Let a - 0 and f be continuous in -a- a- Suppose that f-x exists and f-x - 1 for all x - -a- a- If fa - a and f-a - -a then f0

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Theorems for Differentiability

Let a > 0 a...

Question

Let $a > 0$ and $f$ be continuous in $[- a, a]$ . Suppose that $f^{'} (x)$ exists and $f^{'} (x) \leq 1$ for all $x ϵ (- a, a)$ . If $f (a) = a$ and $f (- a) = - a$ then $f (0)$

equals

0

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equals

\frac{1}{2}

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equals

1

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is not possible to determine

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Solution

The correct option is B equals $0$
As $f$ is continuous in $[- a, a]$ and differential in $(- a, a)$
So using Lagrange's mean value theorem we get some $c ϵ (- a, a)$ such that $f^{^{'}} (c) = \frac{f (- a) - f (a)}{- a - a} = 1$
And $f (c) = c + d$ from $f (a) = a$ we get $d = 0$
Hence $f (c) = c f (0) = 0$
Hence, option 'A' is correct.

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Let a>0 and f be continuous in [−a,a]. Suppose that f′(x) exists and f′(x)≤1 for all xϵ(−a,a). If f(a)=a and f(−a)=−a then f(0)

The correct option is B equals 0As f is continuous in [−a,a] and differential in(−a,a)So using Lagrange's mean value theorem we get some cϵ(−a,a) such that f′(c)=f(−a)−f(a)−a−a=1And f(c)=c+d from f(a)=a we getd=0Hence f(c)=cf(0)=0Hence, option 'A' is correct.

Let $a > 0$ and $f$ be continuous in $[- a, a]$ . Suppose that $f^{'} (x)$ exists and $f^{'} (x) \leq 1$ for all $x ϵ (- a, a)$ . If $f (a) = a$ and $f (- a) = - a$ then $f (0)$