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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

(0.16)^log? (1/3 + 1/3² + . to ∞)
= (4/25)^log? (1/2)
= ( (5/2)? ² )^log? /? (1/2) = (5/2)^ (-2 log? /? (1/2)
= (5/2)^ (log? /? ( (1/2)? ² ) = 4

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

lim_ (x→0) (1/x? ) {1 - cos (x²/2) - cos (x²/4) + cos (x²/2)cos (x²/4)} = 2?
⇒ lim_ (x→0) ( (1 - cos (x²/2) (1 - cos (x²/4) / x? ) = 2?
⇒ 2? = 2? ⇒ k = 8

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

New answer posted

2 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

A = [ x 1 ]
[ 1 0 ]
A² = [ x 1 ] [ x 1 ] = [ x²+1 x ]
[ 1 0 ] [ 1 0 ] [ x 1 ]
A? = [ x²+1 x ] [ x²+1 x ]
[ x 1 ] [ x 1 ]
= [ (x²+1)²+x² x (x²+1)+x ]
[ x (x²+1)+x²+1 ]
a? = (x² + 1)² + x² = 109
⇒ x = ±3
a? = x² + 1 = 10

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

S = (2 . ¹P? - 3 . ²P? + 4 . ³P? upto 51 terms) + (1! - 2! + 3! - . upto 51 terms)
∴ [? ? P_ (n-1) = n!]
= (2! - 3! + 4! + 52!) + (1! - 2! + 3! - 4! + . . + (51)!)
= 1! + 52!

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

0 ≤ y ≤ x² + 1, 0 ≤ y ≤ x + 1, 1/2 ≤ x ≤ 2
Required area
= 19/24 + 5/2 = 79/24

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2

New answer posted

2 months ago

0 Follower 12 Views

V
Vishal Baghel

Contributor-Level 10

2π - (sin? ¹ (4/5) + sin? ¹ (5/13) + sin? ¹ (16/65)
= 2π - (tan? ¹ (4/3) + tan? ¹ (5/12) + tan? ¹ (16/63)
= 2π - (tan? ¹ (63/16) + tan? ¹ (16/63)
= 2π - π/2 = 3π/2

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

f (x) = (3x - 7)x²/³
⇒ f (x) = 3x? /³ - 7x²/³
⇒ f' (x) = 5x²/³ - 14/ (3x¹/³)
= (15x - 14) / (3x¹/³) > 0


∴ f' (x) > 0 ∀x ∈ (-∞, 0) U (14/15, ∞)

New answer posted

2 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

T_r+1 =? C_r (3)^ (n-r)/2) (5)^ (r/8) (n ≥ r)
Clearly r should be a multiple of 8.
∴ there are exactly 33 integral terms
Possible values of r can be
0,8,16, . . .,32 * 8
∴ least value of n = 256

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