Maths NCERT Exemplar Solutions Class 11th Chapter One

$\begin{array}{l}LetthenumberofstudentspassedinMathematicsM,EbeinEnglishandSbeinScience.\\ Thenn\left(U\right)=100,n\left(M\right)=12,n\left(E\right)=15,n\left(S\right)=8\\ n\left(E\cap M\right)=6,n\left(M\cap S\right)=7,n\left(E\cap S\right)=4andn\left(E\cap M\cap S\right)=4\\ Letusdrawavenndiagram.Accordingtothevenndiagram,weget\\ a+b+d+e=15\dots \left(i\right)\\ b+c+e+f=12\dots \left(ii\right)\\ d+e+f+g=8\dots \left(iii\right)\\ n\left(E\cap M\right)=6\therefore b+e=6\dots \left(iv\right)\\ n\left(M\cap S\right)=7\therefore e+f=7\dots \left(v\right)\\ n\left(E\cap S\right)=4\therefore d+e=4\dots \left(vi\right)\\ n\left(E\cap M\cap S\right)=4\therefore e=4\dots \left(vii\right)\\ Fromeqn.\left(iv\right)and\left(vii\right)wegetb+4=6\therefore b=2\\ Fromeqn.\left(v\right)and\left(vii\right)weget4+f=7\therefore f=3\\ Fromeqn.\left(vi\right)and\left(vii\right)wegetd+4=4\therefore d=0\\ Fromeqn.\left(i\right)wegeta+b+d+e=15\\ \Rightarrow a+2+0+4=15\Rightarrow a=9\\ Fromeqn.\left(ii\right)wegetb+c+e+f=12\\ \Rightarrow 2+c+4+3=12\Rightarrow c=3\\ Fromeqn.\left(iii\right)wegetd+e+f+g=8\\ \Rightarrow 0+4+3+g=8\Rightarrow g=1\\ \therefore \left(i\right)NumberofstudentswhopassedinEnglishandMathematicsbutnotinScience,b=2\\ \left(ii\right)NumberofstudentswhopassedinMathematicsandSciencebutnotinEnglish,f=3\\ \left(iii\right)NumberofstudentswhopassedinMathematicsonly,c=3.\end{array}$

$\begin{array}{l}Letx\in A\cap \left(B\cup C\right)\\ \Rightarrow x\in Aand\left(x\in Borx\in C\right)\\ \Rightarrow \left(x\in Aandx\in B\right)or\left(x\in Aandx\in C\right)\\ \Rightarrow \left(x\in A\cap B\right)or\left(x\in A\cap C\right)\\ \Rightarrow \left(x\in A\cap B\right)\cup \left(x\in A\cap C\right)\dots \left(i\right)\\ Lety\in \left(x\in A\cap B\right)\cup \left(x\in A\cap C\right)\\ \Rightarrow y\in \left(x\in A\cap B\right)or\left(x\in A\cap C\right)\\ \Rightarrow \left(y\in Aandy\in B\right)or\left(y\in Aandy\in C\right)\\ \Rightarrow y\in Aand\left(y\in Bory\in C\right)\\ \Rightarrow y\in Aandy\subset \left(B\cup C\right)\\ \Rightarrow y\in A\cap \left(B\cup C\right)\dots \left(ii\right)\\ Fromeq\left(i\right)and\left(ii\right)weget\\ A\cap \left(B\cup C\right)=\left(A\cap B\right)\cup \left(A\cap C\right)\\ L.H.S.=R.H.S.Henceproved.\end{array}$

$\begin{array}{l}Giventhat:T=\left\{x|\frac{x+5}{x-7}-5=\frac{4x-40}{13-x}\right\}\\ \Rightarrow \frac{x+5}{x-7}-5=\frac{4x-40}{13-x}\\ \Rightarrow \frac{x+5}{x-7}-\frac{4x-40}{13-x}=5\\ \Rightarrow \frac{\left(x+5\right)\left(13-x\right)-\left(x-7\right)\left(4x-40\right)}{\left(x-7\right)\left(13-x\right)}=5\\ \Rightarrow \frac{13x-x^2+65-5x-4x^2+40x+28x-280}{13x-x^2-91+7x}=5\\ \Rightarrow -5x^2+76x-215=5\left(13x-x^2+7x-91\right)\\ \Rightarrow -5x^2+76x-215=65x-5x^2+35x-455\\ \Rightarrow 76x-100x=-455+215\\ \Rightarrow -24x=-240\Rightarrow x=\frac{240}{24}=10\\ \therefore T=10\\ Hence,Tisnotanemptyset.\end{array}$

$\begin{array}{l}L.H.S.=\left(A\cup B\right)-B=\left(A\cup B\right)\cap B\text{'}\left[?A-B=A\cap B\text{'}\right]\\ =\left(A\cap B\text{'}\right)\cup \left(B\cap B\text{'}\right)\left[?A\cap A\text{'}=?\right]\\ =\left(A-B\right)\cup ?\\ =A-B=R.H.S.Henceproved.\end{array}$

$\begin{array}{l}L.H.S.=A-\left(A\cap B\right)=A\cap \left(A\cap B\right)\text{'}\\ =A\cap \left(A\text{'}\cup B\text{'}\right)=\left(A\cap A\text{'}\right)\cup \left(A\cap B\text{'}\right)\\ =?\cup \left(A-B\right)\left[?A-B=A\cap B\text{'}\right]\\ =A-B=R.H.S.\\ Hence,thegivenstatementisproved.\end{array}$

$\begin{array}{l}L.H.S.=A-\left(A-B\right)=A-\left(A\cap B\text{'}\right)\left[?A-B=A\cap B\text{'}\right]\\ =A\cap \left(A\cap B\text{'}\right)=A\cap \left[A\text{'}\cup \left(B\text{'}\right)\text{'}\right]\left[?\left(A\cap B\right)\text{'}=A\text{'}\cup B\text{'}\right]\\ =A\cap \left[A\text{'}\cup B\right]\left[?\left(A\text{'}\right)\text{'}=A\right]\\ =\left(A\cap A\text{'}\right)\cup \left(A\cap B\right)=?\cup \left(A\cap B\right)\\ =A\cap B=R.H.S.\\ Hence,thegivenstatementisproved.\end{array}$

$\begin{array}{l}L.H.S.=A\cup \left(B-A\right)=A\cup \left(B\cap A\text{'}\right)\left[?A-B=A\cap B\text{'}\right]\\ =\left(A\cup B\right)\cap \left(A\cup A\text{'}\right)\\ =A\cup B\cap U\left[?A\cup A\text{'}=U\right]\\ =A\cup B=R.H.S.\left[?A\cup U=A\right]\\ Hence,thegivenstatementisproved.\end{array}$

$\begin{array}{l}Letx\in A\cup B\\ \Rightarrow x\in Aorx\in B\\ \Rightarrow x\in Corx\in C\left[?A\subset CandB\subset C\right]\\ \Rightarrow x\in C\Rightarrow A\cup B\subset C\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$

$\begin{array}{l}Letx\in A\cup C\\ \Rightarrow x\in Aorx\in C\\ \Rightarrow x\in Borx\in C\left[?A\subset B\right]\\ \Rightarrow x\in \left(B\cup C\right)\Rightarrow \left(A\cup C\right)\subset \left(B\cup C\right)\\ Hence,thegivenstatementis\text{'}True\text{'}.\end{array}$