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New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

a², b², c² . A.P.
a² + 1, b² + 1, c² + 1 . P.
a² + ab + bc + ca, b² + ab + bc + ca, c² + ab + bc + CA . P.
⇒ (a + b) (a + c), (a + b) (b + c), (b + c) (c + a) . P.
⇒ 1/ (b+c), 1/ (c+a), 1/ (a+b) . A.P.
⇒ b + c, c + a, a + b . P.

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

  5 6 = 5 0 n + 6 0 m m + n

56 n + 56 m = 50 n + 60 m

6n = 4m

n m = 2 / 3            

9 . n m = 9 * 2 3 = 6            

 

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

Let S = z + 2z² + 3z³ + . + 18z¹?
zS = z² + 2z³ + . + 18z¹?
(1 − z)S = z + z² + z³ + . - 18z¹?
(1 − z)S = z (z¹? - 1)/ (z-1) - 18z¹?
Now z = cos 20° + isin 20° ⇒ z¹? = 1
Also |z| = 1
⇒ (1 − z)S = -18z
⇒ S = -18z / (1-z)
|S|? ¹ = | (1-z)/ (-18z)| = |1-z|/18
= (1/18) |1 – cos 20° - isin 20°| = (1/18) |2sin² 10° — 2isin 10°cos 10°|
= (1/18) |2sin 10° (sin 10° – icos 10°)| = (1/9)sin 10°

New answer posted

6 months ago

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Vishal Baghel

Contributor-Level 10

I? = ∫? ¹ (1 − x? )? · 1 dx
= (1 - x? )? x|? ¹ - ∫? ¹ 7 (1 - x? )? (-4x³)xdx = -28∫? ¹ (1 − x? )? (1 − x? − 1)dx
I? = −28∫? ¹ (1 − x? )? dx + 28∫? ¹ (1 − x? )? dx
I? = -28I? + 28I?
29I? = 28I?
I? /I? = 28/29 ⇒ (29/4) * (I? /I? ) = (29/4) * (28/29) = 7

New answer posted

6 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

The two altitudes are

( x 3 y + 1 ) ( x + 3 y 5 ) = 0            

Point of int. of the 2 altitudes is  ( 2 , 3 )  

Let slope of 3rd altitude be 'm'

then   | m 1 3 1 + 1 3 | = 3 3 m 1 3 + m = ± 3

3 m 1 = ± 3 ± 3 m            

  m = , = 2 2 3 = 1 3          

The third altitude is x = 2

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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6 months ago

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Vishal Baghel

Contributor-Level 10

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

  ( λ , 2 λ , 3 λ )

c o s θ = 6 4 2 s i n θ = 6 4 2 (where q is the angle between L1 & L2)

A r e a Δ O A B = 1 2 ( O A ) ( O B ) s i n θ

= 1 2 ( 3 ) | λ | ( 1 4 ) 6 4 2 = 6 λ = ± 2

            

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

dy/dx = (e? – x)? ¹
dx/dy = e? - x
dx/dy + x = e? ⇒ I.F. = e^ (∫dy) = e?
∴ xe? = ∫e? dy + c
xe? = e²? /2 + c
y (0) = 0 ⇒ c = -1/2
∴ x = e? /2 - (1/2)e? ⇒ e? – e? = 2x
e²? – 2xe? - 1 = 0
e? = x ± √ (x² + 1) ⇒ e? = x + √ (x² + 1)
y = ln (x + √ (x² + 1)

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