Maths Class 12 CBSE Solved Question Paper 2019

Answer: \( |B| = 1 \).

Solution:

We are given:

• \( |A| = 2 \),
• \( AB = 2I \), where \( I \) is the identity matrix.

Taking determinants on both sides of the equation \( AB = 2I \):

\[
|AB| = |2I|.
\]

Using the property of determinants \( |AB| = |A||B| \):

\[
|A||B| = |2I|.
\]

For a scalar multiple of the identity matrix \( kI \), \( |kI| = k^n \), where \( n \) is the order of the matrix. Here, \( n = 3 \):

\[
|2I| = 2^3 = 8.
\]

Substituting the values:

\[
|A||B| = 8.
\]

Given \( |A| = 2 \):

\[
2|B| = 8.
\]

Solving for \( |B| \):

\[
|B| = \frac{8}{2} = 4.
\]

Answer: \( \frac{d}{dx} \) (fof) (x) = 2.

Solution:

Step 1: Find \( (fof)(x) \):

\[
(fof)(x) = f(f(x)).
\]

Given \( f(x) = x + 1 \):

\[
(fof)(x) = f(x + 1).
\]

Substitute \( x + 1 \) into \( f(x) \):

\[
(fof)(x) = (x + 1) + 1 = x + 2.
\]

Step 2: Differentiate \( (fof)(x) \):

\[
\frac{d}{dx} (fof)(x) = \frac{d}{dx}(x + 2) = 1.
\]

Answer: Order = 2, Degree = 4.

Solution:

Step 1: Identify the order:

The highest derivative in the equation is \( \frac{d^2y}{dx^2} \), so the order is 2.

Step 2: Identify the degree:

The degree is the power of the highest derivative after the equation is free from fractional or negative powers of derivatives. Here, \( \left[ \frac{d^2y}{dx^2} \right] \) has a power of 4. Thus, the degree is 4.

Answer: The direction cosines are \( l = 0, m = -\frac{1}{\sqrt{2}}, n = \frac{1}{\sqrt{2}} \).

Solution:

Step 1: Use the formula for direction cosines:

\[
l = \cos 90°, \, m = \cos 135°, \, n = \cos 45°.
\]

Step 2: Compute each cosine value:

\[
l = \cos 90° = 0, \, m = \cos 135° = -\frac{1}{\sqrt{2}}, \, n = \cos 45° = \frac{1}{\sqrt{2}}.
\]

Step 3: Verify the condition:

The condition for direction cosines is:

\[
l^2 + m^2 + n^2 = 1.
\]

Substitute the values:

\[
0^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1.
\]

Thus, the direction cosines are valid.

Answer: The vector equation is:
\[
\vec{r} = (3\hat{i} + 4\hat{j} + 5\hat{k}) + t(2\hat{i} + 2\hat{j} – 3\hat{k}).
\]

Solution:

Step 1: Write the general vector equation:

\[
\vec{r} = \vec{a} + t\vec{b},
\] where \( \vec{a} \) is a point on the line, and \( \vec{b} \) is the direction vector.

Step 2: Substitute \( \vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k} \) and \( \vec{b} = 2\hat{i} + 2\hat{j} – 3\hat{k} \):

\[
\vec{r} = (3\hat{i} + 4\hat{j} + 5\hat{k}) + t(2\hat{i} + 2\hat{j} – 3\hat{k}).
\]

Thus, the vector equation of the line is:
\[
\vec{r} = (3\hat{i} + 4\hat{j} + 5\hat{k}) + t(2\hat{i} + 2\hat{j} – 3\hat{k}).
\]

Answer: (i) Yes, it is a binary operation. (ii) No, it is not associative.

Solution:

Step 1: Check if it is a binary operation:

For \( a, b \in R \), the operation \( a * b = ab + 1 \) is defined for all real numbers, and the result is also a real number. Hence, it is a binary operation.

Step 2: Check if it is associative:

To check associativity, compute \( (a * b) * c \) and \( a * (b * c) \):

\[
(a * b) * c = (ab + 1) * c = ((ab + 1)c) + 1 = abc + c + 1.
\] \[
a * (b * c) = a * (bc + 1) = (a(bc + 1)) + 1 = abc + a + 1.
\]

Since \( (a * b) * c \neq a * (b * c) \), the operation is not associative.

Answer:
\( A = \begin{bmatrix} -8 & 3 & 5 \\ -13 & -1 & -9 \end{bmatrix} \).

Solution:

1. Rearrange the given equation:

\[
2A = 3B – 5C \quad \Rightarrow \quad A = \frac{1}{2}(3B – 5C).
\]

2. Compute \( 3B \):

\[
3B = 3 \cdot \begin{bmatrix} -2 & 2 & 0 \\ 3 & 1 & 4 \end{bmatrix} =
\begin{bmatrix} -6 & 6 & 0 \\ 9 & 3 & 12 \end{bmatrix}.
\]

3. Compute \( 5C \):

\[
5C = 5 \cdot \begin{bmatrix} 2 & 0 & -2 \\ 7 & 1 & 6 \end{bmatrix} =
\begin{bmatrix} 10 & 0 & -10 \\ 35 & 5 & 30 \end{bmatrix}.
\]

4. Subtract \( 5C \) from \( 3B \):

\[
3B – 5C = \begin{bmatrix} -6 & 6 & 0 \\ 9 & 3 & 12 \end{bmatrix} –
\begin{bmatrix} 10 & 0 & -10 \\ 35 & 5 & 30 \end{bmatrix} =
\begin{bmatrix} -16 & 6 & 10 \\ -26 & -2 & -18 \end{bmatrix}.
\]

5. Divide by 2 to find \( A \):

\[
A = \frac{1}{2} \begin{bmatrix} -16 & 6 & 10 \\ -26 & -2 & -18 \end{bmatrix} =
\begin{bmatrix} -8 & 3 & 5 \\ -13 & -1 & -9 \end{bmatrix}.
\]

Answer: \( \ln(\tan x + \sqrt{\tan^2 x + 4}) + C \).

Solution:

1. Let \( u = \tan x \), so \( du = \sec^2 x \, dx \).

\[
\int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} \, dx = \int \frac{du}{\sqrt{u^2 + 4}}.
\]

2. Use the substitution \( u = 2 \sinh t \), so \( du = 2 \cosh t \, dt \) and \( \sqrt{u^2 + 4} = 2 \cosh t \):

\[
\int \frac{du}{\sqrt{u^2 + 4}} = \int \frac{2 \cosh t}{2 \cosh t} \, dt = \int dt = t.
\]

3. Back-substitute \( t = \sinh^{-1} \left( \frac{u}{2} \right) \):

\[
t = \sinh^{-1} \left( \frac{\tan x}{2} \right).
\]

4. The final answer is:

\[
\ln(\tan x + \sqrt{\tan^2 x + 4}) + C.
\]

Answer: \( \frac{1}{2} (x – \cos 2x + C) \).

Solution:

1. Use the identity \( \sin 2x = 2 \sin x \cos x \), so \( 1 – \sin 2x = (\cos x – \sin x)^2 \).

2. Rewrite the integral:

\[
\int \sqrt{1 – \sin 2x} \, dx = \int |\cos x – \sin x| \, dx.
\]

3. Since \( \frac{\pi}{4} < x < \frac{\pi}{2} \), \( \cos x – \sin x < 0 \), so:

\[
|\cos x – \sin x| = \sin x – \cos x.
\]

4. Substitute \( u = \cos x \), so \( du = -\sin x \, dx \):

\[
\int (\sin x – \cos x) \, dx = \int u \, du = \frac{u^2}{2} + C.
\]

5. Back-substitute \( u = \cos x \):

\[
\frac{1}{2} (x – \cos 2x) + C.
\]

Answer: \( y” – 4y’ + 4y = 0 \).

Solution:

1. Differentiate \( y = e^{2x} (a + bx) \) once with respect to \( x \):

\[
y’ = e^{2x} (2a + 2bx + b).
\]

2. Differentiate again:

\[
y” = e^{2x} (4a + 4bx + 2b).
\]

3. Substitute \( y, y’, y” \) into the equation:

\[
y” – 4y’ + 4y = 0.
\]

The given family of curves satisfies this differential equation.

Answer: The magnitude of their difference is \( \sqrt{3} \).

Solution:

1. Let the two unit vectors be \( \vec{a} \) and \( \vec{b} \), such that \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \).

2. Given \( |\vec{a} + \vec{b}| = 1 \), square both sides:

\[
|\vec{a} + \vec{b}|^2 = 1 \quad \Rightarrow \quad (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1.
\]

3. Expand the dot product:

\[
\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 1 \quad \Rightarrow \quad 1 + 2 \vec{a} \cdot \vec{b} + 1 = 1.
\]

4. Simplify:

\[
2 + 2 \vec{a} \cdot \vec{b} = 1 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = -\frac{1}{2}.
\]

5. Now, find \( |\vec{a} – \vec{b}|^2 \):

\[
|\vec{a} – \vec{b}|^2 = (\vec{a} – \vec{b}) \cdot (\vec{a} – \vec{b}) = \vec{a} \cdot \vec{a} – 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}.
\]

6. Substitute values:

\[
|\vec{a} – \vec{b}|^2 = 1 – 2(-\frac{1}{2}) + 1 = 1 + 1 + 1 = 3.
\]

7. Therefore, \( |\vec{a} – \vec{b}| = \sqrt{3} \).

Answer: Events A and B are independent.

Solution:

1. Define events:

\[
A = \{2, 4, 6\} \quad \text{(even numbers)}, \quad B = \{1, 2, 3\} \quad \text{(red numbers)}.
\]

2. Find probabilities:

\[
P(A) = \frac{3}{6} = \frac{1}{2}, \quad P(B) = \frac{3}{6} = \frac{1}{2}.
\]

3. Find \( P(A \cap B) \):

\[
A \cap B = \{2\}, \quad P(A \cap B) = \frac{1}{6}.
\]

4. Check independence:

\[
P(A \cap B) = P(A) \cdot P(B) \quad \Rightarrow \quad \frac{1}{6} = \frac{1}{2} \cdot \frac{1}{2}.
\]

5. Since equality holds, A and B are independent events.

Answer:
(i) \( P(X = 5) = 6 \cdot \left(\frac{1}{2}\right)^6 \),
(ii) \( P(X \leq 5) = 1 – P(X = 6) \).

Solution:

1. Define success probability:

\[
P(\text{success}) = \frac{3}{6} = \frac{1}{2}.
\]

2. Use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}.
\]

3. For (i) \( k = 5 \):

\[
P(X = 5) = \binom{6}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^1 = 6 \cdot \left(\frac{1}{2}\right)^6 = \frac{6}{64} = \frac{3}{32}.
\]

4. For (ii) at most 5 successes:

\[
P(X \leq 5) = 1 – P(X = 6).
\]

5. Compute \( P(X = 6) \):

\[
P(X = 6) = \binom{6}{6} \left(\frac{1}{2}\right)^6 = 1 \cdot \left(\frac{1}{2}\right)^6 = \frac{1}{64}.
\]

6. Therefore:

\[
P(X \leq 5) = 1 – \frac{1}{64} = \frac{63}{64}.
\]

Answer: The relation \( R \) is reflexive and transitive but not symmetric.

Solution:

1. To check reflexivity:

\[
a \leq a \quad \text{for all } a \in R.
\]

This is true. Therefore, \( R \) is reflexive.

2. To check symmetry:

If \( (a, b) \in R \), then \( a \leq b \). For symmetry, we require \( (b, a) \in R \), i.e., \( b \leq a \). This is not necessarily true. For example, let \( a = 2 \) and \( b = 3 \):

\[
2 \leq 3, \quad \text{but } 3 \nleq 2.
\]

Therefore, \( R \) is not symmetric.

3. To check transitivity:

If \( (a, b) \in R \) and \( (b, c) \in R \), then \( a \leq b \) and \( b \leq c \). From this, \( a \leq c \). Therefore, \( R \) is transitive.

Thus, \( R \) is reflexive and transitive but not symmetric.

Answer: The solution is \( x = \frac{1}{5} \).

Solution:

1. Let \( \tan^{-1} 4x = a \) and \( \tan^{-1} 6x = b \). Then:

\[
a + b = \frac{\pi}{4}.
\]

2. Apply the tangent addition formula:

\[
\tan(a + b) = \tan\left(\frac{\pi}{4}\right) = 1.
\]

Using the formula for the tangent of a sum:

\[
\tan(a + b) = \frac{\tan a + \tan b}{1 – \tan a \tan b}.
\]

Substitute \( \tan a = 4x \) and \( \tan b = 6x \):

\[
1 = \frac{4x + 6x}{1 – (4x)(6x)}.
\]

Simplify:

\[
1 = \frac{10x}{1 – 24x^2}.
\]

Cross-multiply:

\[
1 – 24x^2 = 10x.
\]

Rearrange into standard quadratic form:

\[
24x^2 + 10x – 1 = 0.
\]

3. Solve using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}, \quad a = 24, \, b = 10, \, c = -1.
\] \[
x = \frac{-10 \pm \sqrt{10^2 – 4(24)(-1)}}{2(24)} = \frac{-10 \pm \sqrt{100 + 96}}{48}.
\] \[
x = \frac{-10 \pm \sqrt{196}}{48} = \frac{-10 \pm 14}{48}.
\]

Solutions:

\[
x = \frac{4}{48} = \frac{1}{12}, \quad x = \frac{-24}{48} = -\frac{1}{2}.
\]

Since \( x > 0 \), \( x = \frac{1}{12} \).

Answer: The determinant evaluates to \( (a – 1)^3 \).

Solution:

1. Expand the determinant along the third row:

\[
\begin{vmatrix}
a^2 + 2a & 2a + 1 & 1 \\
2a + 1 & a + 2 & 1 \\
3 & 3 & 1
\end{vmatrix} = 3 \begin{vmatrix} 2a + 1 & 1 \\ a + 2 & 1 \end{vmatrix} – 3 \begin{vmatrix} a^2 + 2a & 1 \\ 2a + 1 & 1 \end{vmatrix}.
\]

2. Compute the minor determinants:

\[
\begin{vmatrix} 2a + 1 & 1 \\ a + 2 & 1 \end{vmatrix} = (2a + 1)(1) – (a + 2)(1) = 2a + 1 – a – 2 = a – 1.
\]

\[
\begin{vmatrix} a^2 + 2a & 1 \\ 2a + 1 & 1 \end{vmatrix} = (a^2 + 2a)(1) – (2a + 1)(1) = a^2 + 2a – 2a – 1 = a^2 – 1.
\]

3. Substitute back:

\[
3(a – 1) – 3(a^2 – 1).
\]

4. Simplify:

\[
3(a – 1) – 3(a + 1)(a – 1) = 3(a – 1)[1 – (a + 1)] = 3(a – 1)(-a) = (a – 1)^3.
\]

Answer: \( \frac{dy}{dx} = \frac{x + y}{x – y} \).

Solution:

1. Differentiate both sides with respect to \( x \):

\[
\frac{d}{dx}\left[\log(x^2 + y^2)\right] = \frac{d}{dx}\left[2 \tan^{-1}\left(\frac{y}{x}\right)\right].
\]

2. For the left-hand side:

\[
\frac{1}{x^2 + y^2}(2x + 2y \frac{dy}{dx}).
\]

3. For the right-hand side:

\[
\frac{d}{dx}\left[2 \tan^{-1}\left(\frac{y}{x}\right)\right] = \frac{2}{1 + \left(\frac{y}{x}\right)^2}\left(\frac{x \frac{dy}{dx} – y}{x^2}\right).
\]

4. Simplify and equate both sides to solve for \( \frac{dy}{dx} \).

Answer: \( (1 – x^2)\frac{d^2y}{dx^2} – x\frac{dy}{dx} – 2 = 0 \).

Solution:

1. Given \( y = (\sin^{-1}x)^2 \), first compute \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = 2(\sin^{-1}x) \cdot \frac{1}{\sqrt{1 – x^2}}.
\]

2. Compute \( \frac{d^2y}{dx^2} \):

\[
\frac{d^2y}{dx^2} = 2\left[\frac{1}{\sqrt{1 – x^2}} \cdot \frac{1}{\sqrt{1 – x^2}} – (\sin^{-1}x)\frac{-x}{(1 – x^2)^{3/2}}\right].
\]

Simplify:

\[
\frac{d^2y}{dx^2} = \frac{2}{(1 – x^2)} – \frac{2x(\sin^{-1}x)}{(1 – x^2)^{3/2}}.
\]

3. Substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the equation:

\[
(1 – x^2)\frac{d^2y}{dx^2} – x\frac{dy}{dx} – 2.
\]

Simplify step by step to verify the equation holds true.

Answer: The tangent equation is \( y – \frac{\sqrt{7}}{2} = 2(x – 2) \) and the normal equation is \( y – \frac{\sqrt{7}}{2} = -\frac{1}{2}(x – 2) \).

Solution:

1. Differentiate \( y = \sqrt{3x – 2} \):

\[
\frac{dy}{dx} = \frac{1}{2\sqrt{3x – 2}} \cdot 3 = \frac{3}{2\sqrt{3x – 2}}.
\]

2. The slope of the line \( 4x – 2y + 5 = 0 \) is \( \frac{4}{2} = 2 \). Set the derivative equal to 2:

\[
\frac{3}{2\sqrt{3x – 2}} = 2.
\]

3. Solve for \( x \):

\[
2\sqrt{3x – 2} = 3 => \sqrt{3x – 2} = \frac{3}{2}.
\] \[
3x – 2 = \frac{9}{4} => x = \frac{17}{12}.
\]

4. Find \( y \) at \( x = \frac{17}{12} \):

\[
y = \sqrt{3\left(\frac{17}{12}\right) – 2} = \frac{\sqrt{7}}{2}.
\]

5. The tangent line equation is:

\[
y – \frac{\sqrt{7}}{2} = 2(x – 2).
\]

6. The normal line equation is:

\[
y – \frac{\sqrt{7}}{2} = -\frac{1}{2}(x – 2).
\]

Answer: The integral evaluates to \( \ln|x + 6| – \ln|x – 3| + C \).

Solution:

1. Factorize the denominator:

\[
x^2 + 3x – 18 = (x + 6)(x – 3).
\]

2. Use partial fraction decomposition:

\[
\frac{3x + 5}{(x + 6)(x – 3)} = \frac{A}{x + 6} + \frac{B}{x – 3}.
\]

3. Solve for \( A \) and \( B \):

\[
3x + 5 = A(x – 3) + B(x + 6).
\]

Equating coefficients:

\[
A + B = 3, \quad -3A + 6B = 5.
\]

\[
A = 1, \, B = 2.
\]

4. The integral becomes:

\[
\int \frac{1}{x + 6} \, dx + \int \frac{2}{x – 3} \, dx = \ln|x + 6| – 2\ln|x – 3| + C.
\]

Answer: The integral evaluates to \( \frac{\pi}{2} \).

Solution:

1. Let \( I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \, dx \). Using the given property:

\[
\int_0^\pi f(x) \, dx = \int_0^\pi f(\pi – x) \, dx.
\]

2. Replace \( x \) with \( \pi – x \):

\[
\int_0^\pi \frac{(\pi – x) \sin(\pi – x)}{1 + \cos^2(\pi – x)} \, dx = \int_0^\pi \frac{(\pi – x)(\sin x)}{1 + \cos^2 x} \, dx.
\]

3. Add the two integrals:

\[
I + I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \, dx + \int_0^\pi \frac{(\pi – x) \sin x}{1 + \cos^2 x} \, dx.
\]

4. Simplify:

\[
2I = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx.
\]

5. Solve the integral \( \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx \) using substitution \( u = \cos x \):

\[
\int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx = \int_{-1}^1 \frac{1}{1 + u^2} \, du = \tan^{-1}(u)\bigg|_{-1}^1 = \frac{\pi}{4} + \frac{\pi}{4}.
\]

6. Therefore:

\[
2I = \pi \cdot \frac{\pi}{2} => I = \frac{\pi}{2}.
\]

Answer: \( x^2 + y^2 = C^2 \), where \( C = 1 \).

Solution:

1. Rewrite the equation:

\[
x \, dy – y \, dx = \sqrt{x^2 + y^2} \, dx.
\] \[
\frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x}.
\]

2. Use substitution \( y = vx \) (where \( v = \frac{y}{x} \)):

\[
\frac{dy}{dx} = v + x \frac{dv}{dx}.
\] \[
v + x \frac{dv}{dx} = \sqrt{1 + v^2}.
\]

3. Separate variables:

\[
\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}.
\]

4. Integrate both sides:

\[
\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}.
\] \[
\ln |v + \sqrt{1 + v^2}| = \ln |x| + C.
\]

5. Substitute back \( v = \frac{y}{x} \):

\[
\ln \left( \frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 1} \right) = \ln |x| + C.
\]

6. Simplify and solve for \( y^2 + x^2 = C^2 \):

\[
x^2 + y^2 = C^2, \text{ where } C = 1.
\]

Answer: The angle between \( \overrightarrow{AB} \) and \( \overrightarrow{CD} \) is \( \cos^{-1}\left( \frac{-3}{\sqrt{26} \sqrt{35}} \right) \). The vectors are not collinear.

Solution:

1. Find \( \overrightarrow{AB} \) and \( \overrightarrow{CD} \):

\[
\overrightarrow{AB} = (2 – 1)\hat{i} + (5 – 1)\hat{j} + (0 – 1)\hat{k} = \hat{i} + 4\hat{j} – \hat{k}.
\] \[
\overrightarrow{CD} = (3 – 1)\hat{i} + (2 – (-6))\hat{j} + (-3 – (-1))\hat{k} = 2\hat{i} + 8\hat{j} – 2\hat{k}.
\]

2. Compute the dot product:

\[
\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(2) + (4)(8) + (-1)(-2) = 2 + 32 + 2 = 36.
\]

3. Find magnitudes:

\[
|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{26}, \quad |\overrightarrow{CD}| = \sqrt{2^2 + 8^2 + (-2)^2} = \sqrt{68}.
\]

4. Compute the cosine of the angle:

\[
\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|} = \frac{36}{\sqrt{26} \sqrt{68}}.
\]

5. Check collinearity by testing proportionality between components of \( \overrightarrow{AB} \) and \( \overrightarrow{CD} \). They are not collinear.

Answer: \( \lambda = 1 \). The lines are not intersecting.

Solution:

1. Write direction ratios:

For the first line:

\[
\text{Direction ratios: } (3, \lambda, 3).
\]

For the second line:

\[
\text{Direction ratios: } (3\lambda, -7, 5).
\]

2. Use perpendicularity condition:

\[
3(3\lambda) + \lambda(-7) + 3(5) = 0.
\] \[
9\lambda – 7\lambda + 15 = 0 => 2\lambda = -15 => \lambda = -\frac{15}{2}.
\]

3. Test intersection by substituting parametric equations into both lines. No common solution exists.

Answer: The inverse of \( A \) is:
\[
A^{-1} = \begin{bmatrix} \dots \end{bmatrix}.
\] The solution to the system of equations is \( x = \dots, y = \dots, z = \dots \).

Solution:

1. Augment \( A \) with the identity matrix \( I \), forming \( [A | I] \).

\[
[A | I] = \begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & 1 & 0 \\ 3 & 1 & 1 & 0 & 0 & 1 \end{bmatrix}.
\]

2. Perform row operations to reduce \( A \) to \( I \), applying the same operations to \( I \).

Row operations:

(i) \( R_1 \leftrightarrow R_3 \):

\[
\begin{bmatrix} 3 & 1 & 1 & 0 & 0 & 1 \\ 0 & 2 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 \end{bmatrix}.
\]

(ii) \( R_2 \rightarrow R_2 – \frac{1}{3}R_1 \), etc.

Continue these operations until \( A \) becomes \( I \), and the transformed identity matrix becomes \( A^{-1} \).

3. Multiply \( A^{-1} \) with the constant matrix from the system of equations to find \( x, y, z \).

\[
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix}.
\]

Solve for \( x, y, z \).

Answer: The least cost is ₹_____.

Solution:

1. Let the dimensions of the rectangular base be \( x \) and \( y \). Then, the area of the base is \( xy \), and the volume of the tank gives:

\[
xy \cdot 2 = 8 => xy = 4.
\]

2. The surface area of the tank (excluding the top) is:

\[
\text{Surface Area} = xy + 2(x + y) \cdot 2 = xy + 4(x + y).
\]

3. Substitute \( xy = 4 \) and express cost:

\[
\text{Cost} = 70(xy) + 45[4(x + y)] = 280 + 180(x + y).
\]

4. Minimize the cost using \( x + y = \frac{4}{x} \) (from \( xy = 4 \)).

Substitute and solve \( C(x) \):

\[
C(x) = 280 + 180\left(x + \frac{4}{x}\right).
\]

Find the derivative \( \frac{dC}{dx} \) and solve \( \frac{dC}{dx} = 0 \) for \( x \). Then, calculate \( y = \frac{4}{x} \) and the total cost.

Answer: The area of triangle ABC is \( 7 \) square units.

Solution:

1. The formula for the area of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.
\]

Substitute the coordinates:

\[
\text{Area} = \frac{1}{2} \left| 2(7 – 2) + 4(2 – 5) + 6(5 – 7) \right|.
\]

2. Perform the calculations:

\[
\text{Area} = \frac{1}{2} \left| 2(5) + 4(-3) + 6(-2) \right| = \frac{1}{2} \left| 10 – 12 – 12 \right| = \frac{1}{2} \left| -14 \right| = 7.
\]

Thus, the area of the triangle is \( 7 \) square units.

</su_spoiler]

Question 27: Find the vector and Cartesian equations of the plane passing through the points \( (2, 2, -1) \), \( (3, 4, 2) \), and \( (7, 0, 6) \). Also, find the vector equation of a plane passing through \( (4, 3, 1) \) and parallel to the plane obtained above.

Answer:
- The Cartesian equation of the plane is: \( 7x - 5y + 3z = 17 \).
- The vector equation of the plane passing through \( (4, 3, 1) \) is: \( \vec{r} \cdot \vec{n} = d \), where \( \vec{n} = \begin{bmatrix} 7 \\ -5 \\ 3 \end{bmatrix} \) and \( d \) is calculated.

Solution:

1. Calculate two vectors on the plane:

\[
\vec{AB} = (3 - 2, 4 - 2, 2 - (-1)) = (1, 2, 3), \quad \vec{AC} = (7 - 2, 0 - 2, 6 - (-1)) = (5, -2, 7).
\]

2. Find the normal vector \( \vec{n} \) to the plane using the cross product \( \vec{AB} \times \vec{AC} \):

\[
\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 5 & -2 & 7 \end{vmatrix} = \hat{i}(2 \cdot 7 - 3 \cdot (-2)) - \hat{j}(1 \cdot 7 - 3 \cdot 5) + \hat{k}(1 \cdot (-2) - 2 \cdot 5).
\]

3. Simplify the cross product:

\[
\vec{n} = \hat{i}(14 + 6) - \hat{j}(7 - 15) + \hat{k}(-2 - 10) = \hat{i}(20) - \hat{j}(-8) + \hat{k}(-12) = (20, 8, -12).
\]

4. The Cartesian equation of the plane is given by:

\[
20x + 8y - 12z = d.
\]

5. Substitute a point (e.g., \( (2, 2, -1) \)) to find \( d \):

\[
20(2) + 8(2) - 12(-1) = d => d = 40 + 16 + 12 = 68.
\]

6. Therefore, the equation of the plane is:

\[
20x + 8y - 12z = 68.
\]

7. For the second plane passing through \( (4, 3, 1) \):

\[
\text{Normal vector remains the same. Substitute } (4, 3, 1) \text{ to find the equation.}
\]

Answer: The probability that the defective item was produced by A is \( \frac{1}{12} \).

Solution:

1. Let \( P(D|A) = 0.01 \), \( P(D|B) = 0.05 \), and \( P(D|C) = 0.07 \) be the probabilities of producing defective items by A, B, and C respectively.

2. The probabilities of being on the job are \( P(A) = 0.5 \), \( P(B) = 0.3 \), and \( P(C) = 0.2 \).

3. Using the law of total probability, the overall probability of a defective item is:

\[
P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C).
\]

Substitute the values:

\[
P(D) = (0.01)(0.5) + (0.05)(0.3) + (0.07)(0.2) = 0.005 + 0.015 + 0.014 = 0.034.
\]

4. Using Bayes’ theorem, the probability that the defective item was produced by A is:

\[
P(A|D) = \frac{P(D|A)P(A)}{P(D)} = \frac{(0.01)(0.5)}{0.034} = \frac{0.005}{0.034} = \frac{1}{12}.
\]

Thus, the probability that the defective item was produced by A is \( \frac{1}{12} \).

Answer: The maximum profit is ₹ 55, achieved by producing 2 units of model A and 5 units of model B.

Solution:

1. Let \( x \) be the number of units of model A and \( y \) be the number of units of model B.

2. Formulate the constraints based on the available hours:

\[
2x + y \leq 40 \quad \text{(skilled men constraint)}, \quad 2x + 3y \leq 80 \quad \text{(semi-skilled men constraint)}, \quad x \geq 0, \quad y \geq 0.
\]

3. The objective function is:

\[
\text{Maximize } Z = 15x + 10y.
\]

4. Plot the constraints on a graph and identify the feasible region. The corner points are determined by solving the equations of the constraint lines.

5. Calculate \( Z \) at each corner point:

\[
(0, 0): Z = 15(0) + 10(0) = 0, \quad (0, 26.67): Z = 15(0) + 10(26.67) = 266.7,
\] \[
(20, 0): Z = 15(20) + 10(0) = 300.
\]

The maximum profit occurs at the corner points where the objective function \( Z = 15x + 10y \) is maximized:

1. Calculate the values of \( Z \) at all intersection points of the constraints:

• • Intersection of \( 2x + y = 40 \) and \( 2x + 3y = 80 \): Solving these two equations simultaneously,

\[
\text{From } 2x + y = 40, \quad y = 40 – 2x.
\] \[
\text{Substitute } y \text{ into } 2x + 3y = 80:
2x + 3(40 – 2x) = 80 => 2x + 120 – 6x = 80 => -4x + 120 = 80 => 4x = 40 => x = 10.
\] \[
\text{Substitute } x = 10 \text{ into } y = 40 – 2(10) => y = 20.
\] \[
\text{Intersection point is } (x, y) = (10, 20).
\]

• Other points to consider are the boundary points:
  • (0, 0): \( Z = 15(0) + 10(0) = 0. \)
  • (20, 0): From \( 2x + y = 40, y = 0 => x = 20. \quad Z = 15(20) + 10(0) = 300. \)
  • (0, 26.67): From \( 2x + 3y = 80, x = 0 => y = \frac{80}{3} = 26.67. \quad Z = 15(0) + 10(26.67) = 266.7. \)
  • (10, 20): \( Z = 15(10) + 10(20) = 150 + 200 = 350. \)

2. From these calculations, the maximum profit \( Z \) occurs at the point \( (x, y) = (10, 20) \).

3. Therefore, the manufacturer should produce 10 units of model A and 20 units of model B to achieve a maximum profit of \( \mathbf{₹350} \).