Class 12th

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}(i)Giventhat-1\le x\le 1\\ Let{{\displaystyle x}}_1,{{\displaystyle x}}_2\in f(x)\\ f({{\displaystyle x}}_1)=\frac{1}{{{\displaystyle x}}_1}andf({{\displaystyle x}}_2)=\frac{1}{{{\displaystyle x}}_2}\\ f({{\displaystyle x}}_1)=f({{\displaystyle x}}_2)\Rightarrow \frac{1}{{{\displaystyle x}}_1}=\frac{1}{{{\displaystyle x}}_2}\Rightarrow {{\displaystyle x}}_1={{\displaystyle x}}_2\\ So,f(x)isone-onefunction.\\ Letf(x)=y=\frac{x}{2}\Rightarrow x=2y\\ Fory=1,x=2\notin \left[-1,1\right]\\ So,f(x)isnotonto.Hence,f(x)isnotbijectivefunction.\\ (ii)Here,g(x)=\left|x\right|\\ g({{\displaystyle x}}_1)=g({{\displaystyle x}}_2)\Rightarrow \left|{{\displaystyle x}}_1\right|=\left|{{\displaystyle x}}_2\right|\Rightarrow {{\displaystyle x}}_1=\pm {{\displaystyle x}}_2\\ So,g(x)isnotone-onefunction.\\ Letg(x)=y=\left|x\right|\Rightarrow x=\pm y\notin A\forall y\in A\\ So,g(x)isnotontofunction.\\ Hence,g(x)isnotbijectivefunction.\\ (iii)Here,h(x)=x\left|x\right|\\ h({{\displaystyle x}}_1)=h({{\displaystyle x}}_2)\\ \Rightarrow {{\displaystyle x}}_1\left|{{\displaystyle x}}_1\right|={{\displaystyle x}}_2\left|{{\displaystyle x}}_2\right|\Rightarrow {{\displaystyle x}}_1={{\displaystyle x}}_2\\ So,h(x)isone-onefunction.\\ Now,leth(x)=y=x\left|x\right|={{\displaystyle x}}^{2}or-{{\displaystyle x}}^2\\ \Rightarrow \text{\&thi}\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

Na (H₂)PO₂

1+ (2x+1)+ x +2 (−2)=0

x−1=0

x=1

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,A\in R-\{3\},B=R-\{1\}\\ Giventhatf:A\to Bdefinedbyf(x)=\frac{x-2}{x-3}\forall x\in A.\\ Let{{\displaystyle x}}_1,{{\displaystyle x}}_2\in f(x)\\ \therefore f({{\displaystyle x}}_1)=f({{\displaystyle x}}_2)\\ \Rightarrow \frac{{{\displaystyle x}}_1-2}{{{\displaystyle x}}_1-3}=\frac{{{\displaystyle x}}_2-2}{{{\displaystyle x}}_2-3}\\ \Rightarrow ({{\displaystyle x}}_1-2)({{\displaystyle x}}_1-3)=({{\displaystyle x}}_2-2)({{\displaystyle x}}_2-3)\\ \Rightarrow -{{\displaystyle x}}_1=-{{\displaystyle x}}_2\Rightarrow {{\displaystyle x}}_1={{\displaystyle x}}_2\\ So,itisinjectivefunction.\\ Now,Lety=\frac{x-2}{x-3}\\ \Rightarrow xy-3y=x-2\Rightarrow xy-x=3y-2\\ \Rightarrow x(y-1)=3y-2\Rightarrow x=\frac{3y-2}{y-1}\\ f(x)=\frac{x-2}{x-3}=\frac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}\Rightarrow \frac{3y-2-2y+2}{3y-2-3y+3}\Rightarrow y\\ \Rightarrow f(x)=y\in B.\\ So,f(x)issurjectivefunction.\\ Hence,f(x)isbijectivefunction.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (i)

The catalytic oxidation of ammonia produces NO gas, which is used to make HNO₃. 4 moles of NH₃ created 4 moles of NO in the equation below. As a result, the moles of NO produced by oxidising two moles of NH₃ will be two moles.

4NH₃ + 5O₂→4NO + 6H₂O.

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:$\begin{array}{l}\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (i)

We get N₂ in both situations when we heat ammonium dichromate and barium azide separately.

(NH₄)2Cr₂O₇→Cr₂O₃ + 4H₂O + N₂

Ba (N₃)₂→Ba + 3N₂.

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,A=\{2,3,4\}andB=\{2,5,6,7\}\\ \left(i\right)Letf:A\to BbethemappingfromAtoB\\ f=\{\left(x,y\right):y=x+3\}\\ \therefore f=\{\left(2,5\right),\left(3,6\right),\left(4,7\right)\}whichisaninjectivemapping.\\ \left(ii\right)Letg:A\to BbethemappingfromA\to Bsuchthat\\ \therefore g=\{\left(2,5\right),\left(3,5\right),\left(4,2\right)\}whichisnotaninjectivemapping.\\ \left(iii\right)Leth:B\to AbethemappingfromBtoA\\ h=\{\left(y,x\right):x=y-2\}\\ \therefore h=\{\left(5,3\right),\left(6,4\right),\left(7,3\right)\}whichisthemappingfromBtoA.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (ii)

Bismuth is the sole element that has an inert pair effect. Only generates trihalides and has a +3 oxidation state. Florine, on the other hand, is small and has a high electronegativity.

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given\text{?}\text{?}thatx\in N,y\in Nand2x+y=41\\ \therefore DomainofR=\{1,2,3,4,5,\dots ,20\}\\ andRange=\{39,37,35,33,31,\dots ,1\}\\ Here,\left(3,3\right)\notin R\\ as2*3+3\ne 41\\ So,Risnotreflexive.\\ Risnotsymmetricas\left(2,37\right)\in Rbut\left(37,2\right)\notin R\\ Risnottransitiveas\left(11,19\right)\in Rand\left(19,3\right)\in R\\ but\left(11,3\right)\notin R.\\ Hence,Risneitherreflexive,norsymmetricandnortransitive.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (i)

The 'Brown Ring Test' is an identification and conformation test of the nitrate ion  (NO₃^- ) using conc. H₂SO₄ and newly produced FeSO₄. Place the NO₃^-  containing sample in a test tube. Then, with gentle shaking, add conc H₂SO₄ dropwise, heat it briefly on a Bunsen burner, and add freshly made greenish FeSO4 Then, due to the creation of brown coloured  [Fe (H₂O)₅ (NO)]SO₄ complex, a brown ring forms in that test tube.