Question

Let $f\left(x\right)=\frac{3x+2}{\sqrt{x-9}}$ and it is primitive of $F\left(x\right)$ with respect to $x$. If $F\left(10\right)=60$ then twice of sum of digits of the value of $F\left(13\right)$ is.

Answer 1

Consider the given function.

$f\left(x\right)=F^{\prime }\left(x\right)=\frac{3x+2}{\sqrt{x-9}}$

Therefore,

$F\left(x\right)=\int \frac{3x+2}{\sqrt{x-9}}dx$

Let $t=x-9$

$dt=dx$

Put the value of $t$ in above expression,we get

$F\left(x\right)=\int \frac{3(t+9)+2}{\sqrt{t}}dt$

$F\left(x\right)=\int \frac{3t+29}{\sqrt{t}}dt$

$F\left(x\right)=\int (3\sqrt{t}+\frac{29}{\sqrt{t}})dt$

$F\left(x\right)=3\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+29\left(2\sqrt{t}\right)+CF\left(x\right)=2t^{\frac{3}{2}}+58\sqrt{t}+C$

Put the value of $t$ in above expression, we get

$F\left(x\right)=2{(x-9)}^{\frac{3}{2}}+58\sqrt{x-9}+C$

$F\left(x\right)=2(x-9)\times \sqrt{x-9}+58\sqrt{x-9}+C$

$F\left(x\right)=2\sqrt{(x-9)}[x-9+29]+C$

$F\left(x\right)=2\sqrt{(x-9)}[x+20]+C$ ......... (1)

Since, $F\left(10\right)=60$

Therefore,

$60=2\sqrt{(10-9)}[10+20]+C$

$60=60+C$

$C=0$

Now,

$F\left(13\right)=2\sqrt{(13-9)}[13+20]+0$

$F\left(13\right)=2\sqrt{\left(4\right)}[33]$

$F\left(13\right)=4\times 33$

$F\left(13\right)=132$

According to the question,

$=2\times (1+3+2)$

$=2\times 6$

$=12$

Hence, $12$ is the required value.