Functions Test 4

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Functions Test 4

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Weekly Quiz Competition

Question 1
1 / -0

Read the following information and answer the three items that follow :
Let $$f(x) = x^2 + 2x - 5 $$ and $$g(x) = 5x + 30$$
If $$h(x) = 5f(x) - xg (x)$$, then what is the derivative of $$h(x)$$ ?

A
$$-40$$

B
$$-20$$

C
$$-10$$

D
$$0$$

Solution

Given,

$$f(x)=x^2+2x-5$$

$$g(x)=5x+30$$

$$h(x)=5f(x)-xg(x)$$

$$=5(x^2+2x-5)-x(5x+30)$$

$$=5x^2+10x-25-5x^2-30x$$

$$h(x)=-20x-25$$

$$\dfrac{d}{dx}[h(x)]=\dfrac{d}{dx}(-20x-25)$$

$$=-20$$
Question 2
1 / -0

The number of non-surjective mappings that can be defined from $$A = \left \{ 1,4,9,16 \right \} $$ to$$ B=\left \{ 2,8,16,32,64 \right \}$$ is

A
$$1024$$

B
$$20$$

C
$$505$$

D
$$625$$

Solution

Here $$n(B)>n(A)$$, no function can be a surjective.
No of functions $$=$$ No of non-surjectives

Any elements of $$A$$ can be mapped to any of $$5$$ elements in $$B$$
So non surjective mapping from $$A$$ to $$B$$
$$=5\times5\times5\times5$$
$$=625$$
Question 3
1 / -0

The number of one-one functions that can be defined from $$A = \left \{ 1,2,3 \right \} $$ to $$ B = \left \{ a,e,i,o,u \right \}$$ is

A
$$3^{5}$$

B
$$5^{3}$$

C
$${_{}}^{5}P_{3}$$

D
$$5!$$

Solution

You can correlate this to permutation and combination problems.

We have to arrange 3 people on 5 places.

So total no of permutations $$={_{}}^{5}P_{3}$$
Question 4
1 / -0

If $$A = \left \{ 11, 12, 13, 14 \right \} $$ and $$ B = \left \{ 6,8,9,10 \right \} $$ then the number of bijections defined from $$A$$ to $$B$$ is

A
$$256$$

B
$$24$$

C
$$16$$

D
$$64$$

Solution

No of bijections from $$A$$ to $$B$$
$$=4\times3\times2\times1$$
$$=24$$
Question 5
1 / -0

If function $$f$$ has an inverse, then which of the following conditions is necessary and sufficient

A
$$f$$ is a one-one function.

B
$$f$$ is an onto function.

C
$$f$$ is a one-one and onto function.

D
$$f$$ is an identity function.

Solution

To have an inverse for the function $$f$$, $$f$$ should be one-one and onto.
Each point in the domain should have a unique image in range.
So, it should be one-one.
And if it is not onto, then there will be no pre-image of the extra element, so inverse will not be possible.
Question 6
1 / -0

If $$ f:A\rightarrow B $$ is a constant function which is onto then $$B$$ is

A
a singleton set

B
a null set

C
an infinite set

D
a finite set

Solution

$$f(x)$$ is a constant fucntion $$\Rightarrow$$ Range of $$f(x)$$ is a singleton set.
For $$f$$ to be an onto function, Co domain $$B$$ should be equal to Range.
$$\therefore B $$ is a singleton set.
Question 7
1 / -0

The number of bijection that can be defined from $$A = \left \{ 1,2,8,9 \right \} $$ to $$ B = \left \{ 3,4,5,10 \right \}$$ is

A
$$4^{4}$$

B
$$4^{2}$$

C
$$24$$

D
$$18$$

Solution

There are 4 inputs $$\{1, 2, 8, 9\}$$ and 4 outputs $$\{3, 4, 5, 10\}$$. Hence function will be bijective if and only if each output is connected with only one input.
Therefore number of bijective function is $$4! = 24$$
Question 8
1 / -0

If $$ f:A\rightarrow B $$ is a bijection then $$ f^{-1} of = $$

A
$$fof^{-1}$$

B
$$f$$

C
$$f^{-1}$$

D
an identity

Solution

Since $$f$$ is bijective, its inverse will also be bijective.
$$f(x)=y$$
$$x=f^{-1}(y)$$
$$f(x)=f^{-1}(y)$$
$$y=f(f)^{-1}(y)=f(x)=y$$
Hence, it is an identity.
Question 9
1 / -0

Assertion(A): $$ f(x) = \log(x-2)+\log(x-3)$$ and $$ g(x)=\log(x-2)(x-3)$$ then $$ f(x)=g(x)$$

Reason (R): Two functions $$f (x)$$ and $$g (x)$$ are said to be equal if they are defined on the same domain $$A$$ and the co-domain $$B$$ as $$ f(x)=g(x)\forall x \in A$$

A
Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$.

B
Both $$A$$ and $$R$$ are true and $$R$$ is NOT the correct explanation of $$A$$.

C
$$A$$ is correct and $$R$$ is incorrect.

D
$$A$$ is incorrect and $$R$$ is correct.

Solution

Reason is correct.
Assertion:
For f to be defined $$x-2>0 \ \text{and} \ x-3>0\Rightarrow x>3$$
so the domain of f is $$(3,\infty)$$.
Now for g to be defined $$(x-2)(x-3)>0\Rightarrow x\in(-\infty,2)\cup (3,\infty)$$
Thus domain of g is $$(-\infty,2)\cup (3,\infty)$$.
Clearly domain of both the functions are not same, Hence $$f\neq g$$
Question 10
1 / -0

Assertion (A) : $$ f(x)=\dfrac{x^{2}-4}{x-2} $$ and $$ g(x) = x+2$$ are equal.

Reason (R): Two functions $$f$$ and $$g$$ are said to be equal if their domains and ranges are equal and $$f(x)=g(x) \forall x \in $$ domain.

A
Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$

B
Both $$A$$ and $$R$$ are true and $$R $$ is not correct explanation of $$A$$

C
$$A$$ is true but $$R$$ is false

D
$$A$$ is false but $$R$$ is true

Solution

The domain for $$f \left ( x \right )$$ is $$R- \left \{ 2 \right \}$$ and domain of $$g \left ( x \right )$$ is $$R$$. Hence Assertion is false as the domain is not equal.
Reason is true as two functions $$f$$ and $$g$$ are said to be equal if
The domain of $$f$$ = domain of $$g$$.
The co-domain of $$f$$ = co-domain of $$g$$.
The range of $$f$$ = range of $$g$$.

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Re-Attempt Weekly Quiz Competition

Functions Test 4

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Functions Test 4

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SHARING IS CARING

Question 1

$$-40$$

$$-20$$

$$-10$$

$$0$$

Solution

Question 2

$$1024$$

$$20$$

$$505$$

$$625$$

Solution

Question 3

$$3^{5}$$

$$5^{3}$$

$${_{}}^{5}P_{3}$$

$$5!$$

Solution

Question 4

$$256$$

$$24$$

$$16$$

$$64$$

Solution

Question 5

$$f$$ is a one-one function.

$$f$$ is an onto function.

$$f$$ is a one-one and onto function.

$$f$$ is an identity function.

Solution

Question 6

a singleton set

a null set

an infinite set

a finite set

Solution

Question 7

$$4^{4}$$

$$4^{2}$$

$$24$$

$$18$$

Solution

Question 8

$$fof^{-1}$$

$$f$$

$$f^{-1}$$

an identity

Solution

Question 9

Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$.

Both $$A$$ and $$R$$ are true and $$R$$ is NOT the correct explanation of $$A$$.

$$A$$ is correct and $$R$$ is incorrect.

$$A$$ is incorrect and $$R$$ is correct.

Solution

Question 10

Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$

Both $$A$$ and $$R$$ are true and $$R $$ is not correct explanation of $$A$$

$$A$$ is true but $$R$$ is false

$$A$$ is false but $$R$$ is true

Solution

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