Maths Class 12 CBSE Solved Question Paper 2020

Answer: (D) 40

Solution:

We know the property of determinants:
\[
|kA| = k^n |A|,
\] where \( n \) is the order of the square matrix. Here, \( k = 2 \), \( n = 3 \), and \( |A| = 5 \). Therefore:
\[
|2A| = 2^3 \cdot |A| = 8 \cdot 5 = 40.
\] Thus, \( |2A| = 40 \).

Answer: (A) \( I \)

Solution:

Given \( A^2 = A \), \( A \) is an idempotent matrix. To find \( (I – A)^3 + A \), expand \( (I – A)^3 \) using the binomial theorem:
\[
(I – A)^3 = I^3 – 3I^2A + 3IA^2 – A^3.
\] Since \( I^2 = I \), \( I^3 = I \), and \( A^2 = A \), substitute these into the expression:
\[
(I – A)^3 = I – 3A + 3A – A = I – A.
\] Now, add \( A \):
\[
(I – A)^3 + A = (I – A) + A = I.
\] Thus, the result is \( I \).

Answer: (B) \( -\frac{2\pi}{5} \)

Solution:

The principal value of \( \tan^{-1}(x) \) lies in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). Given \( \tan \frac{3\pi}{5} \), note that \( \frac{3\pi}{5} \) lies in the second quadrant where the tangent function is negative. We find the equivalent angle in the principal value range:
\[
\frac{3\pi}{5} – \pi = -\frac{2\pi}{5}.
\] Thus, the principal value of \( \tan^{-1} (\tan \frac{3\pi}{5}) \) is \( -\frac{2\pi}{5} \).

Answer: (C) -2/3

Solution:

The projection of \( \vec{a} \) on \( \vec{b} \) is given by:
\[
\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.
\] For the projection to be zero, \( \vec{a} \cdot \vec{b} = 0 \). Compute \( \vec{a} \cdot \vec{b} \):
\[
\vec{a} \cdot \vec{b} = (1)(2) + (-2)(0) + (3)(\lambda) = 2 + 3\lambda.
\] Setting \( \vec{a} \cdot \vec{b} = 0 \):
\[
2 + 3\lambda = 0 \implies \lambda = -\frac{2}{3}.
\] Thus, the value of \( \lambda \) is \( -\frac{2}{3} \).

Answer: (B) \( \vec{r} = -\hat{i} + 5\hat{j} + (4 + \lambda)\hat{k} \)

Solution:

The equation of a line passing through \((-1, 5, 4)\) and perpendicular to the plane \( z = 0 \) can be written as:
\[
\vec{r} = \vec{a} + \lambda\vec{d},
\] where \( \vec{a} = -\hat{i} + 5\hat{j} + 4\hat{k} \) is a point on the line, and \( \vec{d} = \hat{k} \) is the direction vector (perpendicular to \( z = 0 \)). Therefore, the equation becomes:
\[
\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda\hat{k}.
\] Simplify:
\[
\vec{r} = -\hat{i} + 5\hat{j} + (4 + \lambda)\hat{k}.
\] Thus, the correct option is (B).

Answer: (A) 0

Solution:

The particular solution of a differential equation does not contain any arbitrary constants. Arbitrary constants are present only in the general solution. Since we are given a particular solution, the number of arbitrary constants is \( 0 \).

Answer: (D) 2

Solution:

We know that:
\[
\int \sec^2x \, dx = \tan x + C.
\] Evaluate \( \int_{-\pi/4}^{\pi/4} \sec^2x \, dx \):
\[
\int_{-\pi/4}^{\pi/4} \sec^2x \, dx = \tan\left(\frac{\pi}{4}\right) – \tan\left(-\frac{\pi}{4}\right).
\] Since \( \tan(\pi/4) = 1 \) and \( \tan(-\pi/4) = -1 \), we have:
\[
\tan\left(\frac{\pi}{4}\right) – \tan\left(-\frac{\pi}{4}\right) = 1 – (-1) = 2.
\] Thus, the value of the integral is \( 2 \).

Answer: (C) 5 units

Solution:

The distance of a point \((x, y, z)\) from the \( y \)-axis is given by:
\[
\text{Distance} = \sqrt{x^2 + z^2}
\]

For the point \((4, -7, 3)\), we have:
\[
x = 4, \quad z = 3
\]

Substituting the values:
\[
\text{Distance} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

Answer: (C) \( \frac{3}{4} \)

Solution:

The formula for conditional probability is:
\[
P(B’ \mid A) = \frac{P(A \cap B’)}{P(A)}
\]

Since \( A \) and \( B \) are independent:
\[
P(A \cap B’) = P(A) \cdot P(B’)
\]

Here, \( P(B’) = 1 – P(B) = 1 – \frac{1}{4} = \frac{3}{4} \). Thus:
\[
P(A \cap B’) = \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}
\]

Substituting back into the formula:
\[
P(B’ \mid A) = \frac{\frac{1}{4}}{\frac{1}{3}} = \frac{1}{4} \cdot \frac{3}{1} = \frac{3}{4}
\]

Answer: (A) \( a = 2b \)

Solution:

Let \( z = ax + by \). For the maximum value to occur at \( (2, 4) \) and \( (4, 0) \), \( z \) must have the same value at both points:

At \( (2, 4) \):
\[
z = 2a + 4b
\]

At \( (4, 0) \):
\[
z = 4a + 0b
\]

Equating the two:
\[
2a + 4b = 4a \quad \Rightarrow \quad 4b = 2a \quad \Rightarrow \quad a = 2b
\]

Answer: Symmetric

Answer: \( x = 0, 1, 2, \dots \)

Answer: \( 2 \times 3 \)

OR

A square matrix \( A \) is said to be skew-symmetric if \( A’ = -A \) and the diagonal elements of \( A \) are all zero.

Answer: \( x = 0 \).

Solution:

For the curve \( y^2 = 8x \), differentiate to find the slope of the tangent:
\[
2y \frac{dy}{dx} = 8 \quad => \quad \frac{dy}{dx} = \frac{4}{y}.
\] At the origin \( (0, 0) \), the tangent slope is undefined (as \( y = 0 \)). Hence, the normal is a vertical line \( x = 0 \).

Answer: \( 12\pi \, \text{cm}^2/\text{s} \).

Solution:

Area of a circle is \( A = \pi r^2 \). Differentiate with respect to \( t \):
\[
\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.
\] Substitute \( r = 2 \, \text{cm}, \frac{dr}{dt} = 3 \, \text{cm/s} \):
\[
\frac{dA}{dt} = 2\pi (2)(3) = 12\pi \, \text{cm}^2/\text{s}.
\]

Answer: \( \overrightarrow{OP} = 2\hat{i} – \hat{j} + \frac{1}{3}\hat{k} \).

Solution:

Use the section formula:
\[
\overrightarrow{OP} = \frac{m \overrightarrow{OB} + n \overrightarrow{OA}}{m + n}.
\] Substitute \( m = 2, n = 1 \):
\[
\overrightarrow{OP} = \frac{2(2\hat{i} – \hat{j} + 2\hat{k}) + 1(2\hat{i} – \hat{j} – \hat{k})}{2 + 1}.
\] Simplify:
\[
\overrightarrow{OP} = \frac{4\hat{i} – 2\hat{j} + 4\hat{k} + 2\hat{i} – \hat{j} – \hat{k}}{3} = \frac{6\hat{i} – 3\hat{j} + 3\hat{k}}{3} = 2\hat{i} – \hat{j} + \hat{k}.
\]

Answer: \( A (\text{adj} A) = |A| I = 28 I \).

Solution:

The property of a square matrix states:
\[
A (\text{adj} A) = |A| I,
\] where \( I \) is the identity matrix. First, compute \( |A| \):
\[
|A| = \begin{vmatrix} 2 & 0 & 0 \\ -1 & 2 & 3 \\ 3 & 3 & 5 \end{vmatrix} = 2 \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix} = 2 (10 – 9) = 2(1) = 28.
\] Thus:
\[
A (\text{adj} A) = 28 I.
\]

Answer:
1. For \( \int x^4 \log x \, dx \), the result is:
\[
\frac{x^5}{25} (5 \log x – 1) + C
\] 2. For \( \int \frac{2x}{\sqrt[3]{x^2 + 1}} \, dx \), the result is:
\[
3 \sqrt[3]{(x^2 + 1)^2} + C
\]

Solution:

Part 1: To evaluate \( \int x^4 \log x \, dx \), we use integration by parts:

Let \( u = \log x \) and \( dv = x^4 dx \). Then, \( du = \frac{1}{x} dx \) and \( v = \frac{x^5}{5} \).

Using integration by parts formula:
\[
\int u \, dv = uv – \int v \, du,
\] \[
\int x^4 \log x \, dx = \frac{x^5}{5} \log x – \int \frac{x^5}{5} \cdot \frac{1}{x} dx.
\] \[
= \frac{x^5}{5} \log x – \frac{1}{5} \int x^4 dx.
\] \[
= \frac{x^5}{5} \log x – \frac{1}{5} \cdot \frac{x^5}{5}.
\] \[
= \frac{x^5}{5} \log x – \frac{x^5}{25}.
\] \[
= \frac{x^5}{25} (5 \log x – 1) + C.
\]

Part 2: To evaluate \( \int \frac{2x}{\sqrt[3]{x^2 + 1}} \, dx \):

Let \( t = x^2 + 1 \). Then, \( dt = 2x dx \).

\[
\int \frac{2x}{\sqrt[3]{x^2 + 1}} \, dx = \int \frac{1}{\sqrt[3]{t}} \, dt.
\] \[
= \int t^{-1/3} \, dt = \frac{t^{2/3}}{2/3} + C = 3t^{2/3} + C.
\] \[
= 3 \sqrt[3]{(x^2 + 1)^2} + C.
\]

Answer: \( \int_1^3 |2x – 1| \, dx = 6. \)

Solution:

To solve \( \int_1^3 |2x – 1| \, dx \), note that \( |2x – 1| = 2x – 1 \) since \( 2x – 1 \geq 0 \) for \( x \geq 0.5 \).

\[
\int_1^3 |2x – 1| \, dx = \int_1^3 (2x – 1) \, dx.
\] \[
= \left[ x^2 – x \right]_1^3.
\] \[
= (3^2 – 3) – (1^2 – 1).
\] \[
= (9 – 3) – (1 – 1) = 6.
\] \

Answer: The probability is \( \frac{26}{51} \).

Solution:

There are 26 red cards and 26 black cards in a deck. Probability for the first card:

\[
P(\text{red}) = \frac{26}{52}.
\] If the first card is red, the probability of the second card being black is:
\[
P(\text{black}) = \frac{26}{51}.
\] \[
P(\text{one red, one black}) = P(\text{red first, black second}) + P(\text{black first, red second}).
\] \[
= \frac{26}{52} \cdot \frac{26}{51} + \frac{26}{52} \cdot \frac{26}{51}.
\] \[
= 2 \cdot \frac{26}{52} \cdot \frac{26}{51}.
\] \[
= \frac{2 \cdot 26}{52} \cdot \frac{26}{51} = \frac{26}{51}.
\]

Answer: \( \int \frac{dx}{\sqrt{9 – 4x^2}} = \frac{\arcsin\left(\frac{2x}{3}\right)}{2} + C. \)

Solution:

Let \( x = \frac{3}{2} \sin \theta \). Then \( dx = \frac{3}{2} \cos \theta \, d\theta \) and \( \sqrt{9 – 4x^2} = \sqrt{9 – 4\left(\frac{9}{4} \sin^2 \theta\right)} = \sqrt{9(1 – \sin^2 \theta)} = 3 \cos \theta \).

\[
\int \frac{dx}{\sqrt{9 – 4x^2}} = \int \frac{\frac{3}{2} \cos \theta \, d\theta}{3 \cos \theta}.
\] \[
= \int \frac{1}{2} d\theta = \frac{\theta}{2} + C.
\] \[
\theta = \arcsin\left(\frac{2x}{3}\right).
\] \[
\int \frac{dx}{\sqrt{9 – 4x^2}} = \frac{\arcsin\left(\frac{2x}{3}\right)}{2} + C.
\]

Answer: \( \sin^{-1} \left( 2x \sqrt{1 – x^2} \right) = 2 \cos^{-1} x \).

Solution:

Let \( \theta = \cos^{-1} x \). Then, \( \cos \theta = x \) and \( \sin \theta = \sqrt{1 – x^2} \).

Now, consider:
\[
\sin^{-1} \left( 2x \sqrt{1 – x^2} \right).
\] Substitute \( x = \cos \theta \), so:
\[
2x \sqrt{1 – x^2} = 2 \cos \theta \sin \theta = \sin 2\theta.
\] Thus,
\[
\sin^{-1} \left( 2x \sqrt{1 – x^2} \right) = \sin^{-1} (\sin 2\theta) = 2\theta = 2 \cos^{-1} x.
\]

Answer: \( \frac{d^2y}{dx^2} = \frac{2}{a} \).

Solution:

Given \( x = at^2 \) and \( y = 2at \):

\[
\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 2a.
\] \[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}.
\] Differentiate \( \frac{dy}{dx} \) w.r.t. \( x \):
\[
\frac{d}{dx} \left( \frac{1}{t} \right) = \frac{d}{dt} \left( \frac{1}{t} \right) \cdot \frac{dt}{dx}.
\] \[
\frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}, \quad \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2at}.
\] \[
\frac{d^2y}{dx^2} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3}.
\] Using \( t = \sqrt{\frac{x}{a}} \),
\[
\frac{d^2y}{dx^2} = \frac{2}{a}.
\]

Answer: Points are \( (1, -6) \) and \( (-1, 2) \).

Solution:

The slope of the tangent to the curve is given by:
\[
\frac{dy}{dx} = 3x^2 – 6x – 4.
\] The slope of the line \( 4x + y – 3 = 0 \) is \( -4 \) (compare to \( y = -4x + 3 \)).
Equating slopes:
\[
3x^2 – 6x – 4 = -4.
\] \[
3x^2 – 6x = 0 \quad \Rightarrow \quad 3x(x – 2) = 0.
\] \[
x = 0 \quad \text{or} \quad x = 2.
\] Substitute into \( y = x^3 – 3x^2 – 4x \) to find points:
For \( x = 0 \): \( y = 0^3 – 3(0)^2 – 4(0) = 0 \). Point is \( (0, 0) \).
For \( x = 2 \): \( y = 2^3 – 3(2)^2 – 4(2) = 8 – 12 – 8 = -12 \). Point is \( (2, -12) \).

Answer: Unit vector = \( \frac{\vec{n}}{\|\vec{n}\|} \), where \( \vec{n} = \vec{a} \times \vec{b} \).

Solution:

The cross product \( \vec{a} \times \vec{b} \) is given by:
\[
\vec{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
5 & 6 & -2 \\
7 & 6 & 2
\end{vmatrix}.
\] Expand the determinant:
\[
\vec{n} = \hat{i} \begin{vmatrix} 6 & -2 \\ 6 & 2 \end{vmatrix}
– \hat{j} \begin{vmatrix} 5 & -2 \\ 7 & 2 \end{vmatrix}
+ \hat{k} \begin{vmatrix} 5 & 6 \\ 7 & 6 \end{vmatrix}.
\] \[
\vec{n} = \hat{i} (12 – (-12)) – \hat{j} (10 – (-14)) + \hat{k} (30 – 42).
\] \[
\vec{n} = \hat{i} (24) – \hat{j} (24) – \hat{k} (12).
\] \[
\vec{n} = 24\hat{i} – 24\hat{j} – 12\hat{k}.
\] Normalize \( \vec{n} \) to find the unit vector:
\[
\|\vec{n}\| = \sqrt{24^2 + (-24)^2 + (-12)^2} = \sqrt{576 + 576 + 144} = \sqrt{1296} = 36.
\] \[
\text{Unit vector} = \frac{\vec{n}}{\|\vec{n}\|} = \frac{24\hat{i} – 24\hat{j} – 12\hat{k}}{36} = \frac{2}{3}\hat{i} – \frac{2}{3}\hat{j} – \frac{1}{3}\hat{k}.
\]

Answer: \( k = -1 \).

Solution:

Direction ratios of the first line are \( 1, -1, k \). Direction ratios of the second line are \( 1, \frac{1}{2}, -1 \).

The lines are perpendicular if their dot product is zero:
\[
1 \cdot 1 + (-1) \cdot \frac{1}{2} + k \cdot (-1) = 0.
\] \[
1 – \frac{1}{2} – k = 0 \quad \Rightarrow \quad \frac{1}{2} – k = 0 \quad \Rightarrow \quad k = -1.
\]

Answer: Probability = \( 0.189 \).

Solution:

The probability of finding a green signal on any day is \( P(G) = 0.3 \). The probability of not finding a green signal is \( P(\text{not G}) = 1 – 0.3 = 0.7 \).

We need exactly two green signals in three days. This follows a binomial distribution:
\[
P(X = 2) = \binom{3}{2} (P(G))^2 (P(\text{not G}))^1.
\] \[
P(X = 2) = 3 (0.3)^2 (0.7) = 3 (0.09) (0.7) = 3 \cdot 0.063 = 0.189.
\]

Answer: \( R \) is an equivalence relation.

Solution:

To prove that \( R \) is an equivalence relation, we verify reflexivity, symmetry, and transitivity:

1. **Reflexivity**: For any \( (a, b) \in N \times N \),
\[
(a, b) R (a, b) \iff ab = ab \quad \text{(True)}.
\] Thus, \( R \) is reflexive.

2. **Symmetry**: If \( (a, b) R (c, d) \), then \( ad = bc \). Multiplying both sides by 1 gives \( bc = ad \), so \( (c, d) R (a, b) \). Hence, \( R \) is symmetric.

3. **Transitivity**: If \( (a, b) R (c, d) \) and \( (c, d) R (e, f) \), then \( ad = bc \) and \( cf = de \). Multiplying the two equations gives:
\[
(ad)(cf) = (bc)(de) \quad \Rightarrow \quad af = bf.
\] Thus, \( (a, b) R (e, f) \), proving transitivity.

Since \( R \) satisfies all three properties, it is an equivalence relation.

Answer: \( \frac{dy}{dx} = e^{x^2 \cos x}(2x \cos x – x^2 \sin x) – (\cos x)^x (\ln(\cos x) + x \tan x) \).

Solution:

Given \( y = e^{x^2 \cos x} + (\cos x)^x \), differentiate term by term:

1. For \( e^{x^2 \cos x} \):
\[
\frac{d}{dx} \left( e^{x^2 \cos x} \right) = e^{x^2 \cos x} \cdot \frac{d}{dx}(x^2 \cos x),
\] \[
\frac{d}{dx}(x^2 \cos x) = 2x \cos x – x^2 \sin x.
\] Thus,
\[
\frac{d}{dx} \left( e^{x^2 \cos x} \right) = e^{x^2 \cos x} (2x \cos x – x^2 \sin x).
\]

2. For \( (\cos x)^x \):
\[
\frac{d}{dx} \left( (\cos x)^x \right) = (\cos x)^x \cdot \left( \ln(\cos x) + x \cdot \frac{-\sin x}{\cos x} \right),
\] \[
\frac{d}{dx} \left( (\cos x)^x \right) = (\cos x)^x \cdot (\ln(\cos x) – x \tan x).
\]

Combine the results:
\[
\frac{dy}{dx} = e^{x^2 \cos x} (2x \cos x – x^2 \sin x) – (\cos x)^x (\ln(\cos x) + x \tan x).
\]

Answer: \( \int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| + C \).

Solution:

We use the standard formula for \( \int \sec^3 x \, dx \):
\[
\int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln |\sec x + \tan x| + C.
\] No additional derivation is needed here as this is a well-known result.

Answer: \( x = \frac{y^3}{3} – e^{-y} + C \).

Solution:

Rearranging the given equation:
\[
y e^y \, dx = (y^3 + 2x e^y) \, dy \quad \Rightarrow \quad y e^y \, dx – 2x e^y \, dy = y^3 \, dy.
\] Divide through by \( e^y \):
\[
y \, dx – 2x \, dy = \frac{y^3}{e^y} \, dy.
\]

Separate variables:
\[
\frac{dx}{x} = \frac{2 \, dy}{y} + \frac{y^2}{e^y} \, dy.
\]

Integrate:
1. Integrating \( \int \frac{dx}{x} \), we get:
\[
\ln |x| = \int \frac{2 \, dy}{y} + \int \frac{y^2}{e^y} \, dy.
\]

Solve \( \int \frac{2 \, dy}{y} \):
\[
\int \frac{2 \, dy}{y} = 2 \ln |y|.
\]

Solve \( \int \frac{y^2}{e^y} \, dy \) (use substitution \( u = y^3 \)):
\[
\int \frac{y^2}{e^y} \, dy = -e^{-y} + C_1.
\]

Combine the results:
\[
\ln |x| = 2 \ln |y| – e^{-y} + C.
\]

Simplify:
\[
x = \frac{y^3}{3} – e^{-y} + C.
\]

Answer: The particular solution is \( y = x \tan^{-1}(x) \).

Solution:

Rewriting the given equation:
\[
x \frac{dy}{dx} = y – x \tan \left( \frac{y}{x} \right).
\] Divide through by \( x \):
\[
\frac{dy}{dx} = \frac{y}{x} – \tan \left( \frac{y}{x} \right).
\]

Let \( z = \frac{y}{x} \), so \( y = xz \). Differentiate:
\[
\frac{dy}{dx} = z + x \frac{dz}{dx}.
\]

Substitute:
\[
z + x \frac{dz}{dx} = z – \tan(z).
\]

Simplify:
\[
x \frac{dz}{dx} = -\tan(z).
\]

Separate variables:
\[
\int \frac{dz}{\tan(z)} = -\int \frac{dx}{x}.
\]

Solve:
\[
\ln |\sin(z)| = -\ln |x| + C_1.
\]

Substitute \( z = \tan^{-1} x \):
\[
y = x \tan^{-1} x.
\] At \( x = 1 \), \( y = \frac{\pi}{4} \), so this satisfies the equation.

Answer: Maximize \( Z = 150x + 250y \), subject to constraints \( x + y \leq 35 \), \( 1000x + 2000y \leq 50000 \), and \( x, y \geq 0 \).

Solution:

Let \( x \) be the number of chairs and \( y \) be the number of tables.

1. Objective function (profit):
\[
Z = 150x + 250y.
\]

2. Constraints:
– Investment constraint:
\[
1000x + 2000y \leq 50000 \quad \Rightarrow \quad x + 2y \leq 50.
\] – Space constraint:
\[
x + y \leq 35.
\] – Non-negativity:
\[
x \geq 0, \, y \geq 0.
\]

Graph the constraints and find the feasible region:
– \( x + 2y = 50 \) intersects axes at \( (50, 0) \) and \( (0, 25) \).
– \( x + y = 35 \) intersects axes at \( (35, 0) \) and \( (0, 35) \).

Find corner points of the feasible region and calculate \( Z \) at each:
– At \( (0, 25) \): \( Z = 150(0) + 250(25) = 6250 \).
– At \( (10, 20) \): \( Z = 150(10) + 250(20) = 6500 \).
– At \( (35, 0) \): \( Z = 150(35) + 250(0) = 5250 \).

The maximum profit is \( Z = 6500 \) at \( (10, 20) \), i.e., 10 chairs and 20 tables.

Answer: The probability that the transferred ball is black is \( \frac{7}{10} \).

Solution:

Let \( E_1 \): A black ball is transferred from Bag I to Bag II.
\( E_2 \): A red ball is transferred from Bag I to Bag II.
\( A \): A black ball is drawn from Bag II.

Using Bayes’ theorem:
\[
P(E_1 | A) = \frac{P(E_1) \cdot P(A | E_1)}{P(E_1) \cdot P(A | E_1) + P(E_2) \cdot P(A | E_2)}.
\]

**Step 1: Compute probabilities**
– \( P(E_1) = P(E_2) = \frac{1}{2} \) (equal chances of transferring red or black ball).

**Step 2: Calculate \( P(A | E_1) \):**
If a black ball is transferred to Bag II, Bag II will have \( 4 \) red and \( 4 \) black balls. Probability of drawing a black ball:
\[
P(A | E_1) = \frac{4}{8} = \frac{1}{2}.
\]

**Step 3: Calculate \( P(A | E_2) \):**
If a red ball is transferred to Bag II, Bag II will have \( 5 \) red and \( 3 \) black balls. Probability of drawing a black ball:
\[
P(A | E_2) = \frac{3}{8}.
\]

**Step 4: Substitute into Bayes’ formula:**
\[
P(E_1 | A) = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{3}{8}}.
\]

Simplify:
\[
P(E_1 | A) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{3}{16}} = \frac{\frac{1}{4}}{\frac{7}{16}} = \frac{4}{7}.
\]

Thus, the probability that the transferred ball is black is \( \frac{4}{7} \).

Answer:

Mean: \( 0.6 \). Variance: \( 0.56 \).

Solution:

Let \( X \) be the random variable denoting the number of white balls drawn. \( X \) can take values \( 0, 1, 2 \) with probabilities calculated using hypergeometric distribution:
\[
P(X = x) = \frac{\binom{2}{x} \binom{8}{3-x}}{\binom{10}{3}}.
\]

**Step 1: Calculate probabilities**
– \( P(X = 0) \): No white balls drawn.
\[
P(X = 0) = \frac{\binom{2}{0} \binom{8}{3}}{\binom{10}{3}} = \frac{1 \cdot 56}{120} = \frac{7}{15}.
\]

– \( P(X = 1) \): One white ball drawn.
\[
P(X = 1) = \frac{\binom{2}{1} \binom{8}{2}}{\binom{10}{3}} = \frac{2 \cdot 28}{120} = \frac{14}{15}.
\]

– \( P(X = 2) \): Two white balls drawn.
\[
P(X = 2) = \frac{\binom{2}{2} \binom{8}{1}}{\binom{10}{3}} = \frac{1 \cdot 8}{120} = \frac{2}{15}.
\]

**Step 2: Find mean**
The mean \( \mu = E(X) \) is given by:
\[
E(X) = \sum_{x=0}^{2} x \cdot P(X = x).
\]

\[
E(X) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{14}{15} + 2 \cdot \frac{2}{15} = \frac{14}{15} + \frac{4}{15} = \frac{18}{15} = 1.2.
\]

**Step 3: Find variance**
The variance \( \sigma^2 \) is given by:
\[
\sigma^2 = E(X^2) – (E(X))^2.
\]

First, calculate \( E(X^2) \):
\[
E(X^2) = \sum_{x=0}^{2} x^2 \cdot P(X = x).
\]

\[
E(X^2) = 0^2 \cdot \frac{7}{15} + 1^2 \cdot \frac{14}{15} + 2^2 \cdot \frac{2}{15} = \frac{14}{15} + \frac{8}{15} = \frac{22}{15}.
\]

Now, compute variance:
\[
\sigma^2 = \frac{22}{15} – \left(\frac{18}{15}\right)^2 = \frac{22}{15} – \frac{324}{225} = \frac{110 – 108}{75} = \frac{2}{75}.
\]

Thus, the mean is \( 1.2 \) and the variance is \( \frac{2}{75} \).

Answer: The solution is \( x = 1, y = 2, z = -1 \).

Solution:

Step 1: Find the inverse of \( A \).

To find \( A^{-1} \), calculate the adjoint of \( A \) and divide it by the determinant of \( A \). The determinant is:
\[
\text{det}(A) = 1(2 – (-2)) – 2(3 – 4) + (-3)(-3 – (-4)) = 4 + 2 + 3 = 9.
\]

Now, calculate the adjoint of \( A \) by finding the cofactors of each element of \( A \) and transposing the cofactor matrix.

Finally, compute \( A^{-1} = \frac{\text{Adj}(A)}{\text{det}(A)} \).

The inverse is:
\[
A^{-1} = \frac{1}{9} \begin{bmatrix} 8 & -1 & -5 \\ -3 & 1 & 6 \\ 7 & -2 & -5 \end{bmatrix}.
\]

Step 2: Solve the system of equations.

Write the equations in matrix form:
\[
\begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix}.
\]

Multiply both sides by \( A^{-1} \):
\[
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix}.
\]

Substitute \( A^{-1} \) and solve:
\[
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 8 & -1 & -5 \\ -3 & 1 & 6 \\ 7 & -2 & -5 \end{bmatrix} \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix}.
\]

Perform the matrix multiplication:
\[
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 \\ 18 \\ -9 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}.
\]

Thus, the solution is \( x = 1, y = 2, z = -1 \).

Answer: The area is \( 18 \, \text{units}^2 \).

Solution:

Use the formula for the area of a triangle given vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.
\]

Substitute \( (x_1, y_1) = (2, -2), (x_2, y_2) = (4, 5), (x_3, y_3) = (6, 2) \):
\[
\text{Area} = \frac{1}{2} \left| 2(5 – 2) + 4(2 – (-2)) + 6((-2) – 5) \right|.
\]

Simplify:
\[
\text{Area} = \frac{1}{2} \left| 2(3) + 4(4) + 6(-7) \right| = \frac{1}{2} \left| 6 + 16 – 42 \right| = \frac{1}{2} \left| -20 \right| = 10.
\]

Thus, the area is \( 10 \, \text{units}^2 \).