Maths NCERT Exemplar Solutions Class 11th Chapter Six
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New answer posted
a month agoContributor-Level 9
Let A=[a?]?. tr(AA?)=3.
Σa?² = 3.
Possible cases for non-zero elements are (1,1,1) or (-1,-1,-1) or (1,1,-1) etc.
Total combinations = ?C? * 2³ = 84 * 8 = 672.C? * 2³ = 84 * 8 = 672.
New answer posted
a month agoContributor-Level 9
Sol. Σ? k(k+1)/2 = (1/2)Σ(k²+k) = (1/2)[ (5051101/6) + (5051/2) ]
The solution uses k=1 to 20.
(1/2)[ (202141/6) + (2021/2) ] = (1/2)[2870+210] = 1540
New answer posted
a month agoNew answer posted
a month agoContributor-Level 9
0 Red, 1 Red, 2 Red, 3 Red
Number of ways = ?C? + ?C?.?C? + ?C?.?C? + ?C?.?C? = 35+175+210+70=490
New answer posted
a month agoContributor-Level 9
P=(x?,y?). 2yy'-6x+y'=0 ⇒ y' = 6x/(2y+1)
(y?-0)/(x?-3/2) = (1+2y?)/(6x?)
9-6y? = 1+2y? ⇒ y?=1. x?=±2. Slope = ±12/3 = ±4. |n|=4.
New answer posted
2 months agoContributor-Level 10
This is a Fill in the blanks Type Questions as classified in NCERT Exemplar
New answer posted
2 months agoContributor-Level 10
This is a True or False Type Questions as classified in NCERT Exemplar
New answer posted
3 months agoContributor-Level 10
This is an Objective Type Questions as classified in NCERT Exemplar
New answer posted
3 months agoContributor-Level 10
This is an Objective Type Questions as classified in NCERT Exemplar
New answer posted
3 months agoContributor-Level 10
This is an Objective Type Questions as classified in NCERT Exemplar
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