Question

If the mapping $f\left(x\right)=ax+b,a>0$ maps $[-1,1]$ onto $[0,2]$ then $\cot [{\cot }^{-1}7+{\cot }^{-1}8+{\cot }^{-1}18]$ is equal to

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

Answer 1

The correct option is D $f\left(1\right)+1$

$f(-1)=0\Rightarrow -a+b=0f\left(1\right)=2\Rightarrow a+b=2\Rightarrow b=1,a=1\Rightarrow f\left(x\right)=x+1\cot [{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{18}]=\cot [{\tan }^{-1}\left(\frac{15}{55}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)]=\cot \left[{\tan }^{-1}\left(\frac{65}{195}\right)\right]=\cot \left[{\tan }^{-1}\left(\frac{1}{3}\right)\right]=3=f\left(1\right)+1$