Question

Let $f\left(x\right)=x+1$ where $f$ is defined from $[-1,1]$ onto $[0,2]$ then the value of $\cot \left({\cot }^{-1}\right(1)+{\cot }^{-1}(2)+{\cot }^{-1}(3\left)\right)$ equals

• A. $f(-1)$
• B. $f\left(0\right)$
• C. $f\left(1\right)$
• D. $f\left(1\right)-f\left(0\right)$

Answer 1

Answer: (A)

$f(-1)$

Since $f\left(x\right)$ is a linear function from $[-1,1]$ onto $[0,2]$

Hence $f(-1)=0$ and $f\left(1\right)=2$.

$cot\left(co{t}^{-1}\right(1)+co{t}^{-1}(2)+co{t}^{-1}(3\left)\right)$

$=cot(\frac{\pi }{4}+co{t}^{-1}\left(\frac{\left(3\right)\left(2\right)-1}{3+2}\right))$

$=cot(\frac{\pi }{4}+co{t}^{-1}(1\left)\right)$

$=cot\left(\frac{\pi }{2}\right)$

$=0$$=f(-1)$