Function Test

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Function Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

Let f : R → R be a periodic function such that
f ( T + x) = 1 + [ 1 - 3f (x) + 3f (x) + 3 ( f (x) )² - ( f (x) ³ ]^1/3 where T is a fixed Positive number , then find the period of f (x).

A
2T

B
4T

C
T

D
8T

Solution

f (T + x) = 1 + [ 1 - f (x) }³ ]^1/3

= 1 + (1 - f (x) )

⇒ f (T + x) + f (x) = 2 .......... (1)

⇒ f (2T + x) + f (T + x) = 2 ....... (2)

Using (2) - (1)

⇒ f (2T + x) - f (x) = 0

⇒ f (2T + x) = f (x)

Also T is positive and least therefore period of f(x) =2T
Question 2
1 / -0

Directions For Questions

Let f (x) and g (x) are two distinct continues functions such that f (x) is an odd function and g (x) is an even function ∀ x ∈ R and f ' (x) > g ' (x)  ∀ x ∈ R. If a function h (x)   = f (x) + g (x) is an odd function and

φ (x)   = f (g (x) )   + g (f (x) )

then,

...view full instructions

Function g (x) is

A
a trigonometrical function

B
a polynomial function

C
an absolute value function

D
a constant function

Solution

Since h ( - x)   =   - h (x)

f ( - x) + g ( - x) = - f (x) - g (x)   (^..^. f is odd and g is even)

.^.. 2g (x) = 0

⇒ g (x) = 0 ∀x∈ R

⇒ g (x) is zero function which is constant function.

The correct answer is: a constant function
Question 3
1 / -0

Directions For Questions

Let f (x) and g (x) are two distinct continues functions such that f (x) is an odd function and g (x) is an even function ∀ x ∈ R and f ' (x) > g ' (x)  ∀ x ∈ R. If a function h (x)   = f (x) + g (x) is an odd function and

φ (x)   = f (g (x) )   + g (f (x) )

then,

...view full instructions

Find the number of solutions of f (x) = g (x).

A
1

B
2

C
3

D
4

Solution

g (x) = 0 ⇒ g^' (x)   = 0 ∀ x

So , f ^'(x) > 0 ∀ x ∈ R

⇒ f is increasing function and f is odd.

Let f (x) > 0 for some x (without loss of generality)

f ( - x)   = - f (x) < 0 for that x

Since f (x) and f ( - x) take opposite signs and f is increasing

⇒ f is zero at only one point

.^.. f (x)   = g (x)

⇒ f (x)   = 0

⇒ x = 0

such that f (0)   = 0 (^..^. f ( - x) = - f (x) takes x = 0 then f (0) = f (0)

⇒ f (0) = 0

.^.. f (x)   = g (x) has only one solution

The correct answer is: 1
Question 4
1 / -0

Directions For Questions

Let f (x) and g (x) are two distinct continues functions such that f (x) is an odd function and g (x) is an even function ∀ x ∈ R and f ' (x) > g ' (x)  ∀ x ∈ R. If a function h (x)   = f (x) + g (x) is an odd function and

φ (x)   = f (g (x) )   + g (f (x) )

then,

...view full instructions

Find the number of solutions j (x) = h (x)

A
4

B
1

C
0

D
5

Solution

The number of solutions of ∅ (x) = h (x) is

∅ (x) = f (g (x) ) + g (f (x) )

= f (0) + 0   (^..^. g (x)   = 0 ∀ x ∈ R

⇒ ∅ (x)   = 0 and h (x)   = f (x)

.^.. ∅ (x) = h (x)   = > 0 = f (x)

⇒ f takes 0 at x = 0

.^.. equation has only one solution

The correct answer is: 1
Question 5
1 / -0

If f ( 2x + 3y, 2x - 7y ) = 20 x ,then f (x, y) will be equal to

A
x – y

B
7x – 3y

C
3x – 7y

D
7x + 3y

Solution

7 (2x + 3y) + 3 (2x - 7y) = 20x

.^.. f (2x + 3y, 2x - 7y) = 20x

⇒ f (x, y) = 7x + 3y

The correct answer is: 7x + 3y
Question 6
1 / -0

Let f : R→ R and g : R→ R be two one - one and onto function such

that they are the mirror images of each other about the line y = a.

if h (x) = f (x) + g (x) ,then the function h(x) is

A
one - one onto

B
one - one into

C
many – one onto

D
many – one into

Solution

Since f (x) and g (x) are mirror images of each other about
the line y = a,

f (x) and g (x) are at equal distances from the line y = a .

Let for some particular x₀

f (x₀) = a + k ,then g (x₀) = a - k ,then

h (x₀) = f (x₀) + g (x₀) = 2a

.^.. h (x) = 2a ∀ x ∈ R. So, h(x) must be a constant function,

Which is many-one into

The correct answer is: many – one into

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Re-Attempt Weekly Quiz Competition

Function Test - 5

Result

Function Test - 5

Score

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Rank

-

SHARING IS CARING

Question 1

2T

4T

T

8T

Solution

Question 2

a trigonometrical function

a polynomial function

an absolute value function

a constant function

Solution

Question 3

1

2

3

4

Solution

Question 4

4

1

0

5

Solution

Question 5

x – y

7x – 3y

3x – 7y

7x + 3y

Solution

Question 6

one - one onto

one - one into

many – one onto

many – one into

Solution

User

Question Analysis

My Perfomance

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