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Question: If \[f(x)\] is monotonically increasing function \[\forall x \in R\] , \[{f^{''}}(x) > 0\] and \[{f^...

If f(x)f(x) is monotonically increasing function ∀x∈R\forall x \in R , f′′(x)>0{f^{''}}(x) > 0 and f−1(x){f^{ - 1}}(x) exists, then f−1(x1)+f−1(x2)+f−1(x3)3<f−1(x1+x2+x3)3\dfrac{{{f^{ - 1}}({x_1}) + {f^{ - 1}}({x_2}) + {f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}\dfrac{{({x_1} + {x_2} + {x_3})}}{3} , x1≠x2≠x3{x_1} \ne {x_2} \ne {x_3} is true/false ?

Explanation

Solution

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

Complete step-by-step answer:
Given, f(x)f(x) is monotonically increasing function ∀x∈R\forall x \in R
⇒f(x)\Rightarrow f(x) increases as xx increases ...(i)...(i)
Considering three numbers x1,x2,x3{x_1},{x_2},{x_3} such that x1≠x2≠x3{x_1} \ne {x_2} \ne {x_3}
Since f(x)f(x) is monotonically increasing function ∀x∈R\forall x \in R
Then f(x)f(x) is monotonically increasing function for x1,x2,x3∈R{x_1},{x_2},{x_3} \in R
Monotonically increasing function implies from equation (i)(i) that
x1<x2<x3⇒f(x1)<f(x2)<f(x3){x_1} < {x_2} < {x_3} \Rightarrow f({x_1}) < f({x_2}) < f({x_3}) ...(ii)...(ii)
(Here we are not taking the case of x1⩽x2⩽x3{x_1} \leqslant {x_2} \leqslant {x_3} because we are clearly given x1≠x2≠x3{x_1} \ne {x_2} \ne {x_3}
Clearly x1+x2+x3>x1{x_1} + {x_2} + {x_3} > {x_1}
x1+x2+x3>x2{x_1} + {x_2} + {x_3} > {x_2}
x1+x2+x3>x3{x_1} + {x_2} + {x_3} > {x_3} ...(iii)...(iii)
Therefore, dividing equation (iii)(iii) by 33 on both sides , we get
x1+x2+x33>x13\dfrac{{{x_1} + {x_2} + {x_3}}}{3} > \dfrac{{{x_1}}}{3}
x1+x2+x33>x23\dfrac{{{x_1} + {x_2} + {x_3}}}{3} > \dfrac{{{x_2}}}{3}
x1+x2+x33>x33\dfrac{{{x_1} + {x_2} + {x_3}}}{3} > \dfrac{{{x_3}}}{3} ...(iv)...(iv)
Since, f(x)f(x) is monotonically increasing function ∀x∈R\forall x \in R , also we know f−1(x){f^{ - 1}}(x) exists
Therefore, f−1(x){f^{ - 1}}(x) is also monotonically increasing function ∀x∈R\forall x \in R
i.e. x1<x2<x3⇒f−1(x1)<f−1(x2)<f−1(x3){x_1} < {x_2} < {x_3} \Rightarrow {f^{ - 1}}({x_1}) < {f^{ - 1}}({x_2}) < {f^{ - 1}}({x_3}) ...(v)...(v)
Taking inverse function with the help of equation (v)(v) on both sides of equation (iv)(iv);
f−1(x1)3<f−1(x1+x2+x33)\dfrac{{{f^{ - 1}}({x_1})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)
f−1(x2)3<f−1(x1+x2+x33)\dfrac{{{f^{ - 1}}({x_2})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)
f−1(x3)3<f−1(x1+x2+x33)\dfrac{{{f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right) ...(vi)...(vi)
Also we have
f−1(x1)3<f−1(x1)\dfrac{{{f^{ - 1}}({x_1})}}{3} < {f^{ - 1}}({x_1})
f−1(x2)3<f−1(x2)\dfrac{{{f^{ - 1}}({x_2})}}{3} < {f^{ - 1}}({x_2})
f−1(x3)3<f−1(x3)\dfrac{{{f^{ - 1}}({x_3})}}{3} < {f^{ - 1}}({x_3}) ...(vii)...(vii)
From equation (vi)(vi)and equation (vii)(vii), we get
f−1(x1)3+f−1(x2)3+f−1(x3)3<f−1(x1)+f−1(x2)+f−1(x3)<f−1(x1+x2+x33)+f−1(x1+x2+x33)+f−1(x1+x2+x33)\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(x1)+f−1(x2)+f−1(x3)<3f−1(x1+x2+x33){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. f−1(x1)+f−1(x2)+f−1(x3)3<f−1(x1+x2+x33)\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 xx and f(x)f(x) . Sometimes a common mistake i.e. adding the values of equation (vi)(vi) straight causes a problem
i.e. f−1(x1)3+f−1(x2)3+f−1(x3)3<f−1(x1+x2+x33)+f−1(x1+x2+x33)+f−1(x1+x2+x33)\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. f−1(x1)+f−1(x2)+f−1(x3)3<3×f−1(x1+x2+x33)\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. f−1(x1)+f−1(x2)+f−1(x3)3×3<f−1(x1+x2+x33)\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. f−1(x1)+f−1(x2)+f−1(x3)9<f−1(x1+x2+x33)\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.