Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Since $f(x)=ax+b$ maps $[-1,1]$ onto $[0,2]$,

we have $f(-1)=0$ and $f(1)=2$. Thus, $-a+b=0$ and $a+b=2$, implying $a=1$ and $b=1$. Hence, $f(x)=x+1$.

Now, ${\cot }^{-1}7+{\cot }^{-1}8+{\cot }^{-1}18={\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{18}={\tan }^{-1}\frac{1}{3}$.

Therefore, $\cot [{\cot }^{-1}7+{\cot }^{-1}8+{\cot }^{-1}18]=3=f(2)=f(1)+1$.

Final Answer: The final answer is $\boxed{f(1)+1}$