Let the function be defined by and let be an arbitrary function. Let be the product function defined by . Then number of correct statements among the following: A) If is discontinuous at , then can never be differentiable at . B) If is differen

Question

Let the function $f : R \to R$ be defined by $f (x) = x^{3} - x^{2} + (x - 1) \sin x$ and let $g : R \to R$ be an arbitrary function.

Let $f g : R \to R$ be the product function defined by $(f g) (x) = f (x) g (x)$ .

Then number of correct statements among the following:

A) If $g$ is discontinuous at $x = 1$ , then $f g$ can never be differentiable at $x = 1$ .

B) If $f g$ is differentiable at $x = 1$ , then $g$ is continuous at $x = 1$ .

C) If $f g$ is differentiable at $x = 1$ , then $g$ must be differentiable at $x = 1$ .

Smart Achievers · Accepted Answer

(4)

$Let f : R \to R is defined by f (x) = (x^{2} + \sin x) (x - 1)$

$Then, f (1^{+}) = f (1^{-}) = f (1) = 0$

$Let (f g) : R \to R is defined by (f g) (x) = f (x) \cdot g (x)$

$Let (f g) = h (x) = f (x) \cdot g (x) then h : R \to R$

$h' (x) = f' (x) g (x) + f (x) g' (x)$

$If g is differentiable at x = 1$

$h' (1) = f' (1) g (1) + 0,$ $[∵ f (1) = 0]$

$If g (x) is differentiable then h (x) is also differentiable (true)$

$If g (x) is differentiable at x = 1, then f g is also differentiable at x = 1$

$If g (x) is continuous at x = 1, then g (1^{+}) = g (1^{-}) = g (1)$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{h (1 + h) - h (1)}{h}$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{f (1 + h) g (1 + h) - 0}{h} = f' (1) g (1)$

$h' (1^{-}) = \lim_{h \to 0^{+}} \frac{f (1 - h) g (1 - h) - 0}{- h} = f' (1) g (1)$

$\Rightarrow h (x) = f (x) \cdot g (x) is differentiable at x = 1 (True)$

$So, If g is continuous at x = 1, THEN f g is differentiable at x = 1 .$

$Option (2) (4)$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{h (1 + h) - h (1)}{- h}$

$h' (1^{+}) = \lim_{h \to 0^{+}} \frac{f (1 + h) g (1 + h)}{h} = f' (1) g (1^{+})$

$h' (1^{-}) = \lim_{h \to 0^{+}} \frac{f (1 - h) g (1 - h)}{- h} = f' (1) g (1^{-})$

$\Rightarrow g (1^{+}) = g (1^{-})$

$So, it does not mean that if f g is differentiable at x = 1,$

$then f g is continuous or differentiable at x = 1$

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