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Q.
For $a>\,0$, if $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$, then $\cot \left[\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right]$=
TS EAMCET 2019
Solution:
We have,
$f(x)=a x+b$ 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$
$f(1)=1+1=2$
$\therefore \, \cot \left(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right)=f(1)$