Ncert Solutions Maths class 11th

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a month ago

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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

a month ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a month ago

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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.

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a month ago

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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.

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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)

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a month ago

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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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V
Vishal Baghel

Contributor-Level 10

Circle x²+y²-2x-4y+4=0
⇒ (x-1)²+ (y-2)²=1
Centre: (1,2) radius=1
line 3x+4y-k=0 intersects the circle at two distinct points.
⇒ distance of centre from the line < radius
⇒ |3*1+4*2-k|/√ (3²+4²) < 1
⇒ |11-k|<5
⇒ 6⇒ k∈ {7,8,9, .,15} since k∈I
Number of K is 9.

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820
⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820
⇒ 1+2+.+n = 820
⇒ n (n+1)=2*820
⇒ n (n+1)=40*41
Since n∈N, so n=40

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