Question

The equation ${\int }_0^x(t^2-8t+13)dt=xsin\left(a/x\right)$ has a solution if sin (a/6) is equal to

• A. 0
• B. 1
• C. 1/2
• D. 2/3

Answer 1

The correct option is B 1

After integrating above equation,
$x^3/3-4x^2+13x=xsin\left(a/x\right)$
$\Rightarrow x^2-12x+39=3sin\left(a/x\right)$, since $x=0$ is not in domain
$\Rightarrow (x-6{)}^2+3=3sin(a/x)$
Now maximum possible value of RHS is 3 and minimum possible value of LHS is 3
Therefore, $x=6$ is the only possible value
thus $sin\left(a/6\right)=1$