First, let's determine the parity (even or odd nature) of the function \(G(x)\).


Let \(A(x) = f(x)f(-x)\). Note that \(A(-x) = f(-x)f(x) = A(x)\), so \(A(x)\) is an even function.


Let \(B(x) = f(x) - f(-x)\). Note that \(B(-x) = f(-x) - f(x) = -(f(x) - f(-x)) = -B(x)\), so \(B(x)\) is an odd function.


The determinant is \(G(x) = \begin{vmatrix} A(x) & 0 & x^4
3 & B(x) & \cos x
x^4 & 2x & A(x) \end{vmatrix}\).


Expanding along the second row:

\(G(x) = -3 \begin{vmatrix} 0 & x^4
2x & A(x) \end{vmatrix} + B(x) \begin{vmatrix} A(x) & x^4
x^4 & A(x) \end{vmatrix} - \cos x \begin{vmatrix} A(x) & 0
x^4 & 2x \end{vmatrix}\).


Term 1: \(-3(0 - 2x^5) = 6x^5\). This is an Odd function.


Term 2: \(B(x) ( (A(x))^2 - x^8 )\). Since \(B\) is Odd and \(A^2, x^8\) are Even, Odd \(\times\) Even = Odd function.


Term 3: \(-\cos x ( 2x A(x) - 0 ) = -2x \cos x A(x)\). Since \(x\) is Odd and \(\cos x, A(x)\) are Even, Odd \(\times\) Even = Odd function.


Since \(G(x)\) is the sum of three odd functions, \(G(x)\) is an Odd function.


The integral is \(I = \int_{-2}^{2} x^4 G(x) dx\).


The integrand is \(x^4 \cdot G(x) = (Even) \cdot (Odd) = Odd\).


The definite integral of an odd function over a symmetric interval \([-a, a]\) is always 0.


Therefore, the value is 0.
Quick Tip: Always check for symmetry (even/odd) when integrating over limits like \([-a, a]\). If \(f(-x) = -f(x)\), \(\int_{-a}^a f(x) dx = 0\).

The fourth roots of unity are the roots of the equation \(x^4 - 1 = 0\).


The roots are given as \(1, \alpha_1, \alpha_2, \alpha_3\).


Since \(x^4 - 1 = (x-1)(x+1)(x-i)(x+i)\), the roots are \(1, -1, i, -i\).


So, the set \(\{\alpha_1, \alpha_2, \alpha_3\}\) is equal to \(\{-1, i, -i\}\).


We need to calculate \((1+\alpha_1)(1+\alpha_2)(1+\alpha_3)\).


One of these roots must be \(-1\). Let \(\alpha_1 = -1\).


Then the first factor is \((1 + (-1)) = 0\).


Consequently, the entire product \((0)(1+i)(1-i)\) becomes 0.


Alternatively, using polynomial properties:
\(\frac{x^4-1}{x-1} = x^3 + x^2 + x + 1 = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3)\).


We want the value of \((1+\alpha_1)(1+\alpha_2)(1+\alpha_3)\). This is not directly \(P(1)\) or \(P(-1)\).


However, rewriting the expression: \((1+\alpha_1) = -((-1)-\alpha_1)\).

Product \(= (-1)^3 ((-1)-\alpha_1)((-1)-\alpha_2)((-1)-\alpha_3) = -1 \cdot P(-1)\).

\(P(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0\).


Thus, the value is 0.
Quick Tip: Recognize that if roots are \(n\)-th roots of unity, one of the roots is \(-1\) (if \(n\) is even). Factors involving \((1+root)\) will become zero.

Let \(P(x, y)\) be any point on the conic.


The definition of a conic is \(\frac{SP}{PM} = e\), where \(S\) is the focus, \(M\) is the perpendicular distance to the directrix, and \(e\) is the eccentricity.


Here, \(S = (1, 0)\), Directrix line \(L: x + y - 5 = 0\), and \(e = 2\).

\(SP^2 = e^2 \cdot PM^2\).

\((x-1)^2 + (y-0)^2 = 2^2 \cdot \left( \frac{x + y - 5}{\sqrt{1^2 + 1^2}} \right)^2\).

\((x-1)^2 + y^2 = 4 \cdot \frac{(x + y - 5)^2}{2}\).

\(x^2 - 2x + 1 + y^2 = 2 (x^2 + y^2 + 25 + 2xy - 10x - 10y)\).

\(x^2 + y^2 - 2x + 1 = 2x^2 + 2y^2 + 50 + 4xy - 20x - 20y\).


Rearranging all terms to the right side:
\((2x^2 - x^2) + 4xy + (2y^2 - y^2) + (-20x + 2x) + (-20y) + (50 - 1) = 0\).

\(x^2 + 4xy + y^2 - 18x - 20y + 49 = 0\).
Quick Tip: For any conic section problem given Focus, Directrix, and Eccentricity, directly apply the definition \(SP^2 = e^2 PM^2\).

Given: \(|\vec{u}| = 1, |\vec{v}| = 2, |\vec{w}| = 3\).

\(\vec{v} \perp \vec{w} \implies \vec{v} \cdot \vec{w} = 0\).


Projection of \(\vec{v}\) along \(\vec{u}\) = Projection of \(\vec{w}\) along \(\vec{u}\).
\(\frac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \frac{\vec{w} \cdot \vec{u}}{|\vec{u}|} \implies \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u}\).


Let \(k = \vec{v} \cdot \vec{u} = \vec{w} \cdot \vec{u}\).


We need to find \(|\vec{u} - \vec{v} + \vec{w}|\). Let's calculate its square.
\(|\vec{u} - \vec{v} + \vec{w}|^2 = (\vec{u} - \vec{v} + \vec{w}) \cdot (\vec{u} - \vec{v} + \vec{w})\).

\(= |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) - 2(\vec{v} \cdot \vec{w})\).


Substitute the known values:
\(= 1^2 + 2^2 + 3^2 - 2k + 2k - 2(0)\).

\(= 1 + 4 + 9 = 14\).


Therefore, \(|\vec{u} - \vec{v} + \vec{w}| = \sqrt{14}\).
Quick Tip: Expand the square of the magnitude \(|\vec{A} + \vec{B} + \vec{C}|^2 = A^2 + B^2 + C^2 + 2(\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{C} + \vec{C}\cdot\vec{A})\).

Let the equation of the plane be \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).


The intercepts are \(P(a, 0, 0)\), \(Q(0, b, 0)\), and \(R(0, 0, c)\).


The distance of this plane from the origin \((0,0,0)\) is given as 1.
\(\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1 \implies \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1\).


Let the centroid of \(\Delta PQR\) be \((X, Y, Z)\).
\(X = \frac{a+0+0}{3} \implies a = 3X\).
\(Y = \frac{0+b+0}{3} \implies b = 3Y\).
\(Z = \frac{0+0+c}{3} \implies c = 3Z\).


Substitute these into the distance equation:
\(\frac{1}{(3X)^2} + \frac{1}{(3Y)^2} + \frac{1}{(3Z)^2} = 1\).

\(\frac{1}{9X^2} + \frac{1}{9Y^2} + \frac{1}{9Z^2} = 1\).


Multiply by 9: \(\frac{1}{X^2} + \frac{1}{Y^2} + \frac{1}{Z^2} = 9\).


Comparing with the given equation \(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k\), we get \(k = 9\).
Quick Tip: The centroid of a triangle with vertices \((x_1, y_1, z_1)\) etc., is \((\frac{\Sigma x}{3}, \frac{\Sigma y}{3}, \frac{\Sigma z}{3})\).

Since \(g\) is the inverse function of \(f\), we have \(f(g(x)) = x\).


Differentiating both sides with respect to \(x\) using the chain rule:
\(f'(g(x)) \cdot g'(x) = 1\).


Therefore, \(g'(x) = \frac{1}{f'(g(x))}\).


We are given \(f'(x) = \frac{1}{1+x^5}\).


To find \(f'(g(x))\), substitute \(g(x)\) for \(x\) in the expression for \(f'\):
\(f'(g(x)) = \frac{1}{1+(g(x))^5}\).


Now substitute this back into the equation for \(g'(x)\):
\(g'(x) = \frac{1}{\frac{1}{1+(g(x))^5}} = 1 + (g(x))^5\).
Quick Tip: If \(g = f^{-1}\), then \(g'(x) = \frac{1}{f'(g(x))}\). This is a standard property of derivatives of inverse functions.

The curves are \(C_1: y = |x^3|\) and \(C_2: x = y^3 \implies y = x^{1/3}\).


Let's find the points of intersection.

For \(x \ge 0\), \(y = x^3\). So \(x^3 = x^{1/3} \implies x^9 = x \implies x(x^8 - 1) = 0\).

The real roots are \(x = 0\) and \(x = 1\). The points are \((0,0)\) and \((1,1)\).


In the interval \((0, 1)\), take \(x = 1/8\). \(y_1 = (1/8)^3 = 1/512\). \(y_2 = (1/8)^{1/3} = 1/2\).

So \(y = x^{1/3}\) lies above \(y = x^3\).


For \(x < 0\), \(y = -x^3\) (since \(|x^3| = -x^3\)) and \(y = x^{1/3}\) (which is negative).

Intersection: \(-x^3 = x^{1/3} \implies -x^9 = x \implies x(1 + x^8) = 0\).

The only real root is \(x = 0\). The curves do not enclose a finite area in the 2nd/3rd quadrants as they only meet at the origin and diverge.


Thus, the enclosed area is only in the first quadrant.

Area \(= \int_{0}^{1} (x^{1/3} - x^3) dx\).

\(= \left[ \frac{x^{4/3}}{4/3} - \frac{x^4}{4} \right]_0^1\).

\(= \left[ \frac{3}{4}x^{4/3} - \frac{1}{4}x^4 \right]_0^1\).

\(= (\frac{3}{4} - \frac{1}{4}) - 0 = \frac{2}{4} = \frac{1}{2}\).
Quick Tip: Always visualize or sketch the curves to check which lies above the other and to verify the bounds of integration.

Let \(K = \int_{0}^{2} f(x)dx\). Note that \(K\) is a constant.

The differential equation becomes \(f'(x) = f(x) + K\).
\(\frac{dy}{dx} - y = K\). This is a linear differential equation.

Integrating Factor (I.F.) \(= e^{\int -1 dx} = e^{-x}\).
\(y \cdot e^{-x} = \int K e^{-x} dx = -K e^{-x} + C\).
\(y = f(x) = C e^x - K\).


Given \(f(0) = \frac{4-e^2}{3}\).

Substituting \(x=0\): \(C - K = \frac{4-e^2}{3}\) \quad ...(1)


Now use the definition of K: \(K = \int_{0}^{2} (Ce^x - K) dx\).
\(K = [Ce^x - Kx]_0^2 = (Ce^2 - 2K) - (C - 0)\).
\(K = Ce^2 - 2K - C\).
\(3K = C(e^2 - 1) \implies K = C \frac{e^2 - 1}{3}\).


Substitute \(K\) into (1):
\(C - C \frac{e^2 - 1}{3} = \frac{4-e^2}{3}\).
\(C \left( 1 - \frac{e^2 - 1}{3} \right) = \frac{4-e^2}{3}\).
\(C \left( \frac{3 - e^2 + 1}{3} \right) = \frac{4-e^2}{3}\).
\(C \left( \frac{4 - e^2}{3} \right) = \frac{4-e^2}{3}\).

Since \(e^2 \approx 7.4\), \(4-e^2 \ne 0\), so \(C = 1\).


Then \(K = \frac{e^2 - 1}{3}\).

The function is \(f(x) = 1 \cdot e^x - \frac{e^2 - 1}{3}\).
Quick Tip: When an integral with fixed limits appears in an equation for \(f(x)\), treat it as a constant (\(K\)) and solve for it at the end.

Let \(X\) be the random variable representing the number of heads.

The probability of getting a head in a single toss is \(p = 1/2\). Probability of tails is \(q = 1/2\).

The probability of getting no head in \(n\) tosses is \(P(X=0) = \binom{n}{0} p^0 q^n = (1/2)^n\).


We are given the condition: \(P(X=0) > \frac{1}{16}\).
\(\left(\frac{1}{2}\right)^n > \frac{1}{16}\).
\(\left(\frac{1}{2}\right)^n > \left(\frac{1}{2}\right)^4\).


Since the base \((1/2)\) is less than 1, the inequality sign reverses for the exponents:
\(n < 4\).


Since \(n\) must be a positive integer (number of tosses), the possible values are 1, 2, 3.

The maximum value of \(n\) satisfying the condition is 3.
Quick Tip: Remember that for \(0 < b < 1\), \(b^x > b^y \implies x < y\).

Total numbers formed using digits 1, 2, 3, 4, 5 without repetition = \(5! = 120\).

We need to find the \(100^{th}\) number in ascending order.


Number of numbers starting with 1: \(4! = 24\). (Ranks 1-24)

Number of numbers starting with 2: \(4! = 24\). (Ranks 25-48)

Number of numbers starting with 3: \(4! = 24\). (Ranks 49-72)

Number of numbers starting with 4: \(4! = 24\). (Ranks 73-96)

Total so far = \(24 \times 4 = 96\).


The \(100^{th}\) number must start with 5.

Next, consider the second digit (ascending order of remaining digits {1, 2, 3, 4).

Numbers starting with 51... : Remainder 3 digits ({2,3,4). \(3! = 6\) numbers.

Ranks 97 to \(96+6 = 102\).

So the \(100^{th}\) number lies in the block 51...


Let's list these numbers starting from rank 97:

Digits available: 2, 3, 4. Ascending order: 2, 3, 4.

97: 51234

98: 51243

99: 51324

100: 51342


The \(100^{th}\) number is 51342.
Quick Tip: Break the problem down by fixing the first digit and counting the permutations (\(n!\)) until you get close to the target rank.