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Q.

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

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a

If g is continuous 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

c

If g is differentiable at x=1 , then fg is differentiable at x=1

d

If fg is differentiable at x=1 , then g is differentiable at x=1

answer is A.

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Detailed Solution

f:R→R f(x)=x2+sin⁡x(x−1) f1+=f1−=f(1)=0fg(x):f(x)⋅g(x) fg:R→R let fg⁡(x)=h(x)=f(x)⋅g(x) h:R→R option (c): h′(x)=f′(x)g(x)+f(x)g′(x)                                                              h′(1)=f′(1)g(1)+0                                                         as f(1)=0,g′(x) exists ⇒ if g(x) is differentiable then h(x) is also differentiable (true)  option (A) : If g(x) is continuous at x=1 then g1+=g1−=g1                  h′1+=limh→0+ h(1+h)−h(1)hh′1+=limh→0+ f(1+h)g(1+h)−0h=f′(1)g(1)h′1−=limh→0+ f(1−h)g(1−h)−0−h=f′(1)g(1) So h(x)=f(x)⋅g(x) is differentiable  at x=1  (True)  option (B) (D): h′1+=limh→0+ h(1+h)−h(1)hh′1+=limh→0+ f(1+h)g(1+h)h=f′(1)g1+h′1−=limh→0+ f(1−h)g(1−h)−h=f′(1)⋅g1−⇒g1+=g1−so we cannot comment on the continuity and differentiability of the function gx
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