JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 7)
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?
If g is continuous at x = 1, then fg is differentiable at x = 1
If f g is differentiable at x = 1, then g is continuous at x = 1
If g is differentiable at x = 1, then fg is differentiable at x = 1
If f g is differentiable at x = 1, then g is differentiable at x = 1
Explanation
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
$$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{(fg)(1 - h) - (fg)(1)} \over {1 - h - 1}}$$
$$ \because $$ $$(fg)(x) = f(x)\,.\,g(x)$$ (given)
$$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1 - h) - f(1)\,.\,g(1)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over { - h}}$$
{$$ \because $$ f(1) = 0 and g is continuous at x = 1, so g(1 $$-$$ h) = g(1)}
$$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^2}( - h) + ( - h)\sin (1 - h)} \over { - h}}$$
$$ = (1 + \sin 1)g(1)$$
Similarly, $$\mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over h}$$
$$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 + h)}^2}(h) + h\sin (1 + h)} \over h}$$
$$ = (1 + \sin 1)g(1)$$
$$ \because $$ 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
$$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x)} \over {x - 1}}$$ {$$ \because $$ f(1) = 0}
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,g(1 - h)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{f(1 + h)\,g(1 + h)} \over h}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} {{[{{(1 - h)}^2}( - h) + ( - h)\sin (1 + h)]g(1 + h)} \over { - h}}$$
= $$\mathop {\lim }\limits_{h \to 0} {{[{{(1 + h)}^2}(h) + (h)\sin (1 + h)]g(1 + h)} \over h}$$
$$\mathop {\lim }\limits_{h \to 0} \,[{(1 - h)^2} + \sin (1 - h)]g(1 - h) $$
$$= \mathop {\lim }\limits_{h \to 0} \,[{(1 + h)^2} + \sin (1 + h)]g(1 + h)$$
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.
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
$$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{(fg)(1 - h) - (fg)(1)} \over {1 - h - 1}}$$
$$ \because $$ $$(fg)(x) = f(x)\,.\,g(x)$$ (given)
$$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1 - h) - f(1)\,.\,g(1)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over { - h}}$$
{$$ \because $$ f(1) = 0 and g is continuous at x = 1, so g(1 $$-$$ h) = g(1)}
$$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^2}( - h) + ( - h)\sin (1 - h)} \over { - h}}$$
$$ = (1 + \sin 1)g(1)$$
Similarly, $$\mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over h}$$
$$ = g(1)\mathop {\lim }\limits_{h \to 0} {{{{(1 + h)}^2}(h) + h\sin (1 + h)} \over h}$$
$$ = (1 + \sin 1)g(1)$$
$$ \because $$ 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
$$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x)} \over {x - 1}}$$ {$$ \because $$ f(1) = 0}
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,g(1 - h)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{f(1 + h)\,g(1 + h)} \over h}$$
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{h \to 0} {{[{{(1 - h)}^2}( - h) + ( - h)\sin (1 + h)]g(1 + h)} \over { - h}}$$
= $$\mathop {\lim }\limits_{h \to 0} {{[{{(1 + h)}^2}(h) + (h)\sin (1 + h)]g(1 + h)} \over h}$$
$$\mathop {\lim }\limits_{h \to 0} \,[{(1 - h)^2} + \sin (1 - h)]g(1 - h) $$
$$= \mathop {\lim }\limits_{h \to 0} \,[{(1 + h)^2} + \sin (1 + h)]g(1 + h)$$
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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