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Question: For what values of \(a,m\) and \(b\) Lagrange’s mean value theorem is applicable to the function \(f...

For what values of a,ma,m and bb Lagrange’s mean value theorem is applicable to the function f(x)f\left( x \right) for x∈[0,2]x \in \left[ {0,2} \right]
f\left( x \right) = \left\\{ {\begin{array}{*{20}{c}} 3&{x = 0}&{} \\\ { - {x^2} + a}&{0 < x < 1}&{} \\\ {mx + b}&{1 \leqslant x \leqslant 2}&{} \end{array}} \right.
I. a=3;m=2;b=0a = 3;m = 2;b = 0
II. a=3;m=−2;b=4a = 3;m = - 2;b = 4
III. a=3;m=2;b=1a = 3;m = 2;b = 1
IV. No such a,m,ba,m,b exist

Explanation

Solution

Hint : The given question is related to the concept of continuation and differentiation. Here, in the given question we have to use Lagrange's mean value theorem to find our required solution. We will begin by finding the value of f′(x)f'\left( x \right) and then substituting x=cx = c. Next, we will find the value of f′(c)f'\left( c \right) from Lagrange’s mean value theorem. At last, we will compare the values of f′(c)f'\left( c \right) from both of the equations to find the value of cc.

Complete step by step solution:
According to Lagrange’s mean value theorem, if a function f(x)f\left( x \right) is continuous in the interval [a,b]\left[ {a,b} \right] and is differentiable in (a,b)\left( {a,b} \right) then, there will be at least one point cc in the interval [a,b]\left[ {a,b} \right] such that
f′(c)=f(b)−f(a)b−af'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}
Given is f\left( x \right) = \left\\{ {\begin{array}{*{20}{c}} 3&{x = 0}&{} \\\ { - {x^2} + a}&{0 < x < 1}&{} \\\ {mx + b}&{1 \leqslant x \leqslant 2}&{} \end{array}} \right.
We have to find the values of a,m,ba,m,b.
Using Lagrange’s mean value theorem,
Let f:[a,b]→Rf:\left[ {a,b} \right] \to R be a continuous function and differentiable on (a,b)\left( {a,b} \right). Then, there exists c∈(a,b)c \in \left( {a,b} \right) such that f′(c)=f(b)−f(a)b−af'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}
Hence, f(x)f\left( x \right) is continuous
f(0)=3=a f(1)=2=m+b   f\left( 0 \right) = 3 = a \\\ f\left( 1 \right) = 2 = m + b \;
Hence,b=4b = 4
f(x)f\left( x \right)is differentiable at 11.
So, f′(1)=−2=mf'\left( 1 \right) = - 2 = m
Therefore, a=3;m=−2;b=4a = 3;m = - 2;b = 4
Hence, the correct option is (B).
So, the correct answer is “Option B”.

Note : Before applying the Lagrange’s mean value theorem, the students should check if the given function satisfies all the necessary conditions such as, (i) the function is continuous on (a,b)\left( {a,b} \right) (ii) the function is differentiable on (a,b)\left( {a,b} \right). If all these conditions apply on the given function, then only we should apply the Lagrange’s mean value theorem. The most common mistake which students tend to make is they don’t check these conditions and end up with the wrong solution.