Question

The equation $\int_0^x {\left( {{t^2} - 8t + 13} \right)dt = x\sin \left( {\dfrac{a}{x}} \right)} $ has a solution if $\sin \left( {\dfrac{a}{6}} \right)$ is equal to
1) 0
2) 1
3) 1/2
4) 2/3

Answer 1

Hint: The given equation contains a definite integral which can be calculated by using the properties of integration. On evaluating the definite integral, we can examine the equation so formed to evaluate any points on which a solution exists for the equation. Finally, on putting the solution point, the answer can be formulated.

Complete step by step answer:

The given equation $\int_0^x {\left( {{t^2} - 8t + 13} \right)dt = x\sin \left( {\dfrac{a}{x}} \right)} $ can be simplified by first integrating the L.H.S.
The L.H.S. of the given equation is $\int_0^x {\left( {{t^2} - 8t + 13} \right)dt} $. By using the property of integration, $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ we can integrate the L.H.S. of the equation.
$\int_0^x {\left( {{t^2} - 8t + 13} \right)dt} = \left[ {\dfrac{{{t^3}}}{3} - 8\dfrac{{{t^2}}}{2} + 13t} \right]_0^x$
On putting the limits on the solution of the integration, we can find the definite integration. The solution of integration $\dfrac{{{t^3}}}{3} - 8\dfrac{{{t^2}}}{2} + 13t$ calculated at $t = 0$ is subtracted from the integral value calculated at $t = x$.
$
  \left[ {\dfrac{{{t^3}}}{3} - 4{t^2} + 13t} \right]_0^x = {\left[ {\dfrac{{{t^3}}}{3} - 4{t^2} + 13t} \right]_{t = x}} - {\left[ {\dfrac{{{t^3}}}{3} - 4{t^2} + 13t} \right]_{t = 0}} \\
   = \dfrac{{{x^3}}}{3} - 4{x^2} + 13x \\
$
Thus the given equation becomes
$\dfrac{{{x^3}}}{3} - 4{x^2} + 13x = x\sin \left( {\dfrac{a}{x}} \right)$
$\dfrac{{{x^2}}}{3} - 4x + 13 = \sin \left( {\dfrac{a}{x}} \right)$
On simplifying the equation becomes
${x^2} - 12x + 39 = 3\sin \left( {\dfrac{a}{x}} \right)$
The L.H.S. ${x^2} - 12x + 39$ can be further broken into sum of two non-negative parts, that is
$
  {x^2} - 12x + 39 = {x^2} - 12x + 36 + 3 \\
   = {\left( {x - 6} \right)^2} + 3 \\
 $
Thus the given equation is transformed to
${\left( {x - 6} \right)^2} + 3 = 3\sin \left( {\dfrac{a}{x}} \right)$
On observation we can say that , Since the L.H.S. is made up of two non-negative parts, the minimum value of L.H.S. is 3 for $x = 6$, so that ${\left( {x - 6} \right)^2}$ becomes 0.
Also, the range of the $\sin $function varies from $ - 1$ to 1, the maximum value of the R.H.S. $3\sin \left( {\dfrac{a}{6}} \right)$ will be 3.
Thus the given equation has a solution at $x = 6$. Substituting 6 for $x$ in the equation ${\left( {x - 6} \right)^2} + 3 = 3\sin \left( {\dfrac{a}{x}} \right)$, we get
$
  {\left( {6 - 6} \right)^2} + 3 = 3\sin \left( {\dfrac{a}{6}} \right) \\
  3 = 3\sin \left( {\dfrac{a}{6}} \right) \\
  \sin \left( {\dfrac{a}{6}} \right) = 1 \\
$
Thus the value of \[\sin \left( {\dfrac{a}{6}} \right)\] is 1.
The correct option is (2) which is 1.

Note: Another alternative for the solution can be simply substituting the value 6 for $x$ in the given equation and solving the formed equation for \[\sin \left( {\dfrac{a}{6}} \right)\]. It is important to note that the range of function must be properly understood while formulating the answer.