Relations and Functions Test

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Relations and Functions Test - 44

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Question 1
1 / -0

$$f(x)=(x-1)(x-2)(x-3)(x-4)$$. Then out of three roots of $$f'(x)=0$$

A
three are positive

B
three are negative

C
two are complex

D
three are real, some positive,some negative

Solution
Question 2
1 / -0

If f (x+y).f(y) for all x,y, where f'(0)=3 and f(4)=2 then f'(4) is equal to

A
6

B
10

C
4

D
8

E
12

Solution
Question 3
1 / -0

If $$f \left( \dfrac { x + y } { 2 } \right) = \dfrac { f ( x ) + f ( y ) } { 2 }$$ for all $$x , y \in R$$ and $$f ^ { \prime } ( o ) = - 1 , f ( o ) = 1$$ then $$f(2)=$$

A
$$\dfrac { 1 } { 2 }$$

B
$$1$$

C
$$-1$$

D
$$\dfrac { -1 } { 2 }$$

Solution

let $$f(x)=ax+b$$
$$f(0)=1\implies b=1$$
$$f'(0)=-1 \implies a=-1$$
$$\implies f(x)=1-x$$
$$\implies f(2)=-1$$
Question 4
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If $$z$$, satisfies the condition that $$\dfrac { w-\overline{w}z }{ 1-z }$$ is purely real, then the set of values of $$z$$ is

A
$$\{z:|z|=1\}$$

B
$$\{z:z\neq 1\}$$

C
$$\{z:|z|=1, z\neq 1\}$$

D
None of the above

Solution
Question 5
1 / -0

Let $$R$$ be a relation on $$N$$ defined by $$x+2y=8$$. The domain of $$R$$ is

A
$$\left\{ 2,\ 4,\ 8 \right\} $$

B
$$\left\{ 2,\ 4,\ 6,\ 8 \right\} $$

C
$$\left\{ 2,\ 4,\ 6 \right\} $$

D
$$\left\{ 1,\ 2,\ 3,\ 4 \right\} $$

Solution

R is Relation
$$ R = (x,y) $$
$$ x+2y = 8 $$
$$ R = \left \{ (6,1);(4,2);(2,3) \right \} $$
$$ \therefore $$ Domain of set
$$ = \left \{ 6,4,2 \right \} $$
Question 6
1 / -0

Directions For Questions

$$f:R \rightarrow R$$ defined by, $$f(x) = x^3 + x^2 f'(1) + x.f^n(2) + f^m(3)$$ for all $$x \in R$$

...view full instructions

The value of $$f(1)$$ is

A
$$2$$

B
$$3$$

C
$$-1$$

D
$$4$$

Solution

$$f(x)=x^{3}+x^{2} \cdot f^{\prime}(1)+x \cdot f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$$
Differentiating both sides, $$f^{\prime}(x)=3 x^{2}+2 x f^{\prime \prime}(1)+f^{\prime \prime}(2)$$ (i)
Put $$x=1$$
$$\Rightarrow(1-2 x) f^{\prime}(1)=3 x^{2}+f^{\prime \prime}(2)$$
$$\Rightarrow f^{\prime \prime}(2)=-3 x^{2}+(1-2 x) \cdot f^{\prime}(1) $$
$$\Rightarrow f^{\prime \prime}(2)=-3-f^{\prime}(1)$$

Again, differentiate (i),
$$f^{\prime \prime}(x)=6 x+2 \cdot f^{\prime}(1)$$ (ii)

Put $$x=2$$,
$$f^{\prime \prime}(2)=12+2 \cdot f^{\prime}(1)$$

Substitute $$f^{\prime \prime}(2)=-3-f^{\prime}(1)$$
$$\Rightarrow-3-f^{\prime}(1)=12+2 \cdot f^{\prime}(1)$$
$$\Rightarrow 3 \cdot f^{\prime}(1)=-15$$
$$\Rightarrow f^{\prime}(1)=-5$$

And, $$\quad f^{\prime \prime}(2)=12+2 x-5$$
$$=2$$

Differentiate (ii) again,
$$f^{\prime \prime}(x)=6$$
$$\Rightarrow f^{\prime \prime \prime}(3)=6 \quad(x=3)$$
$$f(x)=x^{3}+x^{2} \cdot(-5)+x \cdot 2+6$$
Put $$x=1,$$
$$f(1)=1-5+2+6$$
$$=4$$
$$\therefore$$ Option $$D$$ is correct.
Question 7
1 / -0

Let $$A\equiv \left\{1,2,3,4\right\},\ B\equiv \left\{a,b,c\right\}$$, then number of function from $$A\rightarrow B$$, which are not onto is

A
$$9$$

B
$$24$$

C
$$45$$

D
$$6$$

Solution

$$A \equiv\{1,2,3,4\} \\$$
$$B \equiv\{a, b, c\}$$
Total function $$(A \rightarrow B)$$

$$\text { Total onto function }(A \rightarrow B)$$

$$=4 c_{3} \times 3 ! \times 3$$
$$=\frac{4 !}{3 ! 1 !} \times 3 ! \times 3$$
$$=41 \times 3=72$$
Hence, no. of $$f^{n}$$ which are not onto
$$=81-72 \\$$
$$=9$$
Question 8
1 / -0

If domain of $$y = f\left( x \right)$$ is $$ \left[ { - 3,\,\,2} \right]$$ then domain of $$y = f\left( {\left| {\left[ x \right]} \right|} \right)$$ is

A
$$[-3, 2]$$

B
$$[-2, 3)$$

C
$$[-3, 3]$$

D
$$[-2, 3]$$

Solution
Question 9
1 / -0

If $$x \epsilon$$ { $$1, 2, 3, ......,9$$} and $$f_n(x)=xxx.....x (n $$digit) then $$f_n ^2(3) + f_n(2)$$ is equal to

A
$$2f_{2n}(1)$$

B
$$f_n ^2(1)$$

C
$$f_{2n}(1)$$

D
$$-f_{2n}(4)$$

Solution
Question 10
1 / -0

Let $$R = \{ ( 3,3 ) , ( 6,6 ) , ( 9,9 ) , ( 6,12 ) , ( 3,9 ) , ( 3,12 ) , ( 3,6 ) \}$$ be a relation on the set $$A=\{ 3,6,9,12\} .$$ Then the relation $$R ^ { - 1 }$$ is

A
reflexive and transitive.

B
not symmetric

C
transitive

D
all the above

Solution

$$\begin{array}{l} We\, have \\ R=\left\{ { \left( { 3,3 } \right) ,\left( { 6,6 } \right) \left( { 9,9 } \right) \left( { 12,12 } \right) ,\, \left( { 6,12 } \right) ,\left( { 3,12 } \right) ,\, \left( { 3,6 } \right) } \right\} \\ and, \\ A=\left\{ { 3,6,9,12 } \right\} \\ Then \\ The\, relation\, is\, reflexive\, and\, transitive\, only.\, because \\ \left\{ { \left( { 3,3 } \right) ,\left( { 6,6 } \right) \left( { 9,9 } \right) \left( { 12,12 } \right) } \right\} \, \, it\, \, is\, { { Re } }flexive \\ and,\, \left( { 6,12 } \right) ,\left( { 3,12 } \right) ,\, \left( { 3,6 } \right) \left[ { it\, is\, transitive. } \right] \\ \end{array}$$
Hence, option $$A$$ is the correct answer.