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New answer posted

a year ago

0 Follower 21 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

The equation of the circle is x²+y²+ax+2ay+c=0, with a<0.
x-intercept = 2√ (g² - c) = 2√ (a/2)² - c) = 2√2. So, a²/4 - c = 2 => a² = 8 + 4c - (i)
y-intercept = 2√ (f² - c) = 2√ (a² - c) = 2√5. So, a² - c = 5 => a² = 5 + c - (ii)
Equating (i) and (ii): 8 + 4c = 5 + c => 3c = -3 => c = -1.
Substituting c in (ii): a² = 5 - 1 = 4. Since a < 0, a = -2.
The equation of the circle is x² + y² - 2x - 4y - 1 = 0.
Completing the square: (x-1)² + (y-2)² = 1+4+1 = 6.
The center is (1,2) and radius is √6.
The tangent is perpendicular to the line x + 2y = 0 (slope -1/2).
So, the slope of the tangent is 2.
Equation of the tangent: (y-2) = 2 (x-1)

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New answer posted

a year ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

Solve sin? ¹ (3x/5) + sin? ¹ (4x/5) = sin? ¹x.
Using the formula sin? ¹a + sin? ¹b = sin? ¹ (a√ (1-b²) + b√ (1-a²):
sin? ¹ ( (3x/5)√ (1 - (4x/5)²) + (4x/5)√ (1 - (3x/5)²) ) = sin? ¹x
(3x/5) * √ (1 - 16x²/25) + (4x/5) * √ (1 - 9x²/25) = x
x * [ (3/5) * √ (25-16x²)/5 + (4/5) * √ (25-9x²)/5 - 1 ] = 0
So, x = 0 is one solution.
For the other part:
3√ (25-16x²) + 4√ (25-9x²) = 25
Let's check integer solutions. If x = 1:
3√ (9) + 4√ (16) = 33 + 44 = 9 + 16 = 25. So x = 1 is a solution.
If x = -1:
3√ (9) + 4√ (16) = 25. So x = -1 is a solution.
The solutions are x = 0, 1, -1.

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

a year ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

The shortest distance D between two skew lines is given by the formula:
D = | (a? - a? ) ⋅ (b? x b? )| / |b? x b? |
Line L? : (x-1)/2 = (y-2)/3 = (z-4)/4
Line L? : (x-2)/3 = (y-4)/4 = (z-5)/5

Here, a? = I + 2j + 4k, b? = 2i + 3j + 4k
a? = 2i + 4j + 5k, b? = 3i + 4j + 5k

a? - a? = I + 2j + k
b? x b? = | I j k |
| 2 3 4 |
| 3 4 5 |
= I (15-16) - j (10-12) + k (8-9) = -i + 2j - k

D = | (i + 2j + k) ⋅ (-i + 2j - k)| / √ (-1)² + 2² + (-1)²)
= |-1 + 4 - 1| / √ (1 + 4 + 1)
= 2 / √6

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

The equation of the plane is given as x + y + z = 42. It is also mentioned that x³ + y³ + z³ = 3xyz.
From the identity, if x³ + y³ + z³ - 3xyz = 0, then x + y + z = 0 or x = y = z.
Given the expression:
3 + (x³ + y³ + z³ - 3xyz) / (xyz)²

Since x³ + y³ + z³ = 3xyz, the expression simplifies to:
3 + 0 = 3

New answer posted

a year ago

0 Follower 23 Views

A
alok kumar singh

Contributor-Level 10

The problem involves a function f (x) defined by a determinant:
f (x) = | sin²x 1+cos²x cos2x |
| 1+sin²x cos²x cos2x |
| sin²x cos²x sin2x |

Applying the row operation R? → R? - R? , we get:
f (x) = | -1 0 |
| 1+sin²x cos²x cos2x |
| sin²x cos²x sin2x |

Expanding the determinant along the first row:
f (x) = -1 (cos²x * sin2x - cos2x * cos²x) - 1 (1+sin²x)sin2x - sin²x * cos2x)
= -cos²x * sin2x + cos2x * cos²x - sin2x - sin²x * sin2x + sin²x * cos2x
= -sin2x (cos²x + sin²x) + cos2x (cos²x + sin²x) - sin2x
= -sin2x + cos2x - sin2x
= cos2x - 2sin2x

To find the maximum value of f (x), we use the form acosθ + bsinθ, where the m

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New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

The problem is to evaluate the integral:
I = ∫? ¹? [x] * e^ [x] / e^ (x-1) dx, where [x] denotes the greatest integer function.

The solution breaks the integral into a sum of integrals over unit intervals:
I = ∑? ∫? ¹ n * e? / e^ (x-1) dx
= ∑? n * e? ∫? ¹ e^ (1-x) dx
= ∑? n * e? [-e^ (1-x)] from n to n+1
= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

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