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Question: Let a > 0 and f be continuous in [-a, a]. suppose that f’(x) exists and f’(x)\( \leqslant \)1 for al...

Let a > 0 and f be continuous in [-a, a]. suppose that f’(x) exists and f’(x)⩽\leqslant1 for all x∈(−a,a)x \in \left( { - a,a} \right). If f(a) = a and f(-a) = -a then f(0)
(A)\left( A \right) Equals 0
(B)\left( B \right) Equals (1/2)
(C)\left( C \right) Equals 1
(D)\left( D \right) It is not possible to determine.

Explanation

Solution

Hint – In this particular type of question use the concept that if a function is continuous in the given interval say [a, b] then there exists a point c between the interval such that f′(c)=f(b)−f(a)b−af'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} according to Lagrange’s mean value theorem so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given data:
a > 0
And f is a continuous function in [-a, a].
Now it is also given that f’(x) exists and f’(x)⩽\leqslant1 for all x∈(−a,a)x \in \left( { - a,a} \right).
And f(a) = a and f(-a) = -a............. (1)
Now we have to find out the value of f(0)
So f is continuous so according to Lagrange’s mean value theorem there exists a point c in between the limits of f such that,
f′(c)=f(b)−f(a)b−af'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}} Where, b and a are the final and initial limits of f.
Now substitute the values we have,
⇒f′(c)=f(−a)−f(a)−a−a\Rightarrow f'\left( c \right) = \dfrac{{f\left( { - a} \right) - f\left( a \right)}}{{ - a - a}}
Now from equation (1) we have,
⇒f′(c)=−a−a−a−a=−2a−2a=1\Rightarrow f'\left( c \right) = \dfrac{{ - a - a}}{{ - a - a}} = \dfrac{{ - 2a}}{{ - 2a}} = 1
Now integrate on both sides w.r.t c we have,
⇒∫f′(c)dc=∫1dc\Rightarrow \int {f'\left( c \right)dc} = \int {1dc}
Now the integration of any differentiation function (f’) gives (f) and integration of constant is given as,
∫1dc=c+K\int {1dc} = c + K, where K is some arbitrary integration constant so we have,
⇒∫f′(c)dc=∫1dc\Rightarrow \int {f'\left( c \right)dc} = \int {1dc}
⇒f(c)=c+K\Rightarrow f\left( c \right) = c + K.................. (2)
Now substitute in place of c, a we have,
⇒f(a)=a+K\Rightarrow f\left( a \right) = a + K
Now from equation (1), f(a) = a so we have,
⇒a=a+K\Rightarrow a = a + K
⇒K=0\Rightarrow K = 0
Now from equation (2) we have,
⇒f(c)=c\Rightarrow f\left( c \right) = c
Now substitute in place of c, 0 we have,
⇒f(0)=0\Rightarrow f\left( 0 \right) = 0
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of question the key concept we have to remember the basic integration property i.e. integration of any differentiation function (f’) gives (f) and integration of constant is given as,∫1dc=c+K\int {1dc} = c + K, where K is some arbitrary integration constant so first find out the function (f’) and then apply integration property and simplify as above we will get the required answer.