Class 12th

Two tangents are drawn from the point P(-1, 1) to the circle x² + y² - 2x – 6y + 6 = 0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to:

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Contributor-Level 10

Circle: (x-1)² + (y-3)² = 4. C= (1,3), r=2.
Length of tangent from P (-1,1) is L = √ (-1)²+1²-2 (-1)-6 (1)+6) = √ (1+1+2-6+6) = √4 = 2.

Area of quadrilateral PACB = 2 * Area (PAC) = 2 * (1/2 * L * r) = 2*2=4.
AB is chord of contact. T=0 => -x+y- (x-1)-3 (y-1)+6=0 => -2x-2y+10=0 => x+y=5.
Distance of C from AB = |1+3-5|/√2 = 1/√2.
Length of AB = 2√ (r²-d²) = 2√ (4-1/2) = 2√ (7/2) = √14.
Area of ABD =?

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7 months ago

If the area of the bounded region R = {(x,y) : max{0, logₑx} ≤ y ≤ 2ˣ, 1/2 ≤ x ≤ 2} is, α(logₑ2)⁻¹ + β(logₑ2) + γ, then the value of (α + β - 2γ)² is equal to:

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Contributor-Level 10

Area = ∫? ² (2? - logx)dx = [2? /ln2 - (xlnx-x)]? ²
= (4/ln2 - (2ln2-2) - (2/ln2 - (0-1) = 2/ln2 - 2ln2 + 1.
α=2, β=-2, γ=1.
(α+β-2γ)² = (2-2-2)² = 4.

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New answer posted

7 months ago

The value of the integral ∫₋₁¹ log(x + √(x² + 1))dx is:

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Class 12th Differential Equations

Contributor-Level 10

log (x + √ (x²+1) is an odd function.
f (-x) = log (-x + √ (x²+1) = log (1/ (x+√ (x²+1) = -log (x+√ (x²+1) = -f (x).
Integral of an odd function over a symmetric interval is 0.

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7 months ago

Let y = y(x) be the solution of the differential equation xdy = (y + x³cosx)dx with y(π) = 0, then y(π/2) is equal to:

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Class 12th Limits and Derivatives

Contributor-Level 10

xdy - ydx = x³cosxdx
(xdy-ydx)/x² = xcosxdx
d (y/x) = xcosxdx
y/x = ∫xcosxdx = xsinx - ∫sinxdx = xsinx + cosx + c
y = x²sinx + xcosx + cx
y (π) = 0 + π (-1) + cπ = 0 ⇒ c = 1
y = x²sinx + xcosx + x
y (π/2) = (π/2)² (1) + 0 + π/2 = π²/4 + π/2

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7 months ago

The value of lim(n→∞) [ Σ(k=1 to n) (2k-1) + 8n ] / [ Σ(k=1 to n) (2k-1) + 4n ] is equal to:

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Contributor-Level 10

lim (n→∞) [n² + 8n] / [n² + 4n] = 1.
The question is likely a Riemann sum.
lim (n→∞) (1/n) Σ [ (2k/n - 1/n) / (2k/n - 1/n + 4) ]
This is too complex. Let's follow the image solution.
lim (n→∞) (1/n) Σ [ 2 (k/n) + 8 ] / [ 2 (k/n) + 4 ]
∫? ¹ (2x+8)/ (2x+4) dx = ∫? ¹ (1 + 4/ (2x+4) dx = [x + 2ln|2x+4|]? ¹
= (1 + 2ln6) - (0 + 2ln4) = 1 + 2ln (6/4) = 1 + 2ln (3/2).

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New answer posted

7 months ago

The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation √13.44, then the standard deviation of the second sample is:

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Contributor-Level 10

1st sample: n? =100, x? =15, σ? =3. Σx? = 1500. Σx? ² = n? (σ? ²+x? ²) = 100 (9+225) = 23400.
Whole group: n=250, x? =15.6, σ²=13.44. Σx = 250*15.6 = 3900.
2nd sample: n? =150. Σy? = 3900 - 1500 = 2400. y? = 2400/150 = 16.
Σ (x+y)² = n (σ²+x? ²) = 250 (13.44+15.6²) = 250 (13.44+243.36) = 64200.
Σy? ² = 64200 - 23400 = 40800.
σ? ² = Σy? ²/n? - y? ² = 40800/150 - 16² = 272 - 256 = 16.
σ? = 4.

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New answer posted

7 months ago

Consider functions f : A → B and g : B → C (A,B,C ⊂ R) such that (gof)⁻¹ exists, then:

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Class 12th Differential Equations

Contributor-Level 10

If (gof)? ¹ exist then gof is a bijective function. For gof to be bijective, f must be one-one and g must be onto.

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7 months ago

A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of any ellipse with eccentricity 1/3 and the distance of the nearer focus from this directrix is 8/√53, then the equation of the other directrix can be:

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Class 12th Chemistry Coordination Compounds

Contributor-Level 10

dy/dx = e^ (3x+4y) = e³? e?
e? dy = e³? dx
∫e? dy = ∫e³? dx
-e? /4 = e³? /3 + C
y (0)=0 ⇒ -1/4 = 1/3 + C ⇒ C = -7/12.
-e? /4 = e³? /3 - 7/12
e? = (7 - 4e³? )/3
y = (-1/4)ln (7-4e³? )/3)
x = -2/3 ln2 = ln (2? ²/³) = ln (1/4¹/³)
e³? = e^ (ln (1/4) = 1/4.
y = (-1/4)ln (7-1)/3) = (-1/4)ln2.
α = -1/4.

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New answer posted

7 months ago

Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl?·3NH?·3H?O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ______.

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Raj Pandey

Contributor-Level 9

CrCl? ·3NH? ·3H? O gives 3 moles of AgCl precipitate. This means all three Cl? are outside the coordination sphere.
The complex is [Cr (NH? )? (H? O)? ]Cl?
The 3 chloride ions satisfy only primary valency.
secondary valency satisfied by chloride ion = 0

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New answer posted

7 months ago

Let X be a random variable such that the probability function of a distribution is given by P(X=0) = 1/2, P(X = j) = (1/3)? (j = 1,2,3,.∞). Then the mean of the distribution and P(X is positive and even) respectively are:

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