CRE 3 - Periodic Functions | Algebra - Functions & Graphs

Answer: 1

Explanation :

For a function to be periodic these should exist a real number p such that

f(p + x) = f(x).

Here, p = 12.5

Hence, 1.

Answer: 1

Explanation :

For a function to be periodic these should exist a real number p such that

f(p + x) = f(x).

Here, p ≈ 12.5

Hence, 1.

Answer: 2

Explanation :

For a function to be periodic these should exist a real number p such that

f(p + x) = f(x).

Here, the function keeps increasing on either side of y axis.

∴ This is not a periodic function.

Hence, 2.

Answer: Option B

Explanation :

We know, Sin(360° + θ) = Sinθ.

∴ The value of f(x) repeats after every 360°.

⇒ Period of f(x) = 360°.

Hence, option (b).

Answer: Option A

Explanation :

Let us draw the graph of (Sinθ)².

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∴ From the graph we can see that the value of f(x) repeats after every π^c i.e., 180°.

⇒ Period of f(x) = 180°.

Hence, option (a).

Answer: Option D

Explanation :

Suppose g(1) = a and g(2) = b

Given, g(x + 1) + g(x - 1) = g(x)

⇒ g(x + 1) = g(x) – g(x – 1)

Put x = 2, g(3) = b - a

Put x = 3, g(4) = (b - a) – b = -a

Put x = 4, g(5) = -a – (b - a) = -b

Put x = 5, g(6) = -b – (-a) = a - b

Put x = 6, g(7) = (a – b) – (-b) = a

Put x = 7, g(8) = a – (a - b) = b

… and so on.

Based on this patter we can notice that the value of f(x) repeats after every 6 terms.

∴ f(6 + x) = f(x).

⇒ p = 6

Hence, option (d).

Answer: Option B

Explanation :

Given, g(x) = g(x - 1) × g(x + 1) and g(0) = 1/3.

⇒ g(x + 1) = g(x)/g(x - 1)

Suppose g(1) = a

Put x = 1, we get g(2) = a/(1/3) = 3a.

Put x = 2, we get g(3) = 3a/a = 3.

Put x = 3, we get g(4) = 3/3a = 1/a

Put x = 4, we get g(5) = (1/a)/3 = 1/3a

Also it is given, g(5) = 1/6

∴ 1/3a = 1/6

⇒ a = 2.

∴ g(4) = 1/2 and g(5) = 1/6

Put x = 5, we get g(6) = (1/6)/(1/2) = 1/3 = g(0)

Put x = 6, we get g(7) = (1/3)/(1/6) = 2 = g(1)

Hence, we can see that g(6 + x) = g(x).

∴ g(1 + 2 + 3 + … + 100) = g(5050) = g(6 × 841 + 4) = g(4) = ½

Hence, option (b).