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4 months ago

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P
Payal Gupta

Contributor-Level 10

Area of shaded region

=(12)032((1x23)32+x)dx+01(1x23)32dx

x=sin3θ

dx=3sin2θcosθdθ

=π4π23sin2θcos4θdθ+(0116)

=9π64+116116=36π256=A

256Aπ=36

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4 months ago

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P
Payal Gupta

Contributor-Level 10

?y(x)=(xx)x

y=xx2

dydx=x2.xx21xx2lnx.2x

dxdy=1xx2+1(1+2lnx) ….(i)

d2xdx=ddx((xx2+1(1+2lnx))1).dxdy

(d2xdy2)x=1=4(d2xdy2)x=1+20=16

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4 months ago

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Payal Gupta

Contributor-Level 10

 f (x)= [1+x]+α2|x|+ {x}+ [x]12 [x]+ {x}

limx0f (x)=α43

limh011+αh111h1=α43

α121α43

32 - 10 + 3 = 0

α=3or1/3

? α in integer, hence = 3

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

4 months ago

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P
Payal Gupta

Contributor-Level 10

= 939

r=4

? 7nnr5r=0

And r = 4 then

n>203

And r should not be 5

n<252

 possible values of n are 7, 8, 9, 10, 11, 12

 Sum of integral value of n = 57

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New question posted

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

( i ) | a z 1 ? b z 2 | 2 + | b z 1 + a z 2 | 2 ? ? ? ? ? ? = | a z 1 | 2 + | b z 2 | 2 ? 2 R e ( a z 1 . b z ¯ 2 ) + | b z 1 | 2 + | a z 2 | 2 + 2 R e ( a z 1 . b z ¯ 2 ) ? ? ? ? ? ? = | a z 1 | 2 + | b z 2 | 2 + | b z 1 | 2 + | a z 2 | 2 ? ? ? ? ? ? = ( a 2 + b 2 ) ( | z 1 | 2 + | z 2 | 2 ) ? ? ? ? ? ? ? ? H e n c e , ? ? t h e ? ? v a l u e ? ? o f ? ? t h e ? ? f i l l e r ? ? i s ? ? ( a 2 + b 2 ) ( | z 1 | 2 + | z 2 | 2 ) . ( i i ) ? 2 5 * ? 9 = ? 1 . 2 5 * ? 1 9 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 5 i * 3 i = 1 5 i 2 = ? 1 5 ? ? ? ? ? ? ? H e n c e , ? ? t h e ? ? v a l u e ? ? o f ? ? t h e ? ? f i l l e r ? ? i s ? ? ? 1 5 .

New answer posted

4 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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