Question

Let $f:[-1,1]\to [0,2]$ be a function defined by $f\left(x\right)=mx+c$ where $m>0.$ If $f$ is onto and $\tan ({\tan }^{-1}\frac{1}{7}+{\cot }^{-1}8+{\cot }^{-1}18)$ equals $f\left(a\right)$ for some $a\in [-1,1],$ then the value of $\left[a\right]+9$ is
($[.]$ denotes the greatest integer function)

Answer 1

Linear function $f\left(x\right)=mx+c$ is an onto function.
$f(-1)=0\Rightarrow c-m=0$
$f\left(1\right)=2\Rightarrow c+m=2$
Solving the above equations, we get
$m=1,c=1$
$\therefore f\left(x\right)=x+1$

Now, $\tan ({\tan }^{-1}\frac{1}{7}+{\cot }^{-1}8+{\cot }^{-1}18)$
$=\tan ({\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{18})$
$=\tan ({\tan }^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\times \frac{1}{8}}\right)+{\tan }^{-1}\frac{1}{18})$
$=\tan ({\tan }^{-1}\frac{3}{11}+{\tan }^{-1}\frac{1}{18})$
$=\tan \left({\tan }^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\times \frac{1}{18}}\right)\right)$
$=\tan \left({\tan }^{-1}\frac{1}{3}\right)=\frac{1}{3}=f\left(a\right)$ (given)

$\frac{1}{3}=a+1$
$\Rightarrow a=-\frac{2}{3}$
$\therefore \left[a\right]+9=-1+9=8$