GO Classes DPP | GATE CS | Discrete Mathematics| Functions

Answer 1

We are given the following functions:

• $f(x) = x^2 - 3x + 1$

• $g(x) = 2x - 1$

• $h(x) = \frac{(x+1)^2}{x^2+1}$

 

Finding the composite function $h \circ g(x):$

The composite function $h \circ g(x)$ means we substitute $g(x)$ into the function $h(x)$.

$$h(g(x)) = h(2x - 1)$$

Now, replace every $x$ in the equation for $h(x)$ with $(2x - 1)$:

$$h(2x - 1) = \frac{((2x - 1) + 1)^2}{(2x - 1)^2 + 1}$$

$$\boxed{h(g(x)) = \frac{4x^2}{(2x - 1)^2 + 1}}$$

 

Finding the composite function $h \circ g \circ f(x):$

This means we substitute $f(x)$ into the composite function $h(g(x))$ that we just found.

Let $k(x) = h(g(x)) = \frac{4x^2}{(2x - 1)^2 + 1}$. We need to find $k(f(x))$.

$$k(f(x)) = k(x^2 - 3x + 1)$$

Replace every $x$ in $k(x)$ with the expression for $f(x)$:

$$h(g(f(x))) = \frac{4(x^2 - 3x + 1)^2}{(2(x^2 - 3x + 1) - 1)^2 + 1}$$

$$\boxed{h(g(f(x))) = \frac{4(x^2 - 3x + 1)^2}{(2x^2 - 6x + 1)^2 + 1}}$$

 

Therefore :

• $h \circ g(x) = \frac{4x^2}{(2x - 1)^2 + 1}$
   

• $h \circ g \circ f(x) = \frac{4(x^2 - 3x + 1)^2}{(2x^2 - 6x + 1)^2 + 1}$
   

The correct answer is A.

Answer 2

 

The correct option is A.

1. $(h \circ g)(x)$

By definition, $(h \circ g)(x) = h(g(x))$. Substitute $g(x) = 2x - 1$ into $h(x)$:

$$h(2x - 1) = \frac{((2x - 1) + 1)^2}{(2x - 1)^2 + 1} = \frac{(2x)^2}{(2x - 1)^2 + 1} = \frac{4x^2}{(2x - 1)^2 + 1}$$

 

2. $(h \circ g \circ f)(x)$

We substitute $f(x) = x^2 - 3x + 1$ directly into our result for $(h \circ g)(x)$:

$$(h \circ g)(f(x)) = \frac{4(f(x))^2}{(2f(x) - 1)^2 + 1}$$

• Numerator: $4(x^2 - 3x + 1)^2$

• Denominator Inner Term: $2(x^2 - 3x + 1) - 1 = 2x^2 - 6x + 2 - 1 = 2x^2 - 6x + 1$

Putting it together:

$$(h \circ g \circ f)(x) = \frac{4(x^2 - 3x + 1)^2}{(2x^2 - 6x + 1)^2 + 1}$$