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5. Continuity and Differentiation
medium
If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is
A
$c = 0$
B
$c = \frac{1}{2}$
C
$c = \frac{1}{4}$
D
$c = 1$
Solution
(b) Here $\frac{{f(b) – f(a)}}{{b – a}} = f'(c)$
==> $\frac{{1 – 0}}{{1 – 0}} = 2 – 2c$ $\left\{ \begin{array}{l} \because b = 1,a = 0\\ \Rightarrow f(1) = 1,\,f(0) = 0\end{array} \right\}$
==> $ – 2c = – 1$ ==> $c = \frac{1}{2}$ $\left\{ \begin{array}{l} \because f'(x) = 2 – 2x\\f'(c) = 2 – 2c\end{array} \right\}$
Standard 12
Mathematics