Test Level 1: Functions - 1 - CAT MCQ

 f(x) = y
g(f(x)) = g(y) = y² + 1
The dependent variable always remains constant, irrespective of the independent variable.

Given f(x) = 
Let us suppose, y = 
Or, (5x – 3)y = 3x + 2
 5xy – 3y = 3x + 2
(5xy – 3x) = 2 + 3y
x(5y – 3) = 2 + 3y
Or, x = 
For finding inverse, we need to interchange x and y.
So, 
So, the function is equal to its inverse.

If y = f (x), f (x) = f (- x).
So the y values are the same for +ve and -ve values of x, and the graph is symmetric about the Y-axis.

Substituting x = 1 and y = -1 in f(x, y), we get 
f(1, -1) = 3 (1)² - 2 × (1) (-1) - (-1)² + 4 = 8

Graph is symmetric about Y - axis;
y = f(x) = I x I
f(-x) = I- x I = I x I
f(-x) = f(x)
therefore it is an even function.

f(x) = + 1 + x + 2x² + 3x³ + 4x⁴ + 5x⁵
f(1/x) = 5x⁵ + 4x⁴ + 3x³ + 2x² + x + 1 + 
f(x) = f(1/x)
Therefore, f(1/2) = f(2) = 260.78.

f(x) = 
Let, y = f(x)

⇒ y² = x - 3
⇒ x = y² + 3
f^-1(x) = x² + 3
Domain of f^-1(x) = R ....(i)
f(x) > 0 ......(ii)
Hence from (i) and (ii)
Range of f(x) = (0, ∞)

f(x) = x², g(x) = x + 3
(fog) (x) = f(x + 3) = (x + 3)² = x² + 6x + 9.

We have, f_n+1(x) = f_n (x) + 3
So, f₃(2) = f₂(2) + 3 = 7
f₄(2) = f₃(2) + 3 = 10
f₅(2) = f₄(2) + 3 = 13
f₆(2) = f₅(2) + 3 = 16

Range of R = {y ∈  A : (x, y) ∈ R for some x ∈ A} = {b, c}