QJEEAdv20Ma1db2: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Ma1db2 Solution: Let the functions f : ( − 1 , 1 ) → R and g : ( − 1 , 1 ) → ( − …

The given functions $$f:(-1,1)\to R$$ and $$g:(-1,1)\to (-1,1)$$ be defined by $$f(x)=|2x-1|+|2x+1|$$ and $$g(x)=x-[x]$$.

As, we know the composite function (fog) (x) is discontinuous at the points, where g(x) is discontinuous for given domain. And, since g(x) is discontinuous at x = 0 lies in interval ($$-$$1, 1), so value of c = 1.

And, since (fog) (x) is not differentiable at the point where g(x) is not differentiable as well as at those points also where g(x) attains the values so that f(g(x)) is non-differentiable.

Since g(x) is not continuous at x = 0$$\in$$ ($$-$$1, 1) so fog(x) is not differentiable and as $$f(x)=|2x-1|+|2x+1|$$ is not differentiable at x = $$-$$1/2 and 1/2, so (fog) (x) is not differentiable for those x, for which g(x) = $$-$$1/2 or 1/2.

But g(x) $$\ge$$ 0, so g(x) can be 1/2 only and for x = $$-$$1/2 and 1/2, g(x) = $$\frac{1}{2}$$.

So, (fog) (x) is not differentiable at x = $$-$$1/2, 0, 1/2, therefore value of d = 3

$$\therefore$$ c + d = 1 + 3 = 4.