Step 1: Understanding the Concept:

We first find the value of \(\lambda\) using the position of vertex A. Then, we find the coordinates of C using properties of the orthocenter and finally compute the tangent length using the standard parabola formula.


Step 2: Key Formula or Approach:

1. A point on the y-axis has \(x = 0\).

2. The altitude from B is perpendicular to AC and passes through the orthocenter H.

3. Length of tangent from \((x_{1}, y_{1})\) to \(y^{2} = 4ax\) is \(\sqrt{y_{1}^{2} - 4ax_{1}}\).


Step 3: Detailed Explanation:

1. Find \(\lambda\): Vertex A lies on both AB and AC and has \(x = 0\).

From AB: \(\lambda y = 4 \Rightarrow y = 4/\lambda\).

From AC: \((1 - \lambda)y + \lambda = 0 \Rightarrow y = -\lambda / (1 - \lambda) = \lambda / (\lambda - 1)\).

Equating: \(4/\lambda = \lambda/(\lambda - 1) \Rightarrow 4\lambda - 4 = \lambda^{2} \Rightarrow (\lambda - 2)^{2} = 0 \Rightarrow \lambda = 2\).

Thus, \(A = (0, 2)\).

2. Equations of sides:

AB: \(3x + 2y = 4\) (Slope \(m_{AB} = -3/2\))

AC: \(2x - y + 2 = 0 \Rightarrow y = 2x + 2\) (Slope \(m_{AC} = 2\))

3. Find C: Let \(H = (1, 2)\) be the orthocenter.

The altitude from C is perpendicular to AB and passes through H.

Slope of altitude \(CH = -1/m_{AB} = 2/3\).

Equation of CH: \(y - 2 = \frac{2}{3}(x - 1) \Rightarrow 2x - 3y = -4\).

Point C is the intersection of AC and CH:
\(y = 2x + 2\) and \(2x - 3(2x + 2) = -4 \Rightarrow -4x = 2 \Rightarrow x = -1/2\).
\(y = 2(-1/2) + 2 = 1\). So \(C = (-1/2, 1)\).

4. Length of tangent: To parabola \(y^{2} - 6x = 0\) from \(C(-1/2, 1)\):
\(L = \sqrt{1^{2} - 6(-1/2)} = \sqrt{1 + 3} = \sqrt{4} = 2\).


Step 4: Final Answer:

The length of the tangent is 2.
Quick Tip: Remember that the vertex A is the intersection of the two given sides. If it lies on the y-axis, simply set \(x=0\) in both equations to solve for the parameter \(\lambda\).

Step 1: Understanding the Concept:

We use logical equivalences and De Morgan's laws to simplify the given expression.


Step 2: Detailed Explanation:

1. Start with the implication: \(p \Rightarrow \sim q \equiv \sim p \vee \sim q\).

2. Substitute into the inner term: \(p \wedge (p \Rightarrow \sim q) \equiv p \wedge (\sim p \vee \sim q)\).

3. Use the Distributive Law: \((p \wedge \sim p) \vee (p \wedge \sim q)\).

4. Since \(p \wedge \sim p\) is a contradiction (False): \(F \vee (p \wedge \sim q) \equiv p \wedge \sim q\).

5. Now apply the outer negation: \(\sim(p \wedge \sim q)\).

6. Use De Morgan's Law: \(\sim p \vee \sim(\sim q) \equiv \sim p \vee q\).


Comparing with options: Option (B) matches perfectly.


Step 3: Final Answer:

The expression is equivalent to \((\sim p) \vee q\).
Quick Tip: A quick shortcut: \(p \wedge (p \Rightarrow r)\) is simply \(p \wedge r\). Here \(r\) is \(\sim q\), so it becomes \(\sim(p \wedge \sim q)\), which is \(\sim p \vee q\).

Step 1: Understanding the Concept:

This is a homogeneous differential equation since the degree of all terms is 2. We can solve it using the substitution \(y = vx\).


Step 2: Key Formula or Approach:

Substitute \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\).


Step 3: Detailed Explanation:

1. Rearrange equation: \(3xy \frac{dy}{dx} = 3y^{2} - x^{2} \Rightarrow \frac{dy}{dx} = \frac{3y^{2} - x^{2}}{3xy}\).

2. Substitute \(y = vx\):
\(v + x \frac{dv}{dx} = \frac{3v^{2}x^{2} - x^{2}}{3vx^{2}} = \frac{3v^{2} - 1}{3v}\).
\(x \frac{dv}{dx} = \frac{3v^{2} - 1}{3v} - v = \frac{3v^{2} - 1 - 3v^{2}}{3v} = \frac{-1}{3v}\).

3. Separate variables: \(3v dv = -\frac{1}{x} dx\).

4. Integrate: \(\frac{3v^{2}}{2} = -\ln |x| + C\).

5. Back-substitute \(v = y/x\): \(\frac{3y^{2}}{2x^{2}} = -\ln |x| + C\).

6. Apply initial condition \(y(1) = 1\): \(\frac{3(1)}{2(1)} = 0 + C \Rightarrow C = \frac{3}{2}\).

Equation: \(\frac{3y^{2}}{2x^{2}} = \frac{3}{2} - \ln |x|\).

7. Find \(6y^{2}(e)\): Multiply entire equation by \(4x^{2}\):
\(6y^{2} = 6x^{2} - 4x^{2} \ln |x|\).

At \(x = e\): \(6y^{2}(e) = 6e^{2} - 4e^{2} \ln(e) = 6e^{2} - 4e^{2}(1) = 2e^{2}\).


Step 4: Final Answer:

The value of \(6y^{2}(e)\) is \(2e^{2}\).
Quick Tip: For homogeneous equations of the form \(M(x,y)dx + N(x,y)dy = 0\), if the substitution \(y=vx\) makes the integration look difficult, try \(x=vy\). In this case, \(y=vx\) led to a very simple separable form.

Step 1: Understanding the Concept:

This is a reciprocal-style quadratic equation. We can simplify it by letting \(t = x + \frac{1}{x}\).


Step 2: Key Formula or Approach:

If \(t = x + \frac{1}{x}\), then \(t^{2} = x^{2} + \frac{1}{x^{2}} + 2 \Rightarrow x^{2} + \frac{1}{x^{2}} = t^{2} - 2\).

Crucially, for real \(x\), \(|x + \frac{1}{x}| \geq 2\).


Step 3: Detailed Explanation:

1. Substitute \(t\):
\(3(t^{2} - 2) - 2t + 5 = 0\)
\(3t^{2} - 6 - 2t + 5 = 0\)
\(3t^{2} - 2t - 1 = 0\).

2. Solve for \(t\):

Factorizing: \(3t^{2} - 3t + t - 1 = 0 \Rightarrow (3t + 1)(t - 1) = 0\).

Possible values: \(t = 1\) or \(t = -1/3\).

3. Verify reality of \(x\):

For \(x\) to be real, the condition \(|t| \geq 2\) must be satisfied.

Check \(t = 1\): \(|1| < 2\). (No real \(x\))

Check \(t = -1/3\): \(|-1/3| < 2\). (No real \(x\))

Since neither value of \(t\) is valid for real \(x\), there are no real solutions.


Step 4: Final Answer:

The number of real solutions is 0.
Quick Tip: In equations involving \(x + \frac{1}{x}\), always check the range of the substituted variable. The property \(|x + \frac{1}{x}| \geq 2\) for real \(x\) (from AM-GM inequality) is a very frequent "trap" in competitive exams.

Step 1: Understanding the Concept:

The given function is of a special form where the sum of function values at symmetric points \(x\) and \(1-x\) often results in a constant. We need to check the value of \(f(x) + f(1-x)\).


Step 2: Key Formula or Approach:

Let's evaluate \(f(1-x)\):
\[ f(1-x) = \frac{2^{2(1-x)}}{2^{2(1-x)} + 2} = \frac{2^2 \cdot 2^{-2x}}{2^2 \cdot 2^{-2x} + 2} = \frac{4/2^{2x}}{4/2^{2x} + 2} \]

Multiplying the numerator and denominator by \(2^{2x}\):
\[ f(1-x) = \frac{4}{4 + 2 \cdot 2^{2x}} = \frac{2}{2 + 2^{2x}} \]

Now, compute \(f(x) + f(1-x)\):
\[ f(x) + f(1-x) = \frac{2^{2x}}{2^{2x} + 2} + \frac{2}{2^{2x} + 2} = \frac{2^{2x} + 2}{2^{2x} + 2} = 1 \]


Step 3: Detailed Explanation:

The given series is \(S = \sum_{k=1}^{2022} f\left(\frac{k}{2023}\right)\).

We can pair the terms from both ends:
\[ S = \left[ f\left(\frac{1}{2023}\right) + f\left(\frac{2022}{2023}\right) \right] + \left[ f\left(\frac{2}{2023}\right) + f\left(\frac{2021}{2023}\right) \right] + \dots \]

Each pair is of the form \(f(x) + f(1-x)\) where \(x = \frac{k}{2023}\), and thus each pair sums to 1.

There are 2022 terms in total, which forms \(\frac{2022}{2} = 1011\) such pairs.
\[ S = 1 \times 1011 = 1011 \]


Step 4: Final Answer:

The value of the sum is 1011.
Quick Tip: Whenever you see a summation of function values like this, always check if \(f(x) + f(1-x)\) or \(f(x) + f(a-x)\) is constant. It simplifies the entire series into pairs.

Step 1: Understanding the Concept:

A system of linear equations has infinitely many solutions if the rank of the augmented matrix is equal to the rank of the coefficient matrix, and both are less than the number of variables. Practically, we can use Gaussian elimination or determinants.


Step 2: Key Formula or Approach:

We can perform row operations on the augmented matrix:
\[ \begin{pmatrix} 1 & 2 & 3 & | & 3
4 & 3 & -4 & | & 4
8 & 4 & -\lambda & | & 9+\mu \end{pmatrix} \]

Perform \(R_2 \to R_2 - 4R_1\) and \(R_3 \to R_3 - 8R_1\):
\[ \begin{pmatrix} 1 & 2 & 3 & | & 3
0 & -5 & -16 & | & -8
0 & -12 & -\lambda-24 & | & \mu-15 \end{pmatrix} \]


Step 3: Detailed Explanation:

For the system to have infinitely many solutions, the third row must be a multiple of the second row:
\[ \frac{-12}{-5} = \frac{-\lambda-24}{-16} = \frac{\mu-15}{-8} \]

From the first equality:
\[ \frac{12}{5} = \frac{\lambda+24}{16} \Rightarrow 192 = 5\lambda + 120 \Rightarrow 5\lambda = 72 \Rightarrow \lambda = \frac{72}{5} \]

From the second equality:
\[ \frac{12}{5} = \frac{\mu-15}{-8} \Rightarrow -96 = 5\mu - 75 \Rightarrow 5\mu = -21 \Rightarrow \mu = -\frac{21}{5} \]


Step 4: Final Answer:

The ordered pair is \((\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5}\right)\).
Quick Tip: For a \(3 \times 3\) system, "infinitely many solutions" means that the determinant \(\Delta = 0\) and all minor determinants \(\Delta_x, \Delta_y, \Delta_z\) are also 0. Row reduction is often faster for finding parameters.

Step 1: Understanding the Concept:

For a chord of a circle with radius \(R\) subtending an angle \(\theta\) at the center, the distance \(d\) of the midpoint of the chord from the center is given by \(d = R \cos\left(\frac{\theta}{2}\right)\). The locus of these midpoints is a concentric circle with radius \(r = d\).


Step 2: Key Formula or Approach:

Given circle \(C_i\) has radius \(R = \sqrt{4} = 2\).

Radius of locus \(r_i = R \cos\left(\frac{\theta_i}{2}\right) = 2 \cos\left(\frac{\theta_i}{2}\right)\).


Step 3: Detailed Explanation:

Using the given angles:

1) For \(\theta_1 = \pi/3\), \(r_1 = 2 \cos(\pi/6) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\).

2) For \(\theta_3 = 2\pi/3\), \(r_3 = 2 \cos(\pi/3) = 2 \cdot \frac{1}{2} = 1\).

Given relation: \(r_1^2 = r_2^2 + r_3^2\).
\[ (\sqrt{3})^2 = r_2^2 + 1^2 \Rightarrow 3 = r_2^2 + 1 \Rightarrow r_2^2 = 2 \Rightarrow r_2 = \sqrt{2} \]

Now find \(\theta_2\):
\[ r_2 = 2 \cos\left(\frac{\theta_2}{2}\right) \Rightarrow \sqrt{2} = 2 \cos\left(\frac{\theta_2}{2}\right) \Rightarrow \cos\left(\frac{\theta_2}{2}\right) = \frac{1}{\sqrt{2}} \]
\[ \frac{\theta_2}{2} = \frac{\pi}{4} \Rightarrow \theta_2 = \frac{\pi}{2} \]

Step 4: Final Answer:

The value of \(\theta_2\) is \(\pi/2\).
Quick Tip: Remember that the midpoint of a chord and the center of the circle form a right-angled triangle with half the chord length and the radius. The distance of the chord from the center is always \(R \cos(\theta/2)\).

Step 1: Understanding the Concept:

The equation of a plane passing through the intersection of two planes \(P_1 = 0\) and \(P_2 = 0\) is given by the family \(P_1 + k P_2 = 0\). We use the given points to find \(\lambda\) and \(k\).


Step 2: Detailed Explanation:

Let the plane be \(L : (x + (\lambda + 4)y + z - 1) + k(2x + y + z - 2) = 0\).

1) Passes through \((0, 1, 0)\):
\[ (\lambda + 4 - 1) + k(1 - 2) = 0 \Rightarrow \lambda + 3 - k = 0 \Rightarrow k = \lambda + 3 \]

2) Passes through \((1, 0, 1)\):
\[ (1 + 1 - 1) + k(2 + 1 - 2) = 0 \Rightarrow 1 + k = 0 \Rightarrow k = -1 \]

Comparing the values of \(k\):
\[ \lambda + 3 = -1 \Rightarrow \lambda = -4 \]

Now, we need to find the distance of point \(A = (2\lambda, \lambda, -\lambda) = (-8, -4, 4)\) from plane \(P_2 : 2x + y + z - 2 = 0\).


Step 3: Key Formula or Approach:

Distance \(d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\).
\[ d = \frac{|2(-8) + (-4) + (4) - 2|}{\sqrt{2^2 + 1^2 + 1^2}} = \frac{|-16 - 2|}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3\sqrt{6} \]

Step 4: Final Answer:

The distance is \(3\sqrt{6}\).
Quick Tip: When dealing with the "Family of Planes", always use the parameters \(k\) and \(\lambda\) systematically. Here, evaluating the points first made the problem straightforward.

Step 1: Understanding the Concept:

We use the properties of an Arithmetic Progression (A.P.) and the formulas for mean and variance of a data set.


Step 2: Key Formula or Approach:

Let the first term be \(a\) and common difference be \(d\).

1) \(a_1 + a_3 = 10 \Rightarrow a + (a + 2d) = 10 \Rightarrow 2a + 2d = 10 \Rightarrow a + d = 5\).

2) Mean \(\bar{x} = \frac{19}{2}\). For 6 terms in A.P., Mean \(= \frac{Sum}{6} = \frac{6/2 [2a + 5d]}{6} = \frac{2a + 5d}{2}\).
\[ \frac{2a + 5d}{2} = \frac{19}{2} \Rightarrow 2a + 5d = 19 \]


Step 3: Detailed Explanation:

Solving the system:
\(a + d = 5 \Rightarrow 2a + 2d = 10\).

Subtracting this from \(2a + 5d = 19\):
\[ 3d = 9 \Rightarrow d = 3 \]
\[ a = 5 - 3 = 2 \]

The terms are: 2, 5, 8, 11, 14, 17.

Variance for A.P. terms: \(\sigma^2 = \frac{d^2(n^2 - 1)}{12}\).
\[ \sigma^2 = \frac{3^2(6^2 - 1)}{12} = \frac{9 \times 35}{12} = \frac{3 \times 35}{4} = \frac{105}{4} \]

We need \(8\sigma^2\):
\[ 8\sigma^2 = 8 \times \frac{105}{4} = 2 \times 105 = 210 \]

Step 4: Final Answer:

The value of \(8\sigma^2\) is 210.
Quick Tip: The formula for the variance of \(n\) terms in A.P. is \(\frac{d^2(n^2 - 1)}{12}\). It is a major time-saver compared to the standard \(\frac{1}{n} \sum (x_i - \bar{x})^2\) method.

Step 1: Understanding the Concept:

For the limit of a floor function expression to be 0, the function must be constant and equal to 0 in a local neighborhood of \(a\). Any point where the floor value jumps (i.e., when the expression inside \([ \dots ]\) is an integer) requires checking left and right limits.


Step 2: Key Formula or Approach:

Let \(f(x) = [x - 5] - [2x + 2] = [x] - 5 - [2x] - 2 = [x] - [2x] - 7\).

We want \(\lim_{x \to a} ([x] - [2x]) = 7\).

Recall \([2x] = [x] + [x + 0.5]\).

Substituting this: \( [x] - ([x] + [x + 0.5]) = 7 \Rightarrow -[x + 0.5] = 7 \Rightarrow [x + 0.5] = -7\).


Step 3: Detailed Explanation:

For \([x + 0.5] = -7\), we must have:
\[ -7 \le x + 0.5 < -6 \Rightarrow -7.5 \le x < -6.5 \]

Now check the endpoints for the existence of the limit:

1) At \(x = -7.5\): L.H.L. involves \([-7.5^- + 0.5] = [-7^-] = -8\). R.H.L. involves \([-7.5^+ + 0.5] = [-7^+] = -7\). Limits are different, so it doesn't exist at \(a = -7.5\).

2) At \(x = -6.5\): L.H.L. involves \([-6.5^- + 0.5] = [-6^-] = -7\). R.H.L. involves \([-6.5^+ + 0.5] = [-6^+] = -6\). Limits are different, so it doesn't exist at \(a = -6.5\).

3) For any \(a \in (-7.5, -6.5)\), the function is locally constant because the only potential jumps for \([x]\) and \([2x]\) in this range are at integers and half-integers. In this specific range \([-7.5, -6.5)\), the values of \([x] - [2x]\) stay constant between the transition points.

Actually, at integers like \(x = -7\), \(\lim_{x \to -7} ([x] - [2x]) = 7\) because both L.H.L and R.H.L are equal to 7. Thus the limit exists for the entire open interval.


Step 4: Final Answer:

The set of values is \((-7.5, -6.5)\).
Quick Tip: For \([x] - [2x] = C\) to have a valid limit at \(a\), \(a\) cannot be a point where the value of the expression jumps between two different integers. Check integers and half-integers carefully.

Step 1: Understanding the Concept:

We are given a polynomial function where the coefficients are its own derivatives evaluated at specific points. We need to solve for these constants to find the exact function.


Step 2: Detailed Explanation:

Let \(f'(1) = A, f''(2) = B, f'''(3) = C\).

Function: \(f(x) = x^3 - Ax^2 + Bx - C\).

Differentiating:
\(f'(x) = 3x^2 - 2Ax + B\)
\(f''(x) = 6x - 2A\)
\(f'''(x) = 6\)

Now, evaluate the points:

1) \(C = f'''(3) = 6\).

2) \(B = f''(2) = 6(2) - 2A = 12 - 2A\).

3) \(A = f'(1) = 3(1)^2 - 2A(1) + B = 3 - 2A + B\).

Substitute \(B\): \(A = 3 - 2A + (12 - 2A) = 15 - 4A \Rightarrow 5A = 15 \Rightarrow A = 3\).

Then, \(B = 12 - 2(3) = 6\).

So, \(f(x) = x^3 - 3x^2 + 6x - 6\).


Step 3: Evaluating Options:

Calculate specific values:
\(f(0) = -6\)
\(f(1) = 1 - 3 + 6 - 6 = -2\)
\(f(2) = 8 - 12 + 12 - 6 = 2\)
\(f(3) = 27 - 27 + 18 - 6 = 12\)

Check option (D): \(2f(0) - f(1) + f(3) = 2(-6) - (-2) + 12 = -12 + 2 + 12 = 2\).

Since \(f(2) = 2\), option (D) is correct.


Step 4: Final Answer:

The relation \(2f(0) - f(1) + f(3) = f(2)\) holds true.
Quick Tip: For problems like \(f(x) = x^n + Ax^{n-1} + \dots\), define the derivative values as variables. It transforms a functional question into a simple system of linear equations.

Step 1: Understanding the Concept:

The functional equation \(f(x + y) = f(x) f(y)\) defines a geometric progression (or exponential function) for natural numbers.


Step 2: Key Formula or Approach:

Since \(f(1) = 3\), we have:
\(f(2) = f(1 + 1) = f(1) \cdot f(1) = 3^2\)
\(f(3) = f(2 + 1) = f(2) \cdot f(1) = 3^2 \cdot 3 = 3^3\)

In general, \(f(k) = 3^k\).


Step 3: Detailed Explanation:

The given sum is:
\[ \sum_{k=1}^n 3^k = 3 + 3^2 + \dots + 3^n = 3279 \]

This is a geometric series with \(a = 3, r = 3\). Sum formula: \(S_n = \frac{a(r^n - 1)}{r - 1}\).
\[ \frac{3(3^n - 1)}{3 - 1} = 3279 \Rightarrow \frac{3}{2}(3^n - 1) = 3279 \]
\[ 3^n - 1 = \frac{3279 \times 2}{3} = 1093 \times 2 = 2186 \]
\[ 3^n = 2187 \]

Powers of 3: \(3^1=3, 3^2=9, 3^3=27, 3^4=81, 3^5=243, 3^6=729, 3^7=2187\).

Thus, \(n = 7\).


Step 4: Final Answer:

The value of \(n\) is 7.
Quick Tip: Common functional equations to remember:
1) \(f(x+y) = f(x) + f(y) \Rightarrow f(x) = cx\)
2) \(f(x+y) = f(x)f(y) \Rightarrow f(x) = a^x\)

Step 1: Understanding the Concept:

This problem involves complex numbers in polar form. The expression inside the brackets is of the form \(\frac{1 + z}{1 + \bar{z}}\).


Step 2: Key Formula or Approach:

Let \(\theta = \frac{2\pi}{9}\).

We know \(\sin\theta + i \cos\theta = \cos(\frac{\pi}{2} - \theta) + i \sin(\frac{\pi}{2} - \theta) = e^{i(\frac{\pi}{2} - \theta)}\).

Let \(z = e^{i\phi}\) where \(\phi = \frac{\pi}{2} - \theta\).

Then the expression is \(\left( \frac{1 + z}{1 + \bar{z}} \right)^3\).

Note that \(\frac{1 + z}{1 + \bar{z}} = \frac{1 + z}{1 + 1/z} = z\).


Step 3: Detailed Explanation:

The value is \(z^3 = (e^{i\phi})^3 = e^{3i\phi}\).

Calculate \(\phi\):
\[ \phi = \frac{\pi}{2} - \frac{2\pi}{9} = \frac{9\pi - 4\pi}{18} = \frac{5\pi}{18} \]

Now find \(3\phi\):
\[ 3\phi = 3 \times \frac{5\pi}{18} = \frac{5\pi}{6} \]

The final value is \(e^{i \frac{5\pi}{6}}\):
\[ e^{i \frac{5\pi}{6}} = \cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{i}{2} = -\frac{1}{2}(\sqrt{3} - i) \]

Step 4: Final Answer:

The value is \(-\frac{1}{2}(\sqrt{3} - i)\).
Quick Tip: For any complex number \(z\) on the unit circle (\(|z|=1\)), the identity \(\frac{1+z}{1+\bar{z}} = z\) is extremely useful. Also, convert \(\sin + i\cos\) to \(\cos + i\sin\) using complement angles.

Step 1: Understanding the Concept:

The numbers can be 4-digit or 5-digit. Since we have 5 distinct digits and repetitions are not allowed, 5-digit numbers will naturally be larger than any 4-digit number.


Step 2: Detailed Explanation:

Case 1: 4-digit numbers greater than 7000

To be greater than 7000, the thousands place must be either 7 or 8.

- Number of choices for thousands place = 2 (7 or 8).

- The remaining 3 places can be filled from the remaining 4 digits in \(P(4, 3)\) ways.

- Total 4-digit numbers = \(2 \times (4 \times 3 \times 2) = 2 \times 24 = 48\).


Case 2: 5-digit numbers

All 5-digit numbers formed from these digits will be greater than 7000.

- Total 5-digit numbers = \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).


Step 3: Summing the Results:

Total integers = \(48 + 120 = 168\).


Step 4: Final Answer:

There are 168 such integers.
Quick Tip: Don't forget to check numbers with more digits than the lower bound. In permutation problems with a threshold, higher-digit count cases are often overlooked.

Step 1: Understanding the Concept:

This is a standard integral of the form \(\int \frac{1}{\sqrt{a^2 - k^2x^2}} dx\), which results in an inverse sine function.


Step 2: Key Formula or Approach:

Evaluate the indefinite integral first:
\[ I = \int \frac{48}{\sqrt{9 - 4x^2}} dx = \int \frac{48}{\sqrt{4((\frac{3}{2})^2 - x^2)}} dx = \int \frac{48}{2\sqrt{(\frac{3}{2})^2 - x^2}} dx \]
\[ I = 24 \int \frac{1}{\sqrt{(\frac{3}{2})^2 - x^2}} dx = 24 \sin^{-1}\left(\frac{x}{3/2}\right) = 24 \sin^{-1}\left(\frac{2x}{3}\right) \]


Step 3: Detailed Explanation:

Apply the limits:
\[ Value = \left[ 24 \sin^{-1}\left(\frac{2x}{3}\right) \right]_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \]

Upper limit: \(24 \sin^{-1}\left(\frac{2}{3} \cdot \frac{3\sqrt{3}}{4}\right) = 24 \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 24 \cdot \frac{\pi}{3} = 8\pi\).

Lower limit: \(24 \sin^{-1}\left(\frac{2}{3} \cdot \frac{3\sqrt{2}}{4}\right) = 24 \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 24 \cdot \frac{\pi}{4} = 6\pi\).

Result \(= 8\pi - 6\pi = 2\pi\).


Step 4: Final Answer:

The integral evaluates to \(2\pi\).
Quick Tip: When integrating \(\frac{1}{\sqrt{a^2 - b^2x^2}}\), you can use \(\frac{1}{b} \sin^{-1}(\frac{bx}{a})\). It saves the step of pulling out the constant from the square root.

Step 1: Understanding the Concept:

Any vector \(\vec{\beta}\) can be decomposed into a component parallel to \(\vec{a}\) (\(\vec{\beta}_1\)) and a component perpendicular to \(\vec{a}\) (\(\vec{\beta}_2\)). \(\vec{\beta}_1\) is the vector projection of \(\vec{\beta}\) on \(\vec{a}\).


Step 2: Key Formula or Approach:
\(\vec{\beta}_1 = \left( \frac{\vec{\beta} \cdot \vec{a}}{|\vec{a}|^2} \right) \vec{a}\)
\(\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1\)


Step 3: Detailed Explanation:

Calculate dot product and magnitude squared:
\(\vec{\beta} \cdot \vec{a} = (1)(4) + (2)(3) + (-4)(5) = 4 + 6 - 20 = -10\).
\(|\vec{a}|^2 = 4^2 + 3^2 + 5^2 = 16 + 9 + 25 = 50\).

Thus, \(\vec{\beta}_1 = \frac{-10}{50} \vec{a} = -\frac{1}{5} \vec{a}\).

Substitute \(\vec{\beta}_1\) into \(\vec{\beta}_2\):
\(\vec{\beta}_2 = \vec{\beta} + \frac{1}{5} \vec{a} \Rightarrow 5\vec{\beta}_2 = 5\vec{\beta} + \vec{a}\).
\(5\vec{\beta}_2 = 5(\hat{i} + 2\hat{j} - 4\hat{k}) + (4\hat{i} + 3\hat{j} + 5\hat{k}) = (5+4)\hat{i} + (10+3)\hat{j} + (-20+5)\hat{k}\).
\(5\vec{\beta}_2 = 9\hat{i} + 13\hat{j} - 15\hat{k}\).

Now, compute the required dot product with \(\vec{v} = \hat{i} + \hat{j} + \hat{k}\):
\(5\vec{\beta}_2 \cdot \vec{v} = (9)(1) + (13)(1) + (-15)(1) = 9 + 13 - 15 = 7\).


Step 4: Final Answer:

The value is 7.
Quick Tip: Instead of calculating \(\vec{\beta}_2\) fully as a vector, working with \(5\vec{\beta}_2 = 5\vec{\beta} + \vec{a}\) avoids fractions and makes the dot product calculation much faster.

Step 1: Understanding the Concept:

We use the determinant properties of adjoint matrices and matrix inverses. For an \(n \times n\) matrix:

1) \(|adj A| = |A|^{n-1}\)

2) \(|adj(adj A)| = |A|^{(n-1)^2}\)


Step 2: Detailed Explanation:

Given \(n = 3\) and \(|adj(adj A)| = 12^4\).

Using the formula: \(|A|^{(3-1)^2} = |A|^4 = 12^4\).

This implies \(|A| = \pm 12\).

Now, evaluate \(|A^{-1} adj A|\):
\[ |A^{-1} adj A| = |A^{-1}| \cdot |adj A| = \frac{1}{|A|} \cdot |A|^{n-1} = |A|^{n-2} \]

For \(n = 3\):
\[ |A|^{3-2} = |A|^1 = |A| \]

Since we need the value and the options are positive, we take \(|12| = 12\).


Step 3: Final Answer:

The value is 12.
Quick Tip: Memorize the generalized property \(|adj^k(A)| = |A|^{(n-1)^k}\). It appears frequently in JEE and other competitive math papers.

Step 1: Understanding the Concept:

The direction of a line parallel to two planes is perpendicular to the normal vectors of both planes. Once we find the direction, we can write the equation of the line and use the property that the vector from the given point to the foot of the perpendicular is orthogonal to the line.


Step 2: Key Formula or Approach:

1. Line direction \(\vec{d} = \vec{n}_1 \times \vec{n}_2\), where \(\vec{n}_1 = (1, 2, 1)\) and \(\vec{n}_2 = (0, 3, -1)\).

2. Foot of perpendicular \(P\) on line \(L(\vec{r}) = \vec{a} + r\vec{d}\) satisfies \(\vec{AP} \cdot \vec{d} = 0\).


Step 3: Detailed Explanation:

Direction of the line:
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 1
0 & 3 & -1 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(-1-0) + \hat{k}(3-0) = -5\hat{i} + \hat{j} + 3\hat{k} \]

Equation of the line passing through \((3, 2, 1)\) with direction \((-5, 1, 3)\):
\[ \frac{x-3}{-5} = \frac{y-2}{1} = \frac{z-1}{3} = r \]

Any point \(P\) (foot of perpendicular) on the line is \((3-5r, 2+r, 1+3r)\).

Let \(A = (1, 9, 7)\). The vector \(\vec{AP} = (3-5r-1, 2+r-9, 1+3r-7) = (2-5r, r-7, 3r-6)\).

Since \(\vec{AP}\) is perpendicular to the line direction \((-5, 1, 3)\):
\[ -5(2-5r) + 1(r-7) + 3(3r-6) = 0 \]
\[ -10 + 25r + r - 7 + 9r - 18 = 0 \Rightarrow 35r - 35 = 0 \Rightarrow r = 1 \]

Substituting \(r=1\) into \(P\):
\((\alpha, \beta, \gamma) = (3-5(1), 2+1, 1+3(1)) = (-2, 3, 4)\).

Sum \(\alpha + \beta + \gamma = -2 + 3 + 4 = 5\).


Step 4: Final Answer:

The sum of the coordinates of the foot of the perpendicular is 5.
Quick Tip: For the foot of the perpendicular from a point \(A\) to a line \(L\), always define a general point \(P\) on the line using a parameter \(r\) and solve the linear equation generated by \(\vec{AP} \cdot \vec{d}_{line} = 0\).

Step 1: Understanding the Concept:

A matrix with entries 0 or 1 where each row and column sums to exactly 1 is known as a permutation matrix. Such a matrix has exactly one '1' in each row and each column.


Step 2: Detailed Explanation:

1. For the first row, there are 5 possible positions to place the single '1'.

2. Once the '1' is placed in the first row, the column it occupies cannot have another '1'.

3. For the second row, there are 4 remaining columns where a '1' can be placed.

4. Continuing this logic, for the third row, there are 3 options; for the fourth, 2 options; and for the fifth, only 1 option.

Total such matrices = \(5 \times 4 \times 3 \times 2 \times 1 = 5!\)
\[ 5! = 120 \]


Step 3: Final Answer:

There are 120 such matrices.
Quick Tip: The number of \(n \times n\) permutation matrices (matrices with entries \(\{0,1\}\) and row/column sum equal to 1) is always \(n!\).

Step 1: Understanding the Concept:

The series involves terms of the form \(k \cdot (^nC_k)^2\). We can use the identity \(k \cdot ^nC_k = n \cdot ^{n-1}C_{k-1}\) and properties of binomial coefficients to simplify the sum.


Step 2: Key Formula or Approach:

1. \(\sum_{k=1}^{n} k (^nC_k)^2 = n \binom{2n-1}{n-1}\).

2. Property: \(\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}\).


Step 3: Detailed Explanation:

Let \(S = \sum_{k=1}^{30} k (^nC_k)^2\), where \(n=30\).

Using the sum identity for \(n=30\):
\[ S = 30 \cdot \binom{59}{29} = 30 \cdot \frac{59!}{29! \cdot 30!} \]

We need to compare this with \(\frac{\alpha 60!}{(30!)^2}\).

Multiply the expression for \(S\) by \(\frac{60}{60}\) and \(\frac{30}{30}\):
\[ S = 30 \cdot \frac{59!}{29! \cdot 30!} \cdot \frac{30}{30} \cdot \frac{60}{60} = \frac{30 \cdot 30 \cdot 60!}{60 \cdot (30 \cdot 29!) \cdot 30!} = \frac{900 \cdot 60!}{60 \cdot 30! \cdot 30!} \]
\[ S = \frac{15 \cdot 60!}{(30!)^2} \]

Comparing with the given form, \(\alpha = 15\).


Step 4: Final Answer:

The value of \(\alpha\) is 15.
Quick Tip: A general useful result is \(\sum_{k=1}^n k \binom{n}{k}^2 = n \binom{2n-1}{n-1}\). Alternatively, the sum of \(\binom{n}{k}^2\) is \(\binom{2n}{n}\). By symmetry, the weighted sum is \(\frac{n}{2} \binom{2n}{n}\).
\(\frac{30}{2} \binom{60}{30} = 15 \binom{60}{30}\), giving \(\alpha = 15\) directly.

Step 1: Understanding the Concept:

We need to calculate the sum of the numerator and the denominator separately using standard summation formulas.


Step 2: Key Formula or Approach:

1. Sum of first \(n\) cubes: \(\sum k^3 = \frac{n^2(n+1)^2}{4}\).

2. Sum of terms in denominator: \(\sum_{k=1}^n k(2k+1) = \sum (2k^2 + k)\).


Step 3: Detailed Explanation:
Numerator: \(S_1 = \frac{n^2(n+1)^2}{4}\).

Denominator: \(S_2 = 2 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6} [2(2n+1) + 3] = \frac{n(n+1)(4n+5)}{6}\).

Given ratio:
\[ \frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(4n+5)}{6}} = \frac{n(n+1) \cdot 6}{4(4n+5)} = \frac{3(n^2+n)}{2(4n+5)} = \frac{9}{5} \]
\[ \frac{n^2+n}{8n+10} = \frac{3}{5} \Rightarrow 5n^2 + 5n = 24n + 30 \]
\[ 5n^2 - 19n - 30 = 0 \Rightarrow 5n^2 - 25n + 6n - 30 = 0 \]
\[ 5n(n-5) + 6(n-5) = 0 \Rightarrow n = 5 \] (since \(n \in \mathbb{N}\)).


Step 4: Final Answer:

The value of \(n\) is 5.
Quick Tip: For integer type questions in series, you can check small values of \(n\). For \(n=5\), Numerator \(= 15^2 = 225\); Denominator \(= 3 + 10 + 21 + 36 + 55 = 125\). Ratio \(225/125 = 9/5\). Verified!

Step 1: Understanding the Concept:

This is an integral equation. We can differentiate both sides with respect to \(x\) using the Leibniz Rule to obtain a differential equation.


Step 2: Detailed Explanation:

Differentiating \(f(x) + \int_{0}^{x} f(t) \sqrt{1 - (\ln f(t))^2} dt = e\):
\[ f'(x) + f(x) \sqrt{1 - (\ln f(x))^2} = 0 \]
\[ \frac{df}{dx} = -f \sqrt{1 - (\ln f)^2} \Rightarrow \int \frac{df}{f \sqrt{1 - (\ln f)^2}} = -\int dx \]

Let \(\ln f = u \Rightarrow \frac{1}{f} df = du\):
\[ \int \frac{du}{\sqrt{1-u^2}} = -x + C \Rightarrow \sin^{-1}(\ln f) = -x + C \]

At \(x=0\), from original equation: \(f(0) + 0 = e \Rightarrow f(0) = e\).
\(\sin^{-1}(\ln e) = -0 + C \Rightarrow \sin^{-1}(1) = C \Rightarrow C = \frac{\pi}{2}\).

So, \(\sin^{-1}(\ln f(x)) = \frac{\pi}{2} - x \Rightarrow \ln f(x) = \sin(\frac{\pi}{2}-x) = \cos x\).

At \(x = \frac{\pi}{6}\): \(\ln f(\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\).

We need \((6 \ln f(\frac{\pi}{6}))^2 = (6 \cdot \frac{\sqrt{3}}{2})^2 = (3\sqrt{3})^2 = 27\).


Step 3: Final Answer:

The final value is 27.
Quick Tip: When a variable upper limit integral appears in an equation, use differentiation (Leibniz rule) to convert it into a solvable differential equation.

Step 1: Understanding the Concept:

The vertices of the triangle are the intersection points of the sides. The centroid formula relates the coordinates of the vertices to the coordinates of the centroid \(P(2, a)\).


Step 2: Detailed Explanation:

Let \(A\) be intersection of \(2x+y=0\) and \(x-y=3\): \(3x=3 \Rightarrow x=1, y=-2\). \(A(1, -2)\).
\(B\) lies on \(2x+y=0\) and \(x+py=21a\). \(C\) lies on \(x-y=3\) and \(x+py=21a\).

Using centroid \(x = \frac{1 + x_B + x_C}{3} = 2 \Rightarrow x_B + x_C = 5\).
\(y = \frac{-2 + y_B + y_C}{3} = a \Rightarrow y_B + y_C = 3a + 2\).

Since \(B\) is on \(2x+y=0\), \(y_B = -2x_B\). Since \(C\) is on \(y=x-3\), \(y_C = x_C - 3\).

Substitute: \(-2x_B + x_C - 3 = 3a + 2 \Rightarrow -2x_B + x_C = 3a + 5\).

From \(x_B + x_C = 5\), subtract equations: \(3x_B = -3a \Rightarrow x_B = -a\).

Then \(x_C = 5 + a\).

Coordinates: \(B(-a, 2a)\), \(C(5+a, a+2)\).

Both lie on \(x+py=21a\). For \(B\): \(-a + p(2a) = 21a \Rightarrow 2p = 22 \Rightarrow p = 11\).

For \(C\): \((5+a) + 11(a+2) = 21a \Rightarrow 5 + a + 11a + 22 = 21a \Rightarrow 9a = 27 \Rightarrow a = 3\).

Vertices: \(B(-3, 6)\), \(C(8, 5)\).
\((BC)^2 = (8 - (-3))^2 + (5 - 6)^2 = 11^2 + (-1)^2 = 121 + 1 = 122\).


Step 3: Final Answer:

The squared distance \((BC)^2\) is 122.
Quick Tip: Solve for parameters \(a\) and \(p\) by expressing vertices in terms of them and applying the centroid constraint. This systematic reduction of variables is key in coordinate geometry.

Step 1: Understanding the Concept:

Shortest distance between two skew lines \(\vec{r} = \vec{a}_1 + t\vec{d}_1\) and \(\vec{r} = \vec{a}_2 + s\vec{d}_2\) is \(d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}\).


Step 2: Detailed Explanation:

Line 1: \(\vec{a}_1 = (-\sqrt{6}, \sqrt{6}, \sqrt{6})\), \(\vec{d}_1 = (2, 3, 4)\).

Line 2: \(\vec{a}_2 = (\lambda, 2\sqrt{6}, -2\sqrt{6})\), \(\vec{d}_2 = (3, 4, 5)\).
\(\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
3 & 4 & 5 \end{vmatrix} = (-1, 2, -1)\). Magnitude \(= \sqrt{6}\).
\(\vec{a}_2 - \vec{a}_1 = (\lambda+\sqrt{6}, \sqrt{6}, -3\sqrt{6})\).

Shortest Distance \(d = \frac{|-1(\lambda+\sqrt{6}) + 2(\sqrt{6}) - 1(-3\sqrt{6})|}{\sqrt{6}} = 6 \Rightarrow |4\sqrt{6} - \lambda| = 6\sqrt{6}\).

Case 1: \(4\sqrt{6} - \lambda = 6\sqrt{6} \Rightarrow \lambda_1 = -2\sqrt{6}\).

Case 2: \(4\sqrt{6} - \lambda = -6\sqrt{6} \Rightarrow \lambda_2 = 10\sqrt{6}\).

Sum of values \(= 8\sqrt{6}\). Square of sum \(= (8\sqrt{6})^2 = 384\).


Step 3: Final Answer:

The result is 384.
Quick Tip: The cross product of directions gives the common perpendicular vector. The shortest distance is simply the projection of the vector joining two points on the lines onto this normal vector.

Step 1: Understanding the Concept:

The equation \(\tan A + \tan B = 0\) implies \(\tan A = \tan(-B)\), which gives \(A = n\pi - B\). This allows us to find the possible values of \(\sin \theta + \cos \theta\).


Step 2: Detailed Explanation:
\(\pi \cos \theta = n\pi - \pi \sin \theta \Rightarrow \sin \theta + \cos \theta = n\).

Since \(-\sqrt{2} \le \sin \theta + \cos \theta \le \sqrt{2}\), integer \(n\) can be \(-1, 0, 1\).

1. \(n = 0\): \(\theta + \frac{\pi}{4} = \pi, 2\pi\) (as \(\sin(\theta+\pi/4)=0\)). \(\sin^2\) values are \(0, 0\).

2. \(n = 1\): \(\sin(\theta + \frac{\pi}{4}) = \frac{1}{\sqrt{2}}\). \(\sin^2\) values are \(\frac{1}{2}, \frac{1}{2}\).

3. \(n = -1\): \(\sin(\theta + \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\). \(\sin^2\) values are \(\frac{1}{2}, \frac{1}{2}\).

Total sum \(= 0 + 0 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2\).


Step 3: Final Answer:

The sum is 2.
Quick Tip: Convert sum/differences of trigonometric functions into a standard phase-shifted form \(R \sin(\theta + \phi)\) to easily solve for \(\theta\) or its functions.

Step 1: Understanding the Concept:

An equivalence relation must be reflexive, symmetric, and transitive. The set of elements connects all \(\{a, b, c, d\}\), meaning they must all belong to the same equivalence class in the smallest equivalence relation containing \(R\).


Step 2: Detailed Explanation:

1. Reflexive: Already given as \(\{(a,a), (b,b), (c,c), (d,d)\}\).

2. Symmetry: Add \((b,a), (c,b), (d,b)\). (3 elements)

3. Transitivity: Since \(a \sim b, b \sim c, b \sim d\), all pairs \((x, y)\) where \(x, y \in \{a, b, c, d\}\) must be in the relation.

Total elements in universal relation on 4 elements \(= 4^2 = 16\).

Current elements in \(R = 7\).

Elements to add \(= 16 - 7 = 9\).


Step 3: Final Answer:

Minimum elements to add is 9.
Quick Tip: If elements of a set are connected via the initial relation, the smallest equivalence relation will always be the partition where those elements form a complete equivalence class.

Step 1: Understanding the Concept:

We use Bayes' Theorem to find \(\lambda\), and then use coordinate geometry for the equilateral triangle properties within a parabola.


Step 2: Detailed Explanation:
\(P(C|R) = \frac{\frac{1}{3} \cdot \frac{\lambda}{\lambda+4}}{\frac{1}{3} [0.4 + 0.5 + \frac{\lambda}{\lambda+4}]} = 0.4 \Rightarrow \frac{\lambda}{\lambda+4} = 0.4 [0.9 + \frac{\lambda}{\lambda+4}]\).
\(0.6 \frac{\lambda}{\lambda+4} = 0.36 \Rightarrow \frac{\lambda}{\lambda+4} = 0.6 \Rightarrow \lambda = 6\).

Parabola: \(y^2 = 6x\). Equilateral triangle vertices: \((0,0), (x, y), (x, -y)\).
\(\tan 30^\circ = \frac{y}{x} = \frac{1}{\sqrt{3}} \Rightarrow y^2 = \frac{x^2}{3}\).

Substitute in parabola: \(\frac{x^2}{3} = 6x \Rightarrow x = 18\).

Side \(L = \sqrt{x^2 + y^2} = \sqrt{x^2 + x^2/3} = \sqrt{4x^2/3} = \frac{2 \times 18}{\sqrt{3}} = 12\sqrt{3}\).
\(L^2 = 144 \times 3 = 432\).


Step 3: Final Answer:

Side squared is 432.
Quick Tip: For an equilateral triangle in \(y^2=4ax\), vertex at origin, the coordinates of the other vertices satisfy \(y/x = \tan 30^\circ\).

Step 1: Understanding the Concept:

The cross product property \((\vec{a} - \vec{b}) \times \vec{c} = 0\) implies \(\vec{c}\) is parallel to \((\vec{a} - \vec{b})\). This allows us to find \(\lambda\).


Step 2: Detailed Explanation:
\(\vec{c} = k(\vec{a} - \vec{b}) = k(-2\hat{i} + 7\hat{j} + 2\lambda\hat{k})\).
\(\vec{a} \cdot \vec{c} = k(1(-2) + 2(7) + \lambda(2\lambda)) = k(12 + 2\lambda^2) = 7\).
\(2\vec{b} \cdot \vec{c} = 2k(3(-2) - 5(7) - \lambda(2\lambda)) = 2k(-41 - 2\lambda^2) = -43 \Rightarrow k(41 + 2\lambda^2) = \frac{43}{2}\).

Ratio: \(\frac{41 + 2\lambda^2}{12 + 2\lambda^2} = \frac{43}{14} \Rightarrow 574 + 28\lambda^2 = 516 + 86\lambda^2 \Rightarrow 58\lambda^2 = 58 \Rightarrow \lambda^2 = 1\).
\(\vec{a} \cdot \vec{b} = 3 - 10 - \lambda^2 = -7 - 1 = -8\).
\(|\vec{a} \cdot \vec{b}| = 8\).


Step 3: Final Answer:

The magnitude of the dot product is 8.
Quick Tip: The relation \(\vec{u} \times \vec{w} = \vec{v} \times \vec{w}\) implies \(\vec{w}\) is collinear with \(\vec{u} - \vec{v}\).

Step 1: Understanding the Concept:

We use the binomial expansion \((a+b)^n = \sum \binom{n}{r} a^{n-r}b^r\) to find the relationship for \(n\).


Step 2: Detailed Explanation:

First three coefficients: \(1, -3n, 9 \frac{n(n-1)}{2}\).
\(1 - 3n + 4.5(n^2 - n) = 376 \Rightarrow 4.5n^2 - 7.5n - 375 = 0 \Rightarrow 3n^2 - 5n - 250 = 0\).
\((n-10)(3n+25) = 0 \Rightarrow n = 10\).

General term: \(T_{r+1} = \binom{10}{r} (x)^{10-r} (-3 x^{-2})^r = \binom{10}{r} (-3)^r x^{10-3r}\).

For \(x^4\), \(10-3r = 4 \Rightarrow r = 2\).

Coefficient \(= \binom{10}{2} (-3)^2 = 45 \times 9 = 405\).


Step 3: Final Answer:

The coefficient is 405.
Quick Tip: Coefficient sum problems are usually solved by setting the variable \(x=1\), but when restricted to "first \(k\) terms", you must explicitly calculate the initial binomial coefficients.

Step 1: Understanding the Concept:

Area between curves is found by integrating the difference \((x_{right} - x_{left})\) with respect to \(y\) between the intersection points.


Step 2: Detailed Explanation:

Curve 1: \(x = 2y - y^2\). Curve 2: \(x = -y\).

Intersections: \(2y - y^2 = -y \Rightarrow y^2 - 3y = 0 \Rightarrow y=0, 3\).

Area \(A = \int_{0}^{3} [(2y - y^2) - (-y)] dy = \int_{0}^{3} (3y - y^2) dy\).
\(A = [\frac{3y^2}{2} - \frac{y^3}{3}]_{0}^{3} = \frac{27}{2} - 9 = 4.5\).
\(8 A = 8 \times 4.5 = 36\).


Step 3: Final Answer:

The final value is 36.
Quick Tip: When a curve is given as a quadratic in \(y\), it is often simpler to integrate with respect to \(dy\) (horizontal strips) rather than \(dx\).