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New answer posted

a year ago

0 Follower 13 Views

P
Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

The C-atom in the CO32-  ion undergoes sp2 hybridization. BF4, NH4+  and SO42-  have a tetrahedral structure, whereas it has a triangular planar structure.

New answer posted

a year ago

0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

(i)  Given  that −1≤x≤1  Let  x1, x2  ∈  f(x)                                         f(x1)  =  1x1     and      f(x2)  =  1x2                                         f(x1)  =  f(x2)       ⇒       1x1  =  1x2        ⇒     x1  =  x2  So,  f(x)  is  one−one  function.  Let                           f(x)  =y  =  x2     ⇒  x  =  2y  For  y=1,  x=2 ∉  [−1,1]   So,f(x)  is  not  onto.  Hence,  f(x)  is  not  bijective  function.  (ii)  Here,                          g(x)=|x|                                                g(x1)=g(x2)         ⇒    |x1|=|x2|       ⇒     x1=±x2  So,  g(x)  is  not  one−one  function.  Let  g(x)=y=|x|     ⇒x=±y ∉A ∀ y∈A  So,  g(x)  is  not  onto  function.  Hence,  g(x)  is  not  bijective  function.  (iii)  Here,            h(x)=x|x|                                   h(x1)=h(x2)  ⇒                          x1|x1|=x2|x2|       ⇒      x1=x2  So,  h(x)  is  one−one  function.  Now,  let             h(x)=y=x|x|=x2  or  −x2  ⇒                            &thi

 

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

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

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

  Here,  A∈ R −{3},  B  = R  −  {1}  Given  that  f: A → B defined byf(x) = x−2x−3  ∀  x  ∈  A.  Let  x1, x2  ∈  f(x)  ∴                                         f(x1)  =  f(x2)  ⇒                                     x1−2x1−3  =  x2−2x2−3  ⇒                              (x1 −2)  (x1 −3)  =   (x2 −2)  (x2 −3)  ⇒                                             −x1  =  −x2         ⇒       x1  =  x2  So,  it  is  injective  function.  Now,  Let              y  =  x−2x−3  ⇒    xy  −3y  =  x  −2      ⇒  xy  −x  =  3y  −2  ⇒  x(y−1)  =  3y  −2       ⇒  x  =  3y−2y−1  f(x)  =  x−2x−3  =  3y−2y−1−23y−2y−1−3       ⇒  3y−2−2y+23y−2−3y+3  ⇒  y  ⇒        f(x)  =  y  ∈  B.  So,  f(x)  is  surjective  function.  Hence,  f(x)  is  bijective  function.

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

    Here,     A  =  {2,  3,  4}        and     B={2,  5,  6,  7}  (i)    Let  f :  A  →  B  be  the  mapping  from  A  to  B           f  =  {(x,  y)  :  y  =  x  +  3}           ∴  f  =  {(2,  5),  (3,  6),  (4,  7)}  which  is  an  injective  mapping.  (ii)    Let  g :  A  →  B  be  the  mapping  from  A  →  B  such  that           ∴  g  =  {(2,  5),  (3,  5),  (4,  2)}  which  is  not  an  injective  mapping.    (iii)  Let  h :  B  →  A  be  the  mapping  from  B  to  A              h  =  {(y,  x)  :  x  =y−2}           ∴  h  =  {(5,  3),  (6,  4),  (7,  3)}  which  is  the  mapping  from  B  to  A.

New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

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.

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

    Given??  that  x∈ N,  y∈N  and  2x + y=41       ∴     Domain  of  R  =  {1, 2, 3, 4, 5,  …,  20}    and                   Range  =  {39, 37, 35, 33, 31,  …,  1}    Here,                   (3,  3)  ∉  R    as                      2* 3 + 3  ≠  41    So,  R  is  not  reflexive.    R  is  not  symmetric  as  (2,  37)  ∈  R  but  (37,  2)  ∉  R    R  is  not  transitive  as  (11,  19)  ∈  R  and  (19,  3)  ∈  R    but  (11,  3)  ∉  R.    Hence,  R  is  neither  reflexive,  nor  symmetric  and  nor  transitive.

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