Maths

The least value of |z| where z is complex number which satisfies the inequality exp( ((|z|+3)(|z|-1))/(|z|+1) logₑ2 ) ≥ log_√2 |5√7 + 9i|, i = √-1, is equal to :

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The inequality is experience ( (|z|+3) (|z|-1) / (|z|+1) * log?2 ) ≥ log√? 16.
The right side is log? (1/2) 16 = log? (2? ¹) 2? = (4/-1)log?2 = -4. This seems incorrect.
Let's assume the base of the log on the right is √2. log√? 16 = log? (1/2) 2? = 2 * log?2? = 8.
The inequality becomes: 2^ (|z|+3) (|z|-1) / (|z|+1) ≥ 8 = 2³.
So, (|z|+3) (|z|-1) / (|z|+1) ≥ 3.
Let |z| = t. (t+3) (t-1) / (t+1) ≥ 3
t² + 2t - 3 ≥ 3t + 3
t² - t - 6 ≥ 0
(t-3) (t+2) ≥ 0
Since t = |z| ≥ 0, we must have t-3 ≥ 0.
So, t ≥ 3, which means |z| ≥ 3.
The minimum value of |z| is 3.

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2 months ago

Let C be the locus of the mirror image of a point on the parabola y² = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is :

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Let the lengths of intercepts on x-axis and y-axis made by the circle x² + y² + ax + 2ay + c = 0, (a < 0) be 22 and 25, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

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2 months ago

Let A = {2,3,4,5,.,30} and '~' be an equivalence relation on A*A, defined by (a,b) ~ (c,d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :

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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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Let α ∈ R be such that the function f(x) = { [cos⁻¹(1-{x}²)sin⁻¹(1-{x})] / ({x}-{x}³), x≠0; α, x=0 } is continuous at x=0, where {x} = x-[x], [x] is the greatest integer less than or equal to x. Then :

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Maths Inverse Trigonometric Functions

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2 months ago

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin⁻¹(3x/5) + sin⁻¹(4x/5) = sin⁻¹x is equal to :

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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.

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2 months ago

Consider a rectangle ABCD having 5,7,6,9 points in the interior of the segments AB, CD, BC, DA respectively. Let α be the number of triangles having these points from different sides as vertices and β be the number of quadrilaterals having these points from different sides as vertices. Then (β-α) is equal to :

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Maths Continuity and Differentiability

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Kindly go through the solution

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2 months ago

Let f be a real valued function, defined on R - {-1, 1} and given by f(x) = 3logₑ|(x-1)/(x+1)| - 2/(x-1). Then in which of the following intervals, function f(x) is increasing?

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Kindly go through the solution

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If the foot of the perpendicular from point (4, 3, 8) on the line L₁ : (x-a)/l = (y-2)/3 = (z-b)/4, l ≠ 0 is (3, 5, 7) then the shortest distance between the line L₁ and line L₂ : (x-2)/3 = (y-4)/4 = (z-5)/5 is equal to:

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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

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2 months ago

If (x,y,z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression
3 + (x-11)/((y-19)²(z-12)²) + (y-9)/((x-11)²(z-12)²) + (z-12)/((x-11)²(y-19)²) - (x+y+z)/(14(x-11)(y-19)(z-12))
is equal to :

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