Question

Left hand derivative and right hand derivative of a function f (x) at a point x = a, are defined as$f'\left( {{a^ - }} \right) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f\left( a \right) - f\left( {a - h} \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f\left( a \right) - f\left( {a - h} \right)}}{h} = \mathop {\lim }\limits_{x \to {a^ + }} \dfrac{{f\left( a \right) - f\left( x \right)}}{{a - x}}$ respectively.
Let f be a twice differentiable function. We also know that the derivative of an even function is an odd function and the derivative of an odd function is an even function.
The statement $\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( { - x} \right) - f\left( { - x - h} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( x \right) - f\left( {x - h} \right)}}{{ - h}}$ implies that for all $x \in R$
$\left( a \right)$ f is odd
$\left( b \right)$ f is even
$\left( c \right)$ f is neither odd nor even
$\left( d \right)$ Nothing can be concluded.

Answer 1

Hint: In this particular question use the concept that for function to be odd f (-x) = -f (x) and for function to be even f (-x) = f (x) and use the concept if f’ (x) i.e. differentiation of f (x) is odd than f (x) is even or vice versa so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given statement
$\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( { - x} \right) - f\left( { - x - h} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( x \right) - f\left( {x - h} \right)}}{{ - h}}$
Let, f’(x) = $\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( { - x} \right) - f\left( { - x - h} \right)}}{h}$ ......................... (1)
And f’(x) = $\mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( x \right) - f\left( {x - h} \right)}}{{ - h}} = - \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( x \right) - f\left( {x - h} \right)}}{h}$................. (2)
Therefore, $f'\left( x \right) = f'\left( x \right)$............... (3)
Now substitute in place of x, -x we have,
$ \Rightarrow f'\left( { - x} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( x \right) - f\left( {x - h} \right)}}{h}$
Now from equation (2) we can say
$ \Rightarrow f'\left( { - x} \right) = - f'\left( x \right)$.............. (4)
Now as we know that for function to be odd f (-x) = -f (x) and for function to be even f (-x) = f (x).
So from equation (3) and (4) f’ (x) is an odd function.
Now as we all know that if f’ (x) i.e. differentiation of f (x) is odd then f (x) is even or vice versa.
So f’(x) is an odd function, therefore f (x) is an even function.
So this is the required answer.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the condition of odd function as well as even function which is stated above, and always recall the property whether it is given or not it is always true, that if a function is odd then the differentiation of the function is an even function or vice versa.