Relations and Functions Test

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Relations and Functions Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

If $$f : R \rightarrow R$$ satisfies $$f(x + y) = f(x) + f(y)$$, $$\forall x, y\, \in\, R$$ and $$f(1) = 7$$, then $$\sum_{r\, =\, 1}^{n}{f(r)}$$ is

A
$$\displaystyle \frac{7n}{2}$$

B
$$\displaystyle \frac{7(n\, +\, 1)}{2}$$

C
$$7n (n + 1)$$

D
$$\displaystyle \frac{7n(n\, +\, 1)}{2}$$

Solution

Given,
$$f(1)=7$$
So, Now
$$f(1+1)=f(1)+f(1)$$ $$ [Since, f(a+b) =f(a) +f(b)]$$
$$i.e. f(2)=7+7$$
$$so, f(2) =2 \times 7$$
Similarly,
$$ f(3) =3 \times 7$$
$$ f(n) =n \times 7$$

Now,
$$\sum \limits _{r=1}^n f(r)= 7(1+2+3+......+n)$$
$$=7\dfrac{n(n+1)}{2}$$
Question 2
1 / -0

If $$A$$ and $$B$$ have $$n$$ elements in common, then the number of elements common to $$A\times B$$ and $$B\times A$$ is

A
$$n$$

B
$$2n$$

C
$$n^2$$

D
$$0$$

Solution

If there are $$m$$ elements in $$A$$ and $$m$$ elements in $$B, A \times B$$ and $$B \times A$$ both will be having $$m^2$$ elements.
Since there are $$n$$ elements common to $$A$$ and $$B$$, there will be $$n^2$$ such pairs in $$A \times B$$ and $$B \times A$$, which will have both the elements same.
Since the elements are same, they are commutative and hence there will be $$n^2$$ elements common to $$A \times B$$ and $$B \times A$$
Question 3
1 / -0

Let $$f(x)=2-|x-3|, 1 \le x \le 5$$ and for rest of the values $$f(x)$$ can be obtained by using the relation $$f(5x)=\alpha\, f(x)\forall\, x \in R$$.
The value of $$f(2007)$$ taking $$\alpha = 5$$, is:

A
118

B
2007

C
1250

D
132

Solution

$$f\left( x \right) =2-\left| x-3 \right| ,1\le x\le 5$$
$$f\left( 1 \right) =0$$
$$f\left( 2 \right) =1$$
$$f\left( 3 \right) =2$$
$$f\left( 4 \right) =1$$
$$f\left( 5 \right) =0$$
$$f\left( 2007 \right) =\alpha f\left( \cfrac { 2007 }{ 5 } \right) $$
$$={ \alpha }^{ 2 }f\left( \cfrac { 2007 }{ 25 } \right) $$
$$={ \alpha }^{ 3 }f\left( \cfrac { 2007 }{ 125 } \right) $$
$$={ \alpha }^{ 4 }f\left( \cfrac { 2007 }{ 625 } \right) $$
$$={ \alpha }^{ 5 }f\left(\cfrac { 2007 }{ 3125 } \right)$$
$$\cfrac { 2007 }{ 3125 } <1\therefore $$ Rejected
$$\quad f\left( 2007 \right) ={ \alpha }^{ 4 }f\left(\cfrac { 2007 }{ 625 } \right)$$
$$\cfrac { 2007 }{ 625 } \approx 3$$
$$f(2007)={ \alpha }^{ 4 }f(3)$$
$$={ 5 }^{ 4 }\times 2$$
$$=1250$$
Question 4
1 / -0

Find the correct statement pertaining to the functions $$\displaystyle f\left( x \right) ={ \left( x-3 \right) }^{ 2 }+2$$ and $$\displaystyle g\left( x \right) =\frac { 1 }{ 2 } x+1$$ graphed above

A
$$\displaystyle f\left( x \right) =g\left( x \right) $$ for exactly $$2$$ values of x

B
$$\displaystyle f\left( x \right) =g\left( x \right) $$ for exactly $$1$$ values of x

C
$$\displaystyle f\left( x \right) $$ is the inverse of $$\displaystyle g(x)$$

D
$$\displaystyle f\left( x \right) >g\left( x \right) $$ for all $$x$$

Solution

If $$f(x)=(x-3)^2+2$$ and $$g(x)=\dfrac { 1 }{ 2 } x+1$$ then $$f(x)=g(x)$$ is as follows:

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

$$x^{ 2 }+9-6x+2=\dfrac { x+2 }{ 2 } \\$$

$$x^{ 2 }-6x+11=\dfrac { x+2 }{ 2 } \\$$

$$2(x^{ 2 }-6x+11)=x+2\\$$

$$2x^{ 2 }-12x+22=x+2\\$$

$$2x^{ 2 }-13x+20=0$$

Now finding the roots of the quadratic equation $$2x^{ 2 }-13x+20=0$$:

$$2x^{ 2 }-13x+20=0\\$$

$$2x^{ 2 }-8x-5x+20=0\\$$

$$2x(x-4)-5(x-4)=0\\$$

$$2x-5=0,\quad x-4=0\\$$

$$x=\dfrac { 5 }{ 2 }, \quad x=4$$

Hence $$f(x)=g(x)$$ for exactly $$2$$ values of $$x$$.
Question 5
1 / -0

Let $$R$$ and $$S$$ be two non-void relations on a set $$A$$. Which of the following statements is false?

A
$$R$$ and $$S$$ are transitive implies $$R\cap S$$ is transitive

B
$$R$$ and $$S$$ are transitive implies $$R\cup S$$ is transitive

C
$$R$$ and $$S$$ are symmetric imples $$R\cup S$$ is symmetric

D
$$R$$ and $$S$$ reflexive implies $$R\cap S$$

Solution

Let $$A=\left\{ 1,2,3 \right\}$$
$$ ;\quad R=\left\{ (1,1),(2,2) \right\} ;\quad S=\left\{ (2,2),(2,3) \right\} $$
be two transitive relations on $$A$$
Thus $$R\cup S=\left\{ (1,1),(1,2),(2,2),(2,3) \right\} $$
then $$\left( 1,2 \right) \in R\cup S\quad $$ and $$\left( 2,3 \right) \in R\cup S\quad $$
but $$\left( 1,3 \right) \notin R\cup S$$
$$R\cup S\quad $$ is not transitive
Question 6
1 / -0

Let $$F_n (\theta) = \displaystyle \sum_{k = 0}^n \frac{1}{4^K} \sin^4 (2^{k} \theta)$$, then which of the following is true

A
$$\displaystyle F_2 \left ( \frac{\pi}{4} \right ) = \frac{1}{\sqrt 2}$$

B
$$\displaystyle F_3 \left ( \frac{\pi}{8} \right ) = \frac{2 + \sqrt 2}{4}$$

C
$$\displaystyle F_4 \left ( \frac{3 \pi}{2} \right ) = 1$$

D
$$F_5 (\pi) = 1$$

Solution

$$F_n (\theta) = \sin^4 \theta + \displaystyle \frac{\sin^4 2 \theta}{4} +

\frac{\sin^4 (2^2 \theta)}{4^2} + \frac{\sin^4 (2^3 \theta)}{4^3}

........$$
$$\displaystyle F_2 \left ( \frac{\pi}{4} \right ) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
$$F_3 \displaystyle \left ( \frac{\pi}{8} \right ) =\left(\frac{\sqrt{2-\sqrt{2}}}{2}\right)^4+\frac{1}{16}+\frac{1}{16}= \frac{2 - \sqrt 2}{4}$$
$$\displaystyle F_4 \left ( \frac{3 \pi}{2} \right ) = 1+0=1$$
$$\displaystyle F_5 \left ( \pi \right ) = 0$$
Question 7
1 / -0

Range of the function $$f(x)=\dfrac{sec^2\,x-tanx}{sec^2\,x+tanx}-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$$, is

A
R

B
$$R-(\dfrac{1}{3},3)$$

C
$$[\dfrac{1}{3},3]$$

D
$$[-1,\dfrac{5}{3}]$$

Solution

Let

$$f(x)=y$$

$$y=\dfrac{sec^2x-\tan x}{sec^2 x+\tan x}$$

$$\Rightarrow y(sec^2 x+\tan x)=sec^2x-\tan x$$

$$\Rightarrow y(1+\tan^2x+\tan x)=1+\tan^2x-\tan x$$

$$\Rightarrow y+y\tan^2x+y\tan x=1\tan^2x-\tan x$$

$$\Rightarrow \tan^2x(y-1)+\tan(y+1)+y-1=0$$

Now it is given $$\dfrac{-\pi}{2}<x< \dfrac{\pi}{2}$$

So $$\tan x$$ is real in $$x\in(\dfrac{-\pi}{2},\dfrac{\pi}{2})$$

So $$D\ge 0$$

$$\Rightarrow (y+1)^2-4(y-1)(y-1)\ge 0$$

$$\Rightarrow y^2+1+2y-4(y-1)^2\ge 0$$

$$\Rightarrow y^2+2y+1-4(y^2+1-2y)\ge 0$$

$$\Rightarrow y^2+2y+1-4y^2-4+8y\ge 0$$

$$\Rightarrow -3y^2+10y-3\ge 0$$

$$\Rightarrow 3y^2-910y+3\le 0$$

$$\Rightarrow 3y^2-9y-y+3\le 0$$

$$\Rightarrow 3y(y-3)-1(y-3)\le 0$$

$$\Rightarrow (3y-1)(y-3)\le 0$$
Question 8
1 / -0

Let $$f:R \to R - \left\{ 3 \right\}$$ be a function such that for some p>0, $$\displaystyle f\left( {x + p} \right) = {{f\left( x \right) - 5} \over {f\left( x \right) - 3}}$$ for all $$x \in R$$. Then, period of $$f$$ is

A
$$2p$$

B
$$3p$$

C
$$4p$$

D
$$5p$$

Solution
Question 9
1 / -0

For a function F, F(0) = 2, F(1) = 3, F(x + 2) = 2 F(x) - F(x + 1) for x $$\geq$$ 0, then F(5) is equal to

A
-7

B
-3

C
17

D
13

Solution
Question 10
1 / -0

If $$f\left( x \right)$$ satisfying the relation $$f\left( x \right) +f\left( x+4 \right) =f\left( x+2 \right) +f\left( x+6 \right) \ \forall x$$, then the period is

A
$$8$$

B
$$6$$

C
$$4$$

D
$$5$$

Solution

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Relations and Functions Test - 51

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Relations and Functions Test - 51

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SHARING IS CARING

Question 1

$$\displaystyle \frac{7n}{2}$$

$$\displaystyle \frac{7(n\, +\, 1)}{2}$$

$$7n (n + 1)$$

$$\displaystyle \frac{7n(n\, +\, 1)}{2}$$

Solution

Question 2

$$n$$

$$2n$$

$$n^2$$

$$0$$

Solution

Question 3

118

2007

1250

132

Solution

Question 4

$$\displaystyle f\left( x \right) =g\left( x \right) $$ for exactly $$2$$ values of x

$$\displaystyle f\left( x \right) =g\left( x \right) $$ for exactly $$1$$ values of x

$$\displaystyle f\left( x \right) $$ is the inverse of $$\displaystyle g(x)$$

$$\displaystyle f\left( x \right) >g\left( x \right) $$ for all $$x$$

Solution

Question 5

$$R$$ and $$S$$ are transitive implies $$R\cap S$$ is transitive

$$R$$ and $$S$$ are transitive implies $$R\cup S$$ is transitive

$$R$$ and $$S$$ are symmetric imples $$R\cup S$$ is symmetric

$$R$$ and $$S$$ reflexive implies $$R\cap S$$

Solution

Question 6

$$\displaystyle F_2 \left ( \frac{\pi}{4} \right ) = \frac{1}{\sqrt 2}$$

$$\displaystyle F_3 \left ( \frac{\pi}{8} \right ) = \frac{2 + \sqrt 2}{4}$$

$$\displaystyle F_4 \left ( \frac{3 \pi}{2} \right ) = 1$$

$$F_5 (\pi) = 1$$

Solution

Question 7

R

$$R-(\dfrac{1}{3},3)$$

$$[\dfrac{1}{3},3]$$

$$[-1,\dfrac{5}{3}]$$

Solution

Question 8

$$2p$$

$$3p$$

$$4p$$

$$5p$$

Solution

Question 9

-7

-3

17

13

Solution

Question 10

$$8$$

$$6$$

$$4$$

$$5$$

Solution

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