Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Given 2x2+3x+410=r=020?arxr

replace x by 2x in above identity :-

2102x2+3x+410x20=r=020??ar2rxr210r=020??arxr=r=020??ar2rx(20-r)( from (i))

now, comparing coefficient of x7 from both sides

(take r=7 in L.H.S. and r=13 in R.H.S.)

210a7=a13213a7a13=23=8

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

xdydx-y=x2(xcos?x+sin?x),x>0

dydx-yx=x(xcos?x+sin?x)dydx+Py=Q

So, I.F. =e-1xdx1|x|=1x(x>0)

Thus, yx=1x(x(xcos?x+sin?x))dx

yx=xsin?x+C

?y(π)=πC=1

So, y=x2sin?x+x(y)π/2=π24+π2

Also, dydx=x2cos?x+2xsin?x+1

d2ydx2=-x2sin?x+4xcos?x+2sin?x

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2 months ago

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A
alok kumar singh

Contributor-Level 10

1+1-22.1+1-42.3+.+1-202.19

=α-220β

=11-221+423+..+20219

=11-22r=110? r2 (2r-1)=11-411022-35*11

=11-220 (103)

α=11, β=103

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2 months ago

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P
Payal Gupta

Contributor-Level 10

R (10+α3, 83)

|mAQ|=|mAP|

|45α|=|32α|

= 7 not possible α=237.7α+3β=23+8=31

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2 months ago

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P
Payal Gupta

Contributor-Level 10

tanπ8=2h80......... (i)

tanθ=h80.......... (ii)

tanπ8=2tanθ

tan2θ=3224

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

x (1x2)dydx+ (3x2yy4x3)=0

x (1x2)dydx+ (3x21)y=4x3

y=2x1 (x3x)y (3)=18

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2 months ago

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P
Payal Gupta

Contributor-Level 10

[xx2y2+ey/x]xdydx=x+[xx2y2+ey/x]y

ey/x[xdyydx]=xdx+xx2y2(ydxxdy)

ey/xd(y/x)=dxxd(y/x)1(y/x)2

Integrating

ey/x=lnxsin1(yx)+c

Passes (1, 0)

1 = c

α=12exp(e1+π6)

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 r=1a+ (r1)d2r=4, 4 (a+d)=?

ar=1 (12)r+dr=1 (r1) (12)r=4

= (12)2k=0k (12)k1

=14.1 (112)2

= 1

a + d = 4

4 (a + d) = 16

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 x¯=15i=120xi=300

i=120 (xi15)220=9i=120 (xi15)2=180

i=120 (xi+α)2=178*20=3560

4680+2α (300)+20α2=3560

α2+30α+234178=0

= 2, 28

αmax2= (2)2=4

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

S = {1,2, 3, …., n, 2022}

HCF (n, 2022) = 1

2022 = 2 * 1011 ->3 * 337

2022 = 2 * 3 * 337 (prime factorization)

Let n (A) = no members divisible by 2 = 1011

Let n (B) = no members divisible by 3 = 674

Let n (C) = no members divisible by 337 = 6

n (ABC)=n (A)+n (B)+n (C)n (AB)n (BC)n (CA)+n (ABC)

= 1011 + 674 + 6 – 337 – 2 – 3 + 1

= 1350

n (AB)=337

n ( (ABC)')=20221350=672

Prob. =6722022=3361011=112337

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