac{167}{840}$
• C. $\frac{17}{38}$
• D. None of these

Answer 1

Answer: (B)

$-\frac{167}{840}$

We have, ${\int }_0^{x}f\left(t\right)dt={\int }_x^1{t}^{2}f\left(t\right)dt+\frac{x^{16}}{8}+\frac{x^6}{3}+k$ ...(1)

For $x=1,{\int }_x^{1}f\left(t\right)dt=0+\frac{1}{8}+\frac{1}{3}+k=\frac{11}{24}+k$ ...(2)

Differentiating both sides of (1), w.r.t $x$, we get

$f\left(x\right)=-x^{2}f\left(x\right)+2x^{15}+2x^5\Rightarrow f\left(x\right)=\frac{2(x^{15}+x^5)}{1+x^2}$

$\therefore {\int }_0^{1}f\left(t\right)dt=2{\int }_0^1\frac{(t^{15}+t^5)}{1+t^2}=\frac{11}{24}+k$ (Using (2))

$\Rightarrow 2{\int }_0^1(t^{13}-t^{11}+t^9-t^7+t^5)dt=\frac{11}{24}+k$

$\Rightarrow 2(\frac{1}{14}-\frac{1}{12}+\frac{1}{10}-\frac{1}{8}+\frac{1}{6})=\frac{11}{24}+k$

$\Rightarrow k=-\frac{167}{840}$