Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=x^{3}-x^{2}+(x-1)\sin x$ and let $g: \mathbb{R} \rightarrow \mathbb{R}$ be an arbitrary function. Let $fg: \mathbb{R} \rightarrow \mathbb{R}$ be the product function defined by $(fg)(x) = f(x)g(x)$. Then which of the following statements is/are TRUE?

(A) If $g$ is continuous at $x=1$, then $fg$ is differentiable at $x=1$; (C) If $g$ is differentiable at $x=1$, then $fg$ is differentiable at $x=1$

(B) If $fg$ is differentiable at $x=1$, then $g$ is continuous at $x=1$

(D) If $fg$ is differentiable at $x=1$, then $g$ is differentiable at $x=1$

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Continuity and Differentiability

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 which of the following statements is/are TRUE?

(A)If $g$ is continuous at $x = 1$ , then $f g$ is differentiable at $x = 1$ ✓ Correct
(B)If $f g$ is differentiable at $x = 1$ , then $g$ is continuous at $x = 1$
(C)If $g$ is differentiable at $x = 1$ , then $f g$ is differentiable at $x = 1$ ✓ Correct
(D)If $f g$ is differentiable at $x = 1$ , then $g$ is differentiable at $x = 1$

Answer: (A), (C)

Note that

f (x) = (x^{2} + sin x) (x - 1)

, so

f (1) = 0

and

f

is continuous at

x = 1

.

Let

h (x) = f (x) g (x)

.

(A) and (C): If

g

is continuous (or differentiable) at

x = 1

, then:

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

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

f g

is differentiable at

x = 1

. Both (A) and (C) are TRUE.

(B) and (D): If

f g

is differentiable at

x = 1

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

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

This gives

g (1^{+}) = g (1^{-})

, but we cannot conclude

g

is continuous at

x = 1

(we need

g (1) = g (1^{\pm})

) or differentiable. Both (B) and (D) are FALSE.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options