JEE Advanced 2020 · Paper 1 · Mathematics Q7

← JEE Advanced 2020 · Paper 1

MathematicsMultiple correct+4 / -2Question 43 of 54

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

g

is continuous at

x=1

, then

fg

is differentiable at

x=1

fg

is differentiable at

x=1

, then

g

is continuous at

x=1

g

is differentiable at

x=1

, then

fg

is differentiable at

x=1

fg

is differentiable at

x=1

, then

g

is differentiable at

x=1

Answer and worked solution

Official answer: A, C

Factor $f$ . $f(x)=x^2(x-1)+(x-1)\sin x=(x-1)(x^2+\sin x)$ , so $f(1)=0$ and
$(fg)'(1)=\lim_{x\to1}\frac{f(x)g(x)}{x-1}=\lim_{x\to1}(x^2+\sin x)\,g(x).$

(A) If $g$ is continuous at 1, the limit exists and equals $(1+\sin1)g(1)$ . True.

(C) A differentiable $g$ is continuous, so this follows from (A). True.

(B), (D) Take $g(x)=0$ for $x\ne1$ and $g(1)=1$ . Then $fg\equiv0$ is differentiable, but $g$ is not continuous (so not differentiable) at 1. Both are false.

(A), (C)

Topics:Differentiability Continuity

← Previous question Next question →

JEE Advanced 2020 · Paper 1, Mathematics question 7