Solution: Given: A × B ⊆ C x D and A ∩ B ∈ ∅ A × B ⊆ C x D implies that A × B is subset of C × D. This means that every element of A × B is in C × D. Similarly, A ∩ B ∈ ∅ implies that A and B does...
Solution: (i) (A ∪ B) x C = (A x C) = (A x C) ∪ (B x C) Suppose that (x, y) is an arbitrary element of (A ∪ B) × C (x, y) ∈ (A ∪ B) C Since, (x, y) are elements of Cartesian product of (A ∪ B) × C...
Solution: We are given that A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6} (i) A × (B ∩ C) $ \left( B~\cap ~C \right)\text{ }=\text{ }\left\{ 4 \right\} $ $ A~\times ~\left( B~\cap ~C \right)\text{...
Solution: We are given that A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} (i) A x C ⊂ B x D Let us first consider LHS A x C $ A\text{ }\times \text{ }C\text{ }=\text{ }\left\{...
Solution: It is given that A = {1, 2, 3}, B = {4} and C = {5} (i) A × (B ∪ C) = (A × B) ∪ (A × C) In order to prove this, we consider the LHS: (B ∪ C) The union of B and C is given by: $(B~\cup...
Solution: We are given that A = {2, 3}, B = {4, 5} and C = {5, 6} We have to determine the values of A x (B ∪ C) and (A x B) ∪ (A x C) For A x (B ∪ C), we first calculate the union of sets B and C...
Solution: It is given that: A = {1, 2, 3} and B = {2, 4} According to the question, we have to find the values of A × B, B × A, A × A, (A × B) ∩ (B × A) (i) A × B = {1, 2, 3} × {2, 4} A × B =...
Solution: We are given that: A = {1, 2, 3} and B = {3, 4} We have to determine the value of A x B A x B = {1, 2, 3} × {3, 4} A x B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} The steps to...
Solution: We are given that A = {1, 2} and B = {1, 3} We have to calculate A x B and B x A Thus, putting values we can write: A × B = {1, 2} × {1, 3} A × B = {(1, 1), (1, 3), (2, 1), (2, 3)}...
Solution: It has been given that a ∈ {2, 4, 6, 9} and b ∈{4, 6, 18, 27} According to the question, we have to form the set of all ordered pairs (a, b) such that a divides b and a<b. So, finding...
Solution: We are given that a ∈ {- 1, 2, 3, 4, 5} and b ∈ {0, 3, 6}, We have to find the ordered pair (a, b) such that a + b = 5. We have to find those values of a and b from the respective sets...
Solution: It has been given that the ordered pairs (x, – 1) and (5, y) belongs to the set {(a, b): b = 2a – 3}. We can solve the given equation by using the definition of equality of ordered pairs....
Solution: It is given that: (a/3 + 1, b – 2/3) = (5/3, 1/3) We can solve the given equation by using the definition of equality of ordered pairs. We can write the above equation as: a/3 + 1 = 5/3...
Solution: It has been given that A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6} We have to determine the value of (A × B) ∩ (B × C) First, we find out the value of A × B, $ \left( A\times B...
(i) We know, A ⊂ {0, 1, 2, 3, 4, 5, 6} B ⊂ {0, 1, 2, 3, 4, 5, 6} So C ⊄ {0, 1, 2, 3, 4, 5, 6} As a result, the set {0, 1, 2, 3, 4, 5, 6} will not be a universal set for the sets A, B, and C. (ii) A...
(i) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} As a result, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set...
(i) The set of triangles or the set of polygons is the universal set among the set of right triangles. (ii) The universal set among isosceles triangles is the set of triangles, polygons, or...
SOLUTION: \[\begin{array}{*{35}{l}} x\text{ }+\text{ }2y\text{ }\le \text{ }3 \\ {} \\ \end{array}\] \[Line:\text{ }x\text{ }+\text{ }2y\text{ }=\text{ }3\] x 3 1 y 0 1 Also, (0, 0) satisfies...
SOLUTION: According to the question, Considering \[3x\text{ }+\text{ }2y\text{ }=\text{ }48,\] The shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy...
As indicated by the inquiry, \[P\left( n \right)\text{ }=\text{ }n^3\text{ }-\text{ }n\] is distinct by 6. Along these lines, subbing various qualities for n, we get, \[P\left( 0 \right)\text{...
(I) In this sentence 'and' is the associating word The part proclamations are as per the following (a) All normal numbers are genuine (b) All genuine numbers are not perplexing (ii) In this...
is an odd natural number} (ii) B = { is an integer, 
}Solution: (i) A = {$x: x$ is an odd natural number} So the elements are A = {1, 3, 5, 7, 9 …..} (ii) B = {$x: x$ is an integer, $-1/2 < x < 9/2$} It known to us that $– 1/2 = – 0.5$ and $9/2 =...
Given, $a–2b+11=0$……. (i) $3a–6b+33=0$……. (ii) From equation (i), ⇒ $b=(a+11)/2$ When $a=-1$, we get $b=(-1+11)/2=5$. When $a=-3$, we get $b=(-3+11)/2=4$. Thus, we have the following table giving...
in 
Solution: Let $y=\sec (\tan \sqrt{x})$ On derivativing both the sides with respect to $x$. $\frac{d y}{d x}=\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec ^{2} \sqrt{x} \frac{d}{d x} \sqrt{x}$...
term are: (iii) 
(iv) 
An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant. Solutions: (iii) ${{a}_{n}}={{3}^{n}}$ Given sequence...
term are: (i) 
(ii) 
An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.Solutions: (i) ${{a}_{n}}=3n+2$ Given sequence whose ${{a}_{n}}=3n+2$ To get the first five terms of given sequence, put $n=1,2,3,4,5$ and we get ${{a}_{1}}=\left( 3\times 1 \right)+2=3+2=5$...
Given, $AP=10cm$ $\angle APB={{60}^{\circ }}$ According to the figure We know that, A line drawn from centre to point from where external tangents are drawn, bisects the angle made by tangents at...
for which the quadratic equation 
has equal roots. Also, find the roots.Quadratic is that type of problem which deals with a variable multiplied by itself – an operation known also as squaring. Solution: The given equation $\left( 3k+1 \right){{x}^{2}}+2\left( k+1...
for which the following equations have real rootsQuadratic is that type of problem which deals with a variable multiplied by itself – an operation known also as squaring. (v) $kx\left( x-3 \right)+9=0$ Solution: Given, $kx\left( x-3 \right)+9=0$...
for which the following equations have real rootsQuadratic is that type of problem which deals with a variable multiplied by itself – an operation known also as squaring. (iii) ${{x}^{2}}-4kx+k=0$ Solution: Given, ${{x}^{2}}-4kx+k=0$ It’s of the...
for which the following equations have real and equal rootsQuadratic is that type of problem which deals with a variable multiplied by itself – an operation known also as squaring. (iii) $\left( k+1 \right){{x}^{2}}-2\left( k-1 \right)x+1=0$ Solution:...
are as follows:(iii) $a=-1.5,d=-0.5$ An arithmetic progression is a number’s sequence such that the difference between the consecutive terms is constant. Formula for this is: $an=d\left( n-1 \right)+c,$ (iii)...
:(iii) $0.3,0.55,0.80,1.05,...$ (iv) $-1.1,-3.1,-5.1,-7.1,...$ An arithmetic progression is a number’s sequence such that the difference between the consecutive terms is constant. Formula for this...
Ages (in years):5 – 1515 – 2525 – 3535 – 4545 – 5555 – 65No of students:6112123145 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. ...
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Solution: It’s given that, \[N\text{ }=\text{ }200\] \[\Rightarrow 46\text{ }+\text{ }x\text{ }+\text{ }y\text{ }+\text{ }25\text{ }+\text{ }10\text{ }+\text{ }5\text{ }=\text{ }200\] \[\Rightarrow...
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Find the median life. Solution: It’s seen that, the cumulative frequency just greater than \[n/2\text{ }\left( 400/2\text{ }=\text{ }200 \right)\text{ }is\text{ }216\]and it belongs to the class...
Calculate the median and interpret the results. Solution: Here, we have \[N\text{ }=\text{ }357,\] So, \[N/2\text{ }=\text{ }357/\text{ }2\text{ }=\text{ }178.5\] The cumulative frequency just...
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Solution: Let the unknown frequency be taken as x, It’s given that Median \[=\text{ }24\] Then, median class = \[20\text{ }\text{ }30;\text{ }L\text{ }=\text{ }20,~h\text{ }=\text{ }30\text{...
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Solution: Here, we have \[N\text{ }=\text{ }100,\] \[So,\text{ }N/2\text{ }=\text{ }100/\text{ }2\text{ }=\text{ }50\] The cumulative frequency just greater than \[N/\text{ }2\text{ }is\text{...
for giving distribution, mean is 
$x:$$3$$5$$7$$9$$11$$13$$f:$$6$$8$$15$$p$$8$$4$ Solution: $x$$f$$fx$$3$$6$$18$$5$$8$$40$$7$$15$$105$$9$$p$$9p$$11$$8$$88$$13$$4$$52$$N=41+p$$\sum{fx=303+9p}$ We know that, Mean $=\sum{fx/N=\left(...
for the given distribution whose 
is mean$x:$$8$$12$$15$$p$$20$$25$$30$$f:$$12$$16$$20$24$16$$8$$4$ Solution: $x$$f$$fx$$8$$12$$96$$12$$16$$192$$15$$20$$300$$P$$24$$24p$$20$$16$$320$$25$$8$$200$$30$$4$$120$$N=100$$\sum{fx=1228+24p}$ We...
. Subsequently, she took out Rs 
on 12.05.2007; deposited a cheque of Rs. 
on 03.06.2007 and paid Rs 
– by cheque on 18.06.2007. 
pa compounded at the end of March and September, find her balance on 1.04.2008From giving data in the question, We have to make the entries in passbook, So, the table have 5 columns. The data in 5 columns are, DateParticulars Withdrawals Depots Balance. Where,...
.
.$$$2x-11\le 7-3x,x\in N$ By transposing we get, $2x+3x\le 7+11$ $5x\le 18$ $5x\le 18/5$ $x\le 3.6$ As per the condition given in the question, x ∈ N. Therefore, solution set $x\in N$x Set can be...
, find the smallest value of x, when(i) 
(ii) 
From the question, (2x + 7)/3 ≤ (5x + 1)/4 So, by cross multiplication we get, $4(2x+7)\le 3(5x+1)$ $8x+28\le 15x+3$ Now transposing we get, $15x-8x\ge 28-3$ $7x\ge 25$ As per the condition given in...
, find the smallest value of x, when:(i) 
<(ii) From the question, $x+17\le 4x+9$ So, by transposing we get, $4x-x\ge 17-9$ $3x\ge 8$ $x\ge 8/3$ As per the condition given in the question, $x\in z$ Therefore, smallest value of $x=\{3\}$ From the...
An equation between two variables that gives a straight line when plotted on a graph. From the question it is given that, $5x+3x<18-3x$ So, by transposing we get, $5x+3x<18+14$ $8x<32$...
Let n be any positive integer. Thus, the three consecutive positive integers are n, n+1 and n+2. We know that any positive integer can be of the form 6q, or...