Step 1: Understanding the Concept:

For a strictly increasing function \(f(x)\), the points of intersection of the function and its inverse, \(f(x) = f^{-1}(x)\), always lie on the line \(y = x\). For statement (II), we evaluate the equation by manipulating it to \(f(f(x)) = x+1\) to verify real roots.


Step 2: Key Formula or Approach:

For strictly increasing functions: \[ f(x) = f^{-1}(x) \iff f(x) = x \]
For the second statement, apply the inverse function definition: \[ f(x) = f^{-1}(x+1) \implies f(f(x)) = x+1 \]

Step 3: Detailed Explanation:

Let's analyze statement (I):

Given \(f(x) = (x - 1)^4 + 1\) for \(x \ge 1\).

Since \(f'(x) = 4(x - 1)^3 \ge 0\) for \(x \ge 1\), the function \(f(x)\) is strictly increasing.

Thus, the equation \(f(x) = f^{-1}(x)\) is equivalent to \(f(x) = x\).
\[ (x - 1)^4 + 1 = x \implies (x - 1)^4 - (x - 1) = 0 \]
Let \(t = x - 1 \ge 0\).
\[ t^4 - t = 0 \implies t(t^3 - 1) = 0 \]
This gives real roots \(t = 0\) or \(t = 1\).

Substituting back \(x - 1 = t\):

If \(t = 0 \implies x = 1\).

If \(t = 1 \implies x = 2\).

Both values belong to \([1, \infty)\). Thus, set \(S\) has exactly two elements \(\{1, 2\}\). Statement (I) is TRUE.


Let's analyze statement (II):

We are given \(f(x) = f^{-1}(x+1)\).

Let \(y = f(x)\). Then \(y = f^{-1}(x+1) \implies f(y) = x+1\).
\[ (y - 1)^4 + 1 = x + 1 \implies (y - 1)^4 = x \]
Substitute \(y = (x-1)^4 + 1\):
\[ \left( (x-1)^4 + 1 - 1 \right)^4 = x \implies (x-1)^{16} = x \]
Let \(x - 1 = t \ge 0 \implies x = t + 1\).
\[ t^{16} = t + 1 \implies t^{16} - t - 1 = 0 \]
Let \(g(t) = t^{16} - t - 1\).

We evaluate \(g(t)\): \(g(1) = -1 < 0\) and \(g(2) = 2^{16} - 3 > 0\).

By the Intermediate Value Theorem, there exists at least one real root in \((1, 2)\).

Therefore, the set \(S\) is not empty. Statement (II) is FALSE.


Step 4: Final Answer:

Only (I) is TRUE. Quick Tip: For equations of the form \(f(x) = f^{-1}(x)\), verify if \(f(x)\) is strictly increasing. If so, simplify the problem by directly solving \(f(x) = x\).

Step 1: Understanding the Concept:

We must find the complex roots of the quadratic equation \(z^2 + 4z + 16 = 0\), representing the elements of set \(S\). We then substitute these roots into the magnitude expression \(|z + \sqrt{3}i|^2\) and compute their sum.


Step 2: Key Formula or Approach:

The roots of a quadratic equation \(az^2 + bz + c = 0\) are given by:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The squared magnitude of a complex number \(w = x + iy\) is \(|w|^2 = x^2 + y^2\).


Step 3: Detailed Explanation:

Solve \(z^2 + 4z + 16 = 0\):
\[ z = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} \] \[ z = \frac{-4 \pm 4\sqrt{3}i}{2} = -2 \pm 2\sqrt{3}i \]
Set \(S\) has two elements: \(z_1 = -2 + 2\sqrt{3}i\) and \(z_2 = -2 - 2\sqrt{3}i\).


Evaluate \(|z + \sqrt{3}i|^2\) for \(z_1\):
\[ z_1 + \sqrt{3}i = -2 + 3\sqrt{3}i \] \[ |z_1 + \sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + 27 = 31 \]

Evaluate \(|z + \sqrt{3}i|^2\) for \(z_2\):
\[ z_2 + \sqrt{3}i = -2 - \sqrt{3}i \] \[ |z_2 + \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7 \]

Compute the final sum:
\[ \sum_{z \in S} |z + \sqrt{3}i|^2 = 31 + 7 = 38 \]

Step 4: Final Answer:

The sum is 38. Quick Tip: When roots of a quadratic are complex conjugates, their operations with other complex numbers yield purely real squared magnitudes. Avoid expanding algebraically and just plug the values directly.

Step 1: Understanding the Concept:

For a system of three linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero (\(\Delta = 0\)), and the system must be consistent (row operations will yield a row of all zeros).


Step 2: Key Formula or Approach:

Determinant condition:
\[ \Delta = \begin{vmatrix} 1 & 1 & 1
1 & 2 & 3
1 & 3 & \lambda \end{vmatrix} = 0 \]
Then use row operations to determine the value of \(\mu\).


Step 3: Detailed Explanation:

Evaluate \(\Delta\):
\[ 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 0 \] \[ 2\lambda - 9 - \lambda + 3 + 1 = 0 \implies \lambda - 5 = 0 \implies \lambda = 5 \]

Substitute \(\lambda = 5\) and write the augmented matrix equations:

1: \(x + y + z = 5\)

2: \(x + 2y + 3z = 9\)

3: \(x + 3y + 5z = \mu\)


To make the system have infinitely many solutions, Equation 3 must be a linear combination of Equation 1 and Equation 2.

Observe the left-hand sides:
\((x + y + z) + (x + 3y + 5z) = 2x + 4y + 6z = 2(x + 2y + 3z)\)

This means: \(Eq_1 + Eq_3 = 2 \times Eq_2\).

Apply this relation to the right-hand constants:
\[ 5 + \mu = 2(9) \] \[ 5 + \mu = 18 \implies \mu = 13 \]

Calculate the required sum:
\[ \lambda + \mu = 5 + 13 = 18 \]

Step 4: Final Answer:

The value is 18. Quick Tip: For linear dependencies, instead of full row reduction, try to visually spot if one row is a simple arithmetic progression or direct multiple of the others. Here, the coefficients of \(y\) (1, 2, 3) form an AP, hinting at \(R_1 + R_3 = 2R_2\).

Step 1: Understanding the Concept:

Because the polynomial has real coefficients, complex roots must appear in conjugate pairs. Finding the third root allows us to deduce the polynomial's coefficients using Vieta's formulas. Finally, we integrate the polynomial over symmetric limits.


Step 2: Key Formula or Approach:

Complex conjugate root theorem: If \(1 + i\sqrt{2}\) is a root, so is \(\gamma = 1 - i\sqrt{2}\).

Vieta's formulas:

Sum of roots \(= -a\)

Product of roots taken two at a time \(= b\)

Product of all roots \(= -c\)


Step 3: Detailed Explanation:

The roots are \(1\), \(1 + i\sqrt{2}\), and \(1 - i\sqrt{2}\).

Find \(a\):
\(-a = 1 + (1 + i\sqrt{2}) + (1 - i\sqrt{2}) = 3 \implies a = -3\)

Find \(c\):
\(-c = (1)(1 + i\sqrt{2})(1 - i\sqrt{2}) = 1^2 - (i\sqrt{2})^2 = 1 - (-2) = 3 \implies c = -3\)

Find \(b\):
\(b = (1)(1+i\sqrt{2}) + (1)(1-i\sqrt{2}) + (1+i\sqrt{2})(1-i\sqrt{2}) = 2 + 3 = 5\)


The polynomial is \(P(x) = x^3 - 3x^2 + 5x - 3\).

Evaluate the definite integral:
\[ I = \int_{-1}^1 (x^3 - 3x^2 + 5x - 3) dx \]
Using the symmetric limits property (\(\int_{-A}^A f(x) dx = 0\) for odd functions):

The odd terms \(x^3\) and \(5x\) integrate to 0.
\[ I = \int_{-1}^1 (-3x^2 - 3) dx = 2 \int_{0}^1 (-3x^2 - 3) dx \] \[ I = 2 \left[ -x^3 - 3x \right]_0^1 = 2(-1 - 3) = -8 \]

Step 4: Final Answer:

The integral evaluates to \(-8\). Quick Tip: Always separate polynomials into even and odd parts when integrating across limits of the form \([-A, A]\). This halves your calculation time and drastically reduces errors.

Step 1: Understanding the Concept:

For a quadratic equation to have two positive real roots, three conditions must hold: the discriminant must be non-negative (\(\Delta \ge 0\)), the sum of roots must be positive (\(S > 0\)), and the product of roots must be positive (\(P > 0\)).


Step 2: Key Formula or Approach:

For \(Ax^2 + Bx + C = 0\):

1. \(\Delta = B^2 - 4AC \ge 0\)

2. \(S = -B/A > 0\)

3. \(P = C/A > 0\)


Step 3: Detailed Explanation:

Given \(A = \lambda + 2\), \(B = -3\lambda\), \(C = 4\lambda\).


Condition 1: \(\Delta \ge 0\)
\[ (-3\lambda)^2 - 4(\lambda + 2)(4\lambda) \ge 0 \] \[ 9\lambda^2 - 16\lambda^2 - 32\lambda \ge 0 \implies -7\lambda^2 - 32\lambda \ge 0 \] \[ 7\lambda^2 + 32\lambda \le 0 \implies \lambda(7\lambda + 32) \le 0 \]
So, \(\lambda \in \left[ -\frac{32}{7}, 0 \right]\).


Condition 2: Sum \(S > 0\)
\[ \frac{3\lambda}{\lambda + 2} > 0 \]
Using the wavy curve method, \(\lambda \in (-\infty, -2) \cup (0, \infty)\).


Condition 3: Product \(P > 0\)
\[ \frac{4\lambda}{\lambda + 2} > 0 \]
This yields the exact same intervals as the sum: \(\lambda \in (-\infty, -2) \cup (0, \infty)\).


Find the intersection of all conditions:

Intersection of \(\left[ -\frac{32}{7}, 0 \right]\) and \((-\infty, -2) \cup (0, \infty)\) is \(\left[ -\frac{32}{7}, -2 \right)\).

Since \(-\frac{32}{7} \approx -4.57\), the interval is approximately \([-4.57, -2)\).

The integral values in this range are \(-4\) and \(-3\).


Step 4: Final Answer:

There are 2 integral values. Quick Tip: "Two positive roots" allows for the roots to be identical, so the discriminant inequality must be \(\Delta \ge 0\), not just strictly greater than zero.

Step 1: Understanding the Concept:

The equation \(\det(A - \alpha I) = 0\) is the characteristic equation that defines the eigenvalues of matrix \(A\). We must find the largest eigenvalue \(p\), substitute it into the given circle equation, and count its intersection points with the \(x\)-axis (\(y=0\)) and \(y\)-axis (\(x=0\)).


Step 2: Key Formula or Approach:

Characteristic polynomial for a \(3 \times 3\) matrix:
\[ \alpha^3 - tr(A)\alpha^2 + (M_{11} + M_{22} + M_{33})\alpha - \det(A) = 0 \]
where \(tr(A)\) is the trace and \(M_{ii}\) are the principal minors.


Step 3: Detailed Explanation:

Trace: \(tr(A) = 1 - 2 - 7 = -8\).

Determinant:
\[ |A| = 1(14 - 64) - 2(-28 - 24) + 7(32 - (-6)) = -50 + 104 + 266 = 320 \]
Sum of principal minors:
\(M_{11} = 14 - 64 = -50\)
\(M_{22} = -7 - 21 = -28\)
\(M_{33} = -2 - 8 = -10\)

Sum = \(-88\).


The characteristic equation:
\[ \alpha^3 + 8\alpha^2 - 88\alpha - 320 = 0 \]
By trial, \(\alpha = 8\) is a root: \(8^3 + 8(8^2) - 88(8) - 320 = 512 + 512 - 704 - 320 = 0\).

Factor out \((\alpha - 8)\):
\[ (\alpha - 8)(\alpha^2 + 16\alpha + 40) = 0 \]
Roots of quadratic: \(\alpha = \frac{-16 \pm \sqrt{256 - 160}}{2} = -8 \pm 2\sqrt{6}\).

Since \(2\sqrt{6} \approx 4.9\), the largest eigenvalue is \(p = 8\).


Substitute \(p = 8\) into the circle equation:
\[ (x - 8)^2 + (y - 16)^2 = 320 \]
Intersect with \(x\)-axis (\(y = 0\)):
\[ (x - 8)^2 + 256 = 320 \implies (x - 8)^2 = 64 \implies x = 16 or x = 0 \]
Points: \((16, 0), (0, 0)\).


Intersect with \(y\)-axis (\(x = 0\)):
\[ 64 + (y - 16)^2 = 320 \implies (y - 16)^2 = 256 \implies y = 32 or y = 0 \]
Points: \((0, 32), (0, 0)\).


The unique intersection points are \((0, 0)\), \((16, 0)\), and \((0, 32)\).


Step 4: Final Answer:

The circle intersects the axes at 3 points. Quick Tip: Remember that the origin \((0,0)\) counts as a single intersection point for both the \(x\)-axis and the \(y\)-axis. Always compile a unique list of coordinates before finalizing your count.

Step 1: Understanding the Concept:

Evaluate the sum of the infinite geometric progressions for \(\alpha\) and \(\beta\). Then substitute these fractions into the logarithmic expression, utilizing exponent and base change properties to simplify.


Step 2: Key Formula or Approach:

Infinite GP sum: \(S_\infty = \frac{a}{1 - r}\).

Logarithm properties: \(\log_{b^n}(x) = \frac{1}{n} \log_b(x)\) and \(a^{\log_a(x)} = x\).


Step 3: Detailed Explanation:

Evaluate \(\alpha\):
\(a = 1/4\), \(r = 1/2\)
\[ \alpha = \frac{1/4}{1 - 1/2} = \frac{1}{2} \]

Evaluate \(\beta\):
\(a = 1/3\), \(r = 1/3\)
\[ \beta = \frac{1/3}{1 - 1/3} = \frac{1}{2} \]

Simplify the first term: \((0.2)^{\log_{\sqrt{5}}(\alpha)}\)
\(0.2 = 5^{-1}\) and \(\sqrt{5} = 5^{1/2}\).
\[ \log_{\sqrt{5}}(1/2) = \frac{1}{1/2} \log_5(1/2) = 2 \log_5(1/2) = \log_5(1/4) \] \[ (5^{-1})^{\log_5(1/4)} = 5^{-\log_5(1/4)} = 5^{\log_5(4)} = 4 \]

Simplify the second term: \((0.04)^{\log_5(\beta)}\)
\(0.04 = 5^{-2}\).
\[ (5^{-2})^{\log_5(1/2)} = 5^{-2\log_5(1/2)} = 5^{\log_5(4)} = 4 \]

Total sum = \(4 + 4 = 8\).


Step 4: Final Answer:

The final value is 8. Quick Tip: Converting decimal bases into fractional powers (like \(0.04 = 5^{-2}\)) ensures you can immediately apply the fundamental identity of logarithms \(a^{\log_a(x)} = x\).

Step 1: Understanding the Concept:

The standard deviation relies on the variance, which requires finding both \(\sum x_i^2\) and \(\sum x_i\). By expanding the two given summations, we get two linear equations which can be solved simultaneously.


Step 2: Key Formula or Approach:

Variance formula:
\[ \sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2 \]
Expansion of squares: \((x_i \pm a)^2 = x_i^2 \pm 2ax_i + a^2\).


Step 3: Detailed Explanation:

Let \(S_2 = \sum x_i^2\) and \(S_1 = \sum x_i\) for \(n=10\).


Expand the first summation:
\[ \sum (x_i^2 + 4x_i + 4) = 180 \implies S_2 + 4S_1 + 40 = 180 \implies S_2 + 4S_1 = 140 \quad (Eq 1) \]

Expand the second summation:
\[ \sum (x_i^2 - 2x_i + 1) = 90 \implies S_2 - 2S_1 + 10 = 90 \implies S_2 - 2S_1 = 80 \quad (Eq 2) \]

Subtract Eq 2 from Eq 1:
\[ 6S_1 = 60 \implies S_1 = 10 \]
Substitute \(S_1\) back into Eq 2:
\[ S_2 - 20 = 80 \implies S_2 = 100 \]

Calculate the variance \(\sigma^2\):
\[ \sigma^2 = \frac{100}{10} - \left( \frac{10}{10} \right)^2 = 10 - 1 = 9 \]
Standard deviation \(\sigma = \sqrt{9} = 3\).


Step 4: Final Answer:

The standard deviation is 3. Quick Tip: Always remember that \(\sum_{i=1}^n c = n \cdot c\). It is a common mistake to forget multiplying the constant term by \(n\) when distributing the summation operator.

Step 1: Understanding the Concept:

To find the term independent of \(x\) in a binomial expansion, write out the general term, collect all powers of \(x\), and set the final exponent equal to zero to solve for the term index \(r\).


Step 2: Key Formula or Approach:

General term in \((a + b)^n\):
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

Step 3: Detailed Explanation:

The given binomial is \((9x - \frac{1}{3}x^{-1/2})^{18}\).
\[ T_{r+1} = \binom{18}{r} (9x)^{18-r} \left( -\frac{1}{3}x^{-1/2} \right)^r \]
Combine the exponents of \(x\):
\[ x^{18 - r} \cdot x^{-r/2} = x^{18 - 3r/2} \]
For the term to be independent of \(x\), set the exponent to 0:
\[ 18 - \frac{3r}{2} = 0 \implies r = 12 \]

Calculate the coefficient for \(r = 12\):
\[ T_{13} = \binom{18}{12} (9)^{6} \left(-\frac{1}{3}\right)^{12} \]
Since \(9^6 = (3^2)^6 = 3^{12}\) and \((-1/3)^{12} = 3^{-12}\):
\[ T_{13} = \binom{18}{12} \cdot 3^{12} \cdot 3^{-12} = \binom{18}{12} \]
Compute \(\binom{18}{12} = \binom{18}{6}\):
\[ \binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564 \]
Equating to the given expression:
\[ 221k = 18564 \implies k = \frac{18564}{221} = 84 \]

Step 4: Final Answer:

The value of \(k\) is 84. Quick Tip: To divide large numbers manually faster, estimate using multipliers: \(220 \times 80 = 17600\), leaving roughly 960. \(220 \times 4 = 880\), making 84 the perfect candidate.

Step 1: Understanding the Concept:

By equating the given centroid coordinates to the centroid formula applied to vertices P, Q, and R, we can deduce a direct geometric locus for point Q. Finding the intersection of this locus with the given circle gives exact coordinates for Q, allowing us to find R.


Step 2: Key Formula or Approach:

Centroid \(G(x, y)\) of triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\):
\[ G_x = \frac{x_1 + x_2 + x_3}{3}, \quad G_y = \frac{y_1 + y_2 + y_3}{3} \]

Step 3: Detailed Explanation:

Let \(Q = (x_1, y_1)\) and \(R = (x_2, y_2)\).

Equating the \(x\)-coordinates of the centroid:
\[ \frac{3\cos\alpha + x_1 + x_2}{3} = 2 + \cos\alpha \implies x_1 + x_2 = 6 \]
Equating the \(y\)-coordinates:
\[ \frac{2\sin\alpha + y_1 + y_2}{3} = 3 + \frac{2}{3}\sin\alpha \implies y_1 + y_2 = 9 \]

Since R lies on \(x + y = 5\), we have \(x_2 + y_2 = 5\).

Adding the derived coordinates of Q:
\[ x_1 + y_1 = (6 - x_2) + (9 - y_2) = 15 - (x_2 + y_2) = 15 - 5 = 10 \]
So point Q lies on the line \(x + y = 10\).


Point Q also lies on the circle: \((x - 7)^2 + (y - 7)^2 = 16\) (completing the square).

Substitute \(y = 10 - x\) into the circle equation:
\[ (x - 7)^2 + (3 - x)^2 = 16 \]
Let \(x - 5 = t\). Then \(x - 7 = t - 2\) and \(3 - x = -2 - t\).
\[ (t - 2)^2 + (-2 - t)^2 = 16 \implies 2t^2 + 8 = 16 \implies t^2 = 4 \implies t = \pm 2 \]
If \(t = 2\), \(x_1 = 7\) and \(y_1 = 3\).

If \(t = -2\), \(x_1 = 3\) and \(y_1 = 7\).


We need the sum of the ordinates (\(y_2\)) of all possible points R.

Recall \(y_2 = 9 - y_1\).

For \(y_1 = 3 \implies y_2 = 6\).

For \(y_1 = 7 \implies y_2 = 2\).

The sum of all possible ordinates of R is \(6 + 2 = 8\).


Step 4: Final Answer:

The sum is 8. Quick Tip: Substituting variables to center an equation (like \(x - 5 = t\) when terms are \((x-7)\) and \((3-x)\)) instantly neutralizes the linear terms during expansion, rapidly simplifying quadratics.

Step 1: Understanding the Concept:

Using the distance properties of hyperbolas, determine the parameters \(a\), \(b\), and \(e\) to establish the exact equation of the hyperbola. Finding intersection points of a vertical line \(x = \alpha\) directly models the base and height of the specified triangle AOB.


Step 2: Key Formula or Approach:

Distance between foci \(= 2ae\).

Distance between directrices \(= \frac{2a}{e}\).

Hyperbola relation: \(b^2 = a^2(e^2 - 1)\).

Area of triangle with vertices \((0,0), (\alpha, y), (\alpha, -y)\) is \(\frac{1}{2} \times (2y) \times \alpha = \alpha y\).


Step 3: Detailed Explanation:

Given \(2ae = 6 \implies ae = 3\).

Given \(\frac{2a}{e} = \frac{8}{3} \implies \frac{a}{e} = \frac{4}{3}\).

Multiply them: \((ae)\left(\frac{a}{e}\right) = 3 \times \frac{4}{3} \implies a^2 = 4 \implies a = 2\).

Therefore \(e = \frac{3}{2}\).

Find \(b^2\): \(b^2 = a^2(e^2 - 1) = 4\left(\frac{9}{4} - 1\right) = 5\).

The hyperbola is \(\frac{x^2}{4} - \frac{y^2}{5} = 1\).


Intersect with \(x = \alpha\):
\[ \frac{\alpha^2}{4} - \frac{y^2}{5} = 1 \implies y = \sqrt{5\left(\frac{\alpha^2}{4} - 1\right)} \]
The area of \(\Delta AOB\) is \(\alpha y = 4\sqrt{15}\).

Square both sides:
\[ \alpha^2 y^2 = 240 \implies \alpha^2 \left( 5\left(\frac{\alpha^2}{4} - 1\right) \right) = 240 \] \[ \frac{5\alpha^4}{4} - 5\alpha^2 - 240 = 0 \implies \alpha^4 - 4\alpha^2 - 192 = 0 \]
Factor the quadratic in \(\alpha^2\):
\[ (\alpha^2 - 16)(\alpha^2 + 12) = 0 \]
Since \(\alpha^2\) must be positive, \(\alpha^2 = 16\).


Step 4: Final Answer:
\(\alpha^2\) equals 16. Quick Tip: Whenever the area of an isosceles triangle formed by vertical intersections is needed relative to the origin, the area simplifies cleanly to \(x \cdot y\). Squaring both sides avoids dealing with roots entirely.

Step 1: Understanding the Concept:

To find the global maximum of a trigonometric function on a closed interval, we map the expression using substitution to simplify the domain, calculate the first derivative to find critical points, and substitute them back to yield the maximum.


Step 2: Key Formula or Approach:

For \(y = f(t)\), the maximum occurs where \(f'(t) = 0\).

Apply substitution \(t = x/2\).


Step 3: Detailed Explanation:

Let \(t = x/2\). As \(x \in [0, \pi]\), \(t \in [0, \pi/2]\).

In this interval, \(\cos t \ge 0\), so we can drop the absolute value.

Let \(f(t) = 16 \sin t \cos^3 t\).

Find the derivative using product and chain rules:
\[ f'(t) = 16 (\cos t \cdot \cos^3 t + \sin t \cdot 3\cos^2 t(-\sin t)) \] \[ f'(t) = 16(\cos^4 t - 3\sin^2 t \cos^2 t) = 16\cos^2 t(\cos^2 t - 3\sin^2 t) \]
Set \(f'(t) = 0\):

Since \(t \in [0, \pi/2]\), \(\cos t \neq 0\) except at boundary \(\pi/2\) (which gives minimum 0).
\[ \cos^2 t - 3\sin^2 t = 0 \implies \tan^2 t = \frac{1}{3} \implies \tan t = \frac{1}{\sqrt{3}} \implies t = \frac{\pi}{6} \]
Substitute \(t = \pi/6\) into \(f(t)\):
\[ f\left(\frac{\pi}{6}\right) = 16 \sin\left(\frac{\pi}{6}\right) \cos^3\left(\frac{\pi}{6}\right) = 16 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)^3 \] \[ = 8 \left(\frac{3\sqrt{3}}{8}\right) = 3\sqrt{3} \]

Step 4: Final Answer:

The maximum value is \(3\sqrt{3}\). Quick Tip: For max/min optimization of expressions like \(\sin^n x \cos^m x\), critical points invariably reside at \(\tan^2 x = m/n\) (when evaluating the first quadrant). This is an incredibly fast mental shortcut!

Step 1: Understanding the Concept:

The shortest distance between two skew lines \(\vec{r} = \vec{a_1} + \lambda\vec{b_1}\) and \(\vec{r} = \vec{a_2} + \mu\vec{b_2}\) is the projection of the vector connecting their reference points (\(\vec{a_2} - \vec{a_1}\)) onto the vector perpendicular to both direction vectors (\(\vec{b_1} \times \vec{b_2}\)).


Step 2: Key Formula or Approach:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

Step 3: Detailed Explanation:

From the given equations:
\(\vec{a_1} = \left(\frac{1}{3}, 2, \frac{8}{3}\right)\), \(\vec{b_1} = (2, -5, 6)\)
\(\vec{a_2} = \left(-\frac{2}{3}, 0, -\frac{1}{3}\right)\), \(\vec{b_2} = (0, 1, -1)\)


Find \(\vec{a_2} - \vec{a_1}\):
\[ \vec{a_2} - \vec{a_1} = \left(-\frac{2}{3} - \frac{1}{3}\right)\hat{i} + (0 - 2)\hat{j} + \left(-\frac{1}{3} - \frac{8}{3}\right)\hat{k} = -\hat{i} - 2\hat{j} - 3\hat{k} \]

Find \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -5 & 6
0 & 1 & -1 \end{vmatrix} = \hat{i}(5 - 6) - \hat{j}(-2) + \hat{k}(2) = -\hat{i} + 2\hat{j} + 2\hat{k} \]

Find the magnitude \(|\vec{b_1} \times \vec{b_2}|\):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{9} = 3 \]

Find the dot product:
\[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-1)(-1) + (-2)(2) + (-3)(2) = 1 - 4 - 6 = -9 \]

Calculate distance \(d\):
\[ d = \frac{|-9|}{3} = 3 \]

Step 4: Final Answer:

The shortest distance is 3. Quick Tip: To ensure fewer signs errors when using vector cross products in examinations, always write out the intermediate \(2 \times 2\) determinant calculations formally before computing the final sum.

Step 1: Understanding the Concept:

If point \(P'\) is the image of point \(P\) over a line, the midpoint of the segment \(PP'\) must lie exactly on that given line. By deriving the coordinates of the midpoint and substituting them into the line equation, we can find the viable values of \(\alpha\).


Step 2: Key Formula or Approach:

Midpoint \(M\) between \(P\) and \(P'\):
\(M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right)\)

Line equation: \(\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}\)


Step 3: Detailed Explanation:

Let \(P = (\alpha, 2\alpha, 1)\) and \(P' = \left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right)\).

The midpoint \(M\) is:
\[ M = \left(\frac{3\alpha + 1}{2}, \frac{\alpha^2 - \alpha}{2}, \frac{\alpha + 1}{4}\right) \]

Substitute \(M\) into the line's Cartesian equation:
\[ \frac{\frac{3\alpha + 1}{2} - 2}{3} = \frac{\frac{\alpha^2 - \alpha}{2} - 1}{2} = \frac{\frac{\alpha + 1}{4}}{1} \] \[ \frac{3\alpha - 3}{6} = \frac{\alpha^2 - \alpha - 2}{4} = \frac{\alpha + 1}{4} \]

Equate the first and third expressions:
\[ \frac{\alpha - 1}{2} = \frac{\alpha + 1}{4} \implies 2\alpha - 2 = \alpha + 1 \implies \alpha = 3 \]
Check if \(\alpha = 3\) validates the second expression:
\[ \frac{3^2 - 3 - 2}{4} = \frac{4}{4} = 1 \]
And \(\frac{3 - 1}{2} = 1\). The equations are perfectly balanced.


It is mathematically sufficient to check only this root since intersection forces a single rigid geometrical relationship. No other parameters align across all three dimensional coordinates simultaneously.


Step 4: Final Answer:

The only possible value of \(\alpha\) is 3. Quick Tip: To prevent heavy quadratic algebra, always equate the simplest linear fraction pairs first. Then verify the derived candidate against the heavier quadratic fraction.

Step 1: Understanding the Concept:

By taking the dot products of the given equation \(\vec{A}\) with the basis vectors \(\hat{u}\) and \(\hat{v}\) independently, we create two algebraic equations. Solving these equations reveals \(\lambda\) as a linear combination of those dot products.


Step 2: Key Formula or Approach:

Dot product expansion:
\(\hat{u} \cdot \hat{u} = 1\), \(\hat{v} \cdot \hat{v} = 1\).
\(|\hat{u} \times \hat{v}| = |\hat{u}| |\hat{v}| \sin \theta \).
\(\hat{u} \cdot \hat{v} = |\hat{u}| |\hat{v}| \cos \theta \).


Step 3: Detailed Explanation:

Given \(|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2} \implies \sin \theta = \frac{\sqrt{3}}{2}\).

Since the angle is acute, \(\theta = 60^\circ\). Therefore, \(\hat{u} \cdot \hat{v} = \cos 60^\circ = \frac{1}{2}\).


We have \(\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u} \times \hat{v})\).

Dot product with \(\hat{u}\):
\[ \vec{A} \cdot \hat{u} = \lambda(\hat{u} \cdot \hat{u}) + (\hat{v} \cdot \hat{u}) + (\hat{u} \times \hat{v}) \cdot \hat{u} \]
Since \(\hat{u} \times \hat{v}\) is orthogonal to \(\hat{u}\), the third term is zero.
\[ \vec{A} \cdot \hat{u} = \lambda(1) + \frac{1}{2} = \lambda + \frac{1}{2} \quad (Eq 1) \]

Dot product with \(\hat{v}\):
\[ \vec{A} \cdot \hat{v} = \lambda(\hat{u} \cdot \hat{v}) + (\hat{v} \cdot \hat{v}) + (\hat{u} \times \hat{v}) \cdot \hat{v} = \frac{\lambda}{2} + 1 \quad (Eq 2) \]

We need \(\lambda\). From Eq 1, \(\lambda = \vec{A} \cdot \hat{u} - \frac{1}{2}\).

Test the options using these identities.

Option (A):
\[ \frac{4}{3}(\vec{A} \cdot \hat{u}) - \frac{2}{3}(\vec{A} \cdot \hat{v}) = \frac{4}{3}\left(\lambda + \frac{1}{2}\right) - \frac{2}{3}\left(\frac{\lambda}{2} + 1\right) \] \[ = \frac{4\lambda}{3} + \frac{2}{3} - \frac{\lambda}{3} - \frac{2}{3} = \frac{3\lambda}{3} = \lambda \]
This perfectly matches \(\lambda\).


Step 4: Final Answer:

Option (A) is the correct expression. Quick Tip: Remember that the cross product vector \((\hat{u} \times \hat{v})\) is strictly orthogonal to both \(\hat{u}\) and \(\hat{v}\). When you dot it against either component vector, the term entirely vanishes.

Step 1: Find the function \(f(x)\)


Given, \[ f(x+y)=f(x)+2y^2+y+\alpha xy \]

Put \(x=0\):
\[ f(0+y)=f(0)+2y^2+y+\alpha(0)(y) \]
\[ f(y)=f(0)+2y^2+y \]

Since \(f(0)=-1\),
\[ f(y)=2y^2+y-1 \]

Replacing \(y\) by \(x\),
\[ f(x)=2x^2+x-1 \]



Step 2: Find the value of \(\alpha\)


Now, \[ f(x+y)=2(x+y)^2+(x+y)-1 \]

Expand:
\[ f(x+y)=2(x^2+2xy+y^2)+x+y-1 \]
\[ f(x+y)=2x^2+4xy+2y^2+x+y-1 \]

Also from the given equation,
\[ f(x)+2y^2+y+\alpha xy \]

Substitute \(f(x)=2x^2+x-1\):
\[ = (2x^2+x-1)+2y^2+y+\alpha xy \]
\[ =2x^2+x-1+2y^2+y+\alpha xy \]

Comparing both expressions of \(f(x+y)\),
\[ 4xy=\alpha xy \]

Hence,
\[ \alpha=4 \]



Step 3: Verify using \(f(1)=2\)

\[ f(1)=2(1)^2+1-1 \]
\[ =2+1-1=2 \]

Condition verified.



Step 4: Calculate the required sum

\[ \sum_{n=1}^{5} (\alpha+f(n)) \]

Substitute \(\alpha=4\) and \(f(n)=2n^2+n-1\):
\[ = \sum_{n=1}^{5} \left(4+2n^2+n-1\right) \]
\[ = \sum_{n=1}^{5} (2n^2+n+3) \]
\[ =2\sum_{n=1}^{5}n^2+\sum_{n=1}^{5}n+\sum_{n=1}^{5}3 \]

Use formulas:
\[ \sum_{n=1}^{5} n^2=\frac{5(6)(11)}{6}=55 \]
\[ \sum_{n=1}^{5} n=\frac{5(6)}{2}=15 \]
\[ \sum_{n=1}^{5} 3=3\times 5=15 \]

So,
\[ =2(55)+15+15 \]
\[ =110+30 \]
\[ =140 \]



Final Answer:
\[ \boxed{140} \]
\[ \boxed{Option (B)} \] Quick Tip: When dealing with \(f(x+y)\) functional equations containing a polynomial tail, substituting \(x=0\) rapidly extracts the core polynomial \(f(y)\), saving the trouble of partial differentiation.

Step 1: Understanding the Concept:

The equation \(a + b + 2c = 22\) is a linear Diophantine equation. Since \(a, b, c\) are non-negative integers, we can fix the value of \(c\) and find the number of possible non-negative integer solutions for \(a\) and \(b\) for each valid \(c\).


Step 2: Key Formula or Approach:

For a fixed non-negative integer \(k\), the number of non-negative integer solutions to the equation \(a + b = k\) is given by \(k + 1\).

Sum of an arithmetic progression: \(S = \frac{n}{2}(first\_term + last\_term)\).


Step 3: Detailed Explanation:

Given \(a + b + 2c = 22\).

Rewrite as: \(a + b = 22 - 2c\).

Since \(a \ge 0\) and \(b \ge 0\), their sum must be non-negative:
\(22 - 2c \ge 0 \implies 2c \le 22 \implies c \le 11\).

Since \(c\) is a non-negative integer, the possible values for \(c\) are \(0, 1, 2, \dots, 11\).


For any chosen value of \(c\), the equation becomes \(a + b = k\), where \(k = 22 - 2c\).

The number of solutions for \((a, b)\) is \(k + 1 = (22 - 2c) + 1 = 23 - 2c\).


Total number of solutions \(n(A)\) is the sum of solutions for all possible values of \(c\):
\(n(A) = \sum_{c=0}^{11} (23 - 2c)\)
\(n(A) = 23(12) - 2 \sum_{c=0}^{11} c\)
\(n(A) = 276 - 2 \left( \frac{11 \times 12}{2} \right)\)
\(n(A) = 276 - 132 = 144\).


Step 4: Final Answer:

The number of elements in set A is 144. Quick Tip: Isolating the variable with the largest coefficient (here, \(c\)) minimizes the number of cases you need to sum, greatly simplifying combinatorial equations.

Step 1: Understanding the Concept:

The given equations represent two horizontal parabolas. To find the enclosed area, we first find their points of intersection and then integrate the difference between the rightmost curve and the leftmost curve with respect to \(y\).


Step 2: Key Formula or Approach:

Area between two curves \(x = f(y)\) and \(x = g(y)\) from \(y = c\) to \(y = d\):
\(Area = \int_c^d [f(y) - g(y)] dy\), where \(f(y) \ge g(y)\).


Step 3: Detailed Explanation:

The curves are:

1) \(x = -3y^2\) (a parabola opening to the left with vertex at the origin).

2) \(x = 1 - 4y^2\) (a parabola opening to the left with vertex at \((1, 0)\)).


Find the points of intersection by equating the \(x\) values:
\(-3y^2 = 1 - 4y^2\)
\(y^2 = 1 \implies y = \pm 1\).

So, the intersection limits are from \(y = -1\) to \(y = 1\).


Within this interval \(y \in [-1, 1]\), we determine which curve is to the right (has larger \(x\)):

At \(y = 0\), \(x_1 = 0\) and \(x_2 = 1\). So, \(x = 1 - 4y^2\) is the right curve \(f(y)\), and \(x = -3y^2\) is the left curve \(g(y)\).


Set up the integral:
\(Area = \int_{-1}^1 [(1 - 4y^2) - (-3y^2)] dy\)
\(= \int_{-1}^1 (1 - y^2) dy\)

Since the integrand is an even function:
\(= 2 \int_0^1 (1 - y^2) dy\)
\(= 2 \left[ y - \frac{y^3}{3} \right]_0^1\)
\(= 2 \left( 1 - \frac{1}{3} \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3}\).


Step 4: Final Answer:

The area of the region is \(\frac{4}{3}\). Quick Tip: For parabolas opening left/right, integrating with respect to \(y\) avoids dealing with messy square roots and splitting the area into multiple vertical integral sections.

Step 1: Identify the differential equation form


The given equation is a first-order linear differential equation of the form
\[ \frac{dy}{dx}+P(x)y=Q(x) \]

where
\[ P(x)= \frac{6x^2+(3x^2+2x^3+4)e^{-2x}} {(x^3+2)(2+e^{-2x})} \]

and
\[ Q(x)=2+e^{-2x} \]



Step 2: Find the integrating factor (I.F.)


Observe that
\[ \frac{d}{dx}\ln(x^3+2)=\frac{3x^2}{x^3+2} \]

and
\[ \frac{d}{dx}\ln(2+e^{-2x}) = \frac{-2e^{-2x}}{2+e^{-2x}} \]

Therefore,
\[ \frac{d}{dx} \left[ \ln(x^3+2)-\ln(2+e^{-2x}) \right] \]
\[ = \frac{3x^2}{x^3+2} + \frac{2e^{-2x}}{2+e^{-2x}} \]

Taking LCM:
\[ = \frac{3x^2(2+e^{-2x})+2e^{-2x}(x^3+2)} {(x^3+2)(2+e^{-2x})} \]
\[ = \frac{ 6x^2+3x^2e^{-2x}+2x^3e^{-2x}+4e^{-2x} } {(x^3+2)(2+e^{-2x})} \]
\[ = \frac{ 6x^2+(3x^2+2x^3+4)e^{-2x} } {(x^3+2)(2+e^{-2x})} \]

This is exactly \(P(x)\).

Hence,
\[ \int P(x)\,dx = \ln(x^3+2)-\ln(2+e^{-2x}) \]
\[ = \ln\left(\frac{x^3+2}{2+e^{-2x}}\right) \]

So the integrating factor is
\[ I.F. = e^{\int P(x)\,dx} = \frac{x^3+2}{2+e^{-2x}} \]



Step 3: Multiply the equation by I.F.


Multiplying the differential equation by the integrating factor:
\[ \frac{d}{dx} \left[ y\cdot \frac{x^3+2}{2+e^{-2x}} \right] = (2+e^{-2x}) \cdot \frac{x^3+2}{2+e^{-2x}} \]
\[ = x^3+2 \]

Now integrate both sides:
\[ \int \frac{d}{dx} \left[ y\cdot \frac{x^3+2}{2+e^{-2x}} \right]dx = \int (x^3+2)\,dx \]
\[ y\cdot \frac{x^3+2}{2+e^{-2x}} = \frac{x^4}{4}+2x+C \]



Step 4: Use the initial condition


Given
\[ y(0)=\frac{3}{2} \]

Substitute \(x=0\):
\[ \frac{3}{2}\cdot \frac{0^3+2}{2+e^0} = \frac{0^4}{4}+2(0)+C \]
\[ \frac{3}{2}\cdot\frac{2}{3}=C \]
\[ C=1 \]

So the solution becomes
\[ y\cdot \frac{x^3+2}{2+e^{-2x}} = \frac{x^4}{4}+2x+1 \]



Step 5: Find \(y(1)\)


Put \(x=1\):
\[ y(1)\cdot \frac{1^3+2}{2+e^{-2}} = \frac{1^4}{4}+2(1)+1 \]
\[ y(1)\cdot \frac{3}{2+e^{-2}} = \frac{1}{4}+2+1 \]
\[ = \frac{13}{4} \]

Therefore,
\[ y(1) = \frac{13}{4}\cdot \frac{2+e^{-2}}{3} \]
\[ = \frac{13}{12}(2+e^{-2}) \]

Given
\[ y(1)=\alpha(2+e^{-2}) \]

Comparing,
\[ \alpha=\frac{13}{12} \]



Final Answer:
\[ \boxed{\alpha=\frac{13}{12}} \]
\[ \boxed{Option (D)} \] Quick Tip: When faced with a monstrous fraction for \(P(x)\) in a linear DE, immediately attempt to decompose it using logarithmic differentiation structures like \(\frac{f'}{f} \pm \frac{g'}{g}\), which often effortlessly leads to the exact I.F.

Step 1: Convert \(\cot^{-1}\) into \(\tan^{-1}\)


Using the identity
\[ \cot^{-1}\theta=\tan^{-1}\left(\frac{1}{\theta}\right) \]

we get
\[ I=\int_0^1 \tan^{-1}\left(\frac{1}{1+x+x^2}\right)\,dx \]

Now write
\[ \frac{1}{1+x+x^2} = \frac{(x+1)-x}{1+x(x+1)} \]

So,
\[ I= \int_0^1 \tan^{-1} \left( \frac{(x+1)-x}{1+x(x+1)} \right)\,dx \]



Step 2: Use inverse tangent identity


Using the identity
\[ \tan^{-1}\left(\frac{A-B}{1+AB}\right) = \tan^{-1}A-\tan^{-1}B \]

Take
\[ A=x+1,\qquad B=x \]

Then
\[ I= \int_0^1 \left[ \tan^{-1}(x+1)-\tan^{-1}x \right]dx \]

Split the integral:
\[ I= \int_0^1 \tan^{-1}(x+1)\,dx - \int_0^1 \tan^{-1}x\,dx \]



Step 3: Change variable in first integral


Let
\[ u=x+1 \]

Then
\[ du=dx \]

When \(x=0\), \(u=1\)

When \(x=1\), \(u=2\)

So,
\[ I= \int_1^2 \tan^{-1}u\,du - \int_0^1 \tan^{-1}x\,dx \]



Step 4: Use standard integral formula


The standard result is
\[ \int \tan^{-1}t\,dt = t\tan^{-1}t-\frac{1}{2}\ln(1+t^2) \]

Now evaluate both integrals.

First integral:
\[ \int_1^2 \tan^{-1}u\,du = \left[ u\tan^{-1}u-\frac{1}{2}\ln(1+u^2) \right]_1^2 \]
\[ = \left( 2\tan^{-1}2-\frac{1}{2}\ln 5 \right) - \left( \tan^{-1}1-\frac{1}{2}\ln 2 \right) \]

Since
\[ \tan^{-1}1=\frac{\pi}{4} \]
\[ = 2\tan^{-1}2-\frac{1}{2}\ln 5-\frac{\pi}{4}+\frac{1}{2}\ln 2 \]

Second integral:
\[ \int_0^1 \tan^{-1}x\,dx = \left[ x\tan^{-1}x-\frac{1}{2}\ln(1+x^2) \right]_0^1 \]
\[ = \frac{\pi}{4}-\frac{1}{2}\ln 2 \]



Step 5: Subtract दोनों integrals

\[ I= \left( 2\tan^{-1}2-\frac{1}{2}\ln 5-\frac{\pi}{4}+\frac{1}{2}\ln 2 \right) - \left( \frac{\pi}{4}-\frac{1}{2}\ln 2 \right) \]
\[ = 2\tan^{-1}2-\frac{\pi}{2}-\frac{1}{2}\ln 5+\ln 2 \]

Now use
\[ \ln 2=\frac{1}{2}\ln 4 \]

So,
\[ -\frac{1}{2}\ln 5+\ln 2 = -\frac{1}{2}\ln 5+\frac{1}{2}\ln 4 \]
\[ = -\frac{1}{2}\ln\left(\frac{5}{4}\right) \]

Hence,
\[ I= 2\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{4}\right) - \frac{\pi}{2} \]



Final Answer:
\[ \boxed{ 2\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{4}\right) - \frac{\pi}{2} } \]
\[ \boxed{Option (D)} \] Quick Tip: Any integrand of the form \(\tan^{-1} \left( \frac{1}{1 + x + x^2} \right)\) or similar quadratic denominators screams for the algebraic manipulation \(\frac{(x+a) - (x+b)}{1 + (x+a)(x+b)}\) to enable telescopic integration.

Step 1: Find total number of ways


Total number of ways to choose \(3\) different dates from \(31\) dates is
\[ \binom{31}{3} \]
\[ = \frac{31\cdot 30\cdot 29}{3\cdot 2\cdot 1} \]
\[ = 4495 \]



Step 2: Condition for 3 numbers to be in A.P.


Let the three dates be
\[ x
For them to be in A.P.,
\[ y-x=z-y \]

which gives
\[ x+z=2y \]

This means the sum \(x+z\) must be even.

So, \(x\) and \(z\) must have the same parity:

- both odd, or
- both even

Once \(x\) and \(z\) are chosen, the middle term \(y\) is uniquely fixed.



Step 3: Count favorable cases


From \(1\) to \(31\):

Number of odd dates
\[ =16 \]
\[ (1,3,5,\dots,31) \]

Number of even dates
\[ =15 \]
\[ (2,4,6,\dots,30) \]

Now choose any \(2\) odd dates:
\[ \binom{16}{2} = \frac{16\cdot 15}{2} = 120 \]

Choose any \(2\) even dates:
\[ \binom{15}{2} = \frac{15\cdot 14}{2} = 105 \]

Hence total favorable cases are
\[ 120+105=225 \]



Step 4: Find probability

\[ P= \frac{225}{4495} \]

Divide numerator and denominator by \(5\):
\[ P= \frac{45}{899} \]

Thus,
\[ a=45,\qquad b=899 \]



Step 5: Find \(a+b\)

\[ a+b=45+899 \]
\[ =944 \]



Final Answer:
\[ \boxed{944} \] Quick Tip: To count 3-term Arithmetic Progressions in a consecutive sequence of integers, you just need to pick the two endpoints. Since the midpoint must be an integer, simply pick two numbers of the same parity!

Step 1: Find \(f(|x|)\)


For \(x<0\), \(|x|=-x>0\)
\[ f(|x|)=(-x)^2-5(-x)+6=x^2+5x+6 \]

For \(x\ge 0\)
\[ f(|x|)=x^2-5x+6 \]



Step 2: Find \(|f(x)|\)


For \(x<0\)
\[ f(x)=e^{x-1}>0 \]

So,
\[ |f(x)|=e^{x-1} \]

For \(x\ge 0\)
\[ f(x)=x^2-5x+6=(x-2)(x-3) \]

Hence,
\[ |f(x)|= \begin{cases} x^2-5x+6, & 0\le x\le 2 or x\ge 3
[4pt] -(x^2-5x+6), & 2


Step 3: Form \(g(x)\)

\[ g(x)= \begin{cases} x^2+5x+6+e^{x-1}, & x<0
[4pt] 2(x^2-5x+6), & 0\le x\le 2
[4pt] 0, & 2[4pt] 2(x^2-5x+6), & x\ge 3 \end{cases} \]



Step 4: Check continuity and differentiability


At \(x=0\)
\[ g(0^-)=6+\frac{1}{e} \]
\[ g(0^+)=12 \]

Since LHL \(\neq\) RHL, discontinuous at \(x=0\)
\[ \alpha=1 \]

This point is also non-differentiable.

At \(x=2\)
\[ g'(x)=4x-10 \quad (0 \[ LHD at x=2=-2 \]
\[ RHD at x=2=0 \]

Not differentiable.

At \(x=3\)
\[ LHD=0 \]
\[ RHD=2 \]

Not differentiable.

Thus
\[ \beta=3 \]
\[ \alpha+\beta=1+3=4 \]



Final Answer:
\[ \boxed{4} \] Quick Tip: Remember that a function containing \(|h(x)|\) typically fails to be differentiable at the roots of \(h(x)=0\) unless the root is a repeated root. Rapidly check the sharp corners at \(x=2, 3\).

he lines
\[ x-\sqrt{3}|y|=\alpha \]

represent two half-lines making angles
\[ 30^\circ and -30^\circ \]

So angle between them is
\[ 60^\circ \]

Given
\[ PA=PB=\alpha \]

Hence \(\triangle PAB\) is equilateral.

Altitude of equilateral triangle:
\[ PQ=\frac{\sqrt{3}}{2}\alpha \]

Given
\[ PQ=\frac{9}{2} \]
\[ \frac{\sqrt{3}}{2}\alpha=\frac{9}{2} \]
\[ \alpha=3\sqrt{3} \]

Circumradius of equilateral triangle:
\[ R=\frac{\alpha}{\sqrt{3}}=3 \]

Now
\[ \frac{\alpha^2}{R} = \frac{(3\sqrt{3})^2}{3} = \frac{27}{3} = 9 \]



Final Answer:
\[ \boxed{9} \] Quick Tip: Recognizing angle geometry from slopes like \(\pm 1/\sqrt{3}\) transforms a messy coordinate geometry calculation directly into elementary equilateral triangle properties.

For parabola
\[ y^2=16x \]

we have
\[ a=4 \]

Parametric point is
\[ (4t^2,8t) \]

Given \(B=(4,8)\) corresponds to
\[ t=1 \]

Let other points be parameters \(t_1,t_2\).

Condition of right angle at \(B\):
\[ m_1m_2=-1 \]

Using slope of chord formula
\[ \frac{2}{1+t_1}\cdot \frac{2}{1+t_2}=-1 \]
\[ (1+t_1)(1+t_2)=-4 \]

After centroid calculation and simplifying, locus becomes
\[ y^2=\frac{16}{3}\left(x-\frac{40}{3}\right) \]

For parabola
\[ y^2=4ax \]

length of latus rectum is
\[ 4a \]

So here
\[ L.R.=\frac{16}{3} \]

Required:
\[ 3\times \frac{16}{3}=16 \]



Final Answer:
\[ \boxed{16} \] Quick Tip: When manipulating parameters for the locus of a centroid, use the identity \((a+b)^2 = a^2+b^2+2ab\) as the unifying bridge to merge your sum, square sum, and product constraints.

Given
\[ f(x)=\int_0^x \tan(t-x)\,dt-\int_0^x f(t)\tan t\,dt \]

Differentiate both sides:
\[ f'(x)= -\tan x-f(x)\tan x \]
\[ f'(x)=-(1+f(x))\tan x \]

This is linear differential equation:
\[ f'(x)+f(x)\tan x=-\tan x \]

Integrating factor:
\[ I.F.=e^{\int \tan x\,dx}=\sec x \]

Multiply throughout:
\[ \frac{d}{dx}(f(x)\sec x)=-\tan x\sec x \]

Integrate:
\[ f(x)\sec x=-\sec x+C \]
\[ f(x)=-1+C\cos x \]

At \(x=0\)
\[ f(0)=0 \]
\[ 0=-1+C \]
\[ C=1 \]

So
\[ f(x)=\cos x-1 \]

Now
\[ f'(x)=-\sin x \]
\[ f''(x)=-\cos x \]

Substitute values:
\[ f''\left(\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2} \]
\[ f'\left(-\frac{\pi}{6}\right)=\frac{1}{2} \]
\[ f\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}-1 \]

Hence
\[ f''\left(\frac{\pi}{6}\right)+12f'\left(-\frac{\pi}{6}\right)+f\left(\frac{\pi}{6}\right) \]
\[ = -\frac{\sqrt{3}}{2}+12\cdot\frac{1}{2}+\frac{\sqrt{3}}{2}-1 \]
\[ = -\frac{\sqrt{3}}{2}+6+\frac{\sqrt{3}}{2}-1 \]
\[ =5 \]



Final Answer:
\[ \boxed{5} \] Quick Tip: Remember to apply the partial derivative to the integrand when using the Leibniz rule on limits that also contain the differentiation variable. Many students forget the \(\int \frac{\partial}{\partial x} g(x,t) dt\) term!