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New answer posted

8 months ago

0 Follower 30 Views

R
Raj Pandey

Contributor-Level 9

f' (x)= (x- (1+x)ln (1+x)/ (x² (1+x). Let h (x)=x- (1+x)ln (1+x).
h' (x)=-ln (1+x). h' (x)>0 for x∈ (-1,0), <0 for x (0, ).
h (0)=0, so h (x)≤0. f' (x)≤0. f is decreasing.

New answer posted

8 months ago

0 Follower 23 Views

R
Raj Pandey

Contributor-Level 9

3+2√-54=3+6√6i= (3+√6i)².
The difference is ± (3+√6i)? (3-√6i).
Possible values are 2√6i, -2√6i, 6, -6.
Imaginary part is ±2√6.

New answer posted

8 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

8 months ago

0 Follower 8 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

8 months ago

0 Follower 12 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

8 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

lim (x?0) (tan (? /4+x)¹/? = e^ (lim (x?0) (tan (? /4+x)-1)/x)
= e^ (lim (x?0) (2tanx/ (1-tanx)/x) = e².

New answer posted

8 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

S? =0 ⇒ (a? +a? ) (11/2)=0 ⇒ a? =-a?
2a? +10d=0 ⇒ a? =-5d.
Sum = a? +a? +.+a? = (a? +a? ) (12/2) = 6 (2a? +22d)
= 6 (2a? +22 (-a? /5) = 6 (2a? -22a? /5) = 6 (-12a? /5)=-72a? /5. k=-72/5.

New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Normal to plane is n= (-4i+5j+7k).
Plane: -4 (x-3)+5 (y-1)+7 (z-1)=0 ⇒ -4x+5y+7z=0.
Passes through (α, -3,5) ⇒ -4α-15+35=0 ⇒ α=5.

New answer posted

8 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

f (x)=a (x-3) (x-α)
f (2)=a (2-α)
f (-1)=a (-4) (-1-α)=4a (1+α)
f (-1)+f (2)=0 ⇒ a (2-α+4+4α)=0 ⇒ a≠0 ⇒ 5α=-2 ⇒ α=-0.4
α ∈ (-1,0)

New answer posted

8 months ago

0 Follower 13 Views

R
Raj Pandey

Contributor-Level 9

λ=- (sin? θ+cos? θ) = - (sin²θ+cos²θ)²-2sin²θcos²θ)
λ = - (1-½sin²2θ) = ½sin²2θ-1
sin²2θ ∈
λ ∈ [-1, -1/2]

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