Question

For $a>0$, if $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$, then $\cot \left[{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{5}\right]$=

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

Answer 1

Answer: (B)

We have,
$f(x)=ax+b$ is onto from $[-1,1]$ to $[0,2]$
$\because f(-1)=-a+b=0$
$f(1)=a+b=2$
From Eqs. (i) and (ii), we get
$a=1,b=1$
$\therefore f(x)=x+1$
Now, $\cot \left({\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{5}\right)$
$\cot \left[{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\left(\frac{\frac{1}{8}+\frac{1}{5}}{1-\frac{1}{8}\times \frac{1}{5}}\right)\right]$
$\cot \left[{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}\right]\Rightarrow \cot \left[{\tan }^{-1}\frac{\frac{1}{7}+\frac{1}{3}}{1-\frac{1}{7}\times \frac{1}{3}}\right]$
$\cot \left[{\tan }^{-1}\frac{1}{2}\right]$
$\Rightarrow \cot \left({\cot }^{-1}2\right)=2$
$f(1)=1+1=2$
$\therefore \cot \left({\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{5}\right)=f(1)$