Ncert Solutions Maths class 12th

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a month ago

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P
Payal Gupta

Contributor-Level 10

(x1) (x2x+1)=0

x=1, x=1±3i2=12+32i, 1232i=eiπ3, eiπ3

Sum of 162th power of roots = 1+ei54π+ei54π=1+1+1=3

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a month ago

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Payal Gupta

Contributor-Level 10

Let A (2, 21, 29), B (1, 16, 23), P (λ, 2, 1), Q (4, 2, 2)

Given ABPQ

AB.PQ=0

(i^+5j^6k^). ( (4λ)i^4j^+k^)=0

4λ206=0

λ=22

(λ11)2+ (4λ11)4=4+84=8

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a month ago

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Payal Gupta

Contributor-Level 10

3cos2x= (31)cosx+1

(3cosx+1) (cosx1)=0

cosx=13 (rejected)

Hence, cos x = 1 x = 0 one solution

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a month ago

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Payal Gupta

Contributor-Level 10

Information missing. The question was droppedby NTA.

 y=||x1|2|

Area bounded region = 12*4*2=4

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a month ago

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Payal Gupta

Contributor-Level 10

R = { (P, Q) |P and Q are at the same distance from the origin}.

at  (1, 1), x2+y2= (1)2+ (1)2=1+1=2

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a month ago

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Payal Gupta

Contributor-Level 10

P 1 : 3 x + 1 5 + 2 1 z = 9           

P2 : x – 3y – z = 5

P3 : 2x + 10y + 14z = 5

Ratio of the direction cosines of P1 and P2

3 2 = 1 5 1 0 = 2 1 1 4

Hence, P1 and P3 are parallel.

New answer posted

a month ago

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Payal Gupta

Contributor-Level 10

 a*(a*b)=(a.b)a|a|2b=0|a|2b(asa.b=0given)

a*(a*(a*b))=|a|2a*b

a*(a*(a*(a*b)))=|a|2a*(a*b)=|a|2(|a|2b)=|a|4b

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a month ago

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Payal Gupta

Contributor-Level 10

Let n be the number of times.

p=12, q=12

According to question,

nC7=nC9n7=9n=16

p (x=2)=16C2 (12)2. (12)14=15213

New answer posted

a month ago

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Payal Gupta

Contributor-Level 10

All the points A (1, 5, 35), B (7, 5,5),  C (1, λ, 7), D (2λ, 1, 2) are coplanar. Hence

AB*AC.AD=|60300λ5282λ1433|=0

5λ244λ+95=0

sumofroots=445

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a month ago

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