Sets, Relations and Functions
The equation x + cosx $$=$$ a has exactly one positive root. Complete set of values of 'a' is
In a combined test in English and Physics; $$36\%$$ candidates failed in English; $$28\%$$ failed in Physics and $$12\%$$ in both; find the total number of candidates appeared, if $$208$$ candidates have failed.
No. of students failed in English $$( n(A) ) = 36 \% $$
No. of students failed in Physics $$( n(B) ) = 28 \% $$
No. of students failed in Both $$( n(C) ) = 12 \% $$
Total percentage of candidates which failed can be found using the formula of union of sets
$$ n(AUB) = n(A) + n (B) - n(C) $$
So, $$ n(AUB) = 36 + 28 - 12 = 52 $$ $$\%$$
Since $$ 52 $$ $$\%$$ of the total candidates failed, which is $$ = 208 $$
$$ \Rightarrow \dfrac {52}{100} \times $$ total candidates $$ = 208 $$
$$\therefore$$ Total candidates $$= 400 $$
The equation x + cosx $$=$$ a has exactly one positive root. Complete set of values of 'a' is
In a combined test in Maths and Chemistry. $$84\%$$candidates passed in Maths, $$76\%$$ in Chemistry and $$8\%$$ failed in both. If $$340$$ candidates passed in the test, then how many appeared ?
Let $$X = \{n \in N : 1 \le n \le 50\}.$$ If $$A = \{n \in X : n$$ is a multiple of $$2\}$$ and $$B = \{n \in X : n$$ is a multiple of $$7\}$$, then the number of elements in the smallest subset of X containing both A and B is _____.
Let $$P = \left \{\theta : \sin \theta - \cos \theta = \sqrt {2} \cos \theta \right \}$$ and $$Q = \left \{\theta : \sin \theta + \cos \theta = \sqrt {2}\sin \theta \right \}$$ be two sets. Then:
If $$ \xi=\{1,2,3, \ldots .9\}, A=\{1,2,3,4,6,7,8\} $$ and $$ B=\{4,6,8\}, $$ then find. (i) $$ A^{\prime} $$ (ii) $$ B^{\prime} $$ (iii) $$ A \cup B $$ (iv) $$ \mathrm{A} \cap \mathrm{B} $$ (v) $$ A-B $$ (vi) $$ B-A $$ (vii) $$ (A \cap B)^{\prime} $$ (viii) $$ A^{\prime} \cup B^{\prime} $$ Also verify that: (a) $$ (A \cap B)^{\prime}=A^{\prime} \cup B^{\prime} $$ (b) $$ n(A)+n\left(A^{\prime}\right)=n(\xi) $$ (c) $$ n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi) $$ (d) $$ n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B) $$
If $$ \xi=\{x: x \in W, x \leq 10\}, A .=\{x: x \geq 5\} $$ and $$ B=\{x: 3 \leq x<8\}, $$ then verify that: $$ A-B=A \cap B^{\prime} $$
If $$ \xi=\{x: x \in W, x \leq 10\}, A .=\{x: x \geq 5\} $$ and $$ B=\{x: 3 \leq x<8\}, $$ then verify that: $$ B-A=B \cap A^{\prime} $$
If $$ n(\xi)=40, n\left(A^{\prime}\right)=15, n(B)=12 $$ and $$ n\left((A \cap B)^{\prime}\right)=32, $$ find : (i) n(A) (ii) $$ n\left(B^{\prime}\right) $$ (iii) $$ n(A \cap B) $$ (iv) $$ n(A \cup B) $$ (v) $$ n(A-B) $$ (vi) $$ n(B-A) $$
$$P\cap Q\cup R$$
$$(P\cap Q)' \cup R$$