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Question: Let f(x) \( = \left( {x + |x|} \right)|x|, \) then for all x? A. f is continuous B. f is differe...

Let f(x) =(x+∣x∣)∣x∣,= \left( {x + |x|} \right)|x|, then for all x?
A. f is continuous
B. f is differentiable for some x
C. f’ is continuous
D. f” is continuous

Explanation

Solution

Hint : We will connect the given information into the form of function say f(x), for x⩾0,x<0,x \geqslant 0,x < 0, we will find f(x) then we will apply left hand limit and right hand side limit. Then we will find f’(x). If LHL=RHL then f(x) or f’(x) are said to be continuous otherwise they are not continuous.

Complete step-by-step answer :
If x⩾0,x \geqslant 0, then f(x) =(x+x)∣x∣= \left( {x + x} \right)|x|
If x<0,x < 0, then f(x) =x−(x)(−x)= x - \left( x \right)\left( { - x} \right)
f(x) =0×(−x)= 0 \times \left( { - x} \right)
f(x)=0f(x)= 0
f(x)=2x2,   x⩾0f(x)= 2{x^2},\,\,\,x \geqslant 0
0,  x<00,\,\,x < 0
Then will take limit both side, separately,
LHL == lim f(x) =0= 0
RHL=x→0+lim⁡f(x) = _{x \to {0^ + }}^{\lim }f\left( x \right)

RHL=x→0+lim⁡f(h) = _{x \to {0^ + }}^{\lim }f\left( h \right)

RHL=x→0+lim⁡2h2 =_{x \to {0^ + }}^{\lim }2{h^2}
RHL =0= 0
Therefore f is continuous.
Now, we will take differentiate of f(x), then
f’(x)=4xf’(x)= 4x for x>0 and
f’(x)=0f’(x)= 0 for x<0
We will take limit both sides, we get
LHL == lim f(x)
LHL =0= 0
RHL=x→0+lim⁡f′(x) = _{x \to {0^ + }}^{\lim }f'\left( x \right)

RHL=x→0+lim⁡f(h) = _{x \to {0^ + }}^{\lim }f\left( h \right)

RHL=h→0lim⁡4h{ = ^{}}_{h \to {0^{}}}^{\lim }4h
RHL =0= 0
Therefor f’ f’ is continuous
Hence, the correct option is (A) and (C).
So, the correct answer is “OptionA AND C”.

Note : Students should carefully solve if the LHL and RHL derivations are not equal at any point then, the function is not differentiable at those points.