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If \[f(x)\] is monotonically increasing function \[\forall x \in R\] , \[{f^{''}}(x) > 0\] and \[{f^{ - 1}}(x)\] exists, then \[\dfrac{{{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}\dfrac{{({x_1} + {x_2} + {x_3})}}{3}\] , \[{x_1} \ne {x_2} \ne {x_3}\] is true/false ?

Answer Verified Verified
Hint: While working with monotonically increasing types of questions, always arrange the elements in increasing order, because \[f(x)\] increases as \[x\] increases in a monotonically increasing function. Also when inverse of a function exists it also follows the same rule i.e. as \[x\] increases \[ \Rightarrow \] \[f(x)\] increases \[ \Rightarrow \]\[{f^{ - 1}}(x)\] increases.

Complete step-by-step answer:
Given, \[f(x)\] is monotonically increasing function \[\forall x \in R\]
\[ \Rightarrow f(x)\] increases as \[x\] increases \[...(i)\]
Considering three numbers \[{x_1},{x_2},{x_3}\] such that \[{x_1} \ne {x_2} \ne {x_3}\]
Since \[f(x)\] is monotonically increasing function \[\forall x \in R\]
Then \[f(x)\] is monotonically increasing function for \[{x_1},{x_2},{x_3} \in R\]
Monotonically increasing function implies from equation \[(i)\] that
\[{x_1} < {x_2} < {x_3} \Rightarrow f({x_1}) < f({x_2}) < f({x_3})\] \[...(ii)\]
(Here we are not taking the case of \[{x_1} \leqslant {x_2} \leqslant {x_3}\] because we are clearly given \[{x_1} \ne {x_2} \ne {x_3}\]
Clearly \[{x_1} + {x_2} + {x_3} > {x_1}\]
               \[{x_1} + {x_2} + {x_3} > {x_2}\]
               \[{x_1} + {x_2} + {x_3} > {x_3}\] \[...(iii)\]
 Therefore, dividing equation \[(iii)\] by \[3\] on both sides , we get
               \[\dfrac{{{x_1} + {x_2} + {x_3}}}{3} > \dfrac{{{x_1}}}{3}\]
               \[\dfrac{{{x_1} + {x_2} + {x_3}}}{3} > \dfrac{{{x_2}}}{3}\]
               \[\dfrac{{{x_1} + {x_2} + {x_3}}}{3} > \dfrac{{{x_3}}}{3}\] \[...(iv)\]
Since, \[f(x)\] is monotonically increasing function \[\forall x \in R\] , also we know \[{f^{ - 1}}(x)\] exists
Therefore, \[{f^{ - 1}}(x)\] is also monotonically increasing function \[\forall x \in R\]
i.e. \[{x_1} < {x_2} < {x_3} \Rightarrow {f^{ - 1}}({x_1}) < {f^{ - 1}}({x_2}) < {f^{ - 1}}({x_3})\] \[...(v)\]
Taking inverse function with the help of equation \[(v)\] on both sides of equation \[(iv)\];
\[\dfrac{{{f^{ - 1}}({x_1})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
\[\dfrac{{{f^{ - 1}}({x_2})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
\[\dfrac{{{f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\] \[...(vi)\]
Also we have
\[\dfrac{{{f^{ - 1}}({x_1})}}{3} < {f^{ - 1}}({x_1})\]
\[\dfrac{{{f^{ - 1}}({x_2})}}{3} < {f^{ - 1}}({x_2})\]
\[\dfrac{{{f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}({x_3})\] \[...(vii)\]
From equation \[(vi)\]and equation \[(vii)\], we get
\[\dfrac{{{f^{ - 1}}({x_1})}}{3} + \dfrac{{{f^{ - 1}}({x_2})}}{3} + \dfrac{{{f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3}) < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right) + {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right) + {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
i.e. \[{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3}) < 3{f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
i.e. \[\dfrac{{{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
Therefore the given value is TRUE.

NOTE: In these types of questions we always try to find a greater than or less than type of relation between \[x\] and \[f(x)\] . Sometimes a common mistake i.e. adding the values of equation \[(vi)\] straight causes a problem
i.e. \[\dfrac{{{f^{ - 1}}({x_1})}}{3} + \dfrac{{{f^{ - 1}}({x_2})}}{3} + \dfrac{{{f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right) + {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right) + {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
i.e. \[\dfrac{{{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3})}}{3} < 3 \times {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
i.e. \[\dfrac{{{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3})}}{{3 \times 3}} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
i.e. \[\dfrac{{{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3})}}{9} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\]
Which is not what we are solving for in the above question.
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