Question

which of the following statements is/are TRUE?

Answer 1

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

solution : let's check differentiability of fg at x = 1

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

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

= $\displaystyle\lim_{h\to 0}\frac{(1+h)^3-(1+h)^2+(1+h-1)sin(1+h)g(1+h)-0}{h}$

=$\displaystyle\lim_{h\to 0}\frac{(h^3+3h^2+3h-h^2-2h+hsin(1+h)g(1+h)}{h}$

= $\displaystyle\lim_{h\to 0}(h^2+2h+1+sin(1+h)g(1+h)$

= $1+sin1g(1)$

if we let g(x) is continuous at x = 1

then, $\displaystyle\lim_{h\to 0}g(1+h)=g(1)$

means, $\displaystyle\lim_{h\to 0}fg'(1)=(1 + sin1)g(1)$ is true.

so it is clear that if g(x) is continuous at x = 1 then fg is differentiable at x = 1.

again, fg(x) : f(x) g(x) , fg : R → R

let fg(x) = h(x) = f(x) g(x) , h : R → R

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

at h'(1) = f'(1) g(1) + g'(1) f(1) = f'(1) g(1) + 0 [ cause f(1) = 0 but g'(x) exists ]

here it is clear that if g(x) is differentiable at x = 1 then fg is also differentiable at x = 1.

but we assume fg is differentiable at x = 1

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

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

so, h(1⁺) = h(1¯)

⇒f'(1)g(1⁺) = f'(1)g(1¯)

⇒g(1⁺) = g(1¯)

so we cannot comment on the continuity and differentiability of the function if we assume fg is differentiable at x = 1

Therefore the correct option is option (A) and (C)