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New answer posted

a year ago

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Payal Gupta

Contributor-Level 10

Circle |z−3|≤1⇒ (x−3)2+y2≤1

and line z (4+3i)+z¯ (4−3i)≤24

⇒4x−3y≤12∴slope=tanθ=43

New answer posted

a year ago

0 Follower 11 Views

P
Payal Gupta

Contributor-Level 10

? fλ (x)=4λx3−36λx2+36x+48

fλ (x)=12 (λx2−6λx+3)

For increasing fλ (x)≥0

fλ* (x)=43x3−12x2+36x+48∴fλ* (1)+fλ* (−1)=7312−112=72

New answer posted

a year ago

0 Follower 10 Views

P
Payal Gupta

Contributor-Level 10

 ? −dxdy=x2xy−x2y2−1

∴dydx=xy−x2y2−1x2

Let   xy=v⇒xdydx+y=dvdx

Put x = 1, y = 1 ⇒tan−1=c⇒c=π4

∴tan−1 (xy)=lnx=π4

e (y (e))=tan (1+π4)=tan1+11−tan1

New answer posted

a year ago

0 Follower 9 Views

P
Payal Gupta

Contributor-Level 10

Let   x3=θ⇒θ2∈ (π4, 3π4)

∴y=tan−1 (secθ−tanθ)

tan−1 (1−sinθcosθ)

⇒dydx=−3x22⇒d2ydx2=−3x

∴x2d2ydx2−6y+3π2=0⇒x2y11−6y+3π2=0

New answer posted

a year ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

|a→+b→+2 (a→*b→)|=2, θ∈ (0, π)

squaring on both sides, we get = 2π3

where θ is angle between a^  and   b^.

2|a^*b^|=3=|a^−b^|, S1 is correct.

and projection of a^  on  a^+b^=|a^. (a^+b^)|a^+b^||=12

So (S2) is correct.

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

Let P (x, y, z) be any point on plane P1 then

(x+4)2+ (y−2)2+ (z−1)2= (x−2)2+ (y+2)2+ (z−3)2

∴cosθ=|6−2+3|14⇒θ=π3

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

S. D. = 13

b1*b2=|i^j^k^23λ145|=i^ (15−4λ)+j^ (λ−10)+k^ (5)

| (−i^−2j^−2k^). { (15−4λ)i^+ (λ−10)j^+5k^}| (15−4λ)2+ (λ−10)2+25=13

⇒λ=16

New answer posted

a year ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

 ?  x is a random variable.

∴k+2k+4k+6k+8k=1∴k=121

∴P ( (1<x<4)|x≤2)=4k7k=47

New answer posted

a year ago

0 Follower 18 Views

P
Payal Gupta

Contributor-Level 10

f (x)=x7−7x−2

f' (x)=7x6−7⇒f' (x)=0, x=±1

∴f (1)=<0   and  f (−1)>0

Hence number of real roots of f (x) = 0 are 3.

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

S = Ltn→∞∑r=1nn2 (n2+r2) (n+r)

π8+14ln2

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