Maths NCERT Exemplar Solutions Class 12th Chapter Three

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2 months ago

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A
alok kumar singh

Contributor-Level 10

A2=cos?2θisin?2θisin?2θcos?2θ

Similarly, A5=cos?5θisin?5θisin?5θcos?5θ=abcd

(1) a2+b2=cos2?5θ-sin2?5θ=cos?10θ=cos?75?

(2) a2-d2=cos2?5θ-cos2?5θ=0

(3) a2-b2=cos2?5θ+sin2?5θ=1

(4) a2-c2=cos2?5θ+sin2?5θ=1

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

03? g (x)-f (x)=03? ||x-2|-2|dx-03? |x-2|dx

=12*2*2+1+12*1*1-12*2*2+12*1*1

=2+1+12-2+12=1

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2 months ago

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P
Payal Gupta

Contributor-Level 10

|A| = 2

||A|adj (5adjA3)|

|AP3||adj (5adj (A3))|

=|A|15.56=215*56=29*106

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2 months ago

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P
Payal Gupta

Contributor-Level 10

System of equation is

(2311111|λ|) (xyz)= [244λ4]

R1 – 2 R2, R3 – R2

(0131102|λ|1) (xyz)= (1044λ8)

System of equation will have no solution for λ = 7.

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 A= [a1a2a3b1b2b3c1c2c3], a2, b2, c2 {0, 1}

S = a1+a2+a3+b1+b2+b3+c1+c2+c3 is prime

0s9

Prime value = 2,3, 5, 7

ForS=2, 1+1+0+0+0+0+0+0+09!2!7!=36

Total number of matrices

= 36 + 84 + 126 + 36 = 282

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

  (pq) (pr)

(pq) (pr)

(A) (q) (pr) (B) (p) (pr)

(C) (pr) (pq) (D) (pq)r

Option (D) is correct

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 A3*3, B3*3&AB=0

00|A|=0&|B|=0 only .

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2 months ago

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A
alok kumar singh

Contributor-Level 10

 r=020?50-rC6=50C6+49C6+48C6+.+30C6

=50C6+49C6+48C6+.+30C6+30C7-30C7=50C6+49C6+48C6+.+31C6+31C7-30C7=50C6+50C7-30C7=51C7-30C7

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

K-2h-11-23-1=-1K=2h

? [? ABC]=55

12 (5) (h-1)2+ (K-2)2=55

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2 months ago

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P
Payal Gupta

Contributor-Level 10

Sum of all entries of matrix A must be prime p such that 2 < p < 8 then sum of entries may be 3, 5 or 7

If sum is 3 then possible entries are

(0, 5), (0, 1, 4), (0, 2, 3), (0, 1, 3)

(0, 1, 2, 2) and (1, 2)

 Total number of matrices = 4 + 12 + 12 + 12 + 12 + 4 = 56

If sum is 7 then possible entries are

(0, 2, 25), (0, 03, 4), (0, 1, 5), (0, 3, 1), (0, 2, 3), (1, 4), (1, 2, 2), (1, 2, 3) and (0, 1, 2, 4)

Total number of matrices with sum 7 = 104

 total number of required matrices

= 20 + 56 = 104

= 108

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