Verified Solution • 2025–26 Session moderate

QJEEAdv20Mafa16: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Mafa16 Solution: Let the function f : R R be defined by f(x) = x 3 x 2 + (x…

Let the function f : R →\to R be defined by f(x) = x3 −- x2 + (x −- 1)sin x and let g : R →\to R be an arbitrary function. Let fg : R →\to R be the product function defined by (fg)(x) = f(x)g(x). Then which of the following statements is/are TRUE?

Multiple Choice Options

  • A If g is continuous at x = 1, then fg is differentiable at x = 1
  • 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 fg is differentiable at x = 1
  • D If f g is differentiable at x = 1, then g is differentiable at x = 1

Step-by-Step Verified Solution

Given functions f : R →\to R be defined

by f(x) = x3 −- x2 + (x + 1) sin x and g : R →\to R be an arbitrary function.

Now, let g is continuous at x = 1, then

lim⁡x→1−(fg)(x)−(fg)(1)x−1\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}}

=lim⁡h→0(fg)(1−h)−(fg)(1)1−h−1= \mathop {\lim }\limits_{h \to 0} {{(fg)(1 - h) - (fg)(1)} \over {1 - h - 1}}

∵\because (fg)(x)=f(x) . g(x)(fg)(x) = f(x)\,.\,g(x) (given)

=lim⁡h→0f(1−h) . g(1−h)−f(1) . g(1)−h= \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1 - h) - f(1)\,.\,g(1)} \over { - h}}

=lim⁡h→0f(1−h) . g(1)−h= \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over { - h}}

{ ∵\because f(1) = 0 and g is continuous at x = 1, so g(1 −- h) = g(1)}

=g(1)lim⁡h→0(1−h)2(−h)+(−h)sin⁡(1−h)−h= g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^2}( - h) + ( - h)\sin (1 - h)} \over { - h}}

=(1+sin⁡1)g(1)= (1 + \sin 1)g(1)

Similarly, lim⁡x→1+(fg)(x)−(fg)(1)x−1\mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}

=lim⁡h→0f(1−h) . g(1)h= \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over h}

=g(1)lim⁡h→0(1+h)2(h)+hsin⁡(1+h)h= g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 + h)}^2}(h) + h\sin (1 + h)} \over h}

=(1+sin⁡1)g(1)= (1 + \sin 1)g(1)

∵\because RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1

Now, let (fg)(x) is differentiable at x = 1, so

lim⁡x→1−(fg)(x)−(fg)(1)x−1=lim⁡x→1+(fg)(x)−(fg)(1)x−1\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}

⇒lim⁡x→1−f(x)g(x)−f(1)g(1)x−1=lim⁡x→1+f(x)g(x)−f(1)g(1)x−1\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}}

⇒\Rightarrow lim⁡x→1−f(x)g(x)x−1=lim⁡x→1+f(x)g(x)x−1\mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x)} \over {x - 1}} { ∵\because f(1) = 0}

⇒\Rightarrow lim⁡h→0f(1−h) g(1−h)−h=lim⁡h→0f(1+h) g(1+h)h\mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,g(1 - h)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{f(1 + h)\,g(1 + h)} \over h}

⇒\Rightarrow lim⁡h→0[(1−h)2(−h)+(−h)sin⁡(1+h)]g(1+h)−h\mathop {\lim }\limits_{h \to 0} {{[{{(1 - h)}^2}( - h) + ( - h)\sin (1 + h)]g(1 + h)} \over { - h}}

= lim⁡h→0[(1+h)2(h)+(h)sin⁡(1+h)]g(1+h)h\mathop {\lim }\limits_{h \to 0} {{[{{(1 + h)}^2}(h) + (h)\sin (1 + h)]g(1 + h)} \over h}

lim⁡h→0 [(1−h)2+sin⁡(1−h)]g(1−h)\mathop {\lim }\limits_{h \to 0} \,[{(1 - h)^2} + \sin (1 - h)]g(1 - h)

=lim⁡h→0 [(1+h)2+sin⁡(1+h)]g(1+h)= \mathop {\lim }\limits_{h \to 0} \,[{(1 + h)^2} + \sin (1 + h)]g(1 + h)

It does not mean that g(x) is continuous or differentiable at x = 1.

But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1.

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