CBSE 11th Standard Maths Subject Sets HOT Questions 2 Mark Questions With Solution 2021
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CBSE 11th Standard Maths Subject Sets HOT Questions 2 Mark Questions With Solution 2021
11th Standard CBSE
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Reg.No. :
Mathematics
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If f: R \(\rightarrow\) R be defined as follows
\(f(x)=\begin{cases} 1,\quad \quad x\quad \in \quad Q \\ -1,\quad x\quad \notin \quad Q \end{cases}\)
Find \(f\left( \frac { 1 }{ 2 } \right) ,f\left( \pi \right) \)(a) -
Show that n {P [P (P ( \(\phi\) ))]}=4
(a) -
Let B be a subset of an A and let P(A: B) = { X \(\in\) P(A): X \(\supset\) B}.
If A = (1,2,3,4} and B ={1,2}. List all the members of the set P(A :B).(a) -
Let A = {1,2,3,4}, B= {1,2,3} and c= (2,4). Find all sets X satisfying each pair of conditions.
X \(\subseteq\) B and X \(\nsubseteq\) C(a) -
Let A = {1,2,3,4}, B= {1,2,3} and c= (2,4). Find all sets X satisfying each pair of conditions
X \(\subseteq\) B, X \(\neq\)B and X \(\nsubseteq\) C(a)
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CBSE 11th Standard Maths Subject Sets HOT Questions 2 Mark Questions With Solution 2021 Answer Keys
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The value of the function for every rational number is 1 and for every irrational number is -1.
\(\therefore \quad \frac { 1 }{ 2 } \in Q\Rightarrow f\left( \frac { 1 }{ 2 } \right) =1\)
\(and \quad \pi \notin Q\Rightarrow f\left( \pi \right) =-1\) -
We have, P ( \(\phi\) )={\(\phi\) }
\(\therefore\) P (P(\(\phi\)))={\(\phi\) ,{\(\phi\) }}
\(\Rightarrow\) P[P(P\(\phi\) ))]={\(\phi\) },{\(\phi\) },{{\(\phi\)}},{\(\phi\) ,{\(\phi\) }}}
Hence, number of elements in P [P(P (\(\phi\) ))] is 4
i.e. n {P [P(P (\(\phi\)))]}=4 -
here, A ={1,2,3,4} ,B={1,2}
Now,P(A:B)= set of all subsets of A which contain B
={{1,2},{1,2,3},{1,2,4},{1,2,3,4} -
Given A={1,2,3,4}, B = {1,2,3} And C={2,4}
Now, P(A)={ \(\phi\), {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}} ......(i)
P(B) ={\(\phi\){1},{2},{3},{1,2},{1,3},{1,3},{2,3},{1,2,3}} ....(ii)
and P(c) ={\(\phi\)),{2},{4},{2,4}} ...(iii)
Given condion is X \(\subseteq\) P(B) and X \(\nsubseteq\) C.
\(\Rightarrow\) X \(\in\) P(B) and X \(\in\) P(c)
\(\therefore\) X= {1},{3},{1,2},{1,3},{2,3}
[using the equation (ii) and (iii)] -
Given A={1,2,3,4}, B = {1,2,3} And C={2,4}
Now, P(A)={ \(\phi\), {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}} ......(i)
P(B) ={\(\phi\){1},{2},{3},{1,2},{1,3},{1,3},{2,3},{1,2,3}} ....(ii)
and P(c) ={\(\phi\)),{2},{4},{2,4}} ...(iii)
Given conditions is X \(\subseteq\) B and X\(\neq\) B and X \(\nsubseteq\) C..
\(\Rightarrow\) X \(\in\) P(B), X \(\neq\) B and X \(\in\) P(c)
\(\therefore\) X ={1},{3},{1,2},{1,3},{2,3}
[using eqs (ii) and (iii)]
