Question

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

• A. $f(-1)$
• B. $f(0)$
• C. $f(1)$
• D. $f(2)$

Answer 1

Answer: (D)

$f(x)=ax+b$
$f^{\prime }(x)=a>0$
$f(x)$ is an increasing function.
$\Rightarrow f(-1)=0$ and $f(1)=2$
or $-a+b=0$
and $a+b=2$
Then, $a=b=1$
$\Rightarrow f(x)=x+1$
Now, $\cot \left[{\cot }^{-1}7+{\cot }^{-1}8+{\cot }^{-1}18\right]$
$=\cot \left\{{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{{\tan }^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\cdot \frac{1}{8}}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{{\tan }^{-1}\left(\frac{15}{55}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{{\tan }^{-1}\left(\frac{3}{11}\right)+{\tan }^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{{\tan }^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{1.1}\cdot \frac{1}{18}}\right)\right\}$
$=\cot \left\{{\tan }^{-1}\left(\frac{65}{195}\right)\right\}$
$=\cot \left\{{\tan }^{-1}\left(\frac{1}{3}\right)\right\}$
$=\cot \left({\cot }^{-1}3\right)=3=1+2=f(2)$