Exemplar Solution

Get insights from 1.3k questions on Exemplar Solution, answered by students, alumni, and experts. You may also ask and answer any question you like about Exemplar Solution

  • 1.3k

    Questions

  • 0

    Discussions

  • 0

    Active Users

  • 0

    Followers

Have a question about Exemplar Solution? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago
S
a year ago

This is a short answer type question as classified in NCERT Exemplar

F i r s t     n     n a t u r a l     n u m b e r s     a r e     1 , 2 , 3 , … , n .     H e r e ,     n     i s     o d d . ∴     M e a n     x ¯ = 1 + 2 + 3 + … + n n = n ( n + 1 ) 2 n = n + 1 2 T h e     d e v i a t i o n s     o f     n u m b e r s     f r o m     m e a n ( n + 1 2 )     a r e 1 − n + 1 2 , 2 − n + 1 2 , 3 − n + 1 2 , … , n − n + 1 2 i . e . , − n − 1 2 , − n − 3 2 , … , − 2 , − 1 , 0 , 1 , 2 , … , n − 1 2 . T h e     a b s o l u t e     v a l u e s     o f     d e v i a t i o n     f r o m     t h e     m e a n     i . e . , | x i − x ¯ |     a r e n − 1 2 , n − 3 2 , … , 2 , 1 , 0 , 1 , 2 , … , n − 1 2 . T h e     s u m     o f     a b s o l u t e     v a l u e s     o f     d e v i a t i o n     f r o m     t h e     m e a n     i . e . , | x i − x ¯ |     a r e = 2 ( 1 + 2 + 3 + … t o     n − 1 2     t e r m s ) = 2 . n − 1 2 ( n − 1 2 + 1 ) 2 = n − 1 2 . n + 1 2 = n 2 − 1 4 . ∴ M e a n     d e v i a t i o n     a b o u t     t h e     m e a n = ∑ | x i − x ¯ | n = n 2 − 1 4 n = n 2 − 1 4 n .

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

L e t     t h e     v e r t i c e s     o f     t h e     Δ A B C     b e A ( x 1 , y 1 , z 1 ) ,     B ( x 2 , y 2 , z 2 )     a n d     C ( x 3 , y 3 , z 3 ) D  is  the  mid-point  of  AB ∴                         1 = x 1 + x 2 2         ⇒ x 1 + x 2 = 2                                                       … ( i )                             2 = y 1 + y 2 2         ⇒ y 1 + y 2 = 4                                                     … ( i i ) a n d     − 3 = z 1 + z 2 2         ⇒ z 1 + z 2 = − 6                                                     … ( i i i ) E  is  the  mid-point  of  BC ∴                         3 = x 2 + x 3 2         ⇒ x 2 + x 3 = 6                                                       … ( i v )                             0 = y 2 + y 3 2         ⇒ y 2 + y 3 = 0                                                       … ( v ) a n d             1 = z 2 + z 3 2         ⇒ z 2 + z 3 = 2                                                             … ( v i ) F  is  the  mid-point  of  AC ∴             − 1 = x 1 + x 3 2         ⇒ x 1 + x 3 = − 2                                                       … ( v i i )                           1 = y 1 + y 3 2         ⇒ y 1 + y 3 = 2                                                           … ( v i i i ) a n d     − 4 = z 1 + z 3 2         ⇒ z 1 + z 3 = − 8                                                         … ( i x ) A d d i n g     e q . ( i ) , ( i v )     a n d     ( v i i )     w e     g e t                       2 ( x 1 + x 2 + x 3 ) = 2 + 6 − 2 ∴                                 x 1 + x 2 + x 3 = 3                                                                                                     … ( x ) A d d i n g     e q . ( i i ) , ( v )     a n d     ( v i i i )     w e     g e t                       2 ( y 1 + y 2 + y 3 ) = 4 + 0 + 2 ∴                                 y 1 + y 2 + y 3 = 3                                                                                                     … ( x i ) A d d i n g     e q . ( i i i ) , ( v i )     a n d     ( i x )     w e     g e t                       2 ( z 1 + z 2 + z 3 ) = − 6 + 2 − 8 ∴                                 z 1 + z 2 + z 3 = − 6                                                                                                   … ( x i i ) S u b t r a c t i n g     ( i )     f r o m     e q . ( x )     w e     g e t                                       x 3 = 3 − 2 = 1             ⇒ x 3 = 1 S u b t r a c t i n g     ( i v )     f r o m     e q . ( x )     w e     g e t                                       x 1 = 3 − 6 = − 3             ⇒ x 1 = − 3 S u b t r a c t i n g     ( v i i )     f r o m     e q . ( x )     w e     g e t                                       x 2 = 3 − ( − 2 ) = 5             ⇒ x 2 = 5

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  given  points  are  (a,2,1)  and  (1,−1,1)∴  Distance=(a−1)2+(2+1)2+(1−1)2                     5=a2+1−2a+9Squaring  both  sides,  we  have                  25=a2+1−2a+9⇒              a2−2a−15=0⇒     a2−5a+3a−15=0⇒a(a−5)+3(a−5)=0⇒            (a+3)(a−5)=0∴    a=−3  or  5Hence,  the  value  of  the  filler  is  5  or  −3.

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  given  dimensions
 are  10,  13  and  8Let  a=10,  b=13  and  c=8∴ Required   length=a2+b2+c2                                        =(10)2+(13)2+(8)2=100+169+64=333Hence,  the  value  of  the  filler  is  333.

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  plane  parallel  to  yz-plane  is  perpendicular  to  x-axis. Hence,   the  value  of  the  filler  is  x-axis.

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

x=a  represents  a  plane  parallel  to  yz-plane.

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

y     a n d     z     c o o r d i n a t e s . H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     y     a n d     z     c o o r d i n a t e s .

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

z-coordinates. Hence,   the  value  of  the  filler  is  z-coordinates.

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  equation  of  z-axis  are,   x=0  and  y=0. H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     x = 0     a n d     y = 0 .

Thumbs Up IconUpvote Thumbs Down Icon
New answer posted•a year ago
S
a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

On  the  z-axis,   x=0  and  y=0. ∴     T h e     r e q u i r e d     c o o r d i n a t e     i s     i n     t h e     f o r m     o f     ( 0 , 0 , z ) . H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     ( 0 , 0 , z ) .

Thumbs Up IconUpvote Thumbs Down Icon