PE 2 - Functions | Algebra - Functions & Graphs

Answer: Option C

Explanation :

f(x²) = x² + $\frac{1}{x^2}$

(f(x))² = ${\left(x+\frac{1}{x}\right)}^2$= x² + $\frac{1}{x^2}$ + 2 = f(x²) + 2

∴ Statement 1 is true.

(f(x))³ = x³ + $\frac{1}{x^3}$ + 3$\left(x+\frac{1}{x}\right)$ = f(x³) + 3f(x)

∴ Statement 2 is true.

Thus both 1 and 2 are correct

Hence, option (c).

Answer: Option B

Explanation :

Given, f(x) × f(y) = f(x × y) and f(3) = 5.

Substitute y = 1, we get

f(x) × f(1) = f(x × 1) 

⇒ f(x) × f(1) = f(x)

⇒ f(1) = 1

Now, substitute x = 3 and y = 1/3, we get

f(3) × f(1/3) = f(3 × 1/3)

⇒ 5 × f(1/3) = f(1) = 1

⇒ f(1/3) = 1/5

Hence, option (b). 

Answer: Option A

Explanation :

f(2) = $\frac{2+1}{2-1}$ = 3

f²(2) = f(f(2)) = f(3) = $\frac{3+1}{3-1}$ = 2

f³(2) = f(f(f(2))) = f(f(3)) = f(2) = $\frac{2+1}{2-1}$ = 3

f⁴(2) = f(f(f(f(2)))) = f(3) = $\frac{3+1}{3-1}$ = 2

f⁵(2) = f(f(f(f(f(2))))) = f(2) = $\frac{2+1}{2-1}$ = 3

∴ f¹⁰⁰(2) = f²(2) = 2

Hence, option (a).

Answer: Option C

Explanation :

Given, f(2x + 1) = 3x³ + 2x² + x + 1

Instead of x if we substitute – x, we get

f(2(-x) + 1) = 3(-x)³ + 2(-x)² + (-x) + 1

⇒ f(-2x + 1) = -3x³ + 2x² - x + 1

Hence, option (c).

Answer: Option C

Explanation :

The function f(x) represents the fractional part of the function as it is the difference between x and the greatest integer function. 
 
For example, if x = 0.499 ⇒ f(x) = 0.499 – 0 = 0.499
 
If x = 1 ⇒ f(x) = 1 – 1 = 0
 
∴ The value of this function will always lie between 0 and 1 for all positive and negative values of x. 
 
The range will include 0 as this will be the value when x is an integer but will never include 1.
 
∴ Range of f(x) = [0, 1)
 
Hence, option (c).

Answer: Option C

Explanation :

Given, f(x − 1) + f(x + 1) = f(x)

We can rewrite the equation as,

f(x + 1) = f(x) – f(x – 1)

Substituting x = 1, we get

f(2) = f(1) - f(0)

⇒ 7 = f(1) – 2

⇒ f(1) = 9

∴ f(0) = 2, f(1) = 9 and f(2) = 7

Now, substituting x = 2, we get

f(3) = f(2) - f(1) = 7 – 9 = -2

Similarly,

f(4) = f(3) - f(2) = -2 – 7 = -9

f(5) = f(4) – f(3) = -9 – (-2) = -7

f(6) = f(5) – f(4) = -7 – (-9) = 2 = f(0)

f(7) = f(6) – f(5) = 2 – (-7) = 9 = f(1)

f(8) = (7) – f(6) = 9 – 2 = 7 = f(2)

Hence, value of f(x) repeats after every 6 integral value of x.

∴ f(6x) = f(0) = 2
f(6x + 1) = f(1) = 9,
f(6x + 2) = f(2) = 7,
f(6x + 3) = f(3) = -2,
f(6x + 4) = f(4) = -9,
f(6x + 5) = f(5) = -7

Now, f(1000) = f(6 × 166 + 4) = f(4) = -9

Hence, option (c).

Answer: 275

Explanation :

Given, f(x + y) = f(x) + f(y)

Substitute y = 0

⇒ f(x + 0) = f(x) + f(0)

⇒ f(0) = 0

Now, substitute x = y = 1, we get

f(1 + 1) = f(1) + f(1)

⇒ f(2) = 2f(1)

Now, substitute x = 2 and y = 1, we get

f(2 + 1) = f(2) + f(1) = 2f(1) + f(1) = 3f(1)

∴ f(3) = 3f(1)

Now, substitute x = 3 and y = 1, we get

f(3 + 1) = f(3) + f(1) = 3f(1) + f(1) = 4f(1)

∴ f(4) = 4f(1)

By observing the pattern, we can say that

f(x) = x × f(1)

∴ f(11) = 11 × f(1) = 55

⇒ f(1) = 5

Now, we have to calculate the value of f(1) + f(2) + f(3) + … + f(10)

= f(1) + 2f(1) + 3f(1) + … + 11f(1)

= f(1) × [1 + 2 + 3 + … + 11]

= 5 × 55

= 275

Hence, 275.

Answer: Option A

Explanation :

fog = f(2x + 3) = a(2x + 3) + 1 = 2ax + 3a + 1

gof = g(ax + 1) = 2(ax + 1) + 3 = 2ax + 5

⇒ 2ax + 3a + 1 = 2ax + 5

⇒ a = 4/3

Hence, option (a).

Answer: Option C

Explanation :

Given, f(x + 2) = 3x – 4

Now substitute x + 2 with x – 2 i.e., substitute x with x – 4

f(x – 4 + 2) = 3(x - 4) – 4

⇒ f(x – 2) = 3x – 16

Hence, option (c).

Answer: Option C

Explanation :

f(x/y) = f(x) − f(y) 
 
Putting x = y, we get,
 
f(1) = f(x) − f(x) = 0
 
Putting x = 1 and y = 1/x, we get,
 
f(x) = f(1) − f(1/x) = 0 − f(1/x)
 
∴ f(1/x) = −f(x) 
 
Also, putting x = x and y = 1/x, we get,
 
f(x²) = f(x) − f(1/x) = f(x) − (−f(x))
 
∴ f(x²) = 2f(x)
 
Now, f(2/72) = f(1/36) = - f(36) = -f(6²) = -2f(6)
 
Hence, option (c).