Relations and Functions Test

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Relations and Functions Test - 69

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Question 1
1 / -0

If $$g(x)={ x }^{ 2 }+x-2$$ and $$\cfrac { 1 }{ 2 } (g\circ f(x))=2{ x }^{ 2 }-5x+2$$, then $$f(x)$$ is equal to

A
$$2x-3$$

B
$$2x+3$$

C
$$2{x}^{2}+3x+1$$

D
$$2{x}^{2}-3x-1$$

Solution

We will solve this problem by the trial and error method.
Let us check option $$A$$ first.
If $$f(x)=2x-3$$
$$g(x)=x^2+x-2$$ [ Given ]
$$\Rightarrow$$ $$\dfrac{1}{2}(g\circ f)(x)=g[f(x)]$$
$$=\dfrac{1}{2}g(2x-3)$$

$$=\dfrac{1}{2}[(2x-3)^2+(2x-3)-2]$$

$$=\dfrac{1}{2}[4x^2+9-12x+2x-3-2]$$

$$=\dfrac{1}{2}[4x^2-10x+4]$$
$$=2x^2-5x+2$$
The given condition is satisfied by $$A.$$
Question 2
1 / -0

Let $$f : x \rightarrow y $$ be such that $$f(1) = 2$$ and $$f(x + y) = f(x) f(y)$$ for all natural numbers x and y. If $$\displaystyle \sum_{k= 1}^n f(a + k) = 16 (2^n - 1)$$ , then a is equal to

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

E
$$7$$

Solution

We have,
$$f(1)=2$$ and $$f(x+y)=f(x).f(y)$$
Now, $$f(2)=f(1+1)=f(1).f(1)=2.2=2^2$$
$$f(3)=f(2+1)+f(2).f(1)=2^2.2=2^3$$
and so on
$$\therefore f(x)=2^n$$......(i)
Now, we have
$$\displaystyle \sum^n_{k=1}f(a+k)=16(2^n-1)$$
$$\Rightarrow f(a+1)+f(a+2)+----f(a+n)=16(2^n-1)$$
$$\Rightarrow f(a).f(1)+f(a).f(2)+-----f(a).f(n)=16(2^n-1)$$
$$\Rightarrow f(a)=[f(1)+f(2)+---f(n)]=16(2^n-1)$$
$$\Rightarrow f(a)[2+2^2+----+2^n]=16(2^n-1)$$

$$\Rightarrow f(a).[2\dfrac{(2^n-1)}{2-1}]=16(2^n-1)$$
$$\Rightarrow 2f(a).(2^n-1)=16.(2^n-1)\Rightarrow f(a)=8$$
$$\Rightarrow 2^a=8[\because f(x)=2^n\Rightarrow f(a)=2^a]$$
$$\Rightarrow 2^a=2^3=a=3$$
Question 3
1 / -0

If $$f(x) = \dfrac{x+1}{x-1}$$, then the valueof $$f(f(x))$$ is equal to

A
$$x$$

B
$$0$$

C
$$-x$$

D
$$1$$

Solution

$$f(x)=\dfrac{x+1}{x-1}$$

$$\therefore f(f(x))=f\left(\dfrac{x+1}{x-1}\right)$$

$$\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}$$

$$=\dfrac{x+1+x-1}{x+1-x+1}$$

$$=\dfrac{2x}{2}$$

$$=x$$
Question 4
1 / -0

If $$f(x)=\dfrac{(4x+3)}{(6x-4)}, x\neq \dfrac{2}{3}$$ then $$(f o f)(x)=?$$

A
$$x$$

B
$$(2x-3)$$

C
$$\dfrac{4x-6}{3x+4}$$

D
None of these

Solution
Question 5
1 / -0

If $$f(x)=\sqrt[3]{3-x^3}$$ then $$(f o f)(x)=?$$

A
$$x^{1/3}$$

B
$$x$$

C
$$(1-x^{1/3})$$

D
None of these

Solution
Question 6
1 / -0

Let $$ f : R \rightarrow R : f(x) =x +1 $$ and $$ g : R \rightarrow R : g(x) = 2x -3 $$.
Find $$(f +g) (x)$$.

A
$$3x -2$$

B
$$4x -5$$

C
$$3x -4$$

D
$$2x -3$$

Solution

Given ,
$$f(x)=x+1,g(x)=2x-3$$

$$\implies (f+g)x=f(x)+g(x)=x+1+2x-3=3x-2$$
Question 7
1 / -0

If $$\displaystyle f(x) = | x - 2 |$$ and $$ g(x) = fof\,(x) $$ , then for $$ x > 20 , {g}\,'(x) = $$

A
$$ 2 $$

B
$$ 1 $$

C
$$ 3 $$

D
None of these

Solution
Question 8
1 / -0

If $$\displaystyle {f}'(x) = g\,(x) $$ and $$\displaystyle {g}'(x) = - f\,(x) $$ for all $$ x $$ and $$ f\,(2) = 4 = {f}'(2) $$ then $$\displaystyle f^{2}\,(19) + g^{2} \,(19) $$ is

A
$$ 16 $$

B
$$ 32 $$

C
$$ 64 $$

D
None of these

Solution

Question 9

1 / -0

Let $$f(n)$$ denote the number of different ways in which the positive integer $$n$$ can be expressed as the sum of $$1s$$ and $$2s$$. For example, $$f(4) = 5$$, since $$4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1$$. Note that order of $$1s$$ and $$2s$$ is important.

$$f : N\rightarrow N$$ is

One-one and onto

One-one and into

Many-one and onto

Many-one and into

Solution

$$6 = 0(2) + 6(1) = 1(2) + 4(1) = 2(2) + 2(1) = 3(2) + 0(1)$$

Number of $$2s$$	Number of $$1s$$	Number of permutations
$$0$$	$$6$$	$$1$$
$$1$$	$$4$$	$$\dfrac {5!}{4!} = 5$$
$$2$$	$$2$$	$$\dfrac {4!}{2!2!} = 6$$
$$3$$	$$0$$	$$\dfrac {3!}{3!} = 1$$
		$$Total = 13$$

$$\therefore f(6) = 13$$

Now, $$f(f(6)) = f(13)$$

Number of $$1s$$	Number of $$2s$$	Number of permutations
$$13$$	$$0$$	$$1$$
$$11$$	$$1$$	$$\dfrac {12!}{11!} = 12$$
$$9$$	$$2$$	$$\dfrac {11!}{9!2!} = 55$$
$$7$$	$$3$$	$$\dfrac {10!}{7!3!} = 120$$
$$5$$	$$4$$	$$\dfrac {9!}{5!4!} = 126$$
$$3$$	$$5$$	$$\dfrac {8!}{3!5!} = 56$$
$$1$$	$$6$$	$$\dfrac {7!}{6!} = 7$$
		$$Total = 377$$

$$\therefore f(f(6)) = f(13) = 377$$

$$f(1) = 1(1)$$

$$f(2) = 2 (1, 1$$ or $$2)$$

$$f(3) = 3(1, 1, 1\ or\ 2, 1\ or\ 1, 2)$$

$$f(4) = 5$$ (explained in the paragraph)

By taking higher value of $$n$$ in $$f(n)$$, we always get more value of $$f(n)$$. Hence, $$f(x)$$ is one-one. Clearly, $$f(x)$$ is into.

Question 10
1 / -0

let $$f(x) = sin^2 x/2 + cos ^2 x/2 $$ and $$g(x) = sec^2 x - tan ^2 x.$$ The two functions are equal over the set

A
$$\phi$$

B
$$R$$

C
$$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$$

D
None of these

Solution

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Relations and Functions Test - 69

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Relations and Functions Test - 69

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SHARING IS CARING

Question 1

$$2x-3$$

$$2x+3$$

$$2{x}^{2}+3x+1$$

$$2{x}^{2}-3x-1$$

Solution

Question 2

$$3$$

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 3

$$x$$

$$0$$

$$-x$$

$$1$$

Solution

Question 4

$$x$$

$$(2x-3)$$

$$\dfrac{4x-6}{3x+4}$$

None of these

Solution

Question 5

$$x^{1/3}$$

$$x$$

$$(1-x^{1/3})$$

None of these

Solution

Question 6

$$3x -2$$

$$4x -5$$

$$3x -4$$

$$2x -3$$

Solution

Question 7

$$ 2 $$

$$ 1 $$

$$ 3 $$

None of these

Solution

Question 8

$$ 16 $$

$$ 32 $$

$$ 64 $$

None of these

Solution

Question 9

One-one and onto

One-one and into

Many-one and onto

Many-one and into

Solution

Question 10

$$\phi$$

$$R$$

$$R-{ x:x (2n+1) \frac{\pi}{2}, n\in1}$$

None of these

Solution

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