Test Level 2: Functions - 1 - CAT MCQ

We have f(x) = 

Since f(x, y) = |x² + y²| and F(x, y) = –f(x, y);
F(x, y) = –|x² + y²|
Again, G(x, y) = –F(x, y)
So, G(x, y) = |x² + y²|
Hence, G(x, y) = f(x, y)

As x is replaced by (x - 2),
The graph function of g(x) = 2^{x - 2} is obtained by shifting the graph of function f(x) = 2^x two units towards right.

Let ( x₁, y₁) be the required point.
Therefore, y₁ = f(x₁) = 2x₁ + 3.....(i)
But x₁ = f(y₁) = 2y₁ + 3.....(ii)
From (i) and (ii), we get
y₁ = 2x₁ + 3
y1 = 
Therefore, x₁ = -3 and y₁ = -3.

(i) f(x) = y = 2^-x2 ; f(- x) = ^-2^x2  = f(x) ; Therefore function is an even function
(ii) f(x) = y = 2^{x - x4}; f(- x) = 2^{x - x4}: Therefore function is neither even nor odd

f(x) = x³ - x² - x + 1 = (x + 1) (x - 1)² and g(x) = x² - 2x + 1 = (x - 1)².
Hence, f(x) = (x + 1) . g(x)

[f(-x)]² = [-x^-x]² = (-x)^-2x = (x)^-2x = 

If f(x) is an odd function, then |f(x)| will become an even function.
An odd function is defined by f(-x) = -f(x)
If the modulus of the function will be taken, it will become positive.
Also the graph in the negative y-axis will shift to the positive y-axis.
|f(-x)| = |-f(x)| = f(x), which is the definition of even function.

Here, f(x) is shifted 3 steps to the left. From the graph, you can see that f(0) = f(3), f(1) = f(4).
This is a horizontal translation of graph, which is obtained when input is increased by 3.
Hence, the only option that satisfies is f(x + 3).

f(x) = 2x - 6
f(-2) = 2 × (-2) - 6 = -10
f(f(-2)) = f(-10) = 2 × (-10) - 6 = -26
f(f(f(-2))) = f(f(-10)) = f(-26)
f(-26) = 2 × (-26) - 6 = -58
f(f(f(-2))) = -58