Relations and Functions Test

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Relations and Functions Test - 42

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Question 1
1 / -0

If $$A$$ and $$B$$ are independent event such that $$P(A \cap B')=\dfrac {3}{25}$$ and $$P(A' \cap B)=\dfrac {8}{25}$$, then $$P(A)=$$

A
$$1/5$$

B
$$3/8$$

C
$$2/5$$

D
$$4/5$$

Solution

$$A$$ & $$B$$ are independent
$$P(A\cap B)=\dfrac {3}{25}$$ and $$P(A'\cap B)=\dfrac {8}{25}\quad P(A)=(?)$$
$$\rightarrow \ P(A)+P(B)=1---(i)$$ ($$A$$& $$B$$ are independent )
$$\rightarrow \ P(A\cap B')=P(A)-P(A\cap B)$$
$$P(A)-P(A\cap B)=\dfrac {3}{25}----(ii)$$
$$\rightarrow \ P(A' \cap B)=P(B)-P(A\cap B)$$
$$P(B)-P(A\cap B)=\dfrac {8}{25}-----(iii)$$
$$\rightarrow \ $$ solving equation $$(ii)$$ and $$(iii)$$
$$\dfrac {\,\,\, P\left( A \right) -P\left( A\cap B \right) =\dfrac { 3 }{ 25 } \\\,\,\, P\left( B \right) -P\left( A\cap B \right) =\dfrac { 8 }{ 25 } \\ -\quad \,\,\,\,\,\,\,+\quad \quad\quad\quad- }{ P\left( A \right) -P\left( B \right) =\dfrac { 3 }{ 25 } -\dfrac { 8 }{ 25 } \\ P\left( A \right) -P\left( B \right) =\dfrac { 3-8 }{ 25 } } $$
$$=\dfrac {-5}{25}$$
$$\rightarrow \ P(A)-P(B)=\dfrac {-1}{5}$$
$$P(A)-(1-P(A))=\dfrac {-1}{5}$$
$$P(A)-I+P(A)=\dfrac {-1}{5}$$
$$\therefore \ 2P(A)=\dfrac {-1}{5}+1$$
$$\therefore \ 2P(A)=\dfrac {4}{5}$$
$$\therefore \ P(A)=\dfrac {4}{5\times 2}$$
$$\therefore \ P(A)=\dfrac {2}{5}$$
Question 2
1 / -0

Let $$S=\{x \in R : x\geq 0$$ and $$2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}$$. Then S?

A
Contains exactly one element

B
Contains exactly two elements

C
Contains exactly four elements

D
Is an empty set

Solution
Question 3
1 / -0

Let $$f ( x ) + f ( 2 x ) + f ( 2 - x ) + f ( 1 + x ) = x , \text { for all } x \in R$$ then $$f ( 0 )$$ equals:

A
$$- \frac { 1 } { 4 }$$

B
$$\frac { 1 } { 4 }$$

C
$$\frac { 1 } { 2 }$$

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

Solution

$$f(x)+f(2x)+f(2-x)+f(1+x)=x$$

$$x=1 \Rightarrow f(1)+f(2)+f(1)+f(2)=1$$

$$\Rightarrow 2f(1)+2f(2)=1$$

$$\Rightarrow f(1)+f(2)=1/2$$ _____ (1)

$$x=0\Rightarrow f(0)+f(0)+f(2)+f(1)=0$$

$$\Rightarrow 2f(0)+f(1)+f(2)=0$$

$$\Rightarrow 2f(0)+\dfrac{1}{2}=0$$ (from (1))

$$\Rightarrow f(0)=\dfrac {-1}{4}$$
Question 4
1 / -0

The domain of $$\dfrac{1}{\sqrt{x-x^{2}}}+\sqrt{3x-1-2x^{2}}$$ is

A
$$(\dfrac{1}{2},11)$$

B
$$(\dfrac{1}{2},31)$$

C
$$(\dfrac{1}{2},17)$$

D
$$(\dfrac{1}{2},41)$$

Solution

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

Domain of $$f(x)$$ : -
$$x (1 - x) > 0$$ and $$3x - 1 - 2x^{2} \geq 0$$

$$2x^{2} - 3x + 1 \leq 0$$

$$2x (x - 1) - 1 (x - 1) \leq 0$$

$$(2x - 1) (x - 1)\leq 0$$

Thus taking the intersection of above $$2$$.

$$x \leftarrow [\dfrac {1}{2}, 1)$$.
Question 5
1 / -0

If $$x<-1$$ then $${ x }^{ 2 }$$ is.

A
=1

B
<1

C
>1

D
None of these

Solution
Question 6
1 / -0

The set of values of '$$b$$' for which the origin and the point $$(1,1)$$ lie on the same side of the straight line $${ a }^{ 2 } x + ab y + 1 = 0 \;\;\forall \;\; a\in R,\; b > 0$$ are:

A
$$b\in (2,4)$$

B
$$b\in (0,2)\quad $$

C
$$b\in (0,4)\quad $$

D
$$b\in (2,6)$$

Solution
Question 7
1 / -0

The minimum value of the sum of real numbers $${ a }^{ -5 },{ a }^{ -4 },{ 3a }^{ -3 },{ 1 },{ a }^{ 8 }$$ and $${ a }^{ 10 }$$ with a > 0 is ______.

A
5

B
6

C
7

D
8

Solution
Question 8
1 / -0

Let $$f\left( x \right)$$ be the primitive of $$\dfrac { 3x+2 }{ \sqrt { x-9 } } $$ with respect to 'x' . if F(10) =60 . them the sum of digit of the value of F (13) i

A
6

B
13

C
24

D
26

Solution
Question 9
1 / -0

If $$a<b<c<d\quad$$ and $$x\epsilon R$$ Then the least value of the functions,
$$f\left( x \right) =\left| x-a \right| +\left| x-b \right| +\left| x-c \right| +\left| x-d \right|$$

A
c-d+b-a

B
c+d+b-a

C
C+D-b-a

D
c-d+b+a

Solution
Question 10
1 / -0

The domain of $$f(x)=\frac{1}{\sqrt{(x-1)(x-2)(x-3)}}$$ is

A
$$(-\infty ,1)\cup (3,\infty )$$

B
$$(1,2)\cup (3,\infty )$$

C
$$(-\infty ,2)$$

D
R

Solution

$${\textbf{Step -1: Put x=0 and change the sign as negative multiply by negative is positive}}$$ $${\textbf{and positive multiply by negative is negative }} $$
$$ x = 0 \Rightarrow - \times - \times - = - $$

$${\textbf{Step -2:Put x=1.5 and change the sign as positive multiply by negative is negative}}$$ $${\textbf{and negative multiply by negative is positive }}$$
$$x = 1.5 \Rightarrow + \times - \times - = + $$

$${\textbf{Step -3:Put x=2.5 and change the sign as positive multiply by positive is positive}}$$ $${\textbf{and positive multiply by negative is negative}}$$
$$x = 2.5 \Rightarrow + \times + \times - = - $$

$${\textbf{Step -4: Put x=4 and change the sign as positive multiply by positive is always positive }}$$
$$x = 4 \Rightarrow + \times + \times + = + $$

$${\text{x is positive between 1 and 2 and between 3 and infinity so }} x \in (1,2) \cup (3,\infty )$$
$$\textbf{Hence option B is correct}$$