Domain and Range

x>=a, so |x-a| = x-a

x<100, so |x-100| = 100-x

f(x) = (x-a) + (100-x) + |x-a-50| =100

or, |x-a-50| = a

From the graph we can can see that when x=a then

|x-a-50|=a

or, a= 50

Similarly when x=a+100

|x-a-50|=a

or, a= 50

So value of a is 50 when f(x) is 100.

Let $$y= \dfrac{2x-3}{2x^{2}+4x-6}$$

$$\Rightarrow 2yx^2 + 4yx - 6y = 2x - 3$$

$$\Rightarrow 2yx^2 + (4y-2)x - 6y + 3 = 0$$ 

Equation (1) is a quadratic in $$x$$, where $$x$$ is real. Therefore, the discriminant of the quadratic has to be greater than or equal to zero.

$$(4y-2)^2 + 4\times 2y\times (6y-3) \geq 0$$

$$16y^2 + 4 - 16y - 24y + 48y^2 \geq 0$$

$$64y^2 - 40y + 4 \geq 0$$

$$16y^2 - 10y + 1 \geq 0$$ 

The roots of the quadratic above will be $$\dfrac{10\pm 6}{32} = \dfrac{1}{2}$$ or $$\dfrac{1}{8}$$

Since the coefficient of $$y^2$$ is positive, the quadratic will be less than zero in the range $$\left(\dfrac{1}{8},\dfrac{1}{2}\right)$$

The quadratic will be greater than or equal to zero otherwise.

Therefore, the domain the quadratic, or possible values of $$y$$, which is the range of $$f(x)$$, will be,

$$\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty \right)$$

Option C is the correct answer.

We check where $$f(x)=\frac{x}{2x-1}$$ or $$g(x)=\frac{x}{x-1}$$, or their compositions, become undefined.

First, f(x) is undefined at $$x=\tfrac12$$ and g(x) is undefined at $$x=1$$.

Next, for $$f(g(x)) = \frac{x}{x+1}$$

Since, the denominator can't be zero, x=-1 must also be excluded.

For $$g(f(x)) = \frac{x}{1-x}$$

$$\Rightarrow x=1$$ is not possible, which is already excluded.

So the values at which $$h(x)=f(g(x))+g(f(x))$$ is undefined are $${-1,\ \tfrac12,\ 1}$$