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New answer posted

9 months ago

0 Follower 20 Views

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

9 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 72 (1+cos2θ)32 (1cos2θ)2cos22θ=2

Put cos 2 θ= t

Equation 2t2– 5t = 0, t (2t – 5) = 0

t=0, 52

cos 20 = 0, 0 < 2 < 4

x1+x2=2 (tan2θ+cot2θ)?

=2 (1+1)+2 (1+1)+2 (1+1)+2 (1+1)=16

New answer posted

9 months ago

0 Follower 18 Views

V
Vishal Baghel

Contributor-Level 10

f (x)=3 (x22)3+4=81.3 (x22)3

f' (x)=81.3 (x22)3.ln3.3 (x22)2.2x

From graph : P, Q, R

New answer posted

9 months ago

0 Follower 5 Views

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

New answer posted

9 months ago

0 Follower 38 Views

V
Vishal Baghel

Contributor-Level 10

 f(x)=|x1|cos|x2|sin|x1|+(x3)|x25x+4|

f(x)=|x1|cos(x2)sin|x1|+(x3)|x1||x4|

x = 1, 4 (doubtful points)

Diff. at x = 1

limx1f(x)f(1)x1=limx1|x1|(sin|x1|cos(x2)+(x3)|x4|)x1

RHD=limx4+3(sin3cos2)0+=+LHD=limx43(sin3cos2)0=} Not diff. at x = 4

New answer posted

9 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

L : y = mx + c, m > 0

y = m (x – 1)

x24y24=1

y=mx±4m24

±4m24=m, m>0

4m24=m

M (5+212, 3+7), N (5212, 37)

=2 (27)=2

New answer posted

9 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

 |x1|y5x2

y=|x1|,y2+x2=5,y0y=5x2

(x1)2+x2=52x22x4=0x2x2=0

(x2)(x+1)=0x=1,2

A=12(5x2|x1|)dx

Area=12(5x2dx|x1|dx)

=125x2dx+11(x1)dx12(x1)dx

=52sin1(1)12=5π412

New answer posted

9 months ago

0 Follower 4 Views

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Vishal Baghel

Contributor-Level 10

 f(α)=1αlog10t1+tdt

f(1α)=11αlog10t1+tdt=1αlog10z1+1z1z2dz

=1αlog10zz(z+1)dz

=1αlog10ttdt=1ln101αlnttdt=1ln10[(lnt)22]1α=(lnα)22ln10

f(e3)+f(e3)=(lne3)22ln10=92ln10

New answer posted

9 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10


(a^+b^).(a^+2b^+2(a^*b^)),a^.b^=12

= 1 + 12+2 + 2 + 0 + 0

=3+32

|a^+2b^+2(a^*b^)2|=1+4+4(12)+22+0=7+22

(2+2)(7+22)cos2θ=9+92+92=27+1822

cos2θ=27+1822*118+112=(27+182)(18112)2(324242)

=486+324229723962*82

164 cos2 (θ)= 90 + 27

New answer posted

9 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

  (a1)2+4= (b1)2+16

= (a1)2+ (b1)2

(b1)2=4& (a1)2=16

b=1±2a=1±4

= 3, 1= 5, 3

x + 2y = 3 …… (i)

3x – y = 8…… (ii)

x + 2y = 3

3x – y = 83 * 2

7x=13x=137

y=397+8=177

k1+k2=x+y=47

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