Test: Operations on Functions - Civil Engineering (CE) MCQ

Given:

g(x) = sin(4x/5)
As we know, period of sin x is 2π.
So, period of g(x) = sin(4x/5) is 

∴ Period of the function g(x) = 

Given,

2f(x) + f(1 - x) = x² ___(i)
Replacing x by (1 - x),
⇒ 2f(1 - x) + f(x) = (1 - x)² ___(ii)
Applying 2 × (i) - (ii),
⇒ 4f(x) + 2f(1 - x) - 2f(1 - x) - f(x) = 2x² - (1 - x)² 
⇒ 3f(x) = 2x² - (1 + x² - 2x)
⇒ 3f(x) = x² - 1 + 2x

f(x) = x + 5 ⇒ f(5) = 5 + 5 = 10.

f(x) = 2x³ + 7x² - 3
∴ f(x - 1) = 2(x - 1)³ + 7(x - 1)² - 3
⇒ 2(x³ - 3x² + 3x - 1) + 7(x² + 1 - 2x) - 3
⇒ 2x³ - 6x² + 6x - 2 + 7x² + 7 - 14x - 3
⇒ 2x³ + x² - 8x + 2

Given: f(x) = x² + 2x – 5 and g(x) = 5x + 30
⇒ g[f(x)] = g(x² + 2x – 5) = 5 (x² + 2x – 5) + 30 = 5x² + 10x + 5 = 0.
⇒ 5x² + 10x + 5 = 0
⇒ x² + 2x + 1 = (x + 1)² = 0
⇒ x = - 1

Given:

We know that period of g(x) is 5x/2 and the period of h(x) is 3π.
As we know, If f and g are two functions with periods p and q, respectively, then the period of the function f + g is LCM (p, q).
So, period of f(x) is 

Given:
h(x) = cos(2x/3)
As we know, period of cos x is 2π.
So, period of

∴ Period of the function h(x) = cos(2x/3) is 3π

Substituting x = tan θ, we get:

Given: f(x) = x² + 2x – 5 and g(x) = 5x + 30
⇒ f[g(x)] = f(5x + 30) = (5x + 30)² + 2(5x + 30) – 5 = 25x² + 310x + 955
⇒ f[g(x)] is a polynomial of degree 2.
Hence statement 1 is false.
⇒ g[g(x)] = g(5x + 30) = 5(5x + 30) + 30 = 25x + 180
⇒ g[g(x)] is a polynomial of degree 1.
Hence statement 2 is wrong.