Relations and Functions Test

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Relations and Functions Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

lf $$f:R\rightarrow R$$ such that $$f(x+y)-Kxy = f(x)+2y^2$$ for all $$x,y \in R$$ and $$f(1)=2,f(2)=8$$ then $$f(20)-f(10)$$

A
$$600$$

B
$$300$$

C
$$60$$

D
$$200$$

Solution

$$f(x+y)-kxy=f(x)+2y^2$$
Substitute $$x=0, y=1$$
$$f(1)-k\times 0=f(0)+2$$
$$f(1)=f(0)+2$$
$$\therefore f(0)=0$$
Now Substitute $$x=0, y=x$$
$$f(x)-k\times 0=f(0)+2x^2$$
$$f(x)=2x^2$$
$$f(10)=200$$
$$f(20)=800$$
$$f(20)-f(10)=600$$
Hence, option 'A' is correct.
Question 2
1 / -0

Vamsi is "X" years old. His sister''s age is $$\tiny \dpi{150} {\color{DarkBlue} 4\frac{1}{2}}$$ times that of Vamsi. Where as his uncle is 30 years older than him. If the total of their ages is 56 years, what is the age of Vamsi?

A
12 years

B
10 years

C
8 years

D
4 years
Question 3
1 / -0

If $$f(x)$$ is a polynomial function of degree $$n$$ satisfying eqation $$f(x) f(2x) = xf(3x)$$ then $$f(x)$$ is

A
one-one

B
many-one

C
bijective

D
surjective

Solution

comparing the degree on both sides, $$2n = n + 1$$
$$\Rightarrow n = 1 \therefore f\left ( x \right ) = ax + b where a \neq 0$$
As $$f\left ( x \right ) f\left ( 2x \right ) = x f\left ( 3x \right ) \Rightarrow \left ( ax + b \right )\left ( 2ax + b \right ) = x\left ( 3ax + b \right )$$
$$\Rightarrow 2a^{2} = 3a, 3ab = b, b^{2} = 0$$
$$\Rightarrow 2a = 3, b = 0$$ $$( \because a \neq 0)$$
$$ \Rightarrow \displaystyle f\left ( x \right ) = \frac{3x}{2}$$
which is an increasing function for all values of $$x$$. Hence, the function is bijective.
Question 4
1 / -0

If $$f(x) = \cos( \log x)$$ then $$f(x^2)f(y^2)-\dfrac{1}{2} \left [ f(x^2y^2)+f\left ( \dfrac{x^2}{y^2} \right) \right]=$$

A
$$0$$

B
$$-1$$

C
$$-2$$

D
None

Solution

Given, $$f(x)=\cos (\log x)$$
Therefore, $$f(x^2)f(y^2) - \dfrac{1}{2} [ f(x^2y^2)+f(x^2 / y^2)]$$
$$=\cos \log x^2 \cos \log y^2 -\dfrac{1}{2} [\cos \log (x^2y^2)+\cos \log (x^2 / y^2)]$$
$$=\cos \log x^2 \cos \log y^2 -\dfrac{1}{2} [\cos (\log x^2 + \log y^2)+\cos (\log x^2 - \log y^2)]$$
Using, $$\cos C + \cos D = 2 \cos \left (\dfrac{C+D}{2}\right) \cos \left (\dfrac{C - D}{2}\right)$$
$$=\cos \log x^2 \cos \log y^2 - \dfrac{1}{2} [2 \cos \log ^2 \cos \log y^2]=0$$
Question 5
1 / -0

Directions For Questions

Consider the function $$y=\frac{x^{2}+x+d}{x^{2}+2x+d}$$

...view full instructions

What will be the range, when d $$=$$ 0

A
R

B
R - {1}

C
$$R-\left \{ 1,\frac{1}{2} \right \}$$

D
R - {0}
Question 6
1 / -0

Let a function $$f(x)$$ satisfies $$f^{2}(x)-f^{2}(y)=4(x-y)$$ and $$f(0)= 2\left ( f\left ( x \right )\geq 0 \right )$$whose domain is $$\left [ a ,\infty \right )$$ and it is differentiable on $$\left (a ,\infty \right )$$
The value of $$f(3)$$ is

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

$$f^{2}(x)-f^{2}(y)=4(x-y)$$

and $$f(0)= 2\left ( f\left ( x \right )\geq 0 \right )$$

$$f^{2}(x)-f^{2}(y)=4(x-y)$$

Put y=0,

$$f^{2}(x)-f^{2}(0)=4x$$

$$f^{2}(x)=4x+4$$

$$f^{2}(3)=16$$

$$f(x) = \pm 4$$

But given $$f(x) \ge 0$$,

Hence$$f(3)=4$$
Question 7
1 / -0

If $$A=\left \{ 1,2,3 \right \} $$ and $$B=\left \{ 4,5,6 \right \}$$ then which of the following sets are relation from $$A$$ to $$B$$
(i) $$\displaystyle R_{1}=\left \{ (4,2) (2,6)(5,1)(2,4)\right \}$$
(ii) $$\displaystyle R_{2}=\left \{ (1,4) (1,5)(3,6)(2,6) (3,4)\right \}$$
(iii) $$\displaystyle R_{3}=\left \{ (1,5) (2,4)(3,6)\right \}$$
(iv) $$\displaystyle R_{4}=\left \{ (1,4) (1,5)(1,6)\right \}$$

A
$$\displaystyle R_{1},R_{2},R_{3}$$

B
$$\displaystyle R_{1},R_{3},R_{4}$$

C
$$\displaystyle R_{2},R_{3},R_{4}$$

D
$$\displaystyle R_{1},R_{2},R_{3},R_{4}$$

Solution

A relation from $$A$$ to $$B$$ will include elements from $$A\times B$$
Elements of $$A\times B$$ are
$$\{(1,4), (1,5) ,(1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}$$
Now the relation $$R_{1}$$ consists of elements form both $$A\times B$$ and $$B\times A$$.
Thus $$R_{1}$$ is not a relation from $$A$$ to $$B.$$
Question 8
1 / -0

The number of solution of the equation $$\displaystyle a^{f\left ( x \right )}+g\left ( x \right )=0,$$ where $$\displaystyle a >0, g\left ( x \right )\neq 0$$ and $$\displaystyle g\left ( x \right )$$ has minimum value $$\dfrac14$$, is

A
one

B
two

C
infinitely many

D
zero

Solution

There are two terms in the question.
The first term is $$a^{f(x)}$$ where 'a' is positive, so this term will always be positive no matter what the value of $$f(x)$$ is.
Since its also mentioned that the minimum value of $$g(x)$$ is $$\dfrac{1}{4}$$, the sum of $$2$$ positive numbers cannot be zero.
Hence, there is no solution.
Question 9
1 / -0

Let $$f$$ be a function satisfying $$\displaystyle f(x+y)=f(x)+f(y)$$ for $$x,y\:\epsilon\:R$$. If $$f(1)=k$$ then $$f(n)$$,$$n\:\in\:N$$, is equal to

A
$$k^{n}$$

B
$$nk$$

C
$$n^{k}$$

D
None of these

Solution

Let $$f(x)=\lambda(x)$$ where $$\lambda$$ is a constant.

Hence

$$f(x+y)$$

$$=\lambda(x+y)$$

$$=\lambda(x)+\lambda(y)$$

$$=f(x)+f(y)$$

Now it has been given that $$f(1)=k$$

Therefore

$$\lambda=k$$

Hence

$$f(x)=k(x)$$

Therefore $$f(n)$$

$$=k(n)$$
Question 10
1 / -0

If the polynomial satisfies $$f\left( x \right) =\dfrac { 1 }{ 2 } \begin{vmatrix} f\left( x \right) & f\left( \dfrac { 1 }{ x } \right) -f\left( x \right) \\ 1 & f\left( \dfrac { 1 }{ x } \right) \end{vmatrix}$$ and $$\displaystyle f \left ( 2 \right ) = 17$$, then the value of $$\displaystyle f\left ( 3 \right )$$ is

A
$$\displaystyle 81$$

B
$$\displaystyle 27$$

C
$$\displaystyle 26$$

D
$$\displaystyle 82$$

Solution

Given that $$f\left( x \right) =\dfrac { 1 }{ 2 } \begin{vmatrix} f\left( x \right) & f\left( \dfrac { 1 }{ x } \right) -f\left( x \right) \\ 1 & f\left( \dfrac { 1 }{ x } \right) \end{vmatrix}$$

$$\displaystyle f\left ( x \right )f\left ( \frac{1}{x} \right )=f\left ( x \right )+f\left ( \frac{1}{x} \right )$$

$$\Rightarrow $$   $$\displaystyle f\left ( x \right )f\left ( \frac{1}{x} \right )-f\left ( x \right )=f\left ( \frac{1}{x} \right )$$

$$\Rightarrow $$   $$\displaystyle f\left ( x \right )=\frac{f\left (\dfrac{1}{x}\right )}{f\left (\dfrac{1}{x}\right )-1}$$ .. (i)

Also, $$\displaystyle f\left ( x \right )f\left ( \frac{1}{x} \right )=f\left ( x \right )+f\left ( \frac{1}{x} \right )$$

$$\Rightarrow $$   $$\displaystyle f\left ( x \right )f\left ( \frac{1}{x} \right )-f\left ( \frac{1}{x} \right )=f\left ( x \right )$$

$$\Rightarrow $$   $$\displaystyle f\left ( \frac{1}{x} \right )=\frac{f\left ( x \right )}{f\left ( x \right )-1}$$ ..(ii)

On multiplying Eqs. (i) and (ii), we get,

$$\displaystyle f\left ( x \right )f\left ( \frac{1}{x} \right )=\frac{f\left (\dfrac{1}{x} \right )f\left ( x \right )}{\left \{ f\left (\dfrac{1}{x} \right )-1 \right \}\left \{ f\left ( x \right )-1 \right \}}$$

$$\Rightarrow $$   $$\displaystyle \left ( f\left ( \frac{1}{x} \right )-1 \right )\left ( f\left ( x \right )-1 \right )=1$$ ..(iii)

Since, $$f(x)$$ is polynomial function, so $$(f(x)-1)$$ and $$\displaystyle f\left ( \frac{1}{x} \right )-1$$ are reciprocals of each other.

Also, $$x$$ and $$\displaystyle \frac{1}{x}$$ are reciprocals of each other.

Thus, Eq. (iii) can hold only when

$$f\left ( x \right )-1=\pm x^{n}$$,          where $$n\in N$$

$$\therefore $$   $$f\left ( x \right )=\pm x^{n}+1$$          but $$f(2)=17$$

$$\Rightarrow $$   $$\pm 2^{n}+1=17$$   $$\Rightarrow $$   $$2^{n}=16$$

$$\Rightarrow $$   $$2^{n}=2^{4}$$          $$\left ( \because 2^{n}> 0 \right )$$

$$\Rightarrow $$   $$n=4$$

So, $$f\left ( x \right )=x^{4}+1$$

Hence, $$f\left ( 3 \right )=3^{4}+1=82$$

Ans: 82

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Relations and Functions Test - 50

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Relations and Functions Test - 50

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SHARING IS CARING

Question 1

$$600$$

$$300$$

$$60$$

$$200$$

Solution

Question 2

12 years

10 years

8 years

4 years

Question 3

one-one

many-one

bijective

surjective

Solution

Question 4

$$0$$

$$-1$$

$$-2$$

None

Solution

Question 5

R

R - {1}

$$R-\left \{ 1,\frac{1}{2} \right \}$$

R - {0}

Question 6

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 7

$$\displaystyle R_{1},R_{2},R_{3}$$

$$\displaystyle R_{1},R_{3},R_{4}$$

$$\displaystyle R_{2},R_{3},R_{4}$$

$$\displaystyle R_{1},R_{2},R_{3},R_{4}$$

Solution

Question 8

one

two

infinitely many

zero

Solution

Question 9

$$k^{n}$$

$$nk$$

$$n^{k}$$

None of these

Solution

Question 10

$$\displaystyle 81$$

$$\displaystyle 27$$

$$\displaystyle 26$$

$$\displaystyle 82$$

Solution

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