Let $f x = x 3 − x 2 + x + 1$ and $g x$ be a function defined by $g x = max f t : 0 ≤ t ≤ x, 0 ≤ x ≤ 1 3 − x, 1 < x ≤ 2$ , then $g x$ is

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continuous and differentiable on

0, 2

continuous but not differentiable on

0, 2

Neither continuous nor differentiable on

0, 2

continuous and differentiable at x=1

answer is B.

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Detailed Solution

we have, $f (x) = x 3 − x 2 + x + 1 ⇒ f ′ (x) = 3 x 2 − 2 x + 1 > 0$ for all x

$∴ g (x) = \{\begin{cases} f (x), 0 \leq x \leq 1 \\ 3 - x, 1 < x \leq 2 \end{cases} a s f (x) i s i n c r e a \sin g f (x) i s m a x i m u m w h e n 0 \leq t \leq x ∴ g (x) = \{\begin{cases} x^{3} - x^{2} + x + 1, 0 \leq x \leq 1 \\ 3 - x, 1 < x \leq 2 \end{cases}$

$\lim_{x \to 1^{-}} g (x) = \lim_{x \to 1^{+}} g (x) = g (1)$ So, it is continuous at x = 1.

We observe that (LHD at x = 1) ${= \frac{d}{d x} (x^{3} - x^{2} + x + 1)\}}_{x = 1} = {(3 x^{2} - 2 x + 1)}_{x = 1} = 2$
and $(R H D a t x = 1) = \frac{d}{d x} {(3 - x)}_{x = 1} = - 1$