Question

Find all the values of ${}^{\prime }{a}^{\prime }(a\ne 0)$ for which the equation : ${\int }_0^x(t^2-8t+13)dt=xsin\frac{a}{x}$ has a solution . Find that solution .

• A. $a=\pi (4n+1),nϵz$
• B. $a=3\pi (4n+1),nϵz$
• C. $a=3\pi (2n+1),nϵz$
• D. $a=3\pi (3n+1),nϵz$

Answer 1

The correct option is B $a=3\pi (4n+1),nϵz$

${\int }_0^x(t^2-8t+13)dt=\frac{x^3}{3}-{4x}^2+13x=x.\sin \left(\frac{a}{x}\right)$
$\Rightarrow x^2-12x+39=3\sin \left(\frac{a}{x}\right)\Rightarrow {(x-6)}^2+3=3\sin \left(\frac{a}{x}\right)$
Max value of $L.H.S=3$ and max of $R.H.S=R$
So solution exists only when
${(x-6)}^2+3=3$ and $3\sin \left(\frac{a}{x}\right)=3$
$\Rightarrow x=6$ and $\sin \left(\frac{a}{x}\right)=1$
$\Rightarrow a=3\pi (4n+1)$