If f(x) = ( ^ - 1 ( ( ^ - 1 x ) ) ) + ( ^ - 1 ( ( ^ - 1 x ) ) ) h… - Vidyadip

MCQ

If $f(x) = \cos \left( {{{\tan }^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}x} \right)} \right)} \right) + \sin \left( {{{\cot }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}x} \right)} \right)} \right)$ has range $\left[ {m,M),} \right.$ then number of solutions of the equation $\operatorname{sgn} \left( {\left| {x - 1} \right| - 2} \right) = \ln \left| {x - 2} \right|$ is (where sgn denotes signum function)

A
$m^2+1$
B
$m^2-M$
C
$M^2+1$
✓
$m^2+M$

✓

Answer

Correct option: D.

$m^2+M$

d
$f(x)=\cos \left(\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)$

$+\sin \left(\cot ^{-1}\left(\cos \left(\frac{\pi}{2}-\cos ^{-1} x\right)\right)\right)$

$=\cos \left(\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)+\sin \left(\cot ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)$

$=\cos \left(\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)+\sin \left(\frac{\pi}{2}-\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)$

$=\cos \left(\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)+\cos \left(\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)$

$=2 \cos \left(\tan ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)\right)=\frac{2}{\sqrt{2-x^{2}}} \in[\sqrt{2}, 2]$

From graph, Number of solutions of

Sgn $\left( {\left| {x - 1} \right| - 2} \right) = \ln \left| {x - 2} \right|$ is $4$

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